College Algebra: Graphs & Models

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Author: John Coburn | J.D. (John) Herdlick

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Graphs and Models John W. Coburn St. Louis Community College at Florissant Valley

J.D. Herdlick St. Louis Community College at Meramec-Kirkwood

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TM

COLLEGE ALGEBRA: GRAPHS AND MODELS Published by McGraw-Hill, a business unit of The McGraw-Hill Companies, Inc., 1221 Avenue of the Americas, New York, NY 10020. Copyright © 2012 by The McGraw-Hill Companies, Inc. All rights reserved. No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written consent of The McGraw-Hill Companies, Inc., including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance learning. Some ancillaries, including electronic and print components, may not be available to customers outside the United States. This book is printed on acid-free paper. 1 2 3 4 5 6 7 8 9 0 DOW/DOW 1 0 9 8 7 6 5 4 3 2 1 ISBN 978–0–07–351954–8 MHID 0–07–351954–5 ISBN 978–0–07–723057–9 (Annotated Instructor’s Edition) MHID 0–07–723057–4 Vice President, Editor-in-Chief: Marty Lange Vice President, EDP: Kimberly Meriwether David Senior Director of Development: Kristine Tibbetts Editorial Director: Stewart K. Mattson Sponsoring Editor: John R. Osgood Developmental Editor: Eve L. Lipton Marketing Manager: Kevin M. Ernzen Senior Project Manager: Vicki Krug

Buyer II: Sherry L. Kane Senior Media Project Manager: Sandra M. Schnee Senior Designer: Laurie B. Janssen Cover Image: © Georgette Douwma and Sami Sarkis / Gettyimages Senior Photo Research Coordinator: John C. Leland Compositor: Aptara, Inc. Typeface: 10.5/12 Times Roman Printer: R. R. Donnelley

All credits appearing on page or at the end of the book are considered to be an extension of the copyright page. Library of Congress Cataloging-in-Publication Data Coburn, John W. College algebra : graphs and models / John W. Coburn, J.D. Herdlick. p. cm. Includes index. ISBN 978–0–07–351954–8 — ISBN 0–07–351954–5 (hard copy : alk. paper) 1. Algebra— Textbooks. 2. Algebra—Graphic methods—Textbooks. I. Herdlick, John D. II. Title. QA154.3.C5953 2012 512.9—dc22 2010035347 www.mhhe.com

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Brief Contents Preface vi Index of Applications

CHAPTER CHAPTER CHAPTER CHAPTER CHAPTER CHAPTER CHAPTER CHAPTER CHAPTER CHAPTER

R 1 2 3 4 5 6 7 8 9

xxxii

A Review of Basic Concepts and Skills 1 Relations, Functions, and Graphs 85 More on Functions 187 Quadratic Functions and Operations on Functions

281

Polynomial and Rational Functions 381 Exponential and Logarithmic Functions 479 Systems of Equations and Inequalities

575

Matrices and Matrix Applications 637 Analytic Geometry and the Conic Sections 707 Additional Topics in Algebra 761

Appendix I

The Language, Notation, and Numbers of Mathematics

Appendix II

Geometry Review with Unit Conversions

Appendix III

More on Synthetic Division

Appendix IV

More on Matrices A-30

Appendix V

Deriving the Equation of a Conic

Appendix VI

Proof Positive—A Selection of Proofs from College Algebra

A-14

A-28 A-32

Student Answer Appendix (SE only)

A-34

SA-1

Instructor Answer Appendix (AIE only) Index

A-1

IA-1

I-1

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About the Authors John Coburn

John Coburn grew up in the Hawaiian Islands, the seventh of sixteen children. He received his Associate of Arts degree in 1977 from Windward Community College, where he graduated with honors. In 1979 he earned a Bachelor’s Degree in Education from the University of Hawaii. After working in the business world for a number of years, he returned to teaching, accepting a position in high school mathematics where he was recognized as Teacher of the Year (1987). Soon afterward, the decision was made to seek a Master's Degree, which he received two years later from the University of Oklahoma. John is now a full professor at the Florissant Valley campus of St. Louis Community College. During his tenure there he has received numerous nominations as an outstanding teacher by the local chapter of Phi Theta Kappa, two nominations to Who’s Who Among America’s Teachers, and was recognized as Post Secondary Teacher of the Year in 2004 by the Mathematics Educators of Greater St. Louis (MEGSL). He has made numerous presentations and local, state, and national conferences on a wide variety of topics and maintains memberships in several mathematics organizations. Some of John’s other interests include body surfing, snorkeling, and beach combing whenever he gets the chance. He is also an avid gamer, enjoying numerous board, card, and party games. His other loves include his family, music, athletics, composition, and the wild outdoors.

J.D. Herdlick

J.D. Herdlick was born and raised in St. Louis, Missouri, very near the Mississippi river. In 1992, he received his bachelor’s degree in mathematics from Santa Clara University (Santa Clara, California). After completing his master’s in mathematics at Washington University (St. Louis, Missouri) in 1994, he felt called to serve as both a campus minister and an aid worker for a number of years in the United States and Honduras. He later returned to education and spent one year teaching high school mathematics, followed by an appointment at Washington University as visiting lecturer, a position he held until 2006. Simultaneously teaching as an adjunct professor at the Meramec campus of St. Louis Community College, he eventually joined the department full time in 2001. While at Santa Clara University, he became a member of the honorary societies Phi Beta Kappa, Pi Mu Epsilon, and Sigma Xi under the tutelage of David Logothetti, Gerald Alexanderson, and Paul Halmos. In addition to the Dean’s Award for Teaching Excellence at Washington University, J.D. has received numerous awards and accolades for his teaching at St. Louis Community College. Outside of the office and classroom, he is likely to be found in the water, on the water, and sometimes above the water, as a passionate wakeboarder and kiteboarder. It is here, in the water and wind, that he finds his inspiration for writing. J.D. and his family currently split their time between the United States and Argentina.

Dedication With boundless gratitude, we dedicate this work to the special people in our lives. To our children, whom we hope were joyfully oblivious to the time, sacrifice, and perseverance required; and to our wives, who were well acquainted with every minute of it.

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About the Cover Most coral reefs in the world are 7000–9000 years old, but new reefs can fully develop in as few as 20 years. In addition to being home to over 4000 species of tropical or reef fish, coral reefs are immensely beneficial to humans and must be carefully preserved. They buffer coastal regions from strong waves and storms, provide millions of people with food and jobs, and prompt advances in modern medicine. Similar to the ancient reefs, a course in College Algebra is based on thousands of years of mathematical curiosity, insight, and wisdom. In this one short course, we study a wealth of important concepts that have taken centuries to mature. Just as the variety of fish in the sea rely on the coral reefs to survive, students in a College Algebra course rely on mastery of this bedrock of concepts to successfully pursue more advanced courses, as well as their career goals.

From the Authors

nges. From the ion has seen some enormo us cha cat edu tics ma the ma s, ade dec In the last two to online homework and and the adv ent of the Intern et, ors ulat calc ng phi gra of n ctio elen ting. intr odu s ago, the cha nges hav e been unr ade dec ut abo am dre only ld cou visual sup plement s we nce tea ching re a combined 40 yea rs of exp erie sha k dlic Her . J.D and urn Cob n of Tog eth er, Joh gies, and hav e dev elop ed a wea lth nolo tech er oth and ors ulat calc ng college alg ebr a wit h gra phi endeav or. firs tha nd exp erience related to the con ver sat iona l style and Models text, we hav e combined the and phs Gra ck rdli /He urn Cob one of In the , wit h this dep th of exp erience. As for wn kno are s text our t tha s the wea lth of application y see functions think visually, to a poin t where the ts den stu help to out set we ls, our primary goa ediately lead to a of gra phs, wit h attr ibutes that imm 2 ily fam a of one as 4x – x = , the nat ure of like f(x) ior, zer oes, solu tions to ineq ualities hav -be end ms, imu min and ums discussion of ma xim an equation that es in con text — instead of mer ely ibut attr se the of tion lica app the le the roots, and the scr een of a calculat or. And whi on ph gra a g etin rpr inte by or off ers much must be solv ed by factor ing nal drudgery, we believe our text atio put com e som eve reli y ma ors gra phing calculat gra phical met hods, wit h ison of algebra ic met hods ver sus par com e -sid -by side ple sim a nua lly. n more tha checking answer s to wor k don e ma ply sim n tha role ant ific sign re mo the calculat or playing a sible wit h pap er and investigate far bey ond what’s pos and k wor to d use are ors ulat Gra phing calc age more applications, and e more tru e-to-lif e equations, eng solv to d use gy nolo tech the h wit text is built on pencil, the end we believe you’ll see this In t. res inte of ns stio que l ntia explore more substa accent uates the visual and dynamic excursion that a ers off t tha one yet als, ent strong fundam use in all areas of their solv ing acumen that studen ts will blem pro and g nnin pla nal atio aniz l tool for the org Gra phs and Models text as an idea ck rdli /He urn Cob the er off we lives. To this end —John Coburn and J.D. Her dlick tics. tea ching and lear ning of mathema

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Making Connections . . . College Algebra tends to be a challenging course for many students. They may not see the connections that College Algebra has to their life or why it is so critical that they succeed in this course. Others may enter into this course underprepared or improperly placed and with very little motivation. Instructors are faced with several challenges as well. They are given the task of improving pass rates and student retention while ensuring the students are adequately prepared for more advanced courses, as a College Algebra course attracts a very diverse audience, with a wide variety of career goals and a large range of prerequisite skills. The goal of this textbook series is to provide both students and instructors with tools to address these challenges, so that both can experience greater success in College Algebra. For instance, the comprehensive exercise sets have a range of difficulty that provides very strong support for weaker students, while advanced students are challenged to reach even further. The rest of this preface further explains the tools that John Coburn, J.D. Herdlick, and McGraw-Hill have developed and how they can be used to connect students to College Algebra and connect instructors to their students.

The Coburn/Herdlick College Algebra Series provides you with strong tools to achieve better outcomes in your College Algebra course as follows:

vi

▶

Making Connections Visually, Symbolically, Numerically, and Verbally

▶

Better Student Preparedness Through Superior Course Management

▶

Increased Student Engagement

▶

Solid Skill Development

▶

Strong Mathematical Connections

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▶


Making Connections Visually, Symbolically, Numerically, and Verbally

In writing their Graphs and Models series, the Coburn/Herdlick team took great care to help students think visually by relating a basic graph to an algebraic equation at every opportunity. This empowers students to see the “Why?” behind many algebraic rules and properties, and offers solid preparation for the connections they’ll need to make in future courses which often depend on these visual skills. ▶

Better Student Preparedness Through Superior Course Management

McGraw-Hill is proud to offer instructors a choice of course management options to accompany Coburn/ Herdlick. If you prefer to assign text-specific problems in a brand new, robust online homework system that contains stepped out and guided solutions for all questions, Connect Math Hosted by ALEKS may be for you. Or perhaps you prefer the diagnostic nature and artificial intelligence engine that is the driving force behind our ALEKS 360 Course product, a true online learning environment, which has been expanded to contain hundreds of new College Algebra & Precalculus topics. We encourage you to take a closer look at each product on preface pages x through xiii and to consult your McGraw-Hill sales representative to setup a demonstration. ▶

Increased Student Engagement

There are many texts that claim they “engage” students, but only the Coburn Series has carefully studied and implemented features and options that make it truly possible. From the on-line support, to the textbook design and a wealth of quality applications, students will remain engaged throughout their studies. ▶

Solid Skill Development

The Coburn/Herdlick series intentionally relates the examples to the exercise sets so there is a strong connection between what students are learning while working through the examples in each section and the homework exercises that they complete. This development of strong mechanical skills is followed closely by a careful development of problem solving skills, with the use of interesting and engaging applications that have been carefully chosen with regard to difficulty and the skills currently under study. There is also an abundance of exercise types to choose from to ensure that homework challenges a wide variety of skills. Furthermore, John and J.D. reconnect students to earlier chapter material with Mid-Chapter Checks; students have praised these exercises for helping them understand what key concepts require additional practice. ▶

Strong Mathematical Connections

John Coburn and J.D. Herdlick’s experience in the classroom and their strong connections to how students comprehend the material are evident in their writing style. This is demonstrated by the way they provide a tight weave from topic to topic and foster an environment that doesn’t just focus on procedures but illustrates the big picture, which is something that so often is sacrificed in this course. Moreover, they employ a clear and supportive writing style, providing the students with a tool they can depend on when the teacher is not available, when they miss a day of class, or simply when working on their own.

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Making Connections . . . Visually, Symbolically, Numerically, and Verbally , the concre te and numer ic “It is widely known that for studen ts to grow stronger algebr aically entations. In this transit ion experiences from their past must give way to more symbolic repres visual connections and verbal from numer ic, to symbolic, to algebr aic thinking, the importance of of rich concep ts or subtle ideas, connections is too often overlooked. To reach a deep understanding concep t or idea using the terms studen ts must develop the ability to menta lly “see” and discuss the seeing the connections that and names needed to describ e it accurately. Only then can they begin . A large part of this involves exist between each new concep t, and concep ts that are already known they’re able to see functio ns like helping our studen ts to begin thinking visuall y, to a point where ical attribu tes that immediately f(x) = x2 – 4x as only one of a large family of functions, with graph , solutions to inequa lities, the lead to a discussion of maximums and minimums, end-behavior, zeroes t. And while it’s important for nature of the roots, and the application of these attribu tes in contex , and that the intersection of students to see that zeroes are x-intercepts and x-intercepts are zeroes g these graphs, these should not two graphs provides a simulta neous solution to the equations formin tions, investigations, connections, remain the sole focus of the tool. Graphing calculators allow explora and we should use the technology and visualizations far beyond what’s possible with paper and pencil, more true-to-life equations, to aid the development of these menta l-visua l skills, in addition to solving ns involving real data, domain engaging more applications, and explor ing the more substa ntial questio tables, and other questions of and range, anticipated graphical behavior, additional uses of lists and be successful in these endeavors.” interest. We believe this text offers instructors the tools they need to —The Authors

EXAMPLE 1

䊳

“I think there is a good balance between technology

Solving a Logarithmic Equation

and paper/pencil techniques. I particularly like how the technology portion does not take the place of paper/pencil, but instead supplements it. I think a lot of departments will like that.

Solve for x and check your answer: log x ⫹ log 1x ⫹ 32 ⫽ 1. 䊲

䊲

Algebraic Solution log x ⫹ log 1x ⫹ 32 ⫽ 1 log 3x 1x ⫹ 32 4 ⫽ 1 x2 ⫹ 3x ⫽ 101 x2 ⫹ 3x ⫺ 10 ⫽ 0 1x ⫹ 52 1x ⫺ 22 ⫽ 0 x ⫽ ⫺5 or x ⫽ 2

original equation product property exponential form, distribute x set equal to 0 factor result

Graphical Solution

Using the intersection-ofgraphs method, we enter Y1 ⫽ log X ⫹ log1X ⫹ 32 and Y2 ⫽ 1. From the domain we know x 7 0, indicating the solution will occur in QI. After graphing both functions using the window shown, the intersection method shows the only solution is x ⫽ 2.

3

”

0

—Daniel Brock, Arkansas State University-Beebe

5

▶ Graphical Examples show students how

⫺3

Check: The “solution” x ⫽ ⫺5 is outside the domain and is ignored. For x ⫽ 2, log x ⫹ log1x ⫹ 32 ⫽ 1 original equation log 2 ⫹ log12 ⫹ 32 ⫽ 1 substitute 2 for x log 2 ⫹ log 5 ⫽ 1 simplify log12 # 52 ⫽ 1 product property log 10 ⫽ 1 Property I

the calculator can be used to supplement their understanding of a problem. EXAMPLE 1A

You could also use a calculator to verify log 2 ⫹ log 5 ⫽ 1 directly. roug g 14 Now try Exercises 7 through

Precalculus: Graphs and Models textbook the best approach ever to the teaching of Precalculus with the inclusion of graphing calculator.

”

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—Alvio Dominguez, Miami-Dade College-Wolfson

Solving an Equation Graphically 1 Solve the equation 21x 32 7 x 2 using 2 a graphing calculator.

䊳

Solution

“I have certainly found the Coburn/Herdlick’s

䊳

䊳

Begin by entering the left-hand expression as Y1 and the right-hand expression as Y2 (Figure 1.74). To find points of intersection, press 2nd TRACE (CALC) and select option 5:intersect, which automatically places you on the graphing window, and asks you to identify the “First curve?.” As discussed, pressing three times in succession will identify each graph, bypass the “Guess?” option, then find and display the point of intersection (Figure 1.75). Here the point of intersection 10 is (2, 3), showing the solution to this equation is x 2 (for which both expressions equal 3). This can be verified by direct substitution or by using the TABLE feature. ENTER

Figure 1.74

Figure 1.75 10

10

10

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▶ Calculator Explanations incorporate the

calculator without sacrificing.

Figure 3.2 Most graphing calculators are programmed to work with imaginary and complex numbers, though for some models the calculator must be placed in complex number mode. After pressing the MODE key (located to the right of the 2nd option key), the screen shown in Figure 3.2 appears and we use the arrow keys to access “a bi” and active this mode (by pressing ). Once active, we can validate our previo previous statements about imaginary numbers (Figure gure 3.3), 3.3 as well as verify our previous calculations like those in Examples 3(a), 3(d), andd 4(a) (F an (Figure 3.4). Note the imaginary unit i is the 2nd option for the decimal point. ENTER

“The technology (graphing calculator) explanations and illustrations are superb. The level of detail is valuable; even an experienced user (myself) learned some new techniques and “tricks” in reading through the text. The text frequently references use of the calculator—yet without sacrificing rigor or mathematical integrity.

Figure 3.3

Figure 3.4

”

—Light Bryant, Arizona Western College

Figure 4.4A

To help illustrate the Intermediate Value Theorem, many graphing calculators offer a useful feature called split screen viewing, that enables us to view a table of values and the graph of a function at the same time. To illustrate, enter the function y x3 9x 6 (from Example 6) as Y1 on the Y= screen, then set the viewing window as shown in Figure 4.4. Set your table in AUTO mode with ¢Tbl 1, then press the MODE key (see Figure 4.4A) and notice the second-to-last entry on this screen reads: Full for full screen viewing, Horiz for splitting the screen horizontally with the graph above a reduced home screen, and G-T, which represents Graph-Table and splits the screen vertically. In the G-T mode, the graph appears on the left and the table of values on the right. Navigate the cursor to the G-T mode and press . Pressing the GRAPH key at this point should give you a screen similar to Figure 4.5. Scrolling downward shows the function also changes sign between x 2 and x 3. For more on this idea, see Exercises 31 and 32. werful yet simple As a final note, while the intermediate value theorem is a powerful tool, it must be used with care. For example, given p1x2 x4 10x2 5, p112 7 0 lly, and p112 7 0, seeming to indicate that no zeroes exist in the intervall (1, 1). Actual Actually, there are two zeroes, as seen in Figure 4.6. ENTER

in every section. I have been using TI calculators for 15 years and I learned a few new tricks while reading this book.

Figure 4.6

Figure 4.5

“The authors give very good uses of the calculator

25

”

B. You’ve just seen how we can use the intermediate value theorem to identify intervals containing a polynomial zero

5

5

—George Hurlburt, Corning Community College

10

▶ Technology Applications show

students how technology can be used to help apply lessons from the classroom to real life.

“I think that the graphing examples, explanations,

and problems are perfect for the average college algebra student who has never touched a graphing calculator. . . . . I think this book would be great to actually have in front of the students.

”

—Dale Duke, Oklahoma City Community College

Use Newton’s law of cooling to complete Exercises 75 and 76: T(x) ⫽ TR ⫹ (T0 ⫺ TR)ekx.

75. Cold party drinks: Janae was late getting ready for the party, and the liters of soft drinks she bought were still at room temperature (73°F) with guests due to arrive in 15 min. If she puts these in her freezer at 10°F, will the drinks be cold enough (35°F) for her guests? Assume k ⬇ 0.031. 76. Warm party drinks: Newton’s law of cooling applies equally well if the “cooling is negative,” meaning the object is taken from a colder medium and placed in a warmer one. If a can of soft drink is taken from a 35°F cooler and placed in a room where the temperature is 75°F, how long will it take the drink to warm to 65°F? Assume k ⬇ 0.031. Photochromat Photochromatic sunglasses: Sunglasses that darken in sunlight sunl unligh ghtt (p (photo (photochromatic sunglasses) contain millions of mole m ole lecules le lec cu of a substance known as silver halide. The molecules molecules m mo o are ttransparent indoors in the absence of ultraviolent (U (UV) light. Outdoors, UV light from the sun causes the mol molecules to change shape, darkening the lenses in respo response to the intensity of the UV light. For certain lenses, the function T1x2 0.85x models the transparency of the lenses (as a percentage) based on a UV index x. Fi Find the transparency (to the nearest percent), if the lenses are exposed to 77 li ht with a UV index of 7 (a high exposure). 77. sunlight 78. sunlight with a UV index of 5.5 (a moderate exposure)

80. Use a trial-and-error process and a graphing calculator to determine the UV index when the lenses are 50% transparent. Modeling inflation: Assuming the rate of inflation is 5% per year, the predicted price of an item can be modeled by the function P1t2 P0 11.052 t, where P0 represents the initial price of the item and t is in years. Use this information to solve Exercises 81 and 82. 81. What will the price of a new car be in the year 2015, if it cost $20,000 in the year 2010? 82. What will the price of a gallon of milk be in the year 2015, if it cost $3.95 in the year 2010? Round to the nearest cent. Modeling radioactive decay: The half-life of a radioactive substance is the time required for half an initial amount of the substance to disappear through decay. The amount of the substance remaining is given t by the formula Q1t2 Q0 1 12 2 h, where h is the half-life, t represents the elapsed time, and Q(t) represents the amount that remains (t and h must have the same unit of time). Use this information to solve Exercises 83 and 84. 83. Some isotopes of the substance known as thorium have a half-life of only 8 min. (a) If 64 grams are initially present, how many grams (g) of the substance remain after 24 min? (b) How many minutes until only 1 gram (g) of the substance remains?

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Efficient Assignment Navigation ▶ Students have access to immediate feedback and help while working through assignments. ▶ Students have direct access ess to a media-rich eBook forr easy referencing. ▶ Students can view detailed, ed, step-by-step solutions written by instructors who teach the course, providing a unique solution on to each and every exercise. e

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Students can easily monitor and track their progress on a given assignment.

Y want a more intuitive and efficient assignment creation process You because of your busy schedule. b

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Built by Math Educators for Math Educators 7

Y want algorithmic content that was developed by math faculty to You ensure the content is pedagogically sound and accurate. e

Digital Content Development Story The development of McGraw-Hill’s Connect Math Hosted by ALEKS Corp. content involved collaboration between McGraw-Hill, experienced instructors, and ALEKS, a company known for its high-quality digital content. The result of this process, outlined below, is accurate content created with your students in mind. It is available in a simple-to-use interface with all the functionality tools needed to manage your course. 1. McGraw-Hill selected experienced instructors to work as Digital Contributors. 2. The Digital Contributors selected the textbook exercises to be included in the algorithmic content to ensure appropriate coverage of the textbook content. 3. The Digital Contributors created detailed, stepped-out solutions for use in the Guided Solution and Show Me features. 4. The Digital Contributors provided detailed instructions for authoring the algorithm specific to each exercise to maintain the original intent and integrity of each unique exercise. 5. Each algorithm was reviewed by the Contributor, went through a detailed quality control process by ALEKS Corporation, and was copyedited prior to being posted live.

Connect Math Hosted by ALEKS Corp. Built by Math Educators for Math Educators Lead Digital Contributors

Tim Chappell Metropolitan Community College, Penn Valley

Digital Contributors Al Bluman, Community College of Allegheny County John Coburn, St. Louis Community College, Florissant Valley Vanessa Coffelt, Blinn College Donna Gerken, Miami-Dade College Kimberly Graham J.D. Herdlick, St. Louis Community College, Meramec

Jeremy Coffelt Blinn College

Nancy Ikeda Fullerton College

Vickie Flanders, Baton Rouge Community College Nic LaHue, Metropolitan Community College, Penn Valley Nicole Lloyd, Lansing Community College Jackie Miller, The Ohio State University Anne Marie Mosher, St. Louis Community College, Florissant Valley Reva Narasimhan, Kean University David Ray, University of Tennessee, Martin

Amy Naughten

Kristin Stoley, Blinn College Stephen Toner, Victor Valley College Paul Vroman, St. Louis Community College, Florissant Valley Michelle Whitmer, Lansing Community College

www.connectmath.com

TM

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Better Student Preparedness . . . College Algebra Enhanced Course Coverage Enables Seamless Integration with Textbooks and Syllabi ALEKS College Algebra features hundreds of new course topics to provide comprehensive course coverage, and ALEKS AI-2, the next generation intelligence engine to dramatically improve student learning outcomes. This enhanced ALEKS course product allows for better curriculum coverage and seamless textbook integration to help students succeed in mathematics, while allowing instructors to customize course content to align with their course syllabi. ALEKS is a Web-based program that uses artificial intelligence and adaptive questioning to assess precisely a student’s knowledge in College Algebra and provide personalized instruction on the exact topics the student is most ready to learn. By providing individualized assessment and learning, ALEKS helps students to master course content quickly and easily.

Topics Added For Comprehensive Coverage: ALEKS College Algebra includes hundreds of new topics for comprehensive coverage of course material. To view College Algebra course content in more detail, please visit:

www.aleks.com/highered/math/course_products The ALEKS Pie summarizes a student’s ▶ current knowledge of course material and provides an individualized learning path with topics each student is most ready to learn.

Robust Graphing Features: ALEKS College Algebra provides more graphing coverage and includes a built-in graphing calculator, an adaptive, open-response environment, and realistic answer input tools to ensure student mastery. The ALEKS Graphing Calculator is accessible via the Student Module and can be turned on or off by the instructor.

▶

◀ Realistic Input Tools provide an adaptive, open-response environment that avoids multiple-choice questions and ensures student mastery.

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ALEKS is a registered trademark of ALEKS Corporation.

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. . .T hrough Superior Course Management New Instructor Module Features for College Algebra Help Students Achieve Success While Saving Instructor Time ALEKS includes an Instructor Module with powerful, assignment-driven features and extensive content flexibility to simplify course management so instructors spend less time with administrative tasks and more time directing student learning. The ALEKS Instructor Module also includes two new features that further simplify course management and provide content flexibility: Partial Credit on Assignments and Supplementary Textbook Integration Topic Coverage.

Partial Credit On Assignments:

▶

With the addition of many more multipart questions to ALEKS College Algebra, instructors now have the option to have ALEKS automatically assign partial credit to students’ responses on multipart questions in an ALEKS Homework, Test, or Quiz. Instructors can also manually adjust scores.

◀

Supplementary Textbook Integration Topic Coverage: Instructors have access to ALL course topics in ALEKS College Algebra, and can include supplementary course topics even if they are not specifically tied to an integrated textbook’s table of contents.

To learn more about how other instructors have successfully implemented ALEKS, please visit:

www.aleks.com/highered/math/implementations

“Overall, both students and I have been very pleased with ALEKS. Students like the flexibility it offers them. I like that students are working where they need to be and can spend as much time reviewing as they need. . . . Students have made such comments as ‘I never liked math in high school but this is kind of fun,’ or ‘I never understood this in high school but now I do.’” —Linda Flanery, Instructor, Sisseton Wahpeton College

For more information about ALEKS, please visit: www.aleks.com/highered/math ALEKS is a registered trademark of ALEKS Corporation.

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Increased Student Engagement . . . Through g Meaningful g Applications pp a con nec tion bet wee n the req uires that student s exp erie nce Ma king mat hematics mea ning ful is the result of a pow erf ul on the wor ld they live in. This text act imp its and y, stud y the s atic ing close ties to the mat hem qua lity, and greates t inte res t, hav est high the of s tion lica app vide commit men t to pro larly ma de an eff ort to sup ply ed leve ls of diff icul ty. We par ticu itor mon lly efu car h wit and les, assignm ent s, and exa mp illus trations, incl uded as hom ework lass in-c for d use be to y ntit qua these in suf ficient ir sup ply premat urely. Ma ny and test s, wit hou t exhaus ting the zes quiz of n ctio stru con the in d cur ious, eve n emp loye nces, wit h oth ers com ing from a erie exp rse dive own our of n bor atics in app lications wer e nts of life, and to see the mat hem eve ay ryd eve the on e seiz to ch tools, wit h visiona ry folly that ena bles one tial libr ary of ref ere nce and resear stan sub a by ted por sup e wer se the backgr oun d. The Aut hor s nts, and modern tren ds. —The an eye toward hist ory, cur ren t eve

▶ Chapter Openers highlight Chapter Connections, an interesting

application exercise from the chapter, and provide a list of other real-world connections to give context for students who wonder how math relates to them.

“I think the book has very modern applications and quite a few of them. The calculator instructions are very well done.”

CHAPTER CONNECTIONS

—Nezam Iraniparast, Western Kentucky University

More on Functions CHAPTER OUTLINE

2.5 Piecewise-Defined Functions 245

power in watts and v is the wind velocity in miles per hour. While the formula enables us to predict the power generated for a given wind speed, the graph offers a visual representation of this relationship, where we note a rapid growth in power output as the wind speed increases. This application appears as Exercise 107 in Section 2.2.

2.6 Variation: The Toolbox Functions in Action 259

Check out these other real-world connections:

2.1 Analyzing the Graph of a Function 188

▶ Examples throughout the text feature word problems, providing

students with a starting point for how to solve these types of problems in their exercise sets.

Viewing a function in terms of an equation, a table of values, and the related graph, often brings a clearer understanding of the relationships involved. For example, the power generated by a wind turbine is often modeled 8v3 by the function P 1v2 , where P is the 125

2.2 The Toolbox Functions and Transformations 202 2.3 Absolute Value Functions, Equations, and Inequalities 218

2.4 Basic Rational Functions and Power Functions; More on the Domain 230

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䊳

䊳

“ The students always want to know ‘When am I ever going to have

䊳

Analyzing the Path of a Projectile (Section 2.1, Exercise 57) Altitude of the Jet Stream (Section 2.3, Exercise 61) Amusement Arcades (Section 2.5, Exercise 42) Volume of Phone Calls (Section 2.6, Exercise 55)

to use algebra anyway?’ Now it will not be hard for them to see for themselves some REAL ways. —Sally Haas, Angelina College

187

”

EXAMPLE 2

▶ Application Exercises at the end of each section are the hallmark of

the Coburn series. Never contrived, always creative, and born out of the author’s life and experiences, each application tells a story and appeals to a variety of teaching styles, disciplines, backgrounds, and interests. The authors have ensured that the applications reflect the most common majors of college algebra students.

“ The amount of technology is great, as are the applications. The quality of the applications is better than my current text.” —Daniel Russow, Arizona Western College–Yuma

▶ Math M th iin Action A ti Applets, A l t llocated t d online, li enable bl students t d t tto work k

collaboratively as they manipulate applets that apply mathematical concepts in real-world contexts. xvi

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Identifying Functions Two relations named f and g are given; f is pointwise-defined (stated as a set of ordered pairs), while g is given as a set of plotted points. Determine whether each is a function. f: 13, 02, 11, 42, 12, 52, 14, 22, 13, 22, 13, 62, 10, 12, (4, 5), and (6, 1)

Solution

䊳

The relation f is not a function, since 3 is paired with two different outputs: 13, 02 and 13, 22 . The relation g shown in the figure is a function. Each input corresponds to exactly one output, otherwise one point would be directly above the other and have the same first coordinate.

g

5

y (0, 5)

(4, 2) (3, 1)

(2, 1) 5

5

x

(4, 1) (1, 3) 5

Now try Exercises 11 through 18 䊳

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Through Timely Examples mp les that set the sta ge to overstate the imp ortance of exa t icul diff be ld wou it , tics ma the was too In ma falt ered due to an exa mp le that e hav s nce erie exp l iona cat edu for lear ning. No t a few a car efu l and a dist rac ting result. In this ser ies, had or ce, uen seq of out fit, r on diff icul t, a poo ely and clea r, wit h a direct foc us tim e wer t tha les mp exa ct sele to deliber ate eff ort was ma de link pre vious e, they wer e fur the r designed to sibl pos e her ryw Eve d. han at l the concep t or skil e. As a tra ined educat or gro undwor k for concep ts to com the lay to and s, idea t ren cur to seq uence of concep ts ore it’s ever asked, and a tim ely bef en oft is n stio que a wer ans knows, the bes t tim e to new idea simply the way in this regard, ma king each long a go can les mp exa ed uct car efu lly constr ity of a studen t grows successfu l, the mathematica l matur en Wh . step d ate icip ant n eve cal, nex t logi . —The Aut hor s it was just sup posed to be that way in unn otic ed incr ements, as tho ugh

“ The authors have succeeded with numerous

calculator examples with easy-to-use instructions to follow along. I truly enjoy seeing plenty of calculator examples throughout the text!!

▶ Side by side graphical and algebraic solutions illustrate the

difference between problem-solving methods, emphasize the connections between algebraic and graphical information, and enable students to understand why one method might be preferable to another for any given problem.

”

—David Bosworth, Huchinson Community College

▶ Titles have been added to examples to

highlight relevant learning objectives and reinforce the importance of speaking mathematically using vocabulary.

EXAMPLE 8

䊳

Analytical Solution

䊳

Solve the inequality x2 6x 9.

▶ Annotations located to the right of the

solution sequence help the student recognize which property or procedure is being applied. ▶ “Now Try” boxes immediately following

examples guide students to specific matched exercises at the end of the section, helping them identify exactly which homework problems coincide with each discussed concept.

Solving a Quadratic Inequality

WORTHY OF NOTE

Begin by writing the inequality in standard form: x2 6x 9 0. Note this is equivalent to g1x2 0 for g1x2 x2 6x 9. Since a 6 0, the graph of g will open downward. The factored form is g1x2 1x 32 2, showing 3 is a zero and a repeated root. Using the x-axis, we plot the point (3, 0) and visualize a parabola opening downward through this point. Figure 3.29 shows the graph is below the x-axis (outputs are negative) for all values of x except x 3. But since this is a less than or equal to inequality, the solution is x 僆⺢.

Since x 3 was a zero of multiplicity 2, the graph “bounced off” the x-axis at this point, with no change of sign for g. The graph is entirely below the x-axis, except at the vertex (3, 0).

Graphical Solution

Figure 3.29 1

0

1

2

3

4

5

6

7

x

a0 䊳

The complete graph of g shown in Figure 3.30 confirms the analytical solution (using the zeroes method). For the intervals of the domain shown in red: 1q, 32 ´ 13, q2 , the graph of g is below the x-axis 3 g1x2 6 04 . The point (3, 0) is on the x-axis 3 g132 04 . As with the analytical solution, the solution to this “less than or equal to” inequality is all real numbers. A calculator check of the original inequality is shown in Figure 3.31. Figure 3.30

Figure 3.31

y

10

2

2

6

x 2

8

g(x)

“The modeling and regression

examples in this text are excellent, and the instructions for using the graphing calculator to investigate these types of problems are great.

3 8

Now try Exercises 121 thro through 132

”

䊳

—Allison Sutton, Austin Community College

“ The examples support the exercises which is very important. The chapter is very well written and is easy to read and understand.

”

—Joseph Lloyd Harris, Gulf Coast Community College

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Solid Skill Development . . . Through g Exercises s. The exe rcise sup por t of eac h sec tion’s ma in idea in es rcis exe of lth wea a d ude We hav e incl t for wea ker stu den ts, e, in an eff ort to pro vide sup por car at gre h wit ed uct str con e of wer set s n fur the r. The qua ntit y and qua lity eve ch rea to ts den stu ed anc adv whi le cha llenging mo re ies to guide eff ort s, and num ero us opp ort unit r’s che tea a for t por sup ong str exe rcis es off ers ing idea s. to illus tra te imp ortant pro blem solv and ions ulat calc t icul diff h oug stu den ts thr —The Aut hor s

Mid-Chapter Checks

MID-CHAPTER CHECK

Mid-Chapter Checks provide students with a good stopping place to assess their knowledge before moving on to the second half of the chapter.

1. Determine whether the following function is even, 冟x冟 odd, or neither. f 1x2 x2 4x

4. Write the equation of the function that has the same graph of f 1x2 2x, shifted left 4 units and up 2 units. 5. For the graph given, (a) identify the function family, (b) describe or identify the end-behavior, inflection point, and x- and y-intercepts, (c) determine the domain and range and

2. Use a graphing calculator to find the maximum and minimum values of f 1x2 1.91x4 2.3x3 2.2x 5.12 . Round to the nearest hundredth. 3 Use interval notation to identify the interval(s)

End-of-Section Exercise Sets

Exercise 5 y 5

f(x)

5

5 x

2.2 EXERCISES

▶ Concepts and Vocabulary exercises to help students

䊳

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.

recall and retain important terms.

1. After a vertical , points on the graph are farther from the x-axis. After a vertical , points on the graph are closer to the x-axis.

2. Transformations that change only the location of a graph and not its shape or form, include and .

3. The vertex of h1x2 31x 52 9 is at and the graph opens .

4. The inflection point of f 1x2 21x 42 3 11 is at and the end-behavior is , .

2

5 Gi

▶ Developing Your Skills exercises to provide

䊳

practice of relevant concepts just learned with increasing levels of difficulty.

th

h f

lf

ti

f ( ) di

/

/E l i

h th

7. f 1x2 x2 4x 5

2

3i

15. r 1x2 3 14 x 3 16. f 1x2 2 1x 1 4 5

y

5

5

r(x)

8. g1x2 x2 2x

y

5

5 x

5

5

5 x

5

f(x)

5 x

5

5

y

y

5

5 x

17. g1x2 2 14 x

formulas and applications bring forward some interesting ideas and problems that are more in depth. These would help hold the students’ interest in the topic.

hift f f 1 2

DEVELOPING YOUR SKILLS

By carefully inspecting each graph given, (a) identify the function family; (b) describe or identify the end-behavior, vertex, intervals where the function is increasing or decreasing, maximum or minimum value(s) and x- and y-intercepts; and (c) determine the domain and range. Assume required features have integer values.

“ The sections in the assignments headed working with

6 Di

18. h1x2 2 1x 1 4

y

5

y

g(x) h(x)

”

5

5

—Sherri Rankin, Huchinson Community College

䊳

WORKING WITH FORMULAS

61. Discriminant of the reduced cubic x3 ⴙ px ⴙ q ⴝ 0: D ⴝ ⴚ14p3 ⴙ 27q2 2 The discriminant of a cubic equation is less well known than that of the quadratic, but serves the same purpose. The discriminant of the reduced cubic is given by the formula shown, where p is the linear coefficient and q is the constant term. If D 7 0, there will be three real and distinct roots. If D 0, there are still three real roots, but one is a repeated root (multiplicity two). If D 6 0, there are one real and two complex roots. Suppose we wish to study the family of cubic equations where q p 1. a. Verify the resulting discriminant is D 14p3 27p2 54p 272. b. Determine the values of p and q for which this family of equations has a repeated real root. In other words, solve the equation 14p3 27p2 54p 272 0 using the rational zeroes theorem and synthetic division to write D in completely factored form.

▶ Working with Formulas exercises to demonstrate

contextual applications of well-known formulas. ▶ Extending the Concept exercises that require

䊳

communication of topics, synthesis of related concepts, and the use of higher-order thinking skills.

EXTENDING THE CONCEPT

59. Use the general solutions from the quadratic formula to show that the average value of the x-intercepts is b . Explain/Discuss why the result is valid even if 2a the roots are complex. b 2b2 4ac

▶ Maintaining Your Skills exercises that address

䊳

skills from previous sections to help students retain previously learning knowledge.

b 2b2 4ac

62. Referring to Exercise 39, discuss the nature (real or complex, rational or irrational) and number of zeroes (0, 1, or 2) given by the vertex/intercept formula if (a) a and k have like signs, (b) a and k k have unlike signs, (c) k is zero, (d) the ratio a is positive and a perfect square and (e) the

MAINTAINING YOUR SKILLS

37. (1.3) Is the graph shown here, the graph of a function? Discuss why or why not.

38. (R.2/R.3) Determine the area of the figure shown 1A LW, A ␲r2 2.

18 cm 24 cm

39. (1.5) Solve for r: A P Prt

“The exercise sets are plentiful. I like having many to

choose from when assigning homework. When there are only one or two exercises of a particular type, it’s hard for the students to get the practice they need.

”

—Sarah Jackson, Pratt Community College

xviii

40

S l

f

(if if

ibl )

“ There seems to be a good selection of easy, moderate, and difficult problems in the exercises.”

—Ed Gallo, Sinclair Community College

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End-of-Chapter Review Material Exercises located at the end of the chapter provide students with the tools they need to prepare for a quiz or test. Each chapter features the following: ▶

Making Connections matching exercises are groups of problems where students must identify graphs based on an equation or description. This feature helps students make the connection between graphical and algebraic information while it enhances students’ ability to read and interpret graphical data.

“ Not only was the algebra rigorously treated, but it

was reinforced throughout the chapters with the MidChapter Check and the Chapter Review and Tests.

”

—Mark Crawford, Waubonsee Community College

MAKING CONNECTIONS Making M ki Connections: C ti G Graphically, hi ll Symbolically, S b li ll Numerically, N i ll and d Verbally V b ll Eight graphs (a) through (h) are given. Match the characteristics shown in 1 through 16 to one of the eight graphs. y

(a)

5 x

5

▶

Chapter Summary and Concept Reviews that present key concepts with corresponding exercises by section in a format easily y used by y students.

5 x

“ The problem sets are really magnificent. I deeply enjoy

”

▶

Practice Tests that give students the opportunity to check their knowledge and prepare for classroom quizzes, tests, and other assessments.

▶

Cumulative Reviews that are presented at the end of each chapter help students retain previously learned skills and concepts by revisiting important ideas from earlier chapters (starting with Chapter 2).

▶

Graphing Calculator icons appear next to exercises where important concepts can be supported by the use of graphing technology.

5 x

5

y

(h)

5

5 x

5

5

5

5 x

5

5

y

(g)

5

5

5 x

5

5

5

y

(f)

5

y

(d)

5

5

y

(e)

5 x

5

y

(c)

5

5

5

and appreciate the many problems that incorporate telescopes, astronomy, reflector design, nuclear cooling tower profiles, charged particle trajectories, and other such examples from science, technology, and engineering. —Light Bryant, Arizona Western College

y

(b)

5

5

5 x

5

5

1 1. ____ y x 1 3

9. ____ f 132 4, f 112 0

2. ____ y x 1

10. ____ f 142 3, f 142 3

SUMMARY AND CONCEPT REVIEW SECTION 1.1 SE

Rectangular Coordinates; Graphing Circles and Other Relations

KEY CONCEPTS KE • A relation is a collection of ordered pairs (x, y) and can be stated as a set or in equation form. • As a set of ordered pairs, we say the relation is pointwise-defined. The domain of the relation is the set of all first coordinates, and the range is the set of all corresponding second coordinates. • A relation can be expressed in mapping notation x S y, indicating an element from the domain is mapped to (corresponds to or is associated with) an element from the range. • The graph of a relation in equation form is the set of all ordered pairs (x, y) that satisfy the equation. We plot a sufficient number of points and connect them with a straight line or smooth curve, depending on the pattern formed. • The x- and y-variables of linear equations and their graphs have implied exponents of 1. • With a relation entered on the Y= screen, a graphing calculator can provide a table of ordered pairs and the related graph. x1 x2 y1 y2 , b. • The midpoint of a line segment with endpoints (x1, y1) and (x2, y2) is a 2 2

• The distance between the points (x1, y1) and (x2, y2) is d 21x2 x1 2 2 1y2 y1 2 2. • The equation of a circle centered at (h, k) with radius r is 1x h2 2 1y k2 2 r2. EXERCISES 1. Represent the relation in mapping notation, then state the domain and range. 517, 32, 14, 22, 15, 12, 17, 02, 13, 22, 10, 826 2

“The authors give very good uses of the calculator in every

section. I have been using TI calculators for 15 years and I learned a few new tricks while reading this book.

”

—George Hurlburt, Corning Community College

Homework Selection Guide A list of suggested homework exercises has been provided for each section of the text (Annotated Instructor’s Edition only). This feature may prove especially useful for departments that encourage consistency among many sections, or those having a large adjunct population. The feature was also designed as a convenience to instructors, enabling them to develop an inventory of exercises that is more in tune with the course as they like to teach it. The guide provides prescreened and preselected p assignments at four different levels: Core, Standard, Extended, and In Depth. 8 10 • Core: These assignments go right to the heart of the material, HOMEWORK SELECTION GUIDE offering a minimal selection of exercises that cover the primary concepts and solution strategies of the section, along with a small selection of the best applications. • Standard: The assignments at this level include the Core exercises, while providing for additional practice without excessive drill. A wider assortment of the possible variations on a theme are included, as well as a greater variety of applications. • Extended: Assignments from the Extended category expand on the Standard exercises to include more applications, as well as some conceptual or theory-based questions. Exercises may include selected items from the Concepts and Vocabulary, Working with Formulas, and the Extending the Concept categories of the exercise sets. • In Depth: The In Depth assignments represent a more comprehensive look at the material from each section, while attempting to keep the assignment manageable for students. These include a selection of the most popular and highest-quality exercises from each category of the exercise set, with an additional emphasis on Maintaining Your Skills. Additional answers can be found in the Instructor Answer Appendix.

Core: 7–91 every other odd, 95–101 odd (26 Exercises) Standard: 1–4, 7–83 every other odd, 85–92 all, 95–101 odd (36 Exercises)

13 3

19 2

Extended: 1–4, 7–31 every other odd, 35–38 all, 39–79 every other odd, 85–92 all, 95–101 odd, 106, 109 (39 Exercises) In Depth: 1–4, 7–31 every other odd, 35–38 all, 39–83 every other odd, 85–92 all, 95, 96, 98, 99, 100, 101, 105, 106, 109 (44 Exercises)

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Strong Mathmatical Connections . . . Through a Conversational Writing Style featur es of a mathematics text, While examples and applications are arguably the most promin ent togeth er. It may be true that it’s the readability and writing style of the author s that bind them when looking for an example some studen ts don’t read the text, and that others open the text only for those studen ts that do (read similar to the exercise they’re working on. But when they do and g concep ts in a form and the text), it’s important they have a text that “speak s to them,” relatin style of this text will help draw at a level they understand and can relate to. We feel the writing and bringing them back a second studen ts in and keep their interest, becoming a positiv e experience begin to see the true value and third time, until it becomes habitual. At this point studen ts might g with any other form of of their text, as it becomes a resour ce for learning on equal footin —The Authors direction. supplementa l instruction. This text represents our best effort s in this

Conversational Writing Style John and J.D.’s experience in the classroom and their strong connections to how students comprehend the material are evident in their writing style. They use a conversational and supportive writing style, providing the students with a tool they can depend on when the teacher is not available, when they miss a day of class, or simply when working on their own. The effort they have put into the writing is representative of John Coburn’s unofficial mantra: “If you want more students to reach the top, you gotta put a few more rungs on the ladder.”

“Coburn strikes a good balance between

providing all of the important information necessary for a certain topic without going too deep.

”

—Barry Monk, Macon State College

“I think the authors have done an excellent job

of interweaving the formal explanations with the ‘plain talk’ descriptions, illustrating with meaningful examples and applications.

”

—Ken Gamber, Hutchinson Community College

Through Student Involvement How do you design a student-friendly textbook? We decided to get students involved by hosting two separate focus groups. During these sessions we asked students to advise us on how they use their books, what pedagogical elements are useful, which elements are distracting and not useful, as well as general feedback on page layout. During this process there were times when we thought, “Now why hasn’t anyone ever thought of that before?” Clearly these student focus groups were invaluable. Taking direct student feedback and incorporating what is feasible and doesn’t detract from instructor use of the text is the best way to design a truly student-friendly text. The next two pages will highlight what we learned from students so you can see for yourself how their feedback played an important role in the development of the Coburn/Herdlick series.

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5.2

Students said that Learning Objectives should clearly define the goals of each section.

Exponential Functions

LEARNING OBJECTIVES In Section 5.2 you will see how we can:

A. Evaluate an exponential

Demographics is the statistical study of human populations. In this section, we introduce the family of exponential functions, which are widely used to model population growth or decline with additional applications in science, engineering, and many other fields. As with other functions, we begin with a study of the graph and its characteristics.

function

A. Evaluating Exponential Functions

B. Graph general exponential functions

C. Graph base-e exponential functions D. Solve exponential equations and applications

In the boomtowns of the old west, it was not uncommon for a town to double in size every year (at least for a time) as the lure of gold drew more and more people westward. When this type of growth is modeled using mathematics, exponents play a lead role. Suppose the town of Goldsboro h d 1000 id t h ld fi t di d

Examples are “boxed” so students can clearly see where they begin and end. Examples are called out in the margins so they are easy for students to spot.

EXAMPLE 4

䊳

Graphing Exponential xponential Functions Using Transformations i transformations f i Graph F1x2 2xx11 2 using off the basic function f 1x2 2x (not by simply plotting points). Clearly state what transformations are applied.

Solution

䊳

Students asked for Check Points throughout each section to alert them when a specific learning objective has been covered and to reinforce the use of correct mathematical terms.

(1, 3) y2

4

4

x

To help sketch a more accurate graph, the point (3, 6) can be used: F132 6. Now try Exercises 15 through 30

䊳

Students told us they liked when the examples were linked to the exercises.

Described by students as one of the most useful features in a math text, Caution Boxes signal a student to stop and take note in order to avoid mistakes in problem solving.

CAUTION

䊳

S d Students told ld us that h the h color l red d should h ld only be used for things that are really important. Also, anything significant should be included in the body of the text; marginal readings imply optional.

Because students spend a lot of time in the exercise section of a text, they said that a white background is hard on their eyes . . . so we used a soft, off-white color for the background.

(0, 2.5)

(3, 6)

F102 21021 2 21 2 1 2 2 2.5 B. You’ve just seen how we can graph general exponential functions

Students said having a lot of icons was confusing. The graphing calculator is the only icon used in the exercise sets; no unnecessary icons are used.

The graph of F is that of the basic function f 1x2 2x with a horizontal shift 1 unit right and a vertical shift 2 units up. With this in mind the horizontal asymptote also shifts from y 0 to y 2 and (0, 1) shifts to (1, 3). The y-intercept of F is at (0, 2.5):

y F(x) = 2x is shifted 1 unit right 2 units up

For equations like those in Example 1, be careful not to treat the absolute value bars as simple grouping symbols. The equation 51x 72 2 13 has only the solution x 10, and “misses” the second solution since it yields x 7 3 in simplified form. The equation 5冟x 7冟 2 13 simplifies to 冟x 7冟 3 and there are actually two solutions. Also note that 5冟x 7冟冟5x 35冟!

Students told us that directions should be in bold so they are easily distinguishable from the problems.

䊳

APPLICATIONS

Use the information given to build a linear equation model, then use the equation to respond. For exercises 71 to 74, develop both an algebraic and a graphical solution.

71. Business depreciation: A business purchases a copier for $8500 and anticipates it will depreciate in value $1250 per year. a. What is the copier’s value after 4 yr of use? b. How many years will it take for this copier’s value to decrease to $2250? 72. Baseball card value: After purchasing an autographed baseball card for $85, its value increases by $1.50 per year. a. What is the card’s value 7 yr after purchase? b. How many years will it take for this card’s value to reach $100?

74. Gas mileage: When empty, a large dump-truck gets about 15 mi per gallon. It is estimated that for each 3 tons of cargo it hauls, gas mileage decreases by 34 mi per gallon. a. If 10 tons of cargo is being carried, what is the truck’s mileage? b. If the truck’s mileage is down to 10 mi per gallon, how much weight is it carrying? 75. Parallel/nonparallel roads: Aberville is 38 mi north and 12 mi west of Boschertown, with a straight “farm and machinery” road (FM 1960) connecting the two cities. In the next county, Crownsburg is 30 mi north and 9.5 mi west of Dower, and these cities are likewise connected by a straight road (FM 830). If the two roads continued indefinitely in both directions, would they intersect at some point?

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Coburn’s Precalculus Series College Algebra: Graphs & Models, First Edition A Review of Basic Concepts and Skills ◆ Functions and Graphs ◆ Relations; More on Functions ◆ Quadratic Functions and Operations on Functions ◆ Polynomial and Rational Functions ◆ Exponential and Logarithmic Functions ◆ Systems of Equations and Inequalities ◆ Matrices and Matrix Applications ◆ Analytic Geometry and the Conic Sections ◆ Additional Topics in Algebra

Precalculus: Graphs & Models, First Edition Functions and Graphs ◆ Relations; More on Functions ◆ Quadratic Functions and Operations on Functions ◆ Polynomial and Rational Functions ◆ Exponential and Logarithmic Functions ◆ Introduction to Trigonometry ◆ trigonometric Identities, Inverses, and Equations ◆ Applications of Trigonometry ◆ Systems of Equations and Inequalities; Matrices ◆ Analytic Geometry; Polar and parametric Equations ◆ Sequences, Series, Counting, and Probability ◆ Bridges to Calculus—An Introduction to Limits

College Algebra Second Edition Review ◆ Equations and Inequalities ◆ Relations, Functions, and Graphs ◆ Polynomial and Rational Functions ◆ Exponential and Logarithmic Functions ◆ Systems of Equations and Inequalities ◆ Matrices ◆ Geometry and Conic Sections ◆ Additional Topics in Algebra MHID 0-07-351941-3, ISBN 978-0-07-351941-8

College Algebra Essentials Second Edition Review ◆ Equations and Inequalities ◆ Relations, Functions, and Graphs ◆ Polynomial and Rational Functions ◆ Exponential and Logarithmic Functions ◆ Systems of Equations and Inequalities MHID 0-07-351968-5, ISBN 978-0-07-351968-5

Algebra and Trigonometry Second Edition Review ◆ Equations and Inequalities ◆ Relations, Functions, and Graphs ◆ Polynomial and Rational Functions ◆ Exponential and Logarithmic Functions ◆ Trigonometric Functions ◆ Trigonometric Identities, Inverses, and Equations ◆ Applications of Trigonometry ◆ Systems of Equations and Inequalities ◆ Matrices ◆ Geometry and Conic Sections ◆ Additional Topics in Algebra MHID 0-07-351952-9, ISBN 978-0-07-351952-4

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Precalculus Second Edition Equations and Inequalities ◆ Relations, Functions, and Graphs ◆ Polynomial and Rational Functions ◆ Exponential and Logarithmic Functions ◆ Trigonometric Functions ◆ Trigonometric Identities, Inverses, and Equations ◆ Applications of Trigonometry ◆ Systems of Equations and Inequalities ◆ Matrices ◆ Geometry and Conic Sections ◆ Additional Topics in Algebra ◆ Limits MHID 0-07-351942-1, ISBN 978-0-07-351942-5

Trigonometry Second Edition Introduction to Trigonometry ◆ Right Triangles and Static Trigonometry ◆ Radian Measure and Dynamic Trigonometry ◆ Trigonometric Graphs and Models ◆ Trigonometric Identities ◆ Inverse Functions and Trigonometric Equations ◆ Applications of Trigonometry ◆ Trigonometric Connections to Algebra MHID 0-07-351948-0, ISBN 978-0-07-351948-7

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/volumes/204/MHDQ234/cob19545_disk1of1/0073519545/cob19545_pagefiles

Making Connections . . . Through 360º Development McGraw-Hill’s 360° Development Process is an ongoing, never-ending, market-oriented approach to building accurate and innovative print and digital products. It is dedicated to continual large-scale and incremental improvement driven by multiple customer feedback loops and checkpoints. This process is initiated during the early planning stages of our new products, intensifies during the development and production stages, and then begins again on publication, in anticipation of the next edition. A key principle in the development of any mathematics text is its ability to adapt to teaching specifications in a

universal way. The only way to do so is by contacting those universal voices—and learning from their suggestions. We are confident that our book has the most current content the industry has to offer, thus pushing our desire for accuracy to the highest standard possible. In order to accomplish this, we have moved through an arduous road to production. Extensive and open-minded advice is critical in the production of a superior text. By investing in this extensive endeavor, McGraw-Hill delivers to you a product suite that has been created, refined, tested, and validated to be a successful tool in your course.

Student Focus Groups Two student focus groups were held at Illinois State University and Southeastern Louisiana University to engage students in the development process and provide feedback as to how the design of a textbook impacts homework and study habits in the College Algebra, Precalculus, and Trigonometry course areas. Francisco Arceo, Illinois State University Candace Banos, Southeastern Louisiana University Dave Cepko, Illinois State University Andrea Connell, Illinois State University Nicholas Curtis, Southeastern Louisiana University M. D. “Boots” Feltenberger, Southeastern Louisiana University Regina Foreman, Southeastern Louisiana University Ashley Lae, Southeastern Louisiana University Brian Lau, Illinois State University Daniel Nathan Mielneczek, Illinois State University Mingaile Orakauskaite, Illinois State University Todd Michael Rapnikas, Illinois State University Bethany Rollet, Illinois State University Teddy Schrishuhn, Illinois State University

Josh Schultz, Illinois State University Jessica Smith, Southeastern Louisiana University Andy Thurman, Illinois State University Ashley Youngblood, Southeastern Louisiana University

Digital Contributors Jeremy Coffelt, Blinn College Vanessa Coffelt, Blinn College Vickie Flanders, Baton Rouge Community College Anne Marie Mosher, Saint Louis Community CollegeFlorissant Valley

Kristen Stoley, Blinn College David Ray, University of Tennessee-Martin Stephen Toner, Victor Valley Community College Paul Vroman, Saint Louis Community College-Florissant Valley

Special Thanks Sherry Meier, Illinois State University Rebecca Muller, Southeastern Louisiana University Anne Schmidt, Illinois State University

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Making Connections . . . Developmental Editing The manuscript has been impacted by numerous developmental reviewers who edited for clarity and consistency. Efforts resulted in cutting length from the manuscript, while retaining a conversational and casual narrative style. Editorial work also ensured the positive visual impact of art and photo placement. Chapter Reviews and Manuscript Reviews Teachers and academics from across the country reviewed the current edition text, the proposed table of contents, and first-draft manuscript to give feedback on reworked narrative, design changes, pedagogical enhancements, and organizational changes. This feedback was summarized by the book team and used to guide the direction of the second-draft manuscript. Betty Anderson, Howard Community College David Bosworth, Hutchinson Community College Daniel Brock, Arkansas State University-Beebe Barry Brunson, Western Kentucky University Light Bryant, Arizona Western College Brenda Burns-Williams, North Carolina State University-Raleigh Charles Cooper, Hutchinson Community College Mark Crawford, Waubonsee Community College Joseph Demaio, Kennesaw State University Alvio Dominguez, Miami-Dade College-Wolfson Dale Duke, Oklahoma City Community College Frank Edwards, Southeastern Louisiana University Caleb Emmons, Pacific University Mike Everett, Santa Ana College Maggie Flint, Northeast State Technical Community College Ed Gallo, Sinclair Community College Ken Gamber, Hutchinson Community College David Gurney, Southeastern Louisiana University

Sally Haas, Angelina College Ben Hill, Lane Community College Jody Hinson, Cape Fear Community College Lynda Hollingsworth, Northwest Missouri State University George Hurlburt, Corning Community College Sarah Jackson, Pratt Community College Laud Kwaku, Owens Community College Kathryn Lavelle, Westchester Community College Joseph Lloyd Harris, Gulf Coast Community College Austin Lovenstein, Pulaski Technical College Rodolfo Maglio, Northeastern Illinois University Barry Monk, Macon State College Camille Moreno, Cosumnes River College Anne Marie Mosher, Saint Louis Community College-Florissant Valley Lilia Orlova, Nassau Community College Susan Pfeifer, Butler Community College Sherri Rankin, Hutchinson Community College Daniel Russow, Arizona Western College-Yuma Rose Shirey, College of the Mainland Joy Shurley, Abraham Baldwin Agricultural College Sean Simpson, Westchester Community College Pam Stogsdill, Bossier Parish Community College Allison Sutton, Austin Community College Linda Tremer, Three Rivers Community Collge Dahlia Vu, Santa Ana College Jackie Wing, Angelina College

Acknowledgments We first want to express a deep appreciation for the guidance, comments, and suggestions offered by all reviewers of the manuscript. We have once again found their collegial exchange of ideas and experience very refreshing and instructive, and always helping to create a better learning tool for our students. Vicki Krug has continued to display an uncanny ability to bring innumerable pieces from all directions into a unified whole, in addition to providing spiritual support during some extremely trying times; Patricia Steele’s skill as a copy editor is as sharp as ever, and her attention to detail continues to pay great dividends; which helps pay the debt we owe Katie White, Michelle Flomenhoft, Christina Lane, and Eve Lipton for their useful suggestions, infinite patience, tireless efforts, and art-counting eyes, which helped in bringing the manuscript to completion. We must also thank John Osgood for his ready wit, creative energies, and ability to step into the flow without missing a beat; Laurie Janssen and our magnificent

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design team, and Dawn Bercier whose influence on this project remains strong although she has moved on, as it was her indefatigable spirit that kept the ship on course through trial and tempest, and her ski-jumper’s vision that brought J.D. on board. In truth, our hats are off to all the fine people at McGraw-Hill for their continuing support and belief in this series. A final word of thanks must go to Rick Armstrong, whose depth of knowledge, experience, and mathematical connections seems endless; Anne Marie Mosher for her contributions to various features of the text, Mitch Levy for his consultation on the exercise sets, Stephen Toner for his work on the videos, Jon Booze and his team for their work on the test bank, Cindy Trimble for her invaluable ability to catch what everyone else misses; and to Rick Pescarino, Kelly Ballard, John Elliot, Jim Frost, Barb Kurt, Lillian Seese, Nate Wilson, and all of our colleagues at St. Louis Community College, whose friendship, encouragement, and love of mathematics makes going to work each day a joy.

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Making Connections . . . Through Supplements *All online supplements are available through the book’s website: www.mhhe.com/coburn.

Instructor Supplements • Computerized Test Bank Online: Utilizing Brownstone Diploma® algorithm-based testing software enables users to create customized exams quickly. • Instructor’s Solutions Manual: Provides comprehensive, worked-out solutions to all exercises in the text. • Annotated Instructor’s Edition: Contains all answers to exercises in the text, which are printed in a second color, adjacent to corresponding exercises, for ease of use by the instructor.

Student Supplements • Student Solutions Manual provides comprehensive, worked-out solutions to all of the odd-numbered exercises. • Graphing Calculator Manual includes detailed instructions for using different calculator models to solve problems throughout the text. Written by the authors to accompany their text, it is designed to match and supplement the text. • Videos • Interactive video lectures are provided for each section in the text, which explain to the students how to do key problem types, as well as highlighting common mistakes to avoid. • Exercise videos provide step-by-step instruction for the key exercises which students will most wish to see worked out. • Graphing calculator videos help students master the most essential calculator skills used in the college algebra course. • The videos are closed-captioned for the hearing impaired, subtitled in Spanish, and meet the Americans with Disabilities Act Standards for Accessible Design.

Connect Math™ Hosted by ALEKS® www.connectmath.com Connect Math Hosted by ALEKS is an exciting, new assessment and assignment platform combining the strengths of McGraw-Hill Higher Education and ALEKS Corporation. Connect Math Hosted by ALEKS is the first platform on the market to combine an artificial-intelligent, diagnostic

assessment with an intuitive ehomework platform designed to meet your needs. Connect Hosted by ALEKS is the culmination of a one-of-a-kind market development process involving math full-time faculty members and adjuncts at every step of the process. This process enables us to provide you with an end product that better meets your needs. Connect Math Hosted by ALEKS is built by mathematicians educators for mathematicians educators!

www.aleks.com ALEKS (Assessment and LEarning in Knowledge Spaces) is a dynamic online learning system for mathematics education, available over the Web 24/7. ALEKS assesses students, accurately determines their knowledge, and then guides them to the material that they are most ready to learn. With a variety of reports, Textbook Integration Plus, quizzes, and homework assignment capabilities, ALEKS offers flexibility and ease of use for instructors. • ALEKS uses artificial intelligence to determine exactly what each student knows and is ready to learn. ALEKS remediates student gaps and provides highly efficient learning and improved learning outcomes • ALEKS is a comprehensive curriculum that aligns with syllabi or specified textbooks. Used in conjunction with McGraw-Hill texts, students also receive links to textspecific videos, multimedia tutorials, and textbook pages. • ALEKS offers a dynamic classroom management system that enables instructors to monitor and direct student progress toward mastery of course objectives.

ALEKS Prep/Remediation: • Helps instructors meet the challenge of remediating underprepared or improperly placed students. • Assesses students on their prerequisite knowledge needed for the course they are entering (i.e., Calculus students are tested on Precalculus knowledge) and prescribes a unique and efficient learning path specifically to address their strengths and weaknesses. • Students can address prerequisite knowledge gaps outside of class freeing the instructor to use class time pursuing course outcomes.

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Making Connections . . . McGraw-Hill Higher Education and Blackboard® have teamed up. Blackboard, the Web-based course-management system, has partnered with McGraw-Hill to better allow students and faculty to use online materials and activities to complement face-to-face teaching. Blackboard features exciting social learning and teaching tools that foster more logical, visually impactful and active learning opportunities for students. You’ll transform your closed-door classrooms into communities where students remain connected to their educational experience 24 hours a day. This partnership allows you and your students access to McGraw-Hill’s Connect™ and Create™ right from within your Blackboard course—all with one single sign-on. Not only do you get single sign-on with Connect and Create, you also get deep integration of McGraw-Hill content and content engines right in Blackboard. Whether you’re choosing a book for your course or building Connect assignments, all the tools you need are right where you want them—inside of Blackboard. Gradebooks are now seamless. When a student completes an integrated Connect assignment, the grade for that assignment automatically (and instantly) feeds your Blackboard grade center. McGraw-Hill and Blackboard can now offer you easy access to industry leading technology and content, whether your campus hosts it, or we do. Be sure to ask your local McGraw-Hill representative for details.

TEGRITY—tegritycampus.mhhe.com McGraw-Hill Tegrity Campus™ is a service that makes class time available all the time by automatically capturing every lecture in a searchable format for students to review when they study and complete assignments. With a simple one-click start and stop process, you capture all computer screens and corresponding audio. Students replay any part of any class with easy-to-use browser-based viewing on a PC or Mac. Educators know that the more students can see, hear, and experience class resources, the better they learn. With Tegrity, students quickly recall key moments by using Tegrity’s unique search feature. This search helps students efficiently find what they need, when they need it across an entire semester of class recordings. Help turn all your students’ study time into learning moments immediately supported by your lecture. To learn more about Tegrity watch a 2-minute Flash demo at tegritycampus.mhhe.com.

Electronic Books: If you or your students are ready for an alternative version of the traditional textbook, McGraw-Hill eBooks offer a cheaper and eco-friendly alternative to traditional textbooks. By purchasing eBooks from McGraw-Hill, students can save as much as 50% on selected titles delivered on the most advanced eBook platform available. Contact your McGraw-Hill sales representative to discuss eBook packaging options.

Create: Craft your teaching resources to match the way you teach! With McGraw-Hill Create, www.mcgrawhillcreate.com, you can easily rearrange chapters, combine material from other content sources, and quickly upload content you have written like your course syllabus or teaching notes. Find the content you need in Create by searching through thousands of leading McGraw-Hill textbooks. Arrange your book to fit your teaching style. Create even allows you to personalize your book’s appearance by selecting the cover and adding your name, school, and course information. Order a Create book and you’ll receive a complimentary print review copy in 3–5 business days or a complimentary electronic review copy (eComp) via email in minutes. Go to www.mcgrawhillcreate.com today and register to experience how McGraw-Hill Create empowers you to teach your students your way. xxvi

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Contents Preface vi Index of Applications

CHAPTER

R

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A Review of Basic Concepts and Skills 1 R.1 R.2 R.3 R.4

Algebraic Expressions and the Properties of Real Numbers 2 Exponents, Scientific Notation, and a Review of Polynomials 11 Solving Linear Equations and Inequalities 25 Factoring Polynomials and Solving Polynomial Equations by Factoring 39 R.5 Rational Expressions and Equations 53 R.6 Radicals, Rational Exponents, and Radical Equations 65 Overview of Chapter R: Prerequisite Definitions, Properties, Formulas, and Relationships 81

CHAPTER

1

Relations, Functions, and Graphs 85 1.1 Rectangular Coordinates; Graphing Circles and Other Relations 86 1.2 Linear Equations and Rates of Change 103 1.3 Functions, Function Notation, and the Graph of a Function 117 Mid-Chapter Check 132 Reinforcing Basic Concepts: Finding the Domain and Range of a Relation from Its Graph 132

1.4 Linear Functions, Special Forms, and More on Rates of Change 134 1.5 Solving Equations and Inequalities Graphically; Formulas and Problem Solving 148 1.6 Linear Function Models and Real Data 163 Making Connections 177 Summary and Concept Review 178 Practice Test 183 Strengthening Core Skills: The Various Forms of a Linear Equation 184 Calculator Exploration and Discovery: Evaluating Expressions and Looking for Patterns 185

CHAPTER

2

More on Functions 187 2.1 Analyzing the Graph of a Function 188 2.2 The Toolbox Functions and Transformations 202 2.3 Absolute Value Functions, Equations, and Inequalities 218 Mid-Chapter Check 228 Reinforcing Basic Concepts: Using Distance to Understand Absolute Value Equations and Inequalities 229

2.4 Basic Rational Functions and Power Functions; More on the Domain

230

2.5 Piecewise-Defined Functions 245

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2.6 Variation: The Toolbox Functions in Action 259 Making Connections 270 Summary and Concept Review 271 Practice Test 275 Calculator Exploration and Discovery: Studying Joint Variations 277 Strengthening Core Skills: Variation and Power Functions: y ⴝ kxp 278 Cumulative Review: Chapters R–2 279

CHAPTER

3

Quadratic Functions and Operations on Functions

281

3.1 Complex Numbers 282 3.2 Solving Quadratic Equations and Inequalities 292 3.3 Quadratic Functions and Applications 313 Mid-Chapter Check 327 Reinforcing Basic Concepts: An Alternative Method for Checking Solutions to Quadratic Equations 327

3.4 Quadratic Models; More on Rates of Change 328 3.5 The Algebra of Functions 340 3.6 The Composition of Functions and the Difference Quotient 352 Making Connections 370 Summary and Concept Review 370 Practice Test 375 Calculator Exploration and Discovery: Residuals, Correlation Coefficients, and Goodness of Fit 376 Strengthening Core Skills: Base Functions and Quadratic Graphs 378 Cumulative Review: Chapters R–3 379

CHAPTER

4

Polynomial and Rational Functions 381 4.1 Synthetic Division; the Remainder and Factor Theorems 382 4.2 The Zeroes of Polynomial Functions 394 4.3 Graphing Polynomial Functions 411 Mid-Chapter Check 428 Reinforcing Basic Concepts: Approximating Real Zeroes

429

4.4 Graphing Rational Functions 430 4.5 Additional Insights into Rational Functions 445 4.6 Polynomial and Rational Inequalities 459 Making Connections 470 Summary and Concept Review 470 Practice Test 474 Calculator Exploration and Discovery: Complex Zeroes, Repeated Zeroes, and Inequalities 475 Strengthening Core Skills: Solving Inequalities Using the Push Principle 476 Cumulative Review: Chapters R–4 477

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CHAPTER

5


Exponential and Logarithmic Functions 479 5.1 5.2 5.3 5.4

One-to-One and Inverse Functions 480 Exponential Functions 492 Logarithms and Logarithmic Functions 503 Properties of Logarithms 516 Mid-Chapter Check 526 Reinforcing Basic Concepts: Understanding Properties of Logarithms 527 5.5 Solving Exponential and Logarithmic Equations 527 5.6 Applications from Business, Finance, and Science 539 5.7 Exponential, Logarithmic, and Logistic Equation Models 552 Making Connections 565 Summary and Concept Review 566 Practice Test 571 Calculator Exploration and Discovery: Investigating Logistic Equations 572 Strengthening Core Skills: The HerdBurn Scale — What’s Hot and What’s Not 573 Cumulative Review: Chapters R–5 574

CHAPTER

6

Systems of Equations and Inequalities

575

6.1 Linear Systems in Two Variables with Applications 576 6.2 Linear Systems in Three Variables with Applications 591 Mid-Chapter Check 603 Reinforcing Basic Concepts: Window Size and Graphing Technology 603 6.3 Nonlinear Systems of Equations and Inequalities 604 6.4 Systems of Inequalities and Linear Programming 615 Making Connections 629 Summary and Concept Review 630 Practice Test 632 Calculator Exploration and Discovery: Optimal Solutions and Linear Programming 633 Strengthening Core Skills: Understanding Why Elimination and Substitution “Work” 634 Cumulative Review: Chapters R–6 635

CHAPTER

7

Matrices and Matrix Applications 637 7.1 Solving Linear Systems Using Matrices and Row Operations 638 7.2 The Algebra of Matrices 650 Mid-Chapter Check 662 Reinforcing Basic Concepts: More on Matrix Multiplication 7.3 Solving Linear Systems Using Matrix Equations 663

663

7.4 Applications of Matrices and Determinants: Cramer’s Rule, Partial Fractions, and More 679 Contents

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7.5 Matrix Applications and Technology Use 692 Making Connections 699 Summary and Concept Review 700 Practice Test 702 Calculator Exploration and Discovery: Cramer’s Rule 704 Strengthening Core Skills: Augmented Matrices and Matrix Inverses Cumulative Review: Chapters R–7 705

CHAPTER

8

704

Analytic Geometry and the Conic Sections 707 8.1 A Brief Introduction to Analytical Geometry 708 8.2 The Circle and the Ellipse 715 Mid-Chapter Check 729 Reinforcing Basic Concepts: More on Completing the Square

730

8.3 The Hyperbola 731 8.4 The Analytic Parabola; More on Nonlinear Systems 744 Making Connections 754 Summary and Concept Review 755 Practice Test 757 Calculator Exploration and Discovery: Elongation and Eccentricity 758 Strengthening Core Skills: Ellipses and Hyperbolas with Rational/Irrational Values of a and b 759 Cumulative Review: Chapters R–8 760

CHAPTER

9

Additional Topics in Algebra 761 9.1 9.2 9.3 9.4

Sequences and Series 762 Arithmetic Sequences 773 Geometric Sequences 782 Mathematical Induction 796 Mid-Chapter Check 803 Reinforcing Basic Concepts: Applications of Summation 803 9.5 Counting Techniques 804 9.6 Introduction to Probability 816 9.7 The Binomial Theorem 829 Making Connections 837 Summary and Concept Review 838 Practice Test 842 Calculator Exploration and Discovery: Infinite Series, Finite Results 844 Strengthening Core Skills: Probability, Quick-Counting, and Card Games 845 Cumulative Review: Chapters R–9 846

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Contents

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Appendix I Appendix II

The Language, Notation, and Numbers of Mathematics Geometry Review with Unit Conversions A-14

A-1

Appendix III

More on Synthetic Division

Appendix IV

More on Matrices A-30

Appendix V

Deriving the Equation of a Conic

Appendix VI

Proof Positive — A Selection of Proofs from College Algebra

A-28 A-32 A-34

Student Answer Appendix (SE only) SA-1 Instructor Answer Appendix (AIE only) IA-1 Index I-1

Contents

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Index of Applications ANATOMY AND PHYSIOLOGY height versus human armspan, 176 height versus male shoe size, 177, 246 length of radial bone, R–38

ARCHITECTURE fountains at foci of elliptical garden, 727 Hall of Mirrors dimensions, 37 heights of selected tall buildings, 590, 630 whispering galleries, 729 window design, 325

ART, FINE ARTS, THEATER actors’ ages, 37 adult and children’s ticket sales, 588 exhibit design for sculpture, 727 heights of selected cartoon characters, 676 prices of paintings at auction, 601 seating arrangements, 812, 815 seating capacity of theater-in-theround, 781 television station programming possibilities, 815

BIOLOGY AND ZOOLOGY animal diets, 678 average animal birth weights, 771 bird wingspan and weight, 246 box turtle life span probability, 844 cat show winner choices, 841 cell division, 570 chimpanzee randomly choosing blocks, 828 concentration of chemical in bloodstream of animal, 475 cricket chirps and temperature, 118 custom feed and cottonseed blend, 163 deep-sea fishing depths, 229 deer and predators, 245 deer population growth, 782 dietary research for pets, 691 elk population growth, 64 fish and shark predators, 245 gestation periods and weights of selected mammals, 246, 601 hawk repopulation, 838 insect population, 426 prairie dog population, 570 predator/prey concentrations, 351, 368, 564

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rabbit populations, 551 species and island area relationship, 246, 525 squirrel populations, 706 stocking a lake, 537, 772 volume of an egg, 270 weight of dog over time, 571 whale weight, 246 wingspans of selected birds, 601 wolf preservation, 772

BUSINESS AND ECONOMICS advertising and sales, 258 applicant combinations for manager positions, 813 appreciation, 537, 843 automated filling of cereal boxes, 230 auto shop repair costs, 10 balance of payments, 426, 475 bicycle sales since 1920, 559 blue-book value of cars, 772 books shipped per box, 52 break even revenue, 589, 613 car rental charges, 132 cost, revenue, profit, 311–312, 350, 374 cost of copper tubing, 270 cost of each business purchased, 706 cost of single can of each vegetable, 632 defective bulb production probability, 844 demand and cost, 269 depreciation, 117, 148, 163, 501, 537, 571, 794, 843 eBay selling and feedback ratings, 270 envelope sizes, 53 equilibrium point, 590, 630 exponential revenue growth, 501 federal surplus of U.S., 202 generator failure probabilities, 824 government expenditures on guns versus butter, 627 hybrid car sales, 178 income depending on number of items sold, 134 inflation, 502, 772 Internet consumer spending, 373 manufacturing costs, 444, 456 market equilibrium price, 630, 633 marketing budgets, 515 marketing choices of table settings, 843 maximizing revenue, 24, 84, 326, 627–628, 744 maximum profit, 323–324, 632, 633, 636

minimizing shipping costs, 628 minimum wage, 10 moving van rental costs, 38 newspaper publishing decline, 258 number of items assembled by employee, 538, 559 number of workers needed to do job, 270 numbers of each item manufactured over time, 676–677, 692 oil imports, 258 overtime wages, 259 packaging costs, 444, 456 paper sizes, 53 patent applications, 176 patents issued, 177 postage costs, 10, 260 power tool rental costs, 38 production cost, 324, 350 production level which will minimize cost of item, 475 production matrix, 660–661 profit and loss, 201 profits, 379, 394, 478 raw material needed to fill orders, 697–698 replacement of aging equipment, 794 salary overtaken by cost of living, 795 sales goals, 781 sales growth, 537 seasonal revenue, 375 service call cost, 132–133 shipping carton length, 162 sizes of Slushies sold, 676 start-up costs, 563 supply and demand, 270, 442, 589–590, 614 thefts of precious metal, 698 tool rental returns on time probabilities, 847 tracking customer service wait time “on hold,” 826 TV repair costs, 10 union workers surveyed, 844 used vehicle sales, 347 wage increases, 771 wages and hours worked, 267 ways for volunteers to replace paid staff, 813

CHEMISTRY carbon-14 dating, 551 chemical mixtures, 163–164, 601, 760

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chlorine levels in swimming pools, 794–795 concentrations of selected chemicals, 443 exponential decay, 501 pH level of solution, 513 photochromatic sunglasses, 502 radioactive decay, 502, 551, 569, 574

COMMUNICATION cell phone charges, 260 cell phone subscribers, 312 customer wait time for cable company service, 843 DVD late rental returns, 836 phone service charges, 259 projected sales of iPhones, 186 radio phone range, 184 radio station broadcast range, 103–104 radio tower cable length, 79

COMPUTERS computer animations, 781 consultant salaries, 229 home computers probabilities, 827 households with Internet connections, 148 number of hours to assemble circuit boards, 115 random generation of numbers, 826 rebound height of ball on screen saver, 795

CONSTRUCTION amounts of wood used to build sail boats, 702 backyard fencing, 325 barrels stacked at storage facility, 847 building height and number of stairs, 267 deflection of a beam, 468 drywall area needed, 39 filling job crew positions, 841 forces on trusses of a roof, 691 height of spines of fireplace door, 727 home ventilation, 515 hours spent rehabbing houses, 703 load-bearing beams and weights carried, 269, 271, 279 load supported by cylindrical post, 538 new home cost, 115 pitch of a roof, 10, 116 plumbing vent pipe in roof, 728


pool dimensions, 614 quality control tests for marble columns, 229 raw material needed to fill orders, 697–698 sewer line slope, 116 sheep pen dimensions, 325 truck rental cost, 277 walkway paving area, 39

CRIMINAL JUSTICE, LEGAL STUDIES estimating time of death, 536 law firm mixture of staff, 825 nonwhite, nonmale Supreme Court justices, 116

DEMOGRAPHICS chicken production in U.S., 561 decrease in smokers in U.S., 118 family farms with milk cows, 560 females/males in workforce, 177 fertility rates in U.S., 338 growth of cell phone use, 563 households owning stocks, 258 households with cable television, 559 logistic growth of populations, 536 military expenditures in U.S., 259 multiple births in U.S., 259 national debt of U.S., 282 number of U.S. post offices, 560 numbers of children homeschooled, 375 percentage of female physicians, 118 population density, 441 population growth, 794–795 population of coastal areas, 561 prison population, 148 reporting of ages by using floor function, 260 research and development expenditures, 562 telephone calls per capita, 560 telephone opinion polls, 836 tourist population at resort, 393 triplets born in U.S., 574 use of debit cards, 561 veterans in civil life, 561

EDUCATION AND TEACHING alumni contributions, 588 applicant combinations for school board positions, 813 average grade computing, 444

college tuition and fees per semester, 118 cost of used and new textbooks, 662 course schedule probabilities, 812 credit hours taught at community college, 840 detentions given out in high school, 703 exam scores, 38 grade point average computing, 444 IQ of selected persons, 649 joining a club, 661 language retention, 443 memory retention, 64, 515 numbers of children homeschooled, 375 passing quiz probabilities, 828 quiz score and minutes spent studying, 184, 561 report card probabilities, 812 scholarship award possibilities, 815 student loan amounts owed, 703, 793 test averages and grade point averages, 444 typing speed, 64 ways for children to line up for lunch, 814 ways to choose committee to attend seminar, 814

ENGINEERING Civil cylindrical vent manufacture, 613 measuring depth of well, 351 traffic lanes on highway, 426 traffic volume at intersections, 229, 426 Electrical impedance calculations, 291 resistance and wire diameter, 245, 270, 629 temperature and electrical resistance, 468–469 voltage calculations, 291 wind turbine energy output, 278 Mechanical commercial flashlights, 753 distance between sides of nuclear cooling towers, 743 horsepower of vehicles, 574 industrial spotlights, 753 parabolic flashlights, 753 solar furnace, 753 wind powered energy, 79, 80, 218–219, 270, 427, 491

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ENVIRONMENTAL STUDIES barrels of toxic waste in storage facility, 803 city-wide recycling, 245 draining of reservoir, 840 energy rationing, 258 forest fire spreading, 367 gold mining depletion rate, 560 hazardous waste disposal, 628 land/crop allocation, 627 number of garbage trucks and volume of garbage, 115 ounces of platinum mined over time, 571 pollution removal, 64, 245, 442, 475 pollution testing, 836 rainfall and cattle per acre, 186 recycling costs, 442 soil acidity, 525 solid and liquid toxic waste containers, 744 stocking a lake, 537 tomato production related to watering amount, 337–338 venting landfill gases, 795 water level in lake, 117 water rationing, 259

FINANCE amount invested at each rate, 591, 601, 632, 649, 677 amounts bequeathed to family members, 627, 781 annuities, 549–550 bills denominations drawn from box, 847 bonds, 588 coin denominations, 588, 631, 677 coins in circulation, 678 compound interest, 366, 548–549, 570, 714, 837 continuously compounded interest, 549 corporate expenditures, 348 credit card usage, 178 currency conversion, 367 debt load, 393 debt-per-capita ratio of U.S., 22 gold coin denominations, 601–602 government deficits, 409 income taxes, 591 inheritance taxes, 551 interest rates, 201 interest rates needed to double investment, 847

xxxiv


investing in coins, 148 investment growth, 549, 636, 760 loan amounts for city renovations, 703 mortgage payments, 550 original purchase price, 615 prices of paintings at auction, 601 return on investments, 691 simple interest, 269, 548, 570, 571, 650 stock prices, 64 stock value, 229 time required for investment to double, 513 time required for investment to triple, 515 title loan interest rate, 548

GEOGRAPHY AND GEOLOGY altitude and atmospheric pressure, 515 altitude and temperature, 537 distance to horizon seen by person, 368 earthquake intensity, 514, 526 earthquake range, 103 gravity and weight, 269 land areas of Tahiti and Tonga, 590, 632 length of Suez Canal, 38 length of Tigris and Euphrates Rivers, 162 predicting tides, 133 river velocities, 246 submarine net depths, 229 temperature and atmospheric pressure, 514–515

HISTORY years Civil War ended and Declaration of Independence signed, 590 years selected U.S. documents signed, 601 years selected wars ended, 601

INTERNATIONAL STUDIES international shoe sizes, 367 license plate probabilities, 812, 814, 843 tourist population at resort, 393 U.S. international trade balance, 312

MATHEMATICS absolute value graphs, 260 angle measurements, 649 arc length of right parabolic segment, 752 area of circle, 22, 491

area of ellipse, 727 area of equilateral triangle, 268 area of inscribed circle, 103 area of oval, 728 area of printed page, 457 area of rectangle, 38, 184, 410, 627 area of right parabolic segment, 752 area of trapezoid, 614 area of triangle, 38, 455, 627, 703 area of triangle inscribed in circle, 103 area of vertical parabola, 614 average rate of change, 338–339, 373, 376 circumference of circle, 176 consecutive integers, 163 constructing and analyzing graphs, 201–203 correlation coefficients, 564 diagonal length of rectangular prism, 84 dimensions of a lawn, 10 dimensions of box, 409, 614 dimensions of closed rectangle, 614 dimensions of cylindrical tank, 614 dimensions of fish tank, 614 dimensions of flag, 614 dimensions of rectangle, 614, 632 dimensions of right triangle, 614 dimensions of triangle, 649 focal chord of hyperbola, 743 functions and rational exponents, 201 graph coordinates probabilities, 827 imaginary numbers, 291 lateral surface area of cone, 80 lateral surface area of frustrum, 80 length of sides of triangle, 633, 757 matric equations, 677 maximum and minimum values, 408 negative exponents, 24 number combination possibilities, 812 numbers made by rearranging digits, 813 Pareto’s principle, 525 perimeter of ellipse, 727 Pick’s theorem for area of polygon, 132 radius of a ripple, 367, 374 radius of circle, 491 radius of sphere, 490 sum of consecutive cubes, 469 sum of consecutive squares, 469 sum of interior angles of decagon, 782 sum of natural numbers, 781 surface area of box, 456 surface area of cube, 268

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surface area of cylinder, 311, 349, 455, 457 surface area of open cylinder, 458 surface area of rectangle with square ends, 323 surface area of sphere, 268 surface area of spherical cap, 458 thickness of folded sheet of paper, 795 variation equations, 269, 279 volume of box, 52 volume of cone, 490 volume of conical shell, 52 volume of cube, 22, 132, 408 volume of cylinder, 132, 176 volume of draining bathtub, 373 volume of grain silo, 39 volume of open box, 392 volume of rectangular solid, 39 volume of sphere, 218, 339 volume of spherical shell, 52 volume of triangular pyramid, 691 volume of water in pool, 393 words made by rearranging letters of given word, 813

MEDICINE, NURSING, NUTRITION, DIETETICS, HEALTH ages of selected famous persons, 603 bacterial growth, 501, 551, 795 blood plasma pH levels, 525 board of directors of hospital selection, 828 calories allotted for lunch, 37 calories in geriatric diet, 602 children and AIDS, 564 coffee bean mixture, 163, 627 cost per pound of selected fruits, 691 days wait time for patients to receive appointment, 842 drug absorption, 537 fetus weight, 338 gastric juice acidity, 525 grams of fat in soup, 602 ground beef mixture, 163 growth rates of children, 564 height of froth on carbonated drinks, 560 ideal weight for males, 132 location of kidney stones for lithotripsy procedure, 727 low birth weight and mother’s age, 563 measures of grain in each bundle, 698


medication concentration in bloodstream, 442, 475 milkfat requirements, 588 milk pH levels, 525 mixtures of selected foods, 163–164, 184 pounds added for each inch of height, 116 prescription drug sales, 148 random selection of dietetics class group, 825 spread of disease, 559 ways to make fruit trays with eight fruits, 814 ways to make hamburger, 814 ways to select recipes for competition, 814 weight loss over time, 559 workout diet, 678

METEOROLOGY altitude at which balloon ruptures, 691 atmospheric pressure at top of mountain, 569 dive depth and water pressure, 281, 562 earthquake intensity, 514, 526 earthquake range, 103 Fahrenheit/Celsius conversion, 38, 132, 588 locating ship/plane using radar, 743 ocean water temperature, 247, 563–564 speed of Caribbean Current, 588 speed of river current, 587, 588 temperature and altitude, 490 temperature drop, 339, 781 thermal conductivity, 677 wind-powered energy, 79, 80, 218–219, 270 wind speed, 589

MUSIC frequencies of musical notes, 562 playing times for selected songs/ movements, 603, 676 product matrix, 660–661

PHYSICS, ASTRONOMY, PLANETARY STUDIES acceleration due to gravity, 219 altitude of jet stream, 229 area illuminated by circle and by ellipse, 728 boiling temperature of water, 116 brightness of star, 514 catapults and projectiles, 201

comparisons of weights, sound, pressure, etc., 571 cooling and warming of liquids, 502 Coulomb’s law, 24 densities of objects, 409 distance between charged particles approaching each other, 743 distance dropped object falls, 490 distance from Earth to Mars, 163 distance from Sun to Mars, 279 distance object rolls due to gravity, 219 distance traveled by bouncing balls, 795 distance traveled on arc of swing, 781, 794 drag resistance on boat, 409 expansion of supernova, 367 force between charged particles, 270 freezing time of water, 537 gravitation between two celestial bodies, 271 height of object thrown downward, 337, 372 height of projectile thrown/shot upward, 311, 338 impedance calculations, 291 intensity of light, 24, 245, 271 intensity of sound, 271, 514, 516 kinetic energy and mass, 269 Lorentz transformation of space-time relationships, 52 orbital velocities of Earth and Jupiter, 728 parabolic sound and radio wave receivers, 753 period of pendulum, 270, 794, 843 planetary orbit times around sun, 79, 80, 246, 563, 728, 757 Poiseuille’s law of tubular liquid flow, 52 projected images and height/distance, 268, 368, 490 radar detection of ships, 103 removing air to create a vacuum, 794 spaceship velocity, 537 space travel costs, 348 speakers chosen randomly at conference, 823 speed of sound depending on temperature, 148 spread of liquid on surface, 278, 571 spring oscillation, 229 strength of a vacuum, 794 temperatures of water mixture, 185

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time for dropped object or projectile to hit ground, 218, 268, 311, 324 time for satellite to orbit Earth time for satellite to reach Jupiter, 22 velocity and times of rocket flight, 426–427 velocity of car depending on drag force, 373 velocity of falling objects, 185, 246, 339, 349 velocity of fluid flowing from tank, 218 weight of object on moon, 270

POLITICS committee composition possibilities, 815 Conservatives and Liberals in senate, 351 ways to choose reconnaissance team, 813

SOCIAL SCIENCE, HUMAN SERVICES applicant combinations for officer positions, 813 area codes possible, 815 arrangement of books on shelf, 843 ciphers to encrypt culinary secrets, 702 ciphers to encrypt switchboard phone numbers, 699 clothing choices and dressing possibilities, 812, 841 cost of cable service, 562 cost of food for retreat, 661 customer count at restaurant, 427 demand for popular songs and singers, 368–369 different meal possibilities, 812 disarming bomb probabilities, 828 European and U.S. shoe sizes, 38 government investment in public works and military, 348 guided tour start-up times and group sizes, 337 household cleaner solutions, 588 increase in restaurant eating, 148 length and width of a book, 281 library fines, 566 lock combination probabilities, 812, 841 long-distance calling-card cost, 706 natural gas pricing for households, 259 new books published in U.S., 376 number of customers in mall, 471 page count and thickness of books, 267

xxxvi


photograph possibilities of multiple-birth families, 813 prices paid for rare books, 649 remote access door opener digit possibilities, 812 retirees surveyed, 828 selecting books to read on vacation, 814 selecting veterans to survey, 825–826 skid marks and speed of car before crash, 79 standardizing e-mail addresses, 815 ticket prices, 337, 372 trash pickup on roads, 276 volume of phone calls, 270

SPORTS AND LEISURE admission prices, 259 amusement arcade scores, 259 amusement park attendance, 393 area of Olympic soccer field, 37 area of race track, 728 arrangements possible in Yahtzee, 814 average weight of football linemen, 38 baseball batting averages, 836 baseball card value, 117, 677 basketball salaries, 562 basketball scores, 649 blanket toss competition, 324–325 bowling strikes probabilities, 842 campers at park in selected months, 847 card game scores, 590, 631 chess tournament probabilities, 813 choosing blocks with numbers, 828 choosing finalists on game show, 824 coin toss probabilities, 229, 812, 814, 823, 826, 829 depth and duration of dive, 375, 474 diameters of sport balls, 229 dice rolling, 824–826, 841 distance jogged at faster pace, 163 distances descended by two cave explorers, 162 dominos numbers, 824 drawing cards at random, 823, 825, 841 drawing colored balls from bag, 813, 827, 843 drawing slips of paper at random, 841 elliptical billiard tables, 729 filling player positions for soccer, 824 frog-jumping contests, 327 height of kicked soccer ball, 602 high jump records, 177

horse-racing finish probabilities, 813, 841 hot-air ballooning, 246 kiteboarding and wind speed, 231 length of kite string let out, 79 marathon training runs, 37 maximum height of arrow shot from bow, 602 medals awarded to sprinters, 814 men’s Olympic freestyle swimming, 203 minimum altitude of stunt plane, 743 motorcycle jumps, 325 numbers of each item in shipment to sports store, 636 orienteering running speeds, 394 Pinochle probabilities, 824 pool table mixture of balls, 825, 826 racing pigeon speeds, 589 random changes in rule book, 823 rebound height of ball, 795, 829 registration for 5-km race, 337 runner first at finish line, 292 shelf size and size of tennis ball cans, 526 sit-ups for training of recruits, 229 snowboard jumping, 325 softball tossing distance, 10 speed of rower in still water, 587, 588 spinner probabilities, 501, 812, 823–824, 826, 843 spin outcomes in Twister, 814 sports promotions, 515 sprinter’s training times, 271 standing broad jump distances, 803 starting line-ups possibilities, 815 swimmers in pool at any time, 24 table tennis scores, 282 tandem bicycle trip finishing probabilities, 843 target-hitting probabilities, 827, 844 tennis court dimensions, 312 ticket pricing, 323 Tic-Tac-Toe ending board possibilities, 816 time arrow is airborne, 602 timed trials in swimming, 38 ways to commit crime in Clue, 815 ways to form female basketball teams, 814 ways to select songs to play at contest, 814 words formed by Scrabble player, 813 words formed from one word, 841

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TRANSPORTATION acceleration of car on highway, 148 aircraft N-Number possibilities, 815 airplane rate of climb, 116 airport moving walkways, 588 air travel distance, 649 bridge length, 163 celebrity airplane arrivals, 37 cruising speed of airliner, 10 cruising speed of ship, 588 distance apart of boats, 117, 292 distance of aircraft carrier from port after departure hour, 115 distribution of car filling stations along route, 350 drag resistance on boat, 409


driving time to work each day, 379, 477 elliptical arch bridge height, 727 fighter pilot training altitudes, 428 fighter plane wing measurements, 839 fines for speeding, 490 gas mileage, 117, 132 gas price per gallon, 588 minimizing transportation costs, 628 parabolic car headlights, 753 probability of responding to call by fire station, 824 road intersections, 117 round-trip average speed, 468 runway takeoff distance, 515 selection of phobics for therapy, 828

spacing of water storage tanks along rail route, 350–351 stopping distance of automobile, 270 stopping distance of boat after cutting engine, 278 time for one vehicle to catch up with another, 163 time spent driving at two different speeds, 163 traffic volume at intersections, 229, 426 trolley cable lengths, 552 tunnel clearances at various points, 613 vehicle fuel economy, 570 velocity and fuel economy, 369 velocity of accelerating car, 185

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College Algebra G&M—

CHAPTER CONNECTIONS

A Review of Basic Concepts and Skills CHAPTER OUTLINE R.1 Algebraic Expressions and the Properties of Real Numbers 2

R.2 Exponents, Scientific Notation, and a Review of Polynomials 11

R.3 Solving Linear Equations and Inequalities 25 R.4 Factoring Polynomials and Solving Polynomial Equations by Factoring 39

R.5 Rational Expressions and Equations 53

One of the primary goals of a college algebra course is to develop the mathematical tools necessary to model, explain, and understand the world around us. Speaking very broadly, this understanding gives us a better perspective of “where we’ve been” and “where we’re going.” For example, you’re likely aware that in 2009, the federal minimum wage got a nice boost. The ability to model increases in the minimum wage over time can give us information of what this wage might become in future years, or remind us of what the wage was long ago. This application is explored in Exercise 96 of Section R.1. Check out these other real-world connections: 䊳

䊳

R.6 Radicals, Rational Exponents, and Radical Equations 65

䊳

䊳

The Cost of Postage Over Time (Section R.1, Exercise 95) Maximizing Revenue of Video Game Sales (Section R.2, Exercise 143) Growth of a New Stock Hitting the Market (Section R.5, Exercise 89) Accident Investigation (Section R.6, Exercise 67)

1

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R.1

Algebraic Expressions and the Properties of Real Numbers

LEARNING OBJECTIVES In Section R.1 you will review how to:

A. Identify terms,

B. C. D.

E.

coefficients, and expressions Create mathematical models Evaluate algebraic expressions Identify and use properties of real numbers Simplify algebraic expressions

EXAMPLE 1

To effectively use mathematics as a problem-solving tool, we must develop the ability to translate written or verbal information into a mathematical model. After obtaining a model, many applications require that you work effectively with algebraic terms and expressions. The basic ideas involved are reviewed here.

A. Terms, Coefficients, and Algebraic Expressions An algebraic term is a collection of factors that may include numbers, variables, or expressions within parentheses. Here are some examples: (a) 3

(b) ⫺6P

(d) ⫺8n2

(c) 5xy

(f ) 21x ⫹ 32

(e) n

If a term consists of a single nonvariable number, it is called a constant term. In (a), 3 is a constant term. Any term that contains a variable is called a variable term. We call the constant factor of a term the numerical coefficient or simply the coefficient. The coefficients for (a), (b), (c), and (d) are 3, ⫺6, 5, and ⫺8, respectively. In (e), the coefficient of n is 1, since 1 # n ⫽ 1n ⫽ n. The term in (f ) has two factors as written, 2 and 1x ⫹ 32. The coefficient is 2. An algebraic expression can be a single term or a sum or difference of terms. To avoid confusion when identifying the coefficient of each term, the expression can be rewritten using algebraic addition if desired: A ⫺ B ⫽ A ⫹ 1⫺B2. For instance, 4 ⫺ 3x ⫽ 4 ⫹ 1⫺3x2 shows the coefficient of x is ⫺3. To identify the coefficient of a rational term, it sometimes helps to decompose the term, rewriting it using a unit fraction as in n ⫺5 2 ⫽ 15 1n ⫺ 22 and 2x ⫽ 12x. 䊳

Identifying Terms and Coefficients State the number of terms in each expression as given, then identify the coefficient of each term. x⫹3 ⫺ 2x a. 2x ⫺ 5y b. c. ⫺1x ⫺ 122 d. ⫺2x2 ⫺ x ⫹ 5 7

Solution

䊳

We can begin by rewriting each subtraction using algebraic addition. Rewritten: Number of terms:

A. You’ve just seen how we can identify terms, coefficients, and expressions

Coefficient(s):

a. 2x ⫹ 1⫺5y2 b. 17 1x ⫹ 32 ⫹ 1⫺2x2 c. ⫺11x ⫺ 122 d. ⫺2x2 ⫹ 1⫺1x2 ⫹ 5 two 2 and ⫺5

two 1 7

and ⫺2

one

three

⫺1

⫺2, ⫺1, and 5

Now try Exercises 7 through 14

䊳

B. Translating Written or Verbal Information into a Mathematical Model The key to solving many applied problems is finding an algebraic expression that accurately models relationships described in context. First, we assign a variable to represent an unknown quantity, then build related expressions using words from the English language that suggest mathematical operations. Variables that remind us of what they represent are often used in the modeling process, such as D ⫽ RT for Distance equals Rate times Time. These are often called descriptive variables. Capital letters are also used due to their widespread appearance in other fields.

2

R–2

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R–3

3

Section R.1 Algebraic Expressions and the Properties of Real Numbers

EXAMPLE 2

䊳

Translating English Phrases into Algebraic Expressions Assign a variable to the unknown number, then translate each phrase into an algebraic expression. a. twice a number, increased by five b. eleven less than eight times the width c. ten less than triple the payment d. two hundred fifty dollars more than double the amount

Solution

䊳

a. Let n represent the number. Then 2n represents twice the number, and 2n ⫹ 5 represents twice the number, increased by five. b. Let W represent the width. Then 8W represents eight times the width, and 8W ⫺ 11 represents 11 less than eight times the width. c. Let p represent the payment. Then 3p represents triple the payment, and 3p ⫺ 10 represents 10 less than triple the payment. d. Let A represent the amount in dollars. Then 2A represents double the amount, and 2A ⫹ 250 represents 250 dollars more than double the amount. Now try Exercises 15 through 28

䊳

Identifying and translating such phrases when they occur in context is an important problem-solving skill. Note how this is done in Example 3. EXAMPLE 3

䊳

Creating a Mathematical Model The cost for a rental car is $35 plus 15 cents per mile. Express the cost of renting a car in terms of the number of miles driven.

Solution

䊳

Let m represent the number of miles driven. Then 0.15m represents the cost for each mile and C ⫽ 35 ⫹ 0.15m represents the total cost for renting the car.

B. You’ve just seen how we can create mathematical models


䊳

C. Evaluating Algebraic Expressions We often need to evaluate expressions to investigate patterns and note relationships. Evaluating a Mathematical Expression 1. Replace each variable with open parentheses ( ). 2. Substitute the values given for each variable. 3. Simplify using the order of operations. In this process, it’s best to use a vertical format, with the original expression written first, the substitutions shown next, followed by the simplified forms and the final result. The numbers substituted or “plugged into” the expression are often called the input values, with the result called the output value.

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4

R–4

CHAPTER R A Review of Basic Concepts and Skills

EXAMPLE 4

䊳

Evaluating an Algebraic Expression Evaluate the expression x3 ⫺ 2x2 ⫹ 5 for x ⫽ ⫺3.

Solution

䊳

WORTHY OF NOTE In Example 4, note the importance of the first step in the evaluation process: replace each variable with open parentheses. Skipping this step could easily lead to confusion as we try to evaluate the squared term, since ⫺32 ⫽ ⫺9, while 1⫺32 2 ⫽ 9. Also see Exercises 55 and 56.

For x ⫽ ⫺3:

x3 ⫺ 2x2 ⫹ 5 ⫽ 1 2 3 ⫺ 21 2 2 ⫹ 5 ⫽ 1⫺32 3 ⫺ 21⫺32 2 ⫹ 5 ⫽ ⫺27 ⫺ 2192 ⫹ 5 ⫽ ⫺27 ⫺ 18 ⫹ 5 ⫽ ⫺40

replace variables with open parentheses substitute ⫺3 for x

simplify: 1⫺32 3 ⫽ ⫺27, 1⫺32 2 ⫽ 9 simplify: 2192 ⫽ 18 result

When the input is ⫺3, the output is ⫺40. Now try Exercises 41 through 60

䊳

If the same expression is evaluated repeatedly, results are often collected and analyzed in a table of values, as shown in Example 5. As a practical matter, the substitutions and simplifications are often done mentally or on scratch paper, with the table showing only the input and output values. EXAMPLE 5

䊳

Evaluating an Algebraic Expression Evaluate x2 ⫺ 2x ⫺ 3 to complete the table shown. Which input value(s) of x cause the expression to have an output of 0?

Solution

䊳

Input x ⫺2 ⫺1

Output x2 ⴚ 2x ⴚ 3

1⫺22 2 ⫺ 21⫺22 ⫺ 3 ⫽ 5 0

0

⫺3

1

⫺4

2

⫺3

3

0

4

5

The expression has an output of 0 when x ⫽ ⫺1 and x ⫽ 3. Now try Exercises 61 through 66 Figure R.1

Graphing calculators provide an efficient means of evaluating many expressions. After entering the expression on the Y= screen (Figure R.1), we can set up the table using the keystrokes 2nd (TBLSET). For this exercise, we’ll put the table in the “Indpnt: Auto Ask” mode, which will have the calculator “automatically” generate the input and output values. In this mode, we can tell the calculator where to start the inputs (we chose TblStart ⫽ ⫺2), and have the calculator produce the input values using any increment desired (we choose ¢Tbl ⫽ 1), as shown in Figure R.2. We access the completed table using 2nd GRAPH (TABLE), and the result for Example 5 is shown in Figure R.3. For exercises that combine the skills from Examples 3 through 5, see Exercises 91 to 98.

Figure R.2

WINDOW

C. You’ve just seen how we can evaluate algebraic expressions

Figure R.3

䊳

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R–5

5


D. Properties of Real Numbers While the phrase, “an unknown number times five,” is accurately modeled by the expression n5 for some number n, in algebra we prefer to have numerical coefficients precede variable factors. When we reorder the factors as 5n, we are using the commutative property of multiplication. A reordering of terms involves the commutative property of addition. The Commutative Properties Given that a and b represent real numbers: ADDITION: a ⫹ b ⫽ b ⫹ a

MULTIPLICATION: a # b ⫽ b # a

Terms can be combined in any order without changing the sum.

Factors can be multiplied in any order without changing the product.

Each property can be extended to include any number of terms or factors. While the commutative property implies a reordering or movement of terms (to commute implies back-and-forth movement), the associative property implies a regrouping or reassociation of terms. For example, the sum 1 34 ⫹ 35 2 ⫹ 25 is easier to compute if we regroup the addends as 34 ⫹ 1 35 ⫹ 25 2 . This illustrates the associative property of addition. Multiplication is also associative. The Associative Properties Given that a, b, and c represent real numbers: ADDITION:

EXAMPLE 6

䊳

MULTIPLICATION:

1a ⫹ b2 ⫹ c ⫽ a ⫹ 1b ⫹ c2

1a # b2 # c ⫽ a # 1b # c2

Terms can be regrouped.

Factors can be regrouped.

Simplifying Expressions Using Properties of Real Numbers Use the commutative and associative properties to simplify each calculation.

Solution

䊳

WORTHY OF NOTE Is subtraction commutative? Consider a situation involving money. If you had $100, you could easily buy an item costing $20: $100 ⫺ $20 leaves you with $80. But if you had $20, could you buy an item costing $100? Obviously $100 ⫺ $20 is not the same as $20 ⫺ $100. Subtraction is not commutative. Likewise, 100 ⫼ 20 is not the same as 20 ⫼ 100, and division is not commutative.

a.

3 8

⫺ 19 ⫹ 58

a.

3 8

⫺ 19 ⫹

5 8

⫽ ⫺19 ⫹

b. 3⫺2.5 # 1⫺1.22 4 # 10

3 5 8 ⫹ 8 1 38 ⫹ 58 2

⫽ ⫺19 ⫹ ⫽ ⫺19 ⫹ 1 ⫽ ⫺18

b. 3⫺2.5 # 1⫺1.22 4 # 10 ⫽ ⫺2.5 # 3 1⫺1.22 # 10 4 ⫽ ⫺2.5 # 1⫺122 ⫽ 30

commutative property (order changes) associative property (grouping changes) simplify result associative property (grouping changes) simplify result

Now try Exercises 67 and 68

䊳

For any real number x, x ⫹ 0 ⫽ x and 0 is called the additive identity since the original number was returned or “identified.” Similarly, 1 is called the multiplicative identity since 1 # x ⫽ x. The identity properties are used extensively in the process of solving equations.

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6

R–6


The Additive and Multiplicative Identities Given that x is a real number, x⫹0⫽x Zero is the identity for addition.

1#x⫽x One is the identity for multiplication.

For any real number x, there is a real number ⫺x such that x ⫹ 1⫺x2 ⫽ 0. The number ⫺x is called the additive inverse of x, since their sum results in the additive identity. Similarly, the multiplicative inverse of any nonzero number x is 1x , since x # 1x ⫽ 1 p q (the multiplicative identity). This property can also be stated as q # p ⫽ 1 1p, q ⫽ 02 for q p p any rational number q. Note that q and p are reciprocals. The Additive and Multiplicative Inverses

Given that p, q, and x represent real numbers 1p, q ⫽ 02: x ⫹ 1⫺x2 ⫽ 0

p q # ⫽1 q p p q

x and ⫺x are additive inverses.

EXAMPLE 7

䊳

Determining Additive and Multiplicative Inverses Replace the box to create a true statement: # ⫺3 x ⫽ 1 # x a. b. x ⫹ 4.7 ⫹ 5

Solution

䊳

q

and p are multiplicative inverses.

a. b.

⫽x

5 5 # ⫺3 , since ⫽1 ⫺3 ⫺3 5 ⫽ ⫺4.7, since 4.7 ⫹ 1⫺4.72 ⫽ 0

⫽


䊳

Note that if no coefficient is indicated, it is assumed to be 1, as in x ⫽ 1x, 1x2 ⫹ 3x2 ⫽ 11x2 ⫹ 3x2 , and ⫺1x3 ⫺ 5x2 2 ⫽ ⫺11x3 ⫺ 5x2 2. The distributive property of multiplication over addition is widely used in a study of algebra, because it enables us to rewrite a product as an equivalent sum and vice versa. The Distributive Property of Multiplication over Addition Given that a, b, and c represent real numbers: a1b ⫹ c2 ⫽ ab ⫹ ac A factor outside a sum can be distributed to each addend in the sum.

ab ⫹ ac ⫽ a1b ⫹ c2 A factor common to each addend in a sum can be “undistributed” and written outside a group.

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EXAMPLE 8

䊳

Simplifying Expressions Using the Distributive Property Apply the distributive property as appropriate. Simplify if possible. a. 71p ⫹ 5.22

Solution

䊳

WORTHY OF NOTE From Example 8b we learn that a negative sign outside a group changes the sign of all terms within the group: ⫺12.5 ⫺ x2 ⫽ ⫺2.5 ⫹ x.

b. ⫺12.5 ⫺ x2

c. 7x3 ⫺ x3

d.

1 5 n⫹ n 2 2

a. 71p ⫹ 5.22 ⫽ 7p ⫹ 715.22 ⫽ 7p ⫹ 36.4

b. ⫺12.5 ⫺ x2 ⫽ ⫺112.5 ⫺ x2 ⫽ ⫺112.52 ⫺ 1⫺121x2 ⫽ ⫺2.5 ⫹ x

c. 7x3 ⫺ x3 ⫽ 7x3 ⫺ 1x3 ⫽ 17 ⫺ 12x3 ⫽ 6x3

d.

1 5 1 5 n ⫹ n ⫽ a ⫹ bn 2 2 2 2 6 ⫽ a bn 2 ⫽ 3n

D. You’ve just seen how we can identify and use properties of real numbers


䊳

E. Simplifying Algebraic Expressions Two terms are like terms only if they have the same variable factors (the coefficient is not used to identify like terms). For instance, 3x2 and ⫺17x2 are like terms, while 5x3 and 5x2 are not. We simplify expressions by combining like terms using the distributive property, along with the commutative and associative properties. Many times the distributive property is used to eliminate grouping symbols and combine like terms within the same expression. EXAMPLE 9

䊳

Simplifying an Algebraic Expression

Solution

䊳

712p2 ⫹ 12 ⫺ 11p2 ⫹ 32

Simplify the expression completely: 712p2 ⫹ 12 ⫺ 1p2 ⫹ 32 . ⫽ 14p ⫹ 7 ⫺ 1p ⫺ 3 ⫽ 114p2 ⫺ 1p2 2 ⫹ 17 ⫺ 32 ⫽ 114 ⫺ 12p2 ⫹ 4 ⫽ 13p2 ⫹ 4 2

2

original expression; note coefficient of ⫺1 distributive property commutative and associative properties (collect like terms) distributive property result


䊳

The steps for simplifying an algebraic expression are summarized here: To Simplify an Expression 1. Eliminate parentheses by applying the distributive property. 2. Use the commutative and associative properties to group like terms. 3. Use the distributive property to combine like terms. E. You’ve just seen how we can simplify algebraic expressions

As you practice with these ideas, many of the steps will become more automatic. At some point, the distributive property, the commutative and associative properties, as well as the use of algebraic addition will all be performed mentally.

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R.1 EXERCISES 䊳


Fill in each blank with the appropriate word or phrase. Carefully reread the section, if necessary.

䊳

1. A term consisting of a single number is called a(n) term.

4. When 3 # 14 # 23 is written as 3 # 23 # 14, the property has been used.

2. A term containing a variable is called a(n) term.

5. Discuss/Explain why the additive inverse of ⫺5 is 5, while the multiplicative inverse of ⫺5 is ⫺15.

3. The constant factor in a variable term is called the .

6. Discuss/Explain how we can rewrite the sum 3x ⫹ 6y as a product, and the product 21x ⫹ 72 as a sum.


Identify the number of terms in each expression and the coefficient of each term.

7. 3x ⫺ 5y 9. 2x ⫹

x⫹3 4

8. ⫺2a ⫺ 3b 10.

n⫺5 ⫹ 7n 3

11. ⫺2x2 ⫹ x ⫺ 5

12. 3n2 ⫹ n ⫺ 7

13. ⫺1x ⫹ 52

14. ⫺1n ⫺ 32

Translate each phrase into an algebraic expression.

15. seven fewer than a number 16. a number decreased by six 17. the sum of a number and four 18. a number increased by nine

Create a mathematical model using descriptive variables.

29. The length of the rectangle is three meters less than twice the width. 30. The height of the triangle is six centimeters less than three times the base. 31. The speed of the car was fifteen miles per hour more than the speed of the bus. 32. It took Romulus three minutes more time than Remus to finish the race. 33. Hovering altitude: The helicopter was hovering 150 ft above the top of the building. Express the altitude of the helicopter in terms of the building’s height.

19. the difference between a number and five is squared 20. the sum of a number and two is cubed 21. thirteen less than twice a number 22. five less than double a number 23. a number squared plus the number doubled 24. a number cubed less the number tripled 25. five fewer than two-thirds of a number 26. fourteen more than one-half of a number 27. three times the sum of a number and five, decreased by seven 28. five times the difference of a number and two, increased by six

34. Stacks on a cruise liner: The smoke stacks of the luxury liner cleared the bridge by 25 ft as it passed beneath it. Express the height of the stacks in terms of the bridge’s height. 35. Dimensions of a city park: The length of a rectangular city park is 20 m more than twice its width. Express the length of the park in terms of the width.

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36. Dimensions of a parking lot: In order to meet the city code while using the available space, a contractor planned to construct a parking lot with a length that was 50 ft less than three times its width. Express the length of the lot in terms of the width. 37. Cost of milk: In 2010, a gallon of milk cost two and one-half times what it did in 1990. Express the cost of a gallon of milk in 2010 in terms of the 1990 cost. 38. Cost of gas: In 2010, a gallon of gasoline cost two and one-half times what it did in 1990. Express the cost of a gallon of gas in 2010 in terms of the 1990 cost. 39. Pest control: In her pest control business, Judy charges $50 per call plus $12.50 per gallon of insecticide for the control of spiders and certain insects. Express the total charge in terms of the number of gallons of insecticide used. 40. Computer repairs: As his reputation and referral business grew, Keith began to charge $75 per service call plus an hourly rate of $50 for the repair and maintenance of home computers. Express the cost of a service call in terms of the number of hours spent on the call. Evaluate each algebraic expression given x ⴝ 2 and y ⴝ ⴚ3.

41. 4x ⫺ 2y

42. 5x ⫺ 3y

43. ⫺2x2 ⫹ 3y2

44. ⫺5x2 ⫹ 4y2

45. 2y2 ⫹ 5y ⫺ 3

46. 3x2 ⫹ 2x ⫺ 5

47. ⫺213y ⫹ 12

48. ⫺312y ⫹ 52

49. 3x2y

50. 6xy2

51. 1⫺3x2 2 ⫺ 4xy ⫺ y2 52. 1⫺2x2 2 ⫺ 5xy ⫺ y2 53. 12x ⫺ 13y

55. 13x ⫺ 2y2 57.

54. 32x ⫺ 12y

2

⫺12y ⫹ 5 ⫺3x ⫹ 1

59. 1⫺12y # 4

56. 12x ⫺ 3y2 58.

2

12x ⫹ 1⫺32 ⫺3y ⫹ 1

60. 7 # 1⫺27y

Evaluate each expression for integers from ⴚ3 to 3 inclusive. Verify results using a graphing calculator. What input(s) give an output of zero?

61. x2 ⫺ 3x ⫺ 4

62. x2 ⫺ 2x ⫺ 3

63. ⫺311 ⫺ x2 ⫺ 6

64. 513 ⫺ x2 ⫺ 10

65. x3 ⫺ 6x ⫹ 4

66. x3 ⫹ 5x ⫹ 18

9

Rewrite each expression using the given property and simplify if possible.

67. Commutative property of addition a. ⫺5 ⫹ 7 b. ⫺2 ⫹ n c. ⫺4.2 ⫹ a ⫹ 13.6 d. 7 ⫹ x ⫺ 7 68. Associative property of multiplication a. 2 # 13 # 62 b. 3 # 14 # b2 # # c. ⫺1.5 16 a2 d. ⫺6 # 1⫺56 # x2 Replace the box so that a true statement results.

69. a. x ⫹ 1⫺3.22 ⫹ b. n ⫺ 56 ⫹ 70. a. b.

⫽x

⫽n

# 23x ⫽ 1x #

n ⫽ 1n ⫺3

Apply the distributive property and simplify if possible.

71. ⫺51x ⫺ 2.62 72. ⫺121v ⫺ 3.22 73. 23 1⫺15p ⫹ 92

2 74. 56 1⫺15 q ⫹ 242

75. 3a ⫹ 1⫺5a2

76. 13m ⫹ 1⫺5m2 77. 23x ⫹ 34x 78.

5 12 y

⫺ 38y

Simplify by removing all grouping symbols (as needed) and combining like terms.

79. 31a2 ⫹ 3a2 ⫺ 15a2 ⫹ 7a2 80. 21b2 ⫹ 5b2 ⫺ 16b2 ⫹ 9b2 81. x2 ⫺ 13x ⫺ 5x2 2

82. n2 ⫺ 15n ⫺ 4n2 2

83. 13a ⫹ 2b ⫺ 5c2 ⫺ 1a ⫺ b ⫺ 7c2

84. 1x ⫺ 4y ⫹ 8z2 ⫺ 18x ⫺ 5y ⫺ 2z2 85. 35 15n ⫺ 42 ⫹ 58 1n ⫹ 162 86. 23 12x ⫺ 92 ⫹ 34 1x ⫹ 122

87. 13a2 ⫺ 5a ⫹ 72 ⫹ 212a2 ⫺ 4a ⫺ 62

88. 213m2 ⫹ 2m ⫺ 72 ⫺ 1m2 ⫺ 5m ⫹ 42

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89. Electrical resistance: R ⴝ

kL d2

The electrical resistance in a wire depends on the length and diameter of the wire. This resistance can be modeled by the formula shown, where R is the resistance in ohms, L is the length in feet, and d is the diameter of the wire in inches. Find the resistance if k ⫽ 0.000025, d ⫽ 0.015 in., and L ⫽ 90 ft

䊳

R–10


k V If temperature remains constant, the pressure of a gas held in a closed container is related to the volume of gas by the formula shown, where P is the pressure in pounds per square inch, V is the volume of gas in cubic inches, and k is a constant that depends on given conditions. Find the pressure exerted by the gas if k ⫽ 440,310 and V ⫽ 22,580 in3.

90. Volume and pressure: P ⴝ

APPLICATIONS

Translate each key phrase into an algebraic expression, then evaluate as indicated.

91. Cruising speed: A turbo-prop airliner has a cruising speed that is one-half the cruising speed of a 767 jet aircraft. (a) Express the speed of the turbo-prop in terms of the speed of the jet, and (b) determine the speed of the airliner if the cruising speed of the jet is 550 mph. 92. Softball toss: Macklyn can throw a softball twothirds as far as her father. (a) Express the distance that Macklyn can throw a softball in terms of the distance her father can throw. (b) If her father can throw the ball 210 ft, how far can Macklyn throw the ball? 93. Dimensions of a lawn: The length of a rectangular lawn is 3 ft more than twice its width. (a) Express the length of the lawn in terms of the width. (b) If the width is 52 ft, what is the length? 94. Pitch of a roof: To obtain the proper pitch, the crossbeam for a roof truss must be 2 ft less than three-halves the rafter. (a) Express the length of the crossbeam in terms of the rafter. (b) If the rafter is 18 ft, how long is the crossbeam? 䊳

95. Postage costs: In 2009, a first class stamp cost 29¢ more than it did in 1978. Express the cost of a 2009 stamp in terms of the 1978 cost. If a stamp cost 15¢ in 1978, what was the cost in 2009? 96. Minimum wage: In 2009, the federal minimum wage was $4.95 per hour more than it was in 1976. Express the 2009 wage in terms of the 1976 wage. If the hourly wage in 1976 was $2.30, what was it in 2009? 97. Repair costs: The TV repair shop charges a flat fee of $43.50 to come to your house and $25 per hour for labor. Express the cost of repairing a TV in terms of the time it takes to repair it. If the repair took 1.5 hr, what was the total cost? 98. Repair costs: At the local car dealership, shop charges are $79.50 to diagnose the problem and $85 per shop hour for labor. Express the cost of a repair in terms of the labor involved. If a repair takes 3.5 hr, how much will it cost?


99. If C must be a positive odd integer and D must be a negative even integer, then C2 ⫹ D2 must be a: a. positive odd integer. b. positive even integer. c. negative odd integer. d. negative even integer. e. cannot be determined.

100. Historically, several attempts have been made to create metric time using factors of 10, but our current system won out. If 1 day was 10 metric hours, 1 metric hour was 10 metric minutes, and 1 metric minute was 10 metric seconds, what time would it really be if a metric clock read 4:3:5? Assume that each new day starts at midnight.

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College Algebra Graphs & Models—

R.2

Exponents, Scientific Notation, and a Review of Polynomials


A. Apply properties of

A. The Properties of Exponents

exponents

B. Perform operations in C. D. E.

An exponent is a superscript number or letter occurring to the upper right of a base number, and indicates how many times the base occurs as a factor. For b # b # b b3, we say b3 is written in exponential form. In some cases, we may refer to b3 as an exponential term. Exponential Notation For any positive integer n, bn b # b # b # . . . # b

b # b # b # . . . # b bn

and

⎞ ⎜ ⎜ ⎜ ⎬ ⎜ ⎜ ⎠

scientific notation Identify and classify polynomial expressions Add and subtract polynomials Compute the product of two polynomials Compute special products: binomial conjugates and binomial squares

⎞ ⎜ ⎜ ⎜ ⎬ ⎜ ⎜ ⎠

F.

In this section, we review basic exponential properties and operations on polynomials. Although there are five to eight exponential properties (depending on how you count them), all can be traced back to the basic definition involving repeated multiplication.

n times

n times

The Product and Power Properties There are two properties that follow immediately from this definition. When b3 is multiplied by b2, we have an uninterrupted string of five factors: b3 # b2 1b # b # b2 # 1b # b2, which can then be written as b5. This is an example of the product property of exponents. Product Property of Exponents

WORTHY OF NOTE In this statement of the product property and the exponential properties that follow, it is assumed that for any expression of the form 0m, m 7 0 (hence 0m 0).

For any base b and positive integers m and n: bm # bn bmn In words, the property says, to multiply exponential terms with the same base, keep the common base and add the exponents. A special application of the product property uses repeated factors of the same exponential term, as in (x2)3. Using the product property, we have 1x2 2 1x2 21x2 2 x6. Notice the same result can be found more quickly by # multiplying the inner exponent by the outer exponent: 1x2 2 3 x2 3 x6. We generalize this idea to state the power property of exponents. In words the property says, to raise an exponential term to a power, keep the same base and multiply the exponents. Power Property of Exponents For any base b and positive integers m and n:

1bm 2 n bm n #

EXAMPLE 1

䊳

Multiplying Terms Using Exponential Properties Compute each product. a. 4x3 # 12x2 b. 1p3 2 2 # 1p4 2 5

Solution

R–11

䊳

a. 4x3 # 12x2 14 # 12 21x3 # x2 2 122 1x32 2 2 x5 # # b. 1p3 2 2 # 1p4 2 5 p3 2 # p4 5 p6 # p20 p620 p26

commutative and associative properties simplify; product property result power property simplify product property result

Now try Exercises 7 through 12 䊳 11

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The power property can easily be extended to include more than one factor within the parentheses. This application of the power property is sometimes called the product to a power property and can be extended to include any number of factors. We can also raise a quotient of exponential terms to a power. The result is called the quotient to a power property. In words the properties say, to raise a product or quotient of exponential terms to a power, multiply every exponent inside the parentheses by the exponent outside the parentheses. Product to a Power Property For any bases a and b, and positive integers m, n, and p: 1ambn 2 p amp # bnp

Quotient to a Power Property For any bases a and b 0, and positive integers m, n, and p: a EXAMPLE 2

䊳

am p amp b np bn b

Simplifying Terms Using the Power Properties Simplify using the power property (if possible): 5a3 2 a. 13a2 2 b. 3a2 c. a b 2b

Solution

䊳

WORTHY OF NOTE Regarding Examples 2a and 2b, note the difference between the expressions 13a2 2 13 # a2 2 and 3a2 3 # a2. In the first, the exponent acts on both the negative 3 and the a; in the second, the exponent acts on only the a and there is no “product to a power.”

a. 13a2 2 132 2 # 1a1 2 2 9a2 152 2 1a3 2 2 5a3 2 b c. a 2b 22b2 25a6 4b2

b. 3a2 is in simplified form

Now try Exercises 13 through 24 䊳 Applications of exponents sometimes involve linking one exponential term with another using a substitution. The result is then simplified using exponential properties.

EXAMPLE 3

䊳

Applying the Power Property after a Substitution The formula for the volume of a cube is V S3, where S is the length of one edge. If the length of each edge is 2x2: a. Find a formula for volume in terms of x. b. Find the volume if x 2.

Solution

䊳

a. V S

3

S

substitute 2x 2 for S

12x2 2 3 8x6

2x2 2x2

2x2

b. For V 8x , V 8122 6 substitute 2 for x # 8 64 or 512 122 6 64 The volume of the cube would be 512 units3. 6

Now try Exercises 25 and 26 䊳

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Section R.2 Exponents, Scientific Notation, and a Review of Polynomials

13

The Quotient Property of Exponents x 1 for x 0, we note a x x5 x # x # x # x # x x3, pattern that helps to simplify a quotient of exponential terms. For 2 # x x x the exponent of the final result appears to be the difference between the exponent in the numerator and the exponent in the denominator. This seems reasonable since the subtraction would indicate a removal of the factors that reduce to 1. Regardless of how many factors are used, we can generalize the idea and state the quotient property of exponents. In words the property says, to divide two exponential terms with the same base, keep the common base and subtract the exponent of the denominator from the exponent of the numerator. By combining exponential notation and the property

Quotient Property of Exponents For any base b 0 and positive integers m and n: bm bmn bn

Zero and Negative Numbers as Exponents If the exponent of the denominator is greater than the exponent in the numerator, the x2 quotient property yields a negative exponent: 5 x25 x3. To help understand x what a negative exponent means, let’s look at the expanded form of the expression: x2 x # x1 1 3 . A negative exponent can literally be interpreted as “write 5 # # # # x x x x x x x the factors as a reciprocal.” A good way to remember this is

!

!

23

three factors of 2 written as a reciprocal

1 1 23 3 1 8 2

Since the result would be similar regardless of the base used, we can generalize this idea and state the property of negative exponents. Property of Negative Exponents For any base b 0 and integer n: bn 1 n 1 b WORTHY OF NOTE The use of zero as an exponent should not strike you as strange or odd; it’s simply a way of saying that no factors of the base remain, since all terms have been reduced to 1. 8 23 For 3 , we have 1, or 8 2 1

1

1

2#2#2 1, or 233 20 1. 2#2#2

1 bn

bn 1

a n b n a b a b ;a0 a b

x3 x3 by division, and 1 x33 x0 using the x3 x3 quotient property, we conclude that x0 1 as long as x 0. We can also generalize this observation and state the meaning of zero as an exponent. In words the property says, any nonzero quantity raised to an exponent of zero is equal to 1. Finally, when we consider that

Zero Exponent Property For any base b 0: b0 1

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EXAMPLE 4

Solution

䊳

䊳

Simplifying Expressions Using Exponential Properties Simplify using exponential properties. Answer using positive exponents only. 2a3 2 a. a 2 b b. 13hk2 2 3 16h2k3 2 2 b 12m2n3 2 5 c. 13x2 0 3x0 32 d. 14mn2 2 3 2a3 2 b2 2 a. a 2 b a 3 b property of negative exponents b 2a 1b2 2 2 2 3 2 power properties 2 1a 2

b4 4a6

result

b. 13hk2 2 3 16h2k3 2 2 33h3 1k2 2 3 # 62 1h2 2 2 1k3 2 2 3 3 6

3hk

2

3

3

#6

2 4 6

#6 hk # h34 # k66

27h k 36

1 32

1 9

Notice in Example 4(c), we have 13x2 0 13 # x2 0 1, while 3x0 3 # x0 3112. This is another example of operations and grouping symbols working together: 13x2 0 1 because any quantity to the zero power is 1. However, for 3x0 there are no grouping symbols, so the exponent 0 acts only on the x and not the 3: 3x0 3 # x0 3112 3.

12m2n3 2 5 14mn2 2 3

122 5 1m2 2 5 1n3 2 5 43m3 1n2 2 3

result

power property

32m10n15 64m3n6

simplify

1m7n9 2

quotient property

m7n9 2

1 b 36

zero exponent property; property of negative exponents

simplify: (3x )0 1, 3x 0 3 1 3

1 37 4 9 9 d.

1 62

result 1k 0 12

c. 13x2 0 3x0 32 1 3112

WORTHY OF NOTE

product property a62

3h7 4

4

simplify

simplify

7 0

power property

result


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15

Summary of Exponential Properties For real numbers a and b, and integers m, n, p (excluding 0 raised to a nonpositive power) Product property: bm # bn bmn # Power property: 1bm 2 n bm n Product to a power: 1ambn 2 p amp # bnp am p amp Quotient to a power: a n b np 1b 02 b b m b Quotient property: bmn 1b 02 bn Zero exponents: b0 11b 02 1 1 a n b n bn Negative exponents: n , n bn, a b a b 1a, b 02 a 1 b b b

A. You’ve just seen how we can apply properties of exponents

B. Exponents and Scientific Notation In many technical and scientific applications, we encounter numbers that are either extremely large or very, very small. For example, the mass of the Moon is over 73 quintillion kilograms (73 followed by 18 zeroes), while the constant for universal gravitation contains 10 zeroes before the first nonzero digit. When computing with numbers of this magnitude, scientific notation has a distinct advantage over the common decimal notation (base-10 place values). WORTHY OF NOTE

Scientific Notation

Recall that multiplying by 10’s (or multiplying by 10k, k 7 02 shifts the decimal point to the right k places, making the number larger. Dividing by 10’s (or multiplying by 10k, k 7 0) shifts the decimal point to the left k places, making the number smaller.

A non-zero number written in scientific notation has the form N 10k

where 1 0N 0 6 10 and k is an integer. To convert a number from decimal notation into scientific notation, we begin by placing the decimal point to the immediate right of the first nonzero digit (creating a number less than 10 but greater than or equal to 1) and multiplying by 10k. Then we determine the power of 10 (the value of k) needed to ensure that the two forms are equivalent. When writing large or small numbers in scientific notation, we sometimes round the value of N to two or three decimal places.

EXAMPLE 5

䊳

Converting from Decimal Notation to Scientific Notation The mass of the Moon is about 73,000,000,000,000,000,000 kg. Write this number in scientific notation.

Solution

䊳

Place decimal to the right of first nonzero digit (7) and multiply by 10k. 73,000,000,000,000,000,000 7.3 10k To return the decimal to its original position would require 19 shifts to the right, so k must be positive 19. 73,000,000,000,000,000,000 7.3 1019 The mass of the Moon is 7.3 1019 kg. Now try Exercises 67 and 68 䊳

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Converting a number from scientific notation to decimal notation is simply an application of multiplication or division with powers of 10. EXAMPLE 6

䊳

Converting from Scientific Notation to Decimal Notation The constant of gravitation is 6.67 1011. Write this number in common decimal form.

Solution

䊳

Since the exponent is negative 11, shift the decimal 11 places to the left, using placeholder zeroes as needed to return the decimal to its original position: 6.67 1011 0.000 000 000 066 7 Now try Exercises 69 and 70 䊳 Computations that involve scientific notation typically use real number properties and the properties of exponents.

EXAMPLE 7

䊳

Storage Space on a Hard Drive A typical 320-gigabyte portable hard drive can hold about 340,000,000,000 bytes of information. A 2-hr DVD movie can take up as much as 8,000,000,000 bytes of storage space. Find the number of movies (to the nearest whole movie) that can be stored on this hard drive.

Solution

䊳

Using the ideas from Example 5, the hard drive holds 3.4 1011 bytes, while the DVD requires 8.0 109 bytes. Divide to find the number of DVDs the hard drive will hold. 3.4 1011 3.4 1011 8.0 8.0 109 109 0.425 102 42.5

rewrite the expression divide; subtract exponents result

The drive will hold approximately 42 DVD movies. A calculator check is shown in the figure. B. You’ve just seen how we can perform operations in scientific notation


C. Identifying and Classifying Polynomial Expressions A monomial is a term using only whole number exponents on variables, with no variables in the denominator. One important characteristic of a monomial is its degree. For a monomial in one variable, the degree is the same as the exponent on the variable. The degree of a monomial in two or more variables is the sum of exponents occurring on variable factors. A polynomial is a monomial or any sum or difference of monomial terms. For instance, 12x2 5x 6 is a polynomial, while 3n2 2n 7 is not (the exponent 2 is not a whole number). Identifying polynomials is an important skill because they represent a very different kind of real-world model than nonpolynomials. In addition, there are different families of polynomials, with each family having different characteristics. We classify polynomials according to their degree and number of terms. The degree of a polynomial in one variable is the largest exponent occurring on the variable. The degree of a polynomial in more than one variable is the largest sum of exponents in any one term. A polynomial with two terms is called a binomial (bi means two) and a polynomial with three terms is called a trinomial (tri means three). There are special names for polynomials with four or more terms, but for these, we simply use the general name polynomial (poly means many).

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EXAMPLE 8

䊳

17

Classifying and Describing Polynomials For each expression: a. Classify as a monomial, binomial, trinomial, or polynomial. b. State the degree of the polynomial. c. Name the coefficient of each term.

Solution

䊳

Expression

Classification

Degree

5x y 2xy

binomial

three

5, 2

x2 0.81

binomial

two

1, 0.81

z3 3z2 9z 27

polynomial (four terms)

three

1, 3, 9, 27

binomial

one

trinomial

two

2

3 4 x 2

5

2x x 3

Coefficients

3 4 ,

5

2, 1, 3

Now try Exercises 73 through 78 䊳 A polynomial expression is in standard form when the terms of the polynomial are written in descending order of degree, beginning with the highest-degree term. The coefficient of the highest-degree term is called the leading coefficient. EXAMPLE 9

䊳

Writing Polynomials in Standard Form Write each polynomial in standard form, then identify the leading coefficient.

Solution

䊳

Polynomial

Standard Form

9x

x 9

5z 7z2 3z3 27

3z3 7z2 5z 27

2

2

1 3 4 2x 2

3 2x x

2

3 4 x 2

2

2x x 3

C. You’ve just seen how we can identify and classify polynomial expressions

Leading Coefficient 1 3 3 4

2


D. Adding and Subtracting Polynomials Adding polynomials simply involves using the distributive, commutative, and associative properties to combine like terms (at this point, the properties are usually applied mentally). As with real numbers, the subtraction of polynomials involves adding the opposite of the second polynomial using algebraic addition. This can be viewed as distributing 1 to the second polynomial and combining like terms. EXAMPLE 10

䊳

Adding and Subtracting Polynomials Perform the indicated operations:

10.7n3 4n2 82 10.5n3 n2 6n2 13n2 7n 102.

Solution

䊳

0.7n3 4n2 8 0.5n3 n2 6n 3n2 7n 10 0.7n3 0.5n3 4n2 1n2 3n2 6n 7n 8 10 1.2n3 13n 18

eliminate parentheses (distributive property) use properties to collect like terms combine like terms


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Sometimes it’s easier to add or subtract polynomials using a vertical format and aligning like terms. Note the use of a placeholder zero in Example 11. EXAMPLE 11

䊳

Subtracting Polynomials Using a Vertical Format Compute the difference of x3 5x 9 and x3 3x2 2x 8 using a vertical format.

Solution

䊳

x3 0x2 5x 9 x3 0x2 5x 9 1x3 3x2 2x 82 ¡ x3 3x2 2x 8 3x2 7x 17 The difference is 3x2 7x 17.

D. You’ve just seen how we can add and subtract polynomials


E. The Product of Two Polynomials Monomial Times Monomial The simplest case of polynomial multiplication is the product of monomials shown in Example 1a. These were computed using exponential properties and the properties of real numbers.

Monomial Times Polynomial To compute the product of a monomial and a polynomial, we use the distributive property. EXAMPLE 12

䊳

Solution

䊳

Multiplying a Monomial by a Polynomial Find the product: 2a2 1a2 2a 12.

2a2 1a2 2a 12 2a2 1a2 2 12a2 212a1 2 12a2 2112 2a4 4a3 2a2

distribute simplify


Binomial Times Polynomial For products involving binomials, we still use a version of the distributive property— this time to distribute one polynomial to each term of the other polynomial factor. Note the distribution can be performed either from the left or from the right. EXAMPLE 13

䊳

Multiplying a Binomial by a Polynomial Multiply as indicated: a. 12z 12 1z 22

Solution

䊳

b. 12v 32 14v2 6v 92

a. 12z 12 1z 22 2z1z 22 11z 22 distribute to every term in the first binomial eliminate parentheses (distribute again) 2z2 4z 1z 2 simplify 2z2 3z 2 2 2 b. 12v 32 14v 6v 92 2v14v 6v 92 314v2 6v 92 distribute 8v3 12v2 18v 12v2 18v 27 simplify 8v3 27 combine like terms


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19

The F-O-I-L Method By observing the product of two binomials in Example 13(a), we note a pattern that can make the process more efficient. The product of two binomials can quickly be computed using the First, Outer, Inner, Last (FOIL) method, an acronym giving the respective position of each term in a product of binomials in relation to the other terms. We illustrate here using the product 12x 12 13x 22.

WORTHY OF NOTE

Consider the product 1x 321x 22 in the context of area. If we view x 3 as the length of a rectangle (an unknown length plus 3 units), and x 2 as its width (the same unknown length plus 2 units), a diagram of the total area would look like the following, with the result x2 5x 6 clearly visible.

Combine like terms 6x2 x 2

Inner Outer

The first term of the result will always be the product of the first terms from each binomial, and the last term of the result is the product of their last terms. We also note that here, the middle term is found by adding the outermost product with the innermost product. As you practice with the F-O-I-L process, much of the work can be done mentally and you can often compute the entire product without writing anything down except the answer. 䊳

Multiplying Binomials Using F-O-I-L Compute each product mentally: a. 15n 121n 22 b. 12b 32 15b 62 a. 15n 12 1n 22:

5n2 9n 2

product of first two terms

S

䊳

S

Solution

10n (1n) 9n

S

(x 3)(x 2) x2 5x 6

EXAMPLE 14

S

6

S

2x

S

2

S

3x

S

x2

S

x

First Outer Inner Last

12x 12 13x 22 S

3

6x2 4x 3x 2

Last First S

x

The F-O-I-L Method for Multiplying Binomials

sum of outer and inner products

product of last two terms

12b 15b 3b

product of first two terms

S

S

S

b. 12b 3215b 62: 10b2 3b 18 sum of outer and inner products

E. You’ve just seen how we can compute the product of two polynomials

product of last two terms


F. Special Polynomial Products Certain polynomial products are considered “special” for two reasons: (1) the product follows a predictable pattern, and (2) the result can be used to simplify expressions, graph functions, solve equations, and/or develop other skills.

Binomial Conjugates Expressions like x 7 and x 7 are called binomial conjugates. For any given binomial, its conjugate is found by using the same two terms with the opposite sign

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between them. Example 15 shows that when we multiply a binomial and its conjugate, the “outers” and “inners” sum to zero and the result is a difference of two squares. EXAMPLE 15

䊳

Multiplying Binomial Conjugates Compute each product mentally: a. 1x 721x 72 b. 12x 5y2 12x 5y2

2 2 c. ax b ax b 5 5

7x 7x 0x

Solution

䊳

a. 1x 72 1x 72 x2 49

difference of squares 1x2 2 172 2

10xy (10xy) 0xy

b. 12x 5y212x 5y2 4x2 25y2

difference of squares: 12x2 2 15y2 2

52 x 25 x 0

2 2 4 c. ax b ax b x2 5 5 25

2 2 difference of squares: x2 a b 5

Now try Exercises 117 through 124 䊳 In summary, we have the following. The Product of a Binomial and Its Conjugate Given any expression that can be written in the form A B, the conjugate of the expression is A B and their product is a difference of two squares: 1A B2 1A B2 A2 B2

Binomial Squares

Expressions like 1x 72 2 are called binomial squares and are useful for solving many equations and sketching a number of basic graphs. Note 1x 72 2 1x 721x 72 x2 14x 49 using the F-O-I-L process. The expression x2 14x 49 is called a perfect square trinomial because it is the result of expanding a binomial square. If we write a binomial square in the more general form 1A B2 2 1A B21A B2 and compute the product, we notice a pattern that helps us write the expanded form more quickly. LOOKING AHEAD Although a binomial square can always be found using repeated factors and F-O-I-L, learning to expand them using the pattern is a valuable skill. Binomial squares occur often in a study of algebra and it helps to find the expanded form quickly.

1A B2 2 1A B2 1A B2

repeated multiplication

A AB AB B

F-O-I-L

A 2AB B

simplify (perfect square trinomial)

2 2

2

2

The first and last terms of the trinomial are squares of the terms A and B. Also, the middle term of the trinomial is twice the product of these two terms: AB AB 2AB. The F-O-I-L process shows us why. Since the outer and inner products are identical, we always end up with two. A similar result holds for 1A B2 2 and the process can be summarized for both cases using the symbol.

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21

The Square of a Binomial

Given any expression that can be written in the form 1A B2 2, 1. 1A B2 2 A2 2AB B2 2. 1A B2 2 A2 2AB B2 䊳

CAUTION

EXAMPLE 16

䊳

Solution

䊳

Note the square of a binomial always results in a trinomial (three terms). In particular, 1A B2 2 A2 B2.

Find each binomial square without using F-O-I-L: a. 1a 92 2

F. You’ve just seen how we can compute special products: binomial conjugates and binomial squares

b. 13x 52 2

a. 1a 92 2 a2 21a # 92 92 a2 18a 81 b. 13x 52 2 13x2 2 213x # 52 52 9x2 30x 25 c. 13 1x2 2 9 213 # 1x2 1 1x2 2 9 6 1x x

c. 13 1x2 2

1A B2 2 A 2 2AB B 2

simplify 1A B2 2 A 2 2AB B 2 simplify 1A B2 2 A 2 2AB B 2 simplify

Now try Exercises 125 through 136 䊳 With practice, you will be able to go directly from the binomial square to the resulting trinomial.

R.2 EXERCISES 䊳



1. The equation 1x2 2 3 x6 is an example of the property of exponents. 2. The equation 1x3 2 2 property of

1 is an example of the x6 exponents.

3. The sum of the “outers” and “inners” for 12x 52 2 is , while the sum of the outers and inners for 12x 52 12x 52 is . 䊳

4. The expression 2x2 3x 10 can be classified as a of degree , with a leading coefficient of . 5. Discuss/Explain why one of the following expressions can be simplified further, while the other cannot: (a) 7n4 3n2; (b) 7n4 # 3n2. 6. Discuss/Explain why the degree of 2x2y3 is greater than the degree of 2x 2 y 3. Include additional examples for contrast and comparison.


Determine each product using the product and/or power properties.

7.

2 2# n 21n5 3

10. 11.5vy2 2 18v4y2

8. 24g5

# 3 g9 8

11. 1a2 2 4 # 1a3 2 2 # b2 # b5

9. 16p2q2 12p3q3 2 12. d 2 # d 4 # 1c5 2 2 # 1c3 2 2

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22

Simplify using the product to a power property.

13. 16pq2 2 3

14. 13p2q2 2

15. 13.2hk2 2 3 17. a

16. 12.5h5k2 2

p 2 b 2q

18. a

19. 10.7c 2 110c d 2 4 2

21.

R–22


1 34x3y2 2

3 2 2

b 3 b 3a

5m2n3 2 b 2r4

45. a

5p2q3r4

2 2 3

22. 1 45x3 2 2

24. 1 23m2n2 2 # 1 12mn2 2

47.

25. Volume of a cube: The 3x2 formula for the volume of a cube is V S3, where S is the length of one edge. If the length of each edge is 3x2, 3x2 a. Find a formula for volume in terms of the variable x. b. Find the volume of the cube if x 2.

49.

23. 138x2 2 116xy2 2

26. Area of a circle: The formula for the area of a circle is A ␲r2, where r is the length of the radius. If the radius is given as 5x3, a. Find a formula for area in terms of the variable x. b. Find the area of the circle if x 2.

3x2

5x3

Simplify using the quotient property or the property of negative exponents. Write answers using positive exponents only.

27.

6w 2w2

28.

8z 16z5

29.

12a3b5 4a2b4

30.

5m3n5 10mn2

5

7

31. 1 23 2 3 33.

2 h3

32. 1 56 2 1 34.

3 m2

35. 122 3

36. 142 2

37.

38.

3 1 1 2 2

2 1 2 3 2

Simplify each expression using the quotient to a power property.

39. a

2p4

41. a

0.2x2 3 b 0.3y3

q3

b

2

40. a

5v4 2 b 7w3

42. a

0.5a3 2 b 0.4b2

44. a 2

b 2pq2r4

46. a

4p3 3x2y

b

3

9p3q2r3 12p5qr

3

b 2

Use properties of exponents to simplify the following. Write the answer using positive exponents only.

20. 12.5a 2 13a b 2 3 2

43. a

51.

9p6q4 12p4q6 20h2 12h5 1a2 2 3 a4 # a5

a3 # b 4 b c2 612x3 2 2

48. 50. 52.

53. a

54.

55.

56.

57.

10x2

14a3bc0 713a2b2c2 3

59. 40 50 61. 21 51 63. 30 31 32 65. 5x0 15x2 0

58.

5m5n2 10m5n 5k3 20k2 153 2 4

59 1p4q8 2 2 p5q2

18n3 813n2 2 3 312x3y4z2 2 18x2yz0

60. 132 0 172 0 62. 41 81

64. 22 21 20 66. 2n0 12n2 0

Convert the following numbers to scientific notation.

67. In mid-2009, the U.S. Census Bureau estimated the world population at nearly 6,770,000,000 people. 68. The mass of a proton is generally given as 0.000 000 000 000 000 000 000 000 001 670 kg. Convert the following numbers to decimal notation.

69. The smallest microprocessors in common use measure 6.5 109 m across. 70. In 2009, the estimated net worth of Bill Gates, the founder of Microsoft, was 5.8 1010 dollars. Compute using scientific notation. Show all work.

71. The average distance between the Earth and the planet Jupiter is 465,000,000 mi. How many hours would it take a satellite to reach the planet if it traveled an average speed of 17,500 mi per hour? How many days? Round to the nearest whole. 72. In fiscal terms, a nation’s debt-per-capita is the ratio of its total debt to its total population. In the year 2009, the total U.S. debt was estimated at $11,300,000,000,000, while the population was estimated at 305,000,000. What was the U.S. debtper-capita ratio for 2009? Round to the nearest whole dollar.

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Identify each expression as a polynomial or nonpolynomial (if a nonpolynomial, state why); classify each as a monomial, binomial, trinomial, or none of these; and state the degree of the polynomial.

73. 35w3 2w2 112w2 14 74. 2x3 23x2 12x 1.2

4 75. 5n2 4n 117 76. 3 2.7r2 r 1 r 2 3 77. p 5 78. q3 2q2 5q Write each polynomial in standard form and name the leading coefficient.

79. 7w 8.2 w3 3w2 80. 2k2 12 k 81. c3 6 2c2 3c

82. 3v3 14 2v2 112v2 83. 12 23x2

84. 8 2n 7n 2

96. 1s 32 15s 42

97. 1x 321x2 3x 92

98. 1z 52 1z2 5z 252

99. 1b2 3b 282 1b 22

100. 12h2 3h 82 1h 12 101. 17v 4213v 52 103. 13 m213 m2

102. 16w 1212w 52 104. 15 n215 n2

105. 1p 2.521p 3.62 106. 1q 4.921q 1.22 107. 1x 12 2 1x 14 2

109. 1m 34 2 1m 34 2

108. 1z 13 21z 56 2

110. 1n 25 21n 25 2

111. 13x 2y212x 5y2 112. 16a b21a 3b2

113. 14c d213c 5d2 114. 15x 3y212x 3y2 115. 12x2 52 1x2 32 116. 13y2 2212y2 12

For each binomial, determine its conjugate and find the product of the binomial with its conjugate.

117. 4m 3

118. 6n 5

85. 13p3 4p2 2p 72 1p2 2p 52

119. 7x 10

120. c 3

121. 6 5k

122. 11 3r

87. 15.75b2 2.6b 1.92 12.1b2 3.2b2

123. x 16

124. p 12

Find the indicated sum or difference.

86. 15q2 3q 42 13q2 3q 42

88. 10.4n2 5n 0.52 10.3n2 2n 0.752

Find each binomial square.

90. 1 59n2 4n 12 2 1 23n2 2n 34 2

127. 14g 32 2

89. 1 34x2 5x 22 1 12x2 3x 42

91. Subtract q5 2q4 q2 2q from q6 2q5 q4 2q3 using a vertical format. 92. Find x4 2x3 x2 2x decreased by x4 3x3 4x2 3x using a vertical format. Compute each product.

93. 3x1x2 x 62

94. 2v2 1v2 2v 152 95. 13r 52 1r 22 䊳

23

125. 1x 42 2

126. 1a 32 2

129. 14p 3q2 2

130. 15c 6d2 2

131. 14 1x2 2

128. 15x 32 2

132. 1 1x 72 2

Compute each product.

133. 1x 321y 22

134. 1a 321b 52

135. 1k 521k 621k 22

136. 1a 621a 121a 52


137. Medication in the bloodstream: M ⴝ 0.5t 4 ⴙ 3t 3 ⴚ 97t 2 ⴙ 348t If 400 mg of a pain medication are taken orally, the number of milligrams in the bloodstream is modeled by the formula shown, where M is the number of milligrams and t is the time in hours, 0 t 6 5. Construct a table of values for t 1 through 5, then answer the following. a. How many milligrams are in the bloodstream after 2 hr? After 3 hr? b. Based on part a, would you expect the number of milligrams in the bloodstream after 4 hr to be less or more? Why? c. Approximately how many hours until the medication wears off (the number of milligrams in the bloodstream is 0)?

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Aa 138. Amount of a mortgage payment: M ⴝ

r n r b a1 ⴙ b 12 12

r n b ⴚ1 12 The monthly mortgage payment required to pay off (or amortize) a loan is given by the formula shown, where M is the monthly payment, A is the original amount of the loan, r is the annual interest rate, and n is the term of the loan in months. Find the monthly payment (to the nearest cent) required to purchase a $198,000 home, if the interest rate is 6.5% and the home is financed over 30 yr.

䊳

a1 ⴙ

APPLICATIONS

139. Attraction between particles: In electrical theory, the force of attraction between two particles P and kPQ Q with opposite charges is modeled by F 2 , d where d is the distance between them and k is a constant that depends on certain conditions. This is known as Coulomb’s law. Rewrite the formula using a negative exponent. 140. Intensity of light: The intensity of illumination from a light source depends on the distance from k the source according to I 2 , where I is the d intensity measured in footcandles, d is the distance from the source in feet, and k is a constant that depends on the conditions. Rewrite the formula using a negative exponent. 141. Rewriting an expression: In advanced mathematics, negative exponents are widely used because they are easier to work with than rational expressions. 5 3 2 Rewrite the expression 3 2 1 4 using x x x negative exponents. 142. Swimming pool hours: A swimming pool opens at 8 A.M. and closes at 6 P.M. In summertime, the

䊳

number of people in the pool at any time can be approximated by the formula S1t2 t 2 10t, where S is the number of swimmers and t is the number of hours the pool has been open (8 A.M.: t 0, 9 A.M.: t 1, 10 A.M.: t 2, etc.). a. How many swimmers are in the pool at 6 P.M.? Why? b. Between what times would you expect the largest number of swimmers? c. Approximately how many swimmers are in the pool at 3 P.M.? d. Create a table of values for t 1, 2, 3, 4, . . . and check your answer to part b. 143. Maximizing revenue: A sporting goods store finds that if they price their video games at $20, they make 200 sales per day. For each decrease of $1, 20 additional video games are sold. This means the store’s revenue can be modeled by the formula R 120 1x2 1200 20x2, where x is the number of $1 decreases. Multiply out the binomials and use a table of values to determine what price will give the most revenue. 144. Maximizing revenue: Due to past experience, a jeweler knows that if they price jade rings at $60, they will sell 120 each day. For each decrease of $2, five additional sales will be made. This means the jeweler’s revenue can be modeled by the formula R 160 2x2 1120 5x2, where x is the number of $2 decreases. Multiply out the binomials and use a table of values to determine what price will give the most revenue.


145. If 13x2 kx 12 1kx2 5x 72 12x2 4x k2 x2 3x 2, what is the value of k?

1 2 1 b 5, then the expression 4x2 2 2x 4x is equal to what number?

146. If a2x

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R.3

Solving Linear Equations and Inequalities

LEARNING OBJECTIVES

In a study of algebra, you will encounter many families of equations, or groups of equations that share common characteristics. Of interest to us here is the family of linear equations in one variable, a study that lays the foundation for understanding more advanced families. This section will also lay the foundation for solving a formula for a specified variable, a practice widely used in science, business, industry, and research.

In Section R.3 you will review how to:

A. Solve linear equations

B.

C. D. E.

F.

using properties of equality Recognize equations that are identities or contradictions Solve linear inequalities Solve compound inequalities Solve basic applications of linear equations and inequalities Solve applications of basic geometry

CAUTION

A. Solving Linear Equations Using Properties of Equality An equation is a statement that two expressions are equal. From the expressions 31x ⫺ 12 ⫹ x and ⫺x ⫹ 7, we can form the equation 31x ⫺ 12 ⫹ x ⫽ ⫺x ⫹ 7,

Table R.1 x

31x ⴚ 12 ⴙ x

ⴚx ⴙ 7

⫺2

⫺11

9

⫺1

⫺7

8

which is a linear equation in one variable (the 0 exponent on any variable is a 1). To solve an equa1 tion, we attempt to find a specific input or x-value 2 that will make the equation true, meaning the left3 hand expression will be equal to the right. Using 4 Table R.1, we find that 31x ⫺ 12 ⫹ x ⫽ ⫺x ⫹ 7 is a true equation when x is replaced by 2, and is a false equation otherwise. Replacement values that make the equation true are called solutions or roots of the equation. 䊳

⫺3

7

1

6

5

5

9

4

13

3

From Section R.1, an algebraic expression is a sum or difference of algebraic terms. Algebraic expressions can be simplified, evaluated or written in an equivalent form, but cannot be “solved,” since we’re not seeking a specific value of the unknown.

Solving equations using a table is too time consuming to be practical. Instead we attempt to write a sequence of equivalent equations, each one simpler than the one before, until we reach a point where the solution is obvious. Equivalent equations are those that have the same solution set, and can be obtained by using the distributive property to simplify the expressions on each side of the equation. The additive and multiplicative properties of equality are then used to obtain an equation of the form x ⫽ constant. The Additive Property of Equality

The Multiplicative Property of Equality

If A, B, and C represent algebraic expressions and A ⫽ B,

If A, B, and C represent algebraic expressions and A ⫽ B,

then A ⫹ C ⫽ B ⫹ C

then AC ⫽ BC and

B A ⫽ , 1C ⫽ 02 C C

In words, the additive property says that like quantities, numbers, or terms can be added to both sides of an equation. A similar statement can be made for the multiplicative property. These properties are combined into a general guide for solving linear equations, which you’ve likely encountered in your previous studies. Note that not all steps in the guide are required to solve every equation.

R–25

25

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Guide to Solving Linear Equations in One Variable • Eliminate parentheses using the distributive property, then combine any like terms. • Use the additive property of equality to write the equation with all variable terms on one side, and all constants on the other. Simplify each side. • Use the multiplicative property of equality to obtain an equation of the form x constant. • For applications, answer in a complete sentence and include any units of measure indicated. For our first example, we’ll use the equation 31x 12 x x 7 from our initial discussion.

EXAMPLE 1

䊳

Solving a Linear Equation Using Properties of Equality Solve for x: 31x 12 x x 7.

Solution

䊳

31x 12 x x 7 3x 3 x x 7 4x 3 x 7 5x 3 7 5x 10 x2

original equation distributive property combine like terms add x to both sides (additive property of equality) add 3 to both sides (additive property of equality) multiply both sides by 15 or divide both sides by 5 (multiplicative property of equality)

As we noted in Table R.1, the solution is x 2. Now try Exercises 7 through 12

䊳

To check a solution by substitution means we substitute the solution back into the original equation (this is sometimes called back-substitution), and verify the lefthand side is equal to the right. For Example 1 we have: 31x 12 x x 7

original equation

312 12 2 2 7

substitute 2 for x

3112 2 5 5 5✓

simplify solution checks

If any coefficients in an equation are fractional, multiply both sides by the least common denominator (LCD) to clear the fractions. Since any decimal number can be written in fraction form, the same idea can be applied to decimal coefficients.

EXAMPLE 2

䊳

Solution

䊳

A. You’ve just seen how we can solve linear equations using properties of equality

Solving a Linear Equation with Fractional Coefficients Solve for n: 14 1n 82 2 12 1n 62. 1 4 1n 82 1 4n 2

2 12 1n 62 2 12n 3 1 1 4n 2n 3 1 41 4n2 41 12n 32 n 2n 12 n 12 n 12

original equation distributive property combine like terms multiply both sides by LCD 4 distributive property subtract 2n multiply by 1

Verify the solution is n 12 using back-substitution. Now try Exercises 13 through 30

䊳

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27

Section R.3 Solving Linear Equations and Inequalities

B. Identities and Contradictions Example 1 illustrates what is called a conditional equation, since the equation is true for x 2, but false for all other values of x. The equation in Example 2 is also conditional. An identity is an equation that is always true, no matter what value is substituted for the variable. For instance, 21x 32 2x 6 is an identity with a solution set of all real numbers, written as 5x|x 僆⺢6, or x 僆 1q, q 2 in interval notation. Contradictions are equations that are never true, no matter what real number is substituted for the variable. The equations x 3 x 1 and 3 1 are contradictions. To state the solution set for a contradiction, we use the symbol “” (the null set) or “{ }” (the empty set). Recognizing these special equations will prevent some surprise and indecision in later chapters.

EXAMPLE 3

䊳

Solving Equations (Special Cases) Solve each equation and state the solution set. a. 21x 42 10x 8 413x 12 b. 8x 16 10x2 24 613x 52

Solution

䊳

a. 21x 42 10x 8 413x 12 2x 8 10x 8 12x 4 12x 8 12x 12 8 12

original equation distributive property combine like terms subtract 12x; contradiction

Since 8 is never equal to 12, the original equation is a contradiction. The solution set is empty: { }

b. 8x 16 10x2 24 613x 52 8x 6 10x 24 18x 30 18x 6 18x 6 6 6

original equation distributive property combine like terms subtract 18x; identity

The result shows that the original equation is an identity, with an infinite number of solutions: 5x | x 僆⺢6 . You may recall this notation is read, “the set of all numbers x, such that x is a real number.” Now try Exercises 31 through 36

B. You’ve just seen how we can recognize equations that are identities or contradictions

䊳

In Example 3(a), our attempt to solve for x ended with all variables being eliminated, leaving an equation that is always false —a contradiction (8 is never equal to 12). There is nothing wrong with the solution process, the result is simply telling us the original equation has no solution. In Example 3(b), all variables were again eliminated but the end result was always true —an identity (6 is always equal to 6). Once again we’ve done nothing wrong mathematically, the result is just telling us that the original equation will be true no matter what value of x we use for an input.

C. Solving Linear Inequalities A linear inequality resembles a linear equality in many respects: Linear Inequality

Related Linear Equation

(1)

x 6 3

x3

(2)

3 p 2 12 8

3 p 2 12 8

A linear inequality in one variable is one that can be written in the form ax b 6 c, where a, b, and c 僆⺢ and a 0. This definition and the following properties also apply when other inequality symbols are used. Solutions to simple inequalities are easy to spot. For instance, x 2 is a solution to x 6 3 since 2 6 3. For more involved inequalities we use the additive property of inequality and the multiplicative property of

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inequality. Similar to solving equations, we solve inequalities by isolating the variable on one side to obtain a solution form such as variable 6 number. The Additive Property of Inequality If A, B, and C represent algebraic expressions and A 6 B, then A C 6 B C Like quantities (numbers or terms) can be added to both sides of an inequality. While there is little difference between the additive property of equality and the additive property of inequality, there is an important difference between the multiplicative property of equality and the multiplicative property of inequality. To illustrate, we begin with 2 6 5. Multiplying both sides by positive three yields 6 6 15, a true inequality. But notice what happens when we multiply both sides by negative three: 2 6 5

original inequality

2132 6 5132 6 6 15

multiply by negative three false

The result is a false inequality, because 6 is to the right of 15 on the number line. Multiplying (or dividing) an inequality by a negative quantity reverses the order relationship between two quantities (we say it changes the sense of the inequality). We must compensate for this by reversing the inequality symbol. 6 7 15

change direction of symbol to maintain a true statement

For this reason, the multiplicative property of inequality is stated in two parts. The Multiplicative Property of Inequality

EXAMPLE 4

䊳

If A, B, and C represent algebraic expressions and A 6 B, then AC 6 BC

If A, B, and C represent algebraic expressions and A 6 B, then AC 7 BC

if C is a positive quantity (inequality symbol remains the same).

if C is a negative quantity (inequality symbol must be reversed).

Solving an Inequality Solve the inequality, then graph the solution set and write it in interval notation: 2 1 5 3 x 2 6.

Solution

WORTHY OF NOTE As an alternative to multiplying or dividing by a negative value, the additive property of inequality can be used to ensure the variable term will be positive. From Example 4, the inequality 4x 2 can be written as 2 4x by adding 4x to both sides and subtracting 2 from both sides. This gives the solution 12 x, which is equivalent to x

12 .

䊳

1 5 2 x 3 2 6 2 1 5 6 a x b 162 3 2 6 4x 3 5 4x 2 4x 2 ⱖ 4 4 1 x 2

original inequality

clear fractions (multiply by LCD) simplify subtract 3 divide by 4, reverse inequality sign result

1 2

• Graph:

3 2 1

[ 0

1

2

3

• Interval notation: x 僆 312, q 2

4


䊳

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To check a linear inequality, you often have an infinite number of choices—any number from the solution set/interval. If a test value from the solution interval results in a true inequality, all numbers in the interval are solutions. For Example 4, using x 0 results in the true statement 12 56 ✓. Some inequalities have all real numbers as the solution set: 5x| x 僆⺢6, while other inequalities have no solutions, with the answer given as the empty set: { }.

EXAMPLE 5

䊳

Solving Inequalities Solve the inequality and write the solution in set notation: a. 7 13x 52 21x 42 5x b. 31x 42 5 6 21x 32 x

Solution

䊳

a. 7 13x 52 21x 42 5x 7 3x 5 2x 8 5x 3x 2 3x 8 2 8

original inequality distributive property combine like terms add 3x

Since the resulting statement is always true, the original inequality is true for all real numbers. The solution is all real numbers ⺢. b. 31x 42 5 3x 12 5 3x 7 7

6 6 6 6

21x 32 x 2x 6 x 3x 6 6

original inequality distribute combine like terms subtract 3x

Since the resulting statement is always false, the original inequality is false for all real numbers. The solution is { }. C. You’ve just seen how we can solve linear inequalities


䊳

D. Solving Compound Inequalities In some applications of inequalities, we must consider more than one solution interval. These are called compound inequalities, and these require us to take a close look at the operations of union “ ´ ” and intersection “ ¨”. The intersection of two sets A and B, written A ¨ B, is the set of all elements common to both sets. The union of two sets A and B, written A ´ B, is the set of all elements that are in either set. When stating the union of two sets, repetitions are unnecessary.

EXAMPLE 6

䊳

Finding the Union and Intersection of Two Sets

Solution

䊳

A ¨ B is the set of all elements in both A and B: A 傽 B 51, 2, 36. A ´ B is the set of all elements in either A or B: A ´ B 52, 1, 0, 1, 2, 3, 4, 56.

WORTHY OF NOTE For the long term, it may help to rephrase the distinction as follows. The intersection is a selection of elements that are common to two sets, while the union is a collection of the elements from two sets (with no repetitions).

For set A 52, 1, 0, 1, 2, 36 and set B 51, 2, 3, 4, 56, determine A ¨ B and A ´ B.


䊳

Notice the intersection of two sets is described using the word “and,” while the union of two sets is described using the word “or.” When compound inequalities are formed using these words, the solution is modeled after the ideas from Example 6. If “and” is used, the solutions must satisfy both inequalities. If “or” is used, the solutions can satisfy either inequality.

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EXAMPLE 7

䊳

Solving a Compound Inequality Solve the compound inequality, then write the solution in interval notation: 3x 1 6 4 or 4x 3 6 6.

Solution

䊳

Begin with the statement as given: 3x 1 6 4 3x 6 3 x 7 1

4x 3 6 6 4x 6 9 9 x 6 4

or or or

original statement isolate variable term solve for x, reverse first inequality symbol

The solution x 7 1 or x 6 94 is better understood by graphing each interval separately, then selecting both intervals (the union). WORTHY OF NOTE

x 1:

The graphs from Example 7 clearly show the solution consists of two disjoint (disconnected) intervals. This is reflected in the “or” statement: x 6 94 or x 7 1, and in the interval notation. Also, note the solution x 6 94 or x 7 1 is not equivalent to 94 7 x 7 1, as there is no single number that is both greater than 1 and less than 94 at the same time.

x 9 : 4

)

8 7 6 5 4 3 2 1

0

1

2

3

4

5

6

9 4

)

8 7 6 5 4 3 2 1

x 9 or x 1: 4

0

1

2

3

4

5

6

9 4

)

)

8 7 6 5 4 3 2 1

0

1

2

3

4

5

6

9 Interval notation: x 僆 aq, b ´ 11, q 2. 4 Now try Exercises 59 and 60

EXAMPLE 8

䊳

䊳

Solving a Compound Inequality Solve the compound inequality, then write the solution in interval notation: 3x 5 7 13 and 3x 5 6 1.

Solution

䊳

Begin with the statement as given: 3x 5 7 13 3x 7 18 x 7 6

3x 5 6 1 3x 6 6 x 6 2

and and and

original statement subtract five divide by 3

The solution x 7 6 and x 6 2 can best be understood by graphing each interval separately, then noting where they intersect. WORTHY OF NOTE x 6: x 2: x 6 and x 2:

)

8 7 6 5 4 3 2 1

)

8 7 6 5 4 3 2 1

)

The inequality a 6 b (a is less than b) can equivalently be written as b 7 a (b is greater than a). In Example 8, the solution is read, “x 7 6 and x 6 2,” but if we rewrite the first inequality as 6 6 x (with the “arrowhead” still pointing at 62, we have 6 6 x and x 6 2 and can clearly see that x must be in the single interval between 6 and 2.

0

1

2

3

4

5

6

0

1

2

3

4

5

6

0

1

2

3

4

5

6

)

8 7 6 5 4 3 2 1

Interval notation: x 僆 16, 22. Now try Exercises 61 through 72

䊳

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The solution from Example 8 consists of the single interval 16, 22, indicating the original inequality could actually be joined and written as 6 6 x 6 2, called a joint inequality. We solve joint inequalities in much the same way as linear inequalities, but must remember they have three parts (left, middle, and right). This means operations must be applied to all three parts in each step of the solution process, to obtain a solution form such as smaller number 6 x 6 larger number. The same ideas apply when other inequality symbols are used.

EXAMPLE 9

䊳

Solving a Joint Inequality Solve the joint inequality, then graph the solution set and write it in interval 2x 5 6. notation: 1 7 3

Solution

䊳

2x 5 6 3 3 6 2x 5 18 8 6 2x 13 13 4 6 x 2 1 7

original inequality multiply all parts by 3; reverse the inequality symbols subtract 5 from all parts divide all parts by 2 13 2

• Graph:

)

5 4 3 2 1

[ 0

• Interval notation: x 僆 14,

D. You’ve just seen how we can solve compound inequalities

1

13 2 4

2

3

4

5

6

7

8


䊳

E. Solving Basic Applications of Linear Equations and Inequalities Applications of linear equations and inequalities come in many forms. In most cases, you are asked to translate written relationships or information given verbally into an equation using words or phrases that indicate mathematical operations or relationships. Here, we’ll practice this skill using ideas that were introduced in Section R.1, where we translated English phrases into mathematical expressions. Very soon these skills will be applied in much more significant ways.

EXAMPLE 10

䊳

Translating Written Information into an Equation Translate the following relationships into equations, then solve: In an effort to lower the outstanding balance on her credit card, Laura paid $10 less than triple her normal payment. If she sent the credit card company $350.75, how much was her normal payment? (See Section R.1, Examples 2 and 3.)

Solution

䊳

Let p represent her normal payment. Then “triple her normal payment” would be 3p, and “ten less than triple” would be 3p 10. Since “she sent the company $350.75,” we have

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3p 10 350.75 3p 360.75 p 120.25

equation form add 10 divide by 3

Laura’s normal payment is $120.25 per month. A calculator check is shown in the figure.


䊳

Inequalities are widely used to help gather information, and to make comparisons that will lead to informed decisions.

EXAMPLE 11

䊳

Using an Inequality to Compute Desired Test Scores Justin earned scores of 78, 72, and 86 on the first three out of four exams. What score must he earn on the fourth exam to have an average of at least 80?

Solution

䊳

The current scores are 78, 72, and 86. An average of at least 80 means A 80. In organized form: Test 1

Test 2

Test 3

Test 4

Computed Average

Minimum

78

72

86

x

78 72 86 x 4

80

Let x represent Justin’s score on the fourth exam, then represents his average score. 78 72 86 x 80 4 78 72 86 x 320 236 x 320 x 84 E. You’ve just seen how we can solve applications of linear equations and inequalities

78 72 86 x 4

average must be greater than or equal to 80 multiply by 4 simplify solve for x (subtract 236)

Justin must score at least an 84 on the last exam to earn an 80 average. Now try Exercises 91 through 100

䊳

F. Solving Applications of Basic Geometry As your translation skills grow, your ability to solve a wider range of more significant applications will grow as well. In many cases, the applications will involve some basic geometry and the most often used figures and formulas appear here. For a more complete review of geometry, see Appendix II Geometry Review with Unit Conversions, which is posted online at www.mhhe.com/coburn.

Perimeter and Area Perimeter is a measure of the distance around a two dimensional figure. As this is a linear measure, results are stated in linear units as in centimeters (cm), feet (ft), kilometers (km), miles (mi), and so on. If no unit is specified, simply write the result as x units. Area is a measure of the surface of a two dimensional figure, with results stated in square units as in x units2. Some of the most common formulas involving perimeter and area are given in Table R.2A.

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Table R.2A Perimeter Formula (linear units or units)

Definition and Diagram

Area Formula (square units or units2)

a three-sided polygon s1

Triangle

s2

P ⫽ s1 ⫹ s2 ⫹ s3

h

s3

A⫽

1 bh 2

b

a quadrilateral with four right angles and opposite sides parallel Rectangle W

P ⫽ 2L ⫹ 2W

A ⫽ LW

P ⫽ 4S

A ⫽ S2

L a rectangle with four equal sides Square

S

a quadrilateral with one pair of parallel sides (called bases b1 and b2) b1

s2

Trapezoid s1

s3 s4

Circle

sum of all sides P ⫽ s1 ⫹ s2 ⫹ s3 ⫹ s4

h

A⫽

h 1b1 ⫹ b2 2 2

b2

the set of all points lying in a plane that are an equal distance (called the radius r) from a given point (called the center C).

r C

C ⫽ 2␲r or C ⫽ ␲d

A ⫽ ␲r2

If an exercise or application uses a formula, begin by stating the formula first. Using the formula as a template for the values substituted will help to prevent many careless errors.

EXAMPLE 12A

䊳

Computing the Area of a Trapezoidal Window A basement window is shaped like an isosceles trapezoid (base angles equal, nonparallel sides equal in length), with a height of 10 in. and bases of 1.5 ft and 2 ft. What is the area of the glass in the window?

1.5 ft

10 in.

2 ft

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Solution

䊳

Before applying the area formula, all measures must use the same unit. In inches, we have 1.5 ft ⫽ 18 in. and 2 ft ⫽ 24 in. h 1b1 ⫹ b2 2 2 10 in. 118 in. ⫹ 24 in.2 A⫽ 2 A ⫽ 15 in.2 142 in.2 A ⫽ 210 in2 A⫽

given formula substitute 10 for h, 18 for b1, and 24 for b2 simplify result

The area of the glass in the window is 210 in2. Now try Exercises 101 and 102

䊳

Volume Volume is a measure of the amount of space occupied by a three dimensional object and is measured in cubic units. Some of the more common formulas are given in Table R.2B. Table R.2B Volume Formula (cubic units or units3)

Definition and Diagram Rectangular solid

Cube

a six-sided, solid figure with opposite faces congruent and adjacent faces meeting at right angles

H

V ⫽ LWH

W

L

a rectangular solid with six congruent, square faces

V ⫽ S3 S

Sphere

the set of all points in space, an equal distance (called the radius) from a given point (called the center)

Right circular cylinder

union of all line segments connecting two congruent circles in parallel planes, meeting each at a right angle

Right circular cone

Right pyramid

union of all line segments connecting a given point (vertex) to a given circle (base) and whose altitude meets the center of the base at a right angle union of all line segments connecting a given point (vertex) to a given square (base) and whose altitude meets the center of the base at a right angle

4 V ⫽ ␲r3 3

r

C

h

V ⫽ ␲r2h

r 1 V ⫽ ␲r2h 3

h r

1 V ⫽ s2h 3 h s

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EXAMPLE 12B

䊳

Computing the Volume of a Composite Figure Sand at a cement factory is being dumped from a conveyor belt into a pile shaped like a right circular cone atop a right circular cylinder (see figure). How many cubic feet of sand are there at the moment the cone is 6 ft high with a diameter of 10 ft?

Solution

䊳

F. You’ve just seen how we can solve applications of basic geometry

Total Volume ⫽ volume of cylinder ⫹ volume of cone 1 V ⫽ ␲r2h1 ⫹ ␲r2h2 3 1 ⫽ ␲152 2 132 ⫹ ␲152 2 162 3 ⫽ 75␲ ⫹ 50␲ ⫽ 125␲

6 ft 3 ft 10 ft

verbal model formula model (note h1 ⫽ h2 2 substitute 5 for r, 3 for h1, and 6 for h2 simplify result (exact form)

There are about 392.7 ft3 of sand in the pile. Now try Exercises 103 and 104

R.3 EXERCISES 䊳



䊳

of sets A and B is written A 傽 B. of sets A and B is written A ´ B.

1. A(n) is an equation that is always true, regardless of the value while a(n) is an equation that is always false, regardless of the value.

4. The The

2. For inequalities, the three ways of writing a solution set are notation, a number line graph, and notation.

5. Discuss/Explain the similarities and differences between the properties of equality for equations and those for inequalities.

3. The mathematical sentence 3x ⫹ 5 6 7 is a(n) inequality, while ⫺2 6 3x ⫹ 5 6 7 is a(n) inequality.

6. Discuss/Explain the use of the words “and” and “or” in the statement of compound inequalities. Include a few examples to illustrate.


Solve each equation. Check your answer by substitution.

7. 4x ⫹ 31x ⫺ 22 ⫽ 18 ⫺ x 8. 15 ⫺ 2x ⫽ ⫺41x ⫹ 12 ⫹ 9

9. 21 ⫺ 12v ⫹ 172 ⫽ ⫺7 ⫺ 3v

10. ⫺12 ⫺ 5w ⫽ ⫺9 ⫺ 16w ⫹ 72

11. 8 ⫺ 13b ⫹ 52 ⫽ ⫺5 ⫹ 21b ⫹ 12 12. 2a ⫹ 41a ⫺ 12 ⫽ 3 ⫺ 12a ⫹ 12

Solve each equation.

13. 15 1b ⫹ 102 ⫺ 7 ⫽ 13 1b ⫺ 92

14. 61 1n ⫺ 122 ⫽ 14 1n ⫹ 82 ⫺ 2 15. 32 1m ⫹ 62 ⫽ ⫺1 2 16. 45 1n ⫺ 102 ⫽

⫺8 9

17. 12x ⫹ 5 ⫽ 13x ⫹ 7 19.

x⫹3 x ⫹ ⫽7 5 3

18. ⫺4 ⫹ 23y ⫽ 12y ⫺ 5 20.

z⫺4 z ⫺2⫽ 6 2

䊳

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21. 15 6

3p 8

22. 15

2q 21 9

Solve each inequality and write the solution in set notation.

23. 0.2124 7.5a2 6.1 4.1

47. 7 21x 32 4x 61x 32

24. 0.4117 4.25b2 3.15 4.16

48. 3 61x 52 217 3x2 1

25. 6.2v 12.1v 52 1.1 3.7v

26. 7.9 2.6w 1.5w 19.1 2.1w2 n 2 n 2 5 3 m 2 m 28. 3 5 4 p p 29. 3p 5 2p 6 4 6 q q 30. 1 3q 2 4q 6 8 27.

Identify the following equations as an identity, a contradiction, or a conditional equation, then state the solution.

31. 314z 52 15z 20 3z 32. 5x 9 2 512 x2 1 33. 8 813n 52 5 611 n2 34. 2a 41a 12 1 312a 12 35. 414x 52 6 218x 72 36. 15x 32 2x 11 41x 22 Write the solution set illustrated on each graph in set notation and interval notation.

37. 38. 39. 40.

[

3 2 1

0

1

0

[

1

2

51. 61p 12 2p 212p 32

52. 91w 12 3w 215 3w2 1 Determine the intersection and union of sets A, B, C, and D as indicated, given A ⴝ 5ⴚ3, ⴚ2, ⴚ1, 0, 1, 2, 36, B ⴝ 52, 4, 6, 86, C ⴝ 5ⴚ 4, ⴚ2, 0, 2, 46, and D ⴝ 54, 5, 6, 76 .

53. A 傽 B and A ´ B 54. A 傽 C and A ´ C 55. A 傽 D and A ´ D 56. B 傽 C and B ´ C

57. B 傽 D and B ´ D 58. C 傽 D and C ´ D Express the compound inequalities graphically and in interval notation.

59. x 6 2 or x 7 1 60. x 6 5 or x 7 5 61. x 6 5 and x 2 62. x 4 and x 6 3 63. x 3 and x 1

Solve the compound inequalities and graph the solution set.

65. 41x 12 20 or x 6 7 9 66. 31x 22 7 15 or x 3 1

69. 35x 12 7

0

1

2

0

1

2

) 4

Solve the inequality and write the solution in set notation. Then graph the solution and write it in interval notation.

41. 5a 11 2a 5 42. 8n 5 7 2n 12 43. 21n 32 4 5n 1 44. 51x 22 3 6 3x 11 3x x 45. 6 4 8 4

3 10

and 4x 7 1

70. 23x 56 0 and 3x 6 2

3

3

64. x 5 and x 7

68. 3x 5 17 and 5x 0

3

[

[

3 2 1

50. 8 16 5m2 7 9m 13 4m2

67. 2x 7 3 and 2x 0

3

)

3 2 1

3 2 1

2

49. 413x 52 18 6 215x 12 2x

2y y 46. 6 2 5 10

71.

x 3x 6 3 or x 1 7 5 8 4

72.

x 2x 6 2 or x 3 7 2 5 10

73. 3 2x 5 6 7 74. 2 6 3x 4 19 75. 0.5 0.3 x 1.7 76. 8.2 6 1.4 x 6 0.9 77. 7 6 34x 1 11 78. 21 23x 9 6 7

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79. Euler’s Polyhedron Formula: V ⴙ F ⴚ E ⴝ 2 Discovered by Leonhard Euler in 1752, this simple but powerful formula states that in any regular polyhedron, the number of vertices V and faces F is always two more than the number of edges E. (a) Verify the formula for a simple cube. (b) Verify the formula for the octahedron shown in the figure. (c) If a dodecahedron has 12 faces and 30 edges, how many vertices does it have? 80. Area of a Regular Polygon: A ⴝ

1 ap 2

The area of any regular polygon can be found a using the formula shown, where a is the apothem of the polygon (perpendicular distance from center to any edge), and p is the perimeter. (a) Verify the formula using a square with sides of length 6 cm. (b) If the hexagon shown has an area of 259.8 cm2 with sides 10 cm in length, what is the length a of the apothem? 䊳

37

81. Body mass index: B ⴝ

704W H2

The U.S. government publishes a body mass index formula to help people consider the risk of heart disease. An index “B” of 27 or more means that a person is at risk. Here W represents weight in pounds and H represents height in inches. If your height is 5¿8– what range of weights will help ensure you remain safe from the risk of heart disease? Source: www.surgeongeneral.gov/topics.

82. Lift capacity: 75S ⴙ 125B ⱕ 750 The capacity in pounds of the lift used by a roofing company to place roofing shingles and buckets of roofing nails on rooftops is modeled by the formula shown, where S represents packs of shingles and B represents buckets of nails. Use the formula to find (a) the largest number of shingle packs that can be lifted, (b) the largest number of nail buckets that can be lifted, and (c) the largest number of shingle packs that can be lifted along with three nail buckets.

APPLICATIONS

Write an equation to model the given information and solve.

83. Celebrity Travel: To avoid paparazzi and overzealous fans, the arrival gates of planes carrying celebrities are often kept secret until the last possible moment. While awaiting the arrival of Angelina Jolie, a large crowd of fans and photographers had gathered at Terminal A, Gate 18. However, the number of fans waiting at Gate 32 was twice that number increased by 5. If there were 73 fans at Gate 32, how many were waiting at Gate 18? (See Section R.1, Example 2a.) 84. Famous Architecture: The Hall of Mirrors is the central gallery of the Palace of Versailles and is one of the most famous rooms in the world. The length of this hall is 11 m less than 8 times the width. If the hall is 73 m long, what is its width? (See Section R.1, Example 2b.) 85. Dietary Goals: At the picnic, Mike abandoned his diet and consumed 13 calories more than twice the number of calories he normally allots for lunch. If he consumed 1467 calories, how many calories are normally allotted for lunch?

86. Marathon Training: While training for the Chicago marathon, Christina’s longest run of the week was 5 mi less than double the shortest. If the longest run was 11.2 mi, how long was the shortest? 87. Actor’s Ages: At the time of this writing, actor Will Smith (Enemy of the State, Seven Pounds, others), was 1 yr older than two-thirds the age of Samuel Jackson (The Negotiator, Die Hard III, others). If Will Smith was 41 at this time, how old was Samuel Jackson? 88. Football versus Fútbol: The area of a regulation field for American football is about 410 square meters (m2) less than three-fifths of an Olympicsized soccer field. If an American football field covers 5350 m2, what is the area of an Olympic soccer field?

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89. Forensic Studies: In forensic studies, skeletal remains are analyzed to determine the height, gender, race, age, and other characteristics of the decedent. For instance, the height of a male individual is approximated as 34 in. more than three and one-third times the length of the radial bone. If a live individual is 74 in. tall, how long is his radial bone?

90. Famous Waterways: The Suez Canal and the Panama Canal are two of the most important waterways in the world, saving ships thousands of miles as they journey from port to destination. The length of the Suez Canal is 39 kilometers (km) less than three times the length of the Panama Canal. If the Egyptian canal is 192 km long, how long is the Central American canal?

Write an inequality to model the given information and solve.

91. Exam scores: Jacques is going to college on an academic scholarship that requires him to maintain at least a 75% average in all of his classes. So far he has scored 82%, 76%, 65%, and 71% on four exams. What scores are possible on his last exam that will enable him to keep his scholarship? 92. Timed trials: In the first three trials of the 100-m butterfly, Johann had times of 50.2, 49.8, and 50.9 sec. How fast must he swim the final timed trial to have an average time of at most 50 sec? 93. Checking account balance: If the average daily balance in a certain checking account drops below $1000, the bank charges the customer a $7.50 service fee. The table Weekday Balance gives the daily balance Monday $1125 for one customer. What Tuesday $850 must the daily balance be Wednesday $625 for Friday to avoid a service charge? Thursday $400 94. Average weight: In the Lineman Weight National Football Left tackle 318 lb League, many consider an offensive line to be Left guard 322 lb “small” if the average Center 326 lb weight of the five down Right guard 315 lb linemen is less than Right tackle ? 325 lb. Using the table, what must the weight of the right tackle be so that the line will not be considered small? 95. Area of a rectangle: Given the rectangle shown, what is the range of values for the width, in order to keep the area less than 150m2? 20 m

w

96. Area of a triangle: Using the triangle shown, find the height that will guarantee an area equal to or greater than 48 in2.

h

12 in.

97. Heating and cooling subsidies: As long as the outside temperature is over 45°F and less than 85°F 145 6 F 6 852, the city does not issue heating or cooling subsidies for low-income families. What is the corresponding range of Celsius temperatures C? Recall that F 95C 32. 98. U.S. and European shoe sizes: To convert a European male shoe size “E” to an American male shoe size “A,” the formula A 0.76E 23 can be used. Lillian has five sons in the U.S. military, with shoe sizes ranging from size 9 to size 14 19 A 142. What is the corresponding range of European sizes? Round to the nearest half-size. 99. Power tool rentals: Sunshine Equipment Co. rents its power tools for a $20 fee, plus $4.50/hr. Kealoha’s Rentals offers the same tools for an $11 fee plus $6.00/hr. How many hours h must a tool be rented to make the cost at Sunshine a better deal? 100. Moving van rentals: Stringer Truck Rentals will rent a moving van for $15.75/day plus $0.35 per mile. Bertz Van Rentals will rent the same van for $25/day plus $0.30 per mile. How many miles m must the van be driven to make the cost at Bertz a better deal?

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101. Cost of drywall: After the studs are up, the 3 ft wall shown in the figure 15 ft must be covered in 10 ft 7 ft drywall. (a) How many square feet of drywall 19 ft are needed? (b) If drywall is sold only in 4-ft by 8-ft sheets, approximately how many sheets are required for this job?

5 in. 103. Trophy bases: The base of a new trophy 7 in. has the form of a cylinder sitting atop a rectangular solid. 2 in. If the base is to be 10 in. cast in a special 10 in. aluminum, determine the volume of aluminum to be used.

102. Paving a walkway: Current plans 104. Grain storage: The dimensions of call for building a circular fountain a grain silo are shown in the figure. 6m 6 m in diameter with a circular If the maximum storage capacity of walkway around it that is 1.5 m the silo is 95% of the total volume wide. (a) What is the approximate of the silo, how many cubic meters area of the walkway? (b) If the 1.5 m of corn can be stored? concrete for the walkway is to be 6 cm deep, what volume of cement must be used 11 cm ⫽ 0.01 m2 ? 䊳

39

Section R.4 Factoring Polynomials and Solving Polynomial Equations by Factoring

16 m

6m


105. Solve for x: ⫺314x2 ⫹ 5x ⫺ 22 ⫹ 7x ⫽ 614 ⫺ x ⫺ 2x2 2 ⫺ 19

106. Solve for n: 553 ⫺ 34 ⫺ 215 ⫺ 9n2 4 6 ⫹ 15 ⫽ ⫺655 ⫹ 2 3n ⫺ 1019 ⫹ n2 4 6 107. Use your local library, the Internet, or another resource to find the highest and lowest point on each of the seven continents. Express the range of altitudes for each continent as a joint inequality. Which continent has the greatest range? 108. The sum of two consecutive even integers is greater than or equal to 12 and less than or equal to 22. List all possible values for the two integers.

R.4

Place the correct inequality symbol in the blank to make the statement true.

109. If m 7 0 and n 6 0, then mn

0.

110. If m 7 n and p 7 0, then mp

np.

111. If m 6 n and p 7 0, then mp

np.

112. If m ⱕ n and p 6 0, then mp

np.

113. If m 7 n, then ⫺m

⫺n.

114. If 0 6 m 6 n, then m1

1 n.

115. If m 7 0 and n 6 0, then m2 116. If m 6 0, then m3

n. 0.

Factoring Polynomials and Solving Polynomial Equations by Factoring

LEARNING OBJECTIVES In Section R.4 you will review:

A. Factoring out the greatest common factor

It is often said that knowing which tool to use is just as important as knowing how to use the tool. In this section, we review the tools needed to factor an expression, an important part of solving polynomial equations. This section will also help us decide which factoring tool is appropriate when many different factorable expressions are presented.

B. Common binomial factors and factoring by grouping C. Factoring quadratic polynomials D. Factoring special forms and quadratic forms E. Solving Polynomial Equations by Factoring

A. The Greatest Common Factor To factor an expression means to rewrite the expression as an equivalent product. The distributive property is an example of factoring in action. To factor 2x2 ⫹ 6x, we might first rewrite each term using the common factor 2x: 2x2 ⫹ 6x ⫽ 2x # x ⫹ 2x # 3, then apply the distributive property to obtain 2x1x ⫹ 32. We commonly say that we have factored out 2x. The greatest common factor (or GCF) is the largest factor common to all terms in the polynomial.

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EXAMPLE 1

䊳

Factoring Polynomials Factor each polynomial: a. 12x2 ⫹ 18xy ⫺ 30y

Solution

䊳

b. x5 ⫹ x2

a. 6 is common to all three terms: 12x2 ⫹ 18xy ⫺ 30y ⫽ 612x2 ⫹ 3xy ⫺ 5y2

mentally: 6 # 2x2 ⫹ 6 # 3xy ⫺ 6 # 5y

b. x2 is common to both terms:

A. You’ve just seen how we can factor out the greatest common factor

x 5 ⫹ x2 ⫽ x2 1x3 ⫹ 12

mentally: x 2 # x 3 ⫹ x 2 # 1


䊳

B. Common Binomial Factors and Factoring by Grouping If the terms of a polynomial have a common binomial factor, it can also be factored out using the distributive property. EXAMPLE 2

䊳

Factoring Out a Common Binomial Factor Factor: a. 1x ⫹ 32x2 ⫹ 1x ⫹ 325

Solution

䊳

b. x2 1x ⫺ 22 ⫺ 31x ⫺ 22

a. 1x ⫹ 32x2 ⫹ 1x ⫹ 325 ⫽ 1x ⫹ 32 1x2 ⫹ 52

b. x2 1x ⫺ 22 ⫺ 31x ⫺ 22 ⫽ 1x ⫺ 22 1x2 ⫺ 32 Now try Exercises 9 and 10

䊳

One application of removing a binomial factor involves factoring by grouping. At first glance, the expression x3 ⫹ 2x2 ⫹ 3x ⫹ 6 appears unfactorable. But by grouping the terms (applying the associative property), we can remove a monomial factor from each subgroup, which then reveals a common binomial factor. x3 ⫹ 2x2 ⫹ 3x ⫹ 6 ⫽ x2 1x ⫹ 22 ⫹ 31x ⫹ 22 ⫽ 1x ⫹ 22 1x2 ⫹ 32

This grouping of terms must take into account any sign changes and common factors, as seen in Example 3. Also, it will be helpful to note that a general four-term polynomial A ⫹ B ⫹ C ⫹ D is factorable by grouping only if AD ⫽ BC. EXAMPLE 3

䊳

Factoring by Grouping Factor 3t3 ⫹ 15t2 ⫺ 6t ⫺ 30.

Solution

䊳

Notice that all four terms have a common factor of 3. Begin by factoring it out. 3t3 ⫹ 15t2 ⫺ 6t ⫺ 30 ⫽ 31t3 ⫹ 5t2 ⫺ 2t ⫺ 102 ⫽ 31t3 ⫹ 5t2 ⫺ 2t ⫺ 102 ⫽ 3 3 t2 1t ⫹ 52 ⫺ 21t ⫹ 52 4 ⫽ 31t ⫹ 52 1t2 ⫺ 22

original polynomial factor out 3 group remaining terms factor common monomial factor common binomial


䊳

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B. You’ve just seen how we can factor common binomial factors and factor by grouping

41

When asked to factor an expression, first look for common factors. The resulting expression will be easier to work with and help ensure the final answer is written in completely factored form. If a four-term polynomial cannot be factored as written, try rearranging the terms to find a combination that enables factoring by grouping.

C. Factoring Quadratic Polynomials A quadratic polynomial is one that can be written in the form ax2 ⫹ bx ⫹ c, where a, b, c 僆⺢ and a ⫽ 0. One common form of factoring involves quadratic trinomials such as x2 ⫹ 7x ⫹ 10 and 2x2 ⫺ 13x ⫹ 15. While we know 1x ⫹ 521x ⫹ 22 ⫽ x2 ⫹ 7x ⫹ 10 and 12x ⫺ 321x ⫺ 52 ⫽ 2x2 ⫺ 13x ⫹ 15 using F-O-I-L, how can we factor these trinomials without seeing the original expression in advance? First, it helps to place the trinomials in two families—those with a leading coefficient of 1 and those with a leading coefficient other than 1.

ax 2 ⴙ bx ⴙ c, where a ⴝ 1

When a ⫽ 1, the only factor pair for x2 (other than 1 # x2 2 is x # x and the first term in each binomial will be x: (x )(x ). The following observation helps guide us to the complete factorization. Consider the product 1x ⫹ b2 1x ⫹ a2: 1x ⫹ b21x ⫹ a2 ⫽ x2 ⫹ ax ⫹ bx ⫹ ab

⫽ x ⫹ 1a ⫹ b2x ⫹ ab 2

F-O-I-L distributive property

Note the last term is the product ab (the lasts), while the coefficient of the middle term is a ⫹ b (the sum of the outers and inners). Since the last term of x2 ⫺ 8x ⫹ 7 is 7 and the coefficient of the middle term is ⫺8, we are seeking two numbers with a product of positive 7 and a sum of negative 8. The numbers are ⫺7 and ⫺1, so the factored form is 1x ⫺ 721x ⫺ 12. It is also helpful to note that if the constant term is positive, the binomials will have like signs, since only the product of like signs is positive. If the constant term is negative, the binomials will have unlike signs, since only the product of unlike signs is negative. This means we can use the sign of the linear term (the term with degree 1) to guide our choice of factors. Factoring Trinomials with a Leading Coefficient of 1 If the constant term is positive, the binomials will have like signs: 1x ⫹ 2 1x ⫹ 2 or 1x ⫺ 2 1x ⫺ 2 ,

to match the sign of the linear (middle) term. If the constant term is negative, the binomials will have unlike signs: 1x ⫹ 2 1x ⫺ 2,

with the larger factor placed in the binomial whose sign matches the linear (middle) term.

EXAMPLE 4

䊳

Factoring Trinomials Factor these expressions: a. ⫺x2 ⫹ 11x ⫺ 24

Solution

䊳

b. x2 ⫺ 10 ⫺ 3x

a. First rewrite the trinomial in standard form as ⫺11x2 ⫺ 11x ⫹ 242. For x2 ⫺ 11x ⫹ 24, the constant term is positive so the binomials will have like signs. Since the linear term is negative, ⫺11x2 ⫺ 11x ⫹ 242 ⫽ ⫺11x ⫺ 21x ⫺ 2 ⫽ ⫺11x ⫺ 821x ⫺ 32

like signs, both negative 1⫺82 1⫺32 ⫽ 24; ⫺8 ⫹ 1⫺32 ⫽ ⫺11

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b. First rewrite the trinomial in standard form as x2 ⫺ 3x ⫺ 10. The constant term is negative so the binomials will have unlike signs. Since the linear term is negative, x2 ⫺ 3x ⫺ 10 ⫽ 1x ⫹ 2 1x ⫺ 2 ⫽ 1x ⫹ 22 1x ⫺ 52

unlike signs, one positive and one negative 5 7 2, 5 is placed in the second binomial; 122 1⫺52 ⫽ ⫺10; 2 ⫹ 1⫺52 ⫽ ⫺3


䊳

Sometimes we encounter prime polynomials, or polynomials that cannot be factored. For x2 ⫹ 9x ⫹ 15, the factor pairs of 15 are 1 # 15 and 3 # 5, with neither pair having a sum of ⫹9. We conclude that x2 ⫹ 9x ⫹ 15 is prime. ax 2 ⴙ bx ⴙ c, where a ⴝ 1 If the leading coefficient is not one, the possible combinations of outers and inners are more numerous. Furthermore, the sum of the outer and inner products will change depending on the position of the possible factors. Note that 12x ⫹ 321x ⫹ 92 ⫽ 2x2 ⫹ 21x ⫹ 27 and 12x ⫹ 921x ⫹ 32 ⫽ 2x2 ⫹ 15x ⫹ 27 result in a different middle term, even though identical numbers were used. To factor 2x2 ⫺ 13x ⫹ 15, note the constant term is positive so the binomials must have like signs. The negative linear term indicates these signs will be negative. We then list possible factors for the first and last terms of each binomial, then sum the outer and inner products. Possible First and Last Terms for 2x2 and 15 1. 12x ⫺ 121x ⫺ 152 2. 12x ⫺ 1521x ⫺ 12 3. 12x ⫺ 321x ⫺ 52 4. 12x ⫺ 521x ⫺ 32 WORTHY OF NOTE The number of trials needed to factor a polynomial can also be reduced by noting that the two terms in any binomial cannot share a common factor (all common factors are removed in a preliminary step).

EXAMPLE 5

䊳

Sum of Outers and Inners ⫺30x ⫺ 1x ⫽ ⫺31x ⫺2x ⫺ 15x ⫽ ⫺17x ⫺10x ⫺ 3x ⫽ ⫺13x

d

⫺6x ⫺ 5x ⫽ ⫺11x

As you can see, only possibility 3 yields a linear term of ⫺13x, and the correct factorization is then 12x ⫺ 321x ⫺ 52. With practice, this trial-and-error process can be completed very quickly. If the constant term is negative, the number of possibilities can be reduced by finding a factor pair with a sum or difference equal to the absolute value of the linear coefficient, as we can then arrange the sign in each binomial to obtain the needed result as shown in Example 5. Factoring a Trinomial Using Trial and Error Factor 6z2 ⫺ 11z ⫺ 35.

Solution

䊳

Note the constant term is negative (binomials will have unlike signs) and 冟⫺11冟 ⫽ 11. The factors of 35 are 1 # 35 and 5 # 7. Two possible first terms are: (6z )(z ) and (3z )(2z ), and we begin with 5 and 7 as factors of 35.

(6z

Outer and Inner Products

)(z

)

Sum

Difference

1. (6z

5)(z

7)

42z ⫹ 5z 47z

42z ⫺ 5z 37z

2. (6z

7)(z

5)

30z ⫹ 7z 37z

30z ⫺ 7z 23z

(3z

Outer and Inner Products

)(2z

)

Sum

Difference

3. (3z

5)(2z

7)

21z ⫹ 10z 31z

21z ⫺ 10z 11z

4. (3z

7)(2z

5)

15z ⫹ 14z 29z

15z ⫺ 14z 1z

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43

Since possibility 3 yields a linear term of 11z, we need not consider other factors of 35 and write the factored form as 6z2 ⫺ 11z ⫺ 35 ⫽ 13z 5212z 72. The signs can then be arranged to obtain a middle term of ⫺11z: 13z ⫹ 5212z ⫺ 72, ⫺21z ⫹ 10z ⫽ ⫺11z ✓. C. You’ve just seen how we can factor quadratic polynomials


䊳

D. Factoring Special Forms and Quadratic Forms Next we consider methods to factor each of the special products we encountered in Section R.2.

The Difference of Two Squares WORTHY OF NOTE In an attempt to factor a sum of two perfect squares, say v2 ⫹ 49, let’s list all possible binomial factors. These are (1) 1v ⫹ 721v ⫹ 72, (2) 1v ⫺ 721v ⫺ 72, and (3) 1v ⫹ 721v ⫺ 72. Note that (1) and (2) are the binomial squares 1v ⫹ 72 2 and 1v ⫺ 72 2, with each product resulting in a “middle” term, whereas (3) is a binomial times its conjugate, resulting in a difference of squares: v2 ⫺ 49. With all possibilities exhausted, we conclude that the sum of two squares is prime!

EXAMPLE 6

䊳

Multiplying and factoring are inverse processes. Since 1x ⫺ 721x ⫹ 72 ⫽ x2 ⫺ 49, we know that x2 ⫺ 49 ⫽ 1x ⫺ 721x ⫹ 72. In words, the difference of two squares will factor into a binomial and its conjugate. To find the terms of the factored form, rewrite each term in the original expression as a square: 1 2 2. Factoring the Difference of Two Perfect Squares Given any expression that can be written in the form A2 ⫺ B2, A2 ⫺ B2 ⫽ 1A ⫹ B21A ⫺ B2

Note that the sum of two perfect squares A2 ⫹ B2 cannot be factored using real numbers (the expression is prime). As a reminder, always check for a common factor first and be sure to write all results in completely factored form. See Example 6(c). Factoring the Difference of Two Perfect Squares Factor each expression completely. a. 4w2 ⫺ 81 b. v2 ⫹ 49 c. ⫺3n2 ⫹ 48

Solution

䊳

a. 4w2 ⫺ 81 ⫽ 12w2 2 ⫺ 92 ⫽ 12w ⫹ 92 12w ⫺ 92 b. v2 ⫹ 49 is prime. c. ⫺3n2 ⫹ 48 ⫽ ⫺31n2 ⫺ 162 ⫽ ⫺33 n2 ⫺ 142 2 4 ⫽ ⫺31n ⫹ 42 1n ⫺ 42 1 d. z4 ⫺ 81 ⫽ 1z2 2 2 ⫺ 1 19 2 2 ⫽ 1z2 ⫹ 19 2 1z2 ⫺ 19 2 ⫽ 1z2 ⫹ 19 2 3 z2 ⫺ 1 13 2 2 4 ⫽ 1z2 ⫹ 19 2 1z ⫹ 13 2 1z ⫺ 13 2 2 e. x ⫺ 7 ⫽ 1x2 2 ⫺ 1 172 2 ⫽ 1x ⫹ 1721x ⫺ 172

1 d. z4 ⫺ 81

e. x2 ⫺ 7

write as a difference of squares A 2 ⫺ B 2 ⫽ 1A ⫹ B21A ⫺ B2 factor out ⫺3 write as a difference of squares A 2 ⫺ B 2 ⫽ 1A ⫹ B21A ⫺ B2 write as a difference of squares A 2 ⫺ B 2 ⫽ 1A ⫹ B21A ⫺ B2

write as a difference of squares (z 2 ⫹ 19 is prime) result write as a difference of squares A 2 ⫺ B 2 ⫽ 1A ⫹ B21A ⫺ B2


Perfect Square Trinomials

䊳

Since 1x ⫹ 72 2 ⫽ x2 ⫹ 14x ⫹ 49, we know that x2 ⫹ 14x ⫹ 49 ⫽ 1x ⫹ 72 2. In words, a perfect square trinomial will factor into a binomial square. To use this idea effectively, we must learn to identify perfect square trinomials. Note that the first and last terms of x2 ⫹ 14x ⫹ 49 are the squares of x and 7, and the middle term is twice the product of these two terms: 217x2 ⫽ 14x. These are the characteristics of a perfect square trinomial.

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Factoring Perfect Square Trinomials Given any expression that can be written in the form A2 ⫾ 2AB ⫹ B2, 1. A2 ⫹ 2AB ⫹ B2 ⫽ 1A ⫹ B2 2 2. A2 ⫺ 2AB ⫹ B2 ⫽ 1A ⫺ B2 2

EXAMPLE 7

䊳

Factoring a Perfect Square Trinomial Factor 12m3 ⫺ 12m2 ⫹ 3m.

Solution

12m3 ⫺ 12m2 ⫹ 3m ⫽ 3m14m2 ⫺ 4m ⫹ 12

䊳

check for common factors: GCF ⫽ 3m factor out 3m

For the remaining trinomial 4m2 ⫺ 4m ⫹ 1 p 1. Are the first and last terms perfect squares?

4m2 ⫽ 12m2 2 and 1 ⫽ 112 2 ✓ Yes.

2. Is the linear term twice the product of 2m and 1? 2 # 2m # 1 ⫽ 4m ✓ Yes.

Factor as a binomial square: 4m2 ⫺ 4m ⫹ 1 ⫽ 12m ⫺ 12 2 This shows 12m3 ⫺ 12m2 ⫹ 3m ⫽ 3m12m ⫺ 12 2. Now try Exercises 19 and 20

CAUTION

䊳

䊳

As shown in Example 7, be sure to include the GCF in your final answer. It is a common error to “leave the GCF behind.”

In actual practice, these calculations can be performed mentally, making the process much more efficient.

Sum or Difference of Two Perfect Cubes Recall that the difference of two perfect squares is factorable, but the sum of two perfect squares is prime. In contrast, both the sum and difference of two perfect cubes are factorable. For either A3 ⫹ B3 or A3 ⫺ B3 we have the following: 1. Each will factor into the product of a binomial and a trinomial: 2. The terms of the binomial are the quantities being cubed: 3. The terms of the trinomial are the square of A, the product AB, and the square of B, respectively: 4. The binomial takes the same sign as the original expression 5. The middle term of the trinomial takes the opposite sign of the original expression (the last term is always positive):

(

)(

binomial

) trinomial

(A

B)(

)

(A

B)(A2

AB

B 2)

(A ⫾ B)(A2

AB

B 2)

(A ⫾ B)(A2 ⫿ AB ⫹ B 2)

Factoring the Sum or Difference of Two Perfect Cubes: A3 ⫾ B3 1. A3 ⫹ B3 ⫽ 1A ⫹ B2 1A2 ⫺ AB ⫹ B2 2 2. A3 ⫺ B3 ⫽ 1A ⫺ B21A2 ⫹ AB ⫹ B2 2

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EXAMPLE 8

䊳

Factoring the Sum and Difference of Two Perfect Cubes Factor completely: a. x3 ⫹ 125

Solution

䊳

45

a.

b.

b. ⫺5m3n ⫹ 40n4

x3 ⫹ 125 ⫽ x3 ⫹ 53 Use A3 ⫹ B3 ⫽ 1A ⫹ B21A2 ⫺ AB ⫹ B2 2 x3 ⫹ 53 ⫽ 1x ⫹ 52 1x2 ⫺ 5x ⫹ 252

write terms as perfect cubes factoring template A S x and B S 5

⫺5m3n ⫹ 40n4 ⫽ ⫺5n1m3 ⫺ 8n3 2 ⫽ ⫺5n 3m3 ⫺ 12n2 3 4 Use A3 ⫺ B3 ⫽ 1A ⫺ B21A2 ⫹ AB ⫹ B2 2 m3 ⫺ 12n2 3 ⫽ 1m ⫺ 2n2 3 m2 ⫹ m12n2 ⫹ 12n2 2 4 ⫽ 1m ⫺ 2n21m2 ⫹ 2mn ⫹ 4n2 2 3 4 1 ⫺5m n ⫹ 40n ⫽ ⫺5n1m ⫺ 2n21m2 ⫹ 2mn ⫹ 4n2 2.

check for common factors 1GCF ⫽ ⫺5n2 write terms as perfect cubes factoring template A S m and B S 2n simplify factored form

The results for parts (a) and (b) can be checked using multiplication. Now try Exercises 21 and 22

䊳

Quadratic Forms and u-Substitution For any quadratic expression ax2 ⫹ bx ⫹ c in standard form, the degree of the leading term is twice the degree of the middle term. Generally, a trinomial is in quadratic form if it can be written as a1 __ 2 2 ⫹ b1 __ 2 ⫹ c, where the parentheses “hold” the same factors. The equation x4 ⫺ 13x2 ⫹ 36 ⫽ 0 is in quadratic form since 1x2 2 2 ⫺ 131x2 2 ⫹ 36 ⫽ 0. In many cases, we can factor these expressions using a placeholder substitution that transforms them into a more recognizable form. In a study of algebra, the letter “u” often plays this role. If we let u represent x2, the expression 1x2 2 2 ⫺ 131x2 2 ⫹ 36 becomes u2 ⫺ 13u ⫹ 36, which can be factored into 1u ⫺ 92 1u ⫺ 42. After “unsubstituting” (replace u with x2), we have 1x2 ⫺ 92 1x2 ⫺ 42 ⫽ 1x ⫹ 321x ⫺ 321x ⫹ 22 1x ⫺ 22. EXAMPLE 9

䊳

Factoring a Quadratic Form

Solution

䊳

Expanding the binomials would produce a fourth-degree polynomial that would be very difficult to factor. Instead we note the expression is in quadratic form. Letting u represent x2 ⫺ 2x (the variable part of the “middle” term), 1x2 ⫺ 2x2 2 ⫺ 21x2 ⫺ 2x2 ⫺ 3 becomes u2 ⫺ 2u ⫺ 3.

Write in completely factored form: 1x2 ⫺ 2x2 2 ⫺ 21x2 ⫺ 2x2 ⫺ 3.

u2 ⫺ 2u ⫺ 3 ⫽ 1u ⫺ 32 1u ⫹ 12

factor

To finish up, write the expression in terms of x, substituting x2 ⫺ 2x for u. ⫽ 1x2 ⫺ 2x ⫺ 32 1x2 ⫺ 2x ⫹ 12

substitute x2 ⫺ 2x for u

The resulting trinomials can be further factored. ⫽ 1x ⫺ 32 1x ⫹ 12 1x ⫺ 12 2

x2 ⫺ 2x ⫹ 1 ⫽ 1x ⫺ 12 2


D. You’ve just seen how we can factor special forms and quadratic forms

䊳

It is well known that information is retained longer and used more effectively when it’s placed in an organized form. The “factoring flowchart” provided in Figure R.4 offers a streamlined and systematic approach to factoring and the concepts involved. However, with some practice the process tends to “flow” more naturally than following a chart, with many of the decisions becoming automatic.

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Factoring Polynomials

Standard Form: decreasing order of degree

Greatest Common Factor (positive leading coefficient)

Number of Terms

Two

Difference of squares

Difference of cubes

Four

Three

Sum of cubes

Trinomials (a ⫽ 1)

• Can any result be factored further?

Trinomials (a ⫽ 1)

Factor by grouping

Advanced methods (Section 4.2)

• Polynomials that cannot be factored are said to be prime.

Figure R.4

For additional practice with these ideas, see Exercises 25 through 52.

E. Polynomial Equations and the Zero Product Property The ability to solve linear and quadratic equations is the foundation on which a large percentage of our future studies are built. Both are closely linked to the solution of other equation types, as well as to the graphs of these equations. In standard form, linear and quadratic equations have a known number of terms, so we commonly represent their coefficients using the early letters of the alphabet, as in ax2 ⫹ bx ⫹ c ⫽ 0. However, these equations belong to the larger family of polynomial equations. To write a general polynomial, where the number of terms is unknown, we often represent the coefficients using subscripts on a single variable, such as a1, a2, a3, and so on. Polynomial Equations A polynomial equation of degree n is one of the form anxn ⫹ an⫺1xn⫺1 ⫹ p ⫹ a1x1 ⫹ a0 ⫽ 0 where an, an⫺1, p , a1, a0 are real numbers and an ⫽ 0. As a prelude to solving polynomial equations of higher degree, we’ll first look at quadratic equations and the zero product property. As before, a quadratic equation is one that can be written in the form ax2 ⫹ bx ⫹ c ⫽ 0, where a, b, and c are real numbers and a ⫽ 0. As written, the equation is in standard form, meaning the terms are in decreasing order of degree and the equation is set equal to zero.

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Quadratic Equations A quadratic equation can be written in the form ax2 ⫹ bx ⫹ c ⫽ 0, with a, b, c 僆⺢, and a ⫽ 0. Notice that a is the leading coefficient, b is the coefficient of the linear (first degree) term, and c is a constant. All quadratic equations have degree two, but can have one, two, or three terms. The equation n2 ⫺ 81 ⫽ 0 is a quadratic equation with two terms, where a ⫽ 1, b ⫽ 0, and c ⫽ ⫺81. EXAMPLE 10

䊳

Determining Whether an Equation Is Quadratic State whether the given equation is quadratic. If yes, identify coefficients a, b, and c. ⫺3 a. 2x2 ⫺ 18 ⫽ 0 b. z ⫺ 12 ⫺ 3z2 ⫽ 0 c. x⫹5⫽0 4 d. z3 ⫺ 2z2 ⫹ 7z ⫽ 8 e. 0.8x2 ⫽ 0

Solution WORTHY OF NOTE The word quadratic comes from the Latin word quadratum, meaning square. The word historically refers to the “four sidedness” of a square, but mathematically to the area of a square. Hence its application to polynomials of the form ax2 ⫹ bx ⫹ c, where the variable of the leading term is squared.

䊳

Standard Form

Quadratic

a.

2x ⫺ 18 ⫽ 0

yes, deg 2

a⫽2

b.

⫺3z ⫹ z ⫺ 12 ⫽ 0

yes, deg 2

a ⫽ ⫺3

c.

⫺3 x⫹5⫽0 4

no, deg 1

(linear equation)

d.

z3 ⫺ 2z2 ⫹ 7z ⫺ 8 ⫽ 0

no, deg 3

(cubic equation)

e.

0.8x ⫽ 0

yes, deg 2

2

2

2

Coefficients b⫽0

a ⫽ 0.8

c ⫽ ⫺18

b⫽1

b⫽0

c ⫽ ⫺12

c⫽0


䊳

With quadratic and other polynomial equations, we generally cannot isolate the variable on one side using only properties of equality, because the variable is raised to different powers. Instead we attempt to solve the equation by factoring and applying the zero product property. Zero Product Property If A and B represent real numbers or real-valued expressions and A # B ⫽ 0, then A ⫽ 0 or B ⫽ 0. In words, the property says, If the product of any two (or more) factors is equal to zero, then at least one of the factors must be equal to zero. We can use this property to solve higher degree equations after rewriting them in terms of equations with lesser degree. As with linear equations, values that make the original equation true are called solutions or roots of the equation.

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EXAMPLE 11

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Solving Equations Using the Zero Product Property Solve by writing the equations in factored form and applying the zero product property. a. 3x2 ⫽ 5x b. ⫺5x ⫹ 2x2 ⫽ 3 c. 4x2 ⫽ 12x ⫺ 9

Solution

3x2 ⫽ 5x 3x ⫺ 5x ⫽ 0 x13x ⫺ 52 ⫽ 0 x ⫽ 0 or 3x ⫺ 5 ⫽ 0 5 x ⫽ 0 or x⫽ 3 b. ⫺5x ⫹ 2x2 ⫽ 3 2x2 ⫺ 5x ⫺ 3 ⫽ 0 12x ⫹ 121x ⫺ 32 ⫽ 0 or x ⫺ 3 ⫽ 0 2x ⫹ 1 ⫽ 0 1 x⫽⫺ or x⫽3 2 c. 4x2 ⫽ 12x ⫺ 9 2 4x ⫺ 12x ⫹ 9 ⫽ 0 12x ⫺ 3212x ⫺ 32 ⫽ 0 2x ⫺ 3 ⫽ 0 or 2x ⫺ 3 ⫽ 0 3 3 x⫽ or x⫽ 2 2

䊳

a.

2

given equation standard form factor set factors equal to zero (zero product property) result given equation standard form factor set factors equal to zero (zero product property) result given equation standard form factor set factors equal to zero (zero product property) result

3 This equation has only the solution x ⫽ , which we call a repeated root. 2 Now try Exercises 65 through 88

䊳

CAUTION

EXAMPLE 12

䊳

䊳

Consider the equation x2 ⫺ 2x ⫺ 3 ⫽ 12. While the left-hand side is factorable, the result is 1x ⫺ 321x ⫹ 12 ⫽ 12 and finding a solution becomes a “guessing game” because the equation is not set equal to zero. If you misapply the zero factor property and say that x ⫺ 3 ⫽ 12 or x ⫹ 1 ⫽ 12, the “solutions” are x ⫽ 15 or x ⫽ 11, which are both incorrect! After subtracting 12 from both sides, x2 ⫺ 2x ⫺ 3 ⫽ 12 becomes x2 ⫺ 2x ⫺ 15 ⫽ 0 giving 1x ⫺ 521x ⫹ 32 ⫽ 0 with solutions x ⫽ 5 or x ⫽ ⫺3.

Solving Polynomials by Factoring Solve by factoring: 4x3 ⫺ 40x ⫽ 6x2.

Solution

䊳

4x3 ⫺ 40x ⫽ 6x2 4x ⫺ 6x2 ⫺ 40x ⫽ 0 2x 12x2 ⫺ 3x ⫺ 202 ⫽ 0 2x12x ⫹ 521x ⫺ 42 ⫽ 0 2x ⫽ 0 or 2x ⫹ 5 ⫽ 0 or x ⫺ 4 ⫽ 0 x ⫽ 0 or x ⫽ ⫺5 or x⫽4 2 3

given equation standard form common factor is 2x factored form zero product property result — solve for x

Substituting these values into the original equation verifies they are solutions. Now try Exercises 89 through 92

䊳

Example 12 reminds us that in the process of factoring polynomials, there may be a common monomial factor. This factor is also set equal to zero in the solution process (if the monomial is a constant, no solution is generated).

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EXAMPLE 13

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Solving Higher Degree Equations Solve each equation by factoring. a. x3 ⫺ 4x ⫹ 20 ⫽ 5x2

Solution

䊳

49

a.

b. x4 ⫺ 10x2 ⫹ 9 ⫽ 0

x3 ⫺ 4x ⫹ 20 ⫽ 5x2 x ⫺ 5x2 ⫺ 4x ⫹ 20 ⫽ 0

original equation

3

x 1x ⫺ 52 ⫺ 41x ⫺ 52 ⫽ 0 1x ⫺ 521x2 ⫺ 42 ⫽ 0 1x ⫺ 52 1x ⫹ 221x ⫺ 22 ⫽ 0 or x ⫺ 2 ⫽ 0 x ⫺ 5 ⫽ 0 or x ⫹ 2 ⫽ 0 x ⫽ ⫺2 or x⫽2 x ⫽ 5 or 2

standard form; factor by grouping remove common factors from each group factor common binomial factored form zero product property solve

The solutions are x ⫽ 5, x ⫽ ⫺2, and x ⫽ 2. b. The equation appears to be in quadratic form and we begin by substituting u for x2 and u2 for x4. original equation x4 ⫺ 10x2 ⫹ 9 ⫽ 0 substitute u for x 2 and u 2 for x 4 u2 ⫺ 10u ⫹ 9 ⫽ 0 factored form 1u ⫺ 921u ⫺ 12 ⫽ 0 zero product property u ⫺ 9 ⫽ 0 or u ⫺ 1 ⫽ 0 2 2 substitute x 2 for u x ⫺ 9 ⫽ 0 or x ⫺ 1 ⫽ 0 1x ⫹ 321x ⫺ 32 ⫽ 0 or 1x ⫹ 12 1x ⫺ 12 ⫽ 0 factor x ⫽ ⫺1 or x ⫽ 1 zero product property x ⫽ ⫺3 or x ⫽ 3 or The solutions are x ⫽ ⫺3, x ⫽ 3, x ⫽ ⫺1, and x ⫽ 1. Now try Exercises 93 through 100 E. You’ve just seen how we can solve polynomial equations by factoring

䊳

In Examples 12 and 13, we were able to solve higher degree polynomial equations by “breaking them down” into linear and quadratic forms. This basic idea can be applied to other kinds of equations as well.

R.4 EXERCISES 䊳



1. To factor an expression means to rewrite the expression as an equivalent .

2. If a polynomial will not factor, it is said to be a(n) polynomial.

3. The difference of two perfect squares always factors into the product of a(n) and its .

4. The expression x2 ⫹ 6x ⫹ 9 is said to be a(n) trinomial, since its factored form is a perfect (binomial) square.

5. Discuss/Explain why 4x2 ⫺ 36 ⫽ 12x ⫺ 62 12x ⫹ 62 is not written in completely factored form, then rewrite it so it is factored completely.

6. Discuss/Explain why a3 ⫹ b3 is factorable, but a2 ⫹ b2 is not. Demonstrate by writing x3 ⫹ 64 in factored form, and by exhausting all possibilities for x2 ⫹ 64 to show it is prime.

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Factor each expression using the method indicated. Greatest Common Factor

7. a. ⫺17x2 ⫹ 51 c. ⫺3a4 ⫹ 9a2 ⫺ 6a3

b. 21b3 ⫺ 14b2 ⫹ 56b

8. a. ⫺13n2 ⫺ 52 b. 9p2 ⫹ 27p3 ⫺ 18p4 c. ⫺6g5 ⫹ 12g4 ⫺ 9g3 Common Binomial Factor

9. a. 2a1a ⫹ 22 ⫹ 31a ⫹ 22 b. 1b2 ⫹ 323b ⫹ 1b2 ⫹ 322 c. 4m1n ⫹ 72 ⫺ 111n ⫹ 72 10. a. 5x1x ⫺ 32 ⫺ 21x ⫺ 32 b. 1v ⫺ 522v ⫹ 1v ⫺ 523 c. 3p1q2 ⫹ 52 ⫹ 71q2 ⫹ 52

b. z2 ⫺ 18z ⫹ 81 d. 16q2 ⫹ 40q ⫹ 25

Sum/Difference of Perfect Cubes

21. a. 8p3 ⫺ 27 c. g3 ⫺ 0.027

b. m3 ⫹ 18 d. ⫺2t 4 ⫹ 54t

22. a. 27q3 ⫺ 125 c. b3 ⫺ 0.125

8 b. n3 ⫹ 27 d. 3r4 ⫺ 24r

b. x4 ⫹ 13x2 ⫹ 36

25. Completely factor each of the following (recall that “1” is its own perfect square and perfect cube). a. n2 ⫺ 1 b. n3 ⫺ 1 3 c. n ⫹ 1 d. 28x3 ⫺ 7x

2

Trinomial Factoring where 円a円 ⴝ 1

13. a. ⫺p2 ⫹ 5p ⫹ 14 c. n2 ⫹ 20 ⫺ 9n

b. q2 ⫺ 4q ⫹ 12

14. a. ⫺m2 ⫹ 13m ⫺ 42 c. v2 ⫹ 10v ⫹ 15

b. x2 ⫹ 12 ⫹ 13x

Trinomial Factoring where a ⴝ 1

15. a. 3p2 ⫺ 13p ⫺ 10 c. 10u2 ⫺ 19u ⫺ 15

b. 4q2 ⫹ 7q ⫺ 15

16. a. 6v ⫹ v ⫺ 35 c. 15z2 ⫺ 22z ⫺ 48

b. 20x ⫹ 53x ⫹ 18 2

Difference of Perfect Squares

17. a. 4s2 ⫺ 25 c. 50x2 ⫺ 72 e. b2 ⫺ 5

b. 9x2 ⫺ 49 d. 121h2 ⫺ 144

18. a. 9v ⫺ c. v4 ⫺ 1 e. x2 ⫺ 17

b. 25w ⫺ d. 16z4 ⫺ 81

1 25

20. a. x2 ⫹ 12x ⫹ 36 c. 25p2 ⫺ 60p ⫹ 36

24. a. x6 ⫺ 26x3 ⫺ 27 b. 31n ⫹ 52 2 ⫹ 21n ⫹ 52 ⫺ 21 c. 21z ⫹ 32 2 ⫹ 31z ⫹ 32 ⫺ 54

12. a. 6h ⫺ 9h ⫺ 2h ⫹ 3 b. 4k3 ⫹ 6k2 ⫺ 2k ⫺ 3 c. 3x2 ⫺ xy ⫺ 6x ⫹ 2y

2

b. b2 ⫹ 10b ⫹ 25 d. 9n2 ⫺ 42n ⫹ 49

23. a. x4 ⫺ 10x2 ⫹ 9 c. x6 ⫺ 7x3 ⫺ 8

11. a. 9q3 ⫹ 6q2 ⫹ 15q ⫹ 10 b. h5 ⫺ 12h4 ⫺ 3h ⫹ 36 c. k5 ⫺ 7k3 ⫺ 5k2 ⫹ 35

2

19. a. a2 ⫺ 6a ⫹ 9 c. 4m2 ⫺ 20m ⫹ 25

u-Substitution

Grouping

3

Perfect Square Trinomials

2

1 49

26. Carefully factor each of the following trinomials, if possible. Note differences and similarities. a. x2 ⫺ x ⫹ 6 b. x2 ⫹ x ⫺ 6 2 c. x ⫹ x ⫹ 6 d. x2 ⫺ x ⫺ 6 e. x2 ⫺ 5x ⫹ 6 f. x2 ⫹ 5x ⫺ 6 Factor each expression completely, if possible. Rewrite the expression in standard form (factor out “⫺1” if needed) and factor out the GCF if one exists. If you believe the expression will not factor, write “prime.”

27. a2 ⫹ 7a ⫹ 10

28. b2 ⫹ 9b ⫹ 20

29. 2x2 ⫺ 24x ⫹ 40

30. 10z2 ⫺ 140z ⫹ 450

31. 64 ⫺ 9m2

32. 25 ⫺ 16n2

33. ⫺9r ⫹ r2 ⫹ 18

34. 28 ⫹ s2 ⫺ 11s

35. 2h2 ⫹ 7h ⫹ 6

36. 3k2 ⫹ 10k ⫹ 8

37. 9k2 ⫺ 24k ⫹ 16

38. 4p2 ⫺ 20p ⫹ 25

39. ⫺6x3 ⫹ 39x2 ⫺ 63x

40. ⫺28z3 ⫹ 16z2 ⫹ 80z

41. 12m2 ⫺ 40m ⫹ 4m3

42. ⫺30n ⫺ 4n2 ⫹ 2n3

43. a2 ⫺ 7a ⫺ 60

44. b2 ⫺ 9b ⫺ 36

45. 8x3 ⫺ 125

46. 27r3 ⫹ 64

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47. m2 ⫹ 9m ⫺ 24

48. n2 ⫺ 14n ⫺ 36

49. x3 ⫺ 5x2 ⫺ 9x ⫹ 45

50. x3 ⫹ 3x2 ⫺ 4x ⫺ 12

51. Match each expression with the description that fits best. a. prime polynomial b. standard trinomial a ⫽ 1 c. perfect square trinomial d. difference of cubes e. binomial square f. sum of cubes g. binomial conjugates h. difference of squares i. standard trinomial a ⫽ 1 3 A. x ⫹ 27 B. 1x ⫹ 32 2 C. x2 ⫺ 10x ⫹ 25 D. x2 ⫺ 144 2 E. x ⫺ 3x ⫺ 10 F. 8s3 ⫺ 125t3 G. 2x2 ⫺ x ⫺ 3 H. x2 ⫹ 9 I. 1x ⫺ 72 and 1x ⫹ 72 52. Match each polynomial to its factored form. Two of them are prime. a. 4x2 ⫺ 9 b. 4x2 ⫺ 28x ⫹ 49 c. x3 ⫺ 125 d. 8x3 ⫹ 27 e. x2 ⫺ 3x ⫺ 10 f. x2 ⫹ 3x ⫹ 10 g. 2x2 ⫺ x ⫺ 3 h. 2x2 ⫹ x ⫺ 3 i. x2 ⫹ 25 A. 1x ⫺ 521x2 ⫹ 5x ⫹ 252 B. 12x ⫺ 321x ⫹ 12 C. 12x ⫹ 3212x ⫺ 32 D. 12x ⫺ 72 2 E. prime trinomial F. prime binomial G. 12x ⫹ 321x ⫺ 12 2 H. 12x ⫹ 3214x ⫺ 6x ⫹ 92 I. 1x ⫺ 521x ⫹ 22 Determine whether each equation is quadratic. If so, identify the coefficients a, b, and c. If not, discuss why.

53. 2x ⫺ 15 ⫺ x2 ⫽ 0 55.

2 x⫺7⫽0 3

54. 21 ⫹ x2 ⫺ 4x ⫽ 0 56. 12 ⫺ 4x ⫽ 9

63. 1x ⫺ 12 2 ⫹ 1x ⫺ 12 ⫹ 4 ⫽ 9

64. 1x ⫹ 52 2 ⫺ 1x ⫹ 52 ⫹ 4 ⫽ 17 Solve using the zero factor property. Be sure each equation is in standard form and factor out any common factors before attempting to solve. Check all answers in the original equation.

65. x2 ⫺ 15 ⫽ 2x

66. z2 ⫺ 10z ⫽ ⫺21

67. m2 ⫽ 8m ⫺ 16

68. ⫺10n ⫽ n2 ⫹ 25

69. 5p2 ⫺ 10p ⫽ 0

70. 6q2 ⫺ 18q ⫽ 0

71. ⫺14h2 ⫽ 7h

72. 9w ⫽ ⫺6w2

73. a2 ⫺ 17 ⫽ ⫺8

74. b2 ⫹ 8 ⫽ 12

75. g2 ⫹ 18g ⫹ 70 ⫽ ⫺11 76. h2 ⫹ 14h ⫺ 2 ⫽ ⫺51 77. m3 ⫹ 5m2 ⫺ 9m ⫺ 45 ⫽ 0 78. n3 ⫺ 3n2 ⫺ 4n ⫹ 12 ⫽ 0 79. 1c ⫺ 122c ⫺ 15 ⫽ 30

80. 1d ⫺ 102d ⫹ 10 ⫽ ⫺6 81. 9 ⫹ 1r ⫺ 52r ⫽ 33 82. 7 ⫹ 1s ⫺ 42s ⫽ 28

83. 1t ⫹ 421t ⫹ 72 ⫽ 54

84. 1g ⫹ 1721g ⫺ 22 ⫽ 20 85. 2x2 ⫺ 4x ⫺ 30 ⫽ 0

86. ⫺3z2 ⫹ 12z ⫹ 36 ⫽ 0 87. 2w2 ⫺ 5w ⫽ 3 88. ⫺3v2 ⫽ ⫺v ⫺ 2 89. 22x ⫽ x3 ⫺ 9x2 90. x3 ⫽ 13x2 ⫺ 42x 91. 3x3 ⫽ ⫺7x2 ⫹ 6x 92. 7x2 ⫹ 15x ⫽ 2x3 93. p3 ⫹ 7p2 ⫺ 63 ⫽ 9p 94. q3 ⫺ 4q ⫹ 24 ⫽ 6q2 95. x3 ⫺ 25x ⫽ 2x2 ⫺ 50 96. 3c2 ⫹ c ⫽ c3 ⫹ 3

1 57. x2 ⫽ 6x 4

58. 0.5x ⫽ 0.25x

59. 2x2 ⫹ 7 ⫽ 0

60. 5 ⫽ ⫺4x2

2

61. ⫺3x2 ⫹ 9x ⫺ 5 ⫹ 2x3 ⫽ 0

62. z2 ⫺ 6z ⫹ 9 ⫺ z3 ⫽ 0

97. x4 ⫺ 29x2 ⫹ 100 ⫽ 0 98. z4 ⫺ 20z2 ⫹ 64 ⫽ 0

99. 1b2 ⫺ 3b2 2 ⫺ 141b2 ⫺ 3b2 ⫹ 40 ⫽ 0

100. 1d2 ⫺ d 2 2 ⫺ 81d2 ⫺ d2 ⫹ 12 ⫽ 0

51

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101. Surface area of a cylinder: 2␲r2 ⴙ 2␲rh

102. Volume of a cylindrical shell: ␲R2h ⴚ ␲r2h

The surface area of a cylinder is given by the formula shown, where h is the height of the cylinder and r is the radius. Factor out the GCF and use the result to find the surface area of a cylinder where r ⫽ 35 cm and h ⫽ 65 cm. Answer in exact form and in approximate form rounded to the nearest whole number.

r The volume of a cylindrical shell (a larger cylinder with a smaller cylinder removed) can be found using the formula shown, where R is the radius of the larger cylinder and r is the radius of the smaller. Factor the expression R completely and use the result to find the volume of a shell where R ⫽ 9 cm, r ⫽ 3 cm, and h ⫽ 10 cm. Answer in exact form and in approximate form rounded to the nearest whole number.

䊳

APPLICATIONS

In many cases, factoring an expression can make it easier to evaluate as in the following applications.

103. Conical shells: The volume of a conical shell (like the shell of an ice cream cone) is given by the 1 1 formula V ⫽ ␲R2h ⫺ ␲r2h, where R is the outer 3 3 radius and r is the inner radius of the cone. Write the formula in completely factored form, then find the volume of a shell when R ⫽ 5.1 cm, r ⫽ 4.9 cm, and h ⫽ 9 cm. Answer in exact form and in approximate form rounded to the nearest tenth. 104. Spherical shells: The volume of a spherical shell (like the outer r shell of a cherry cordial) is given R 4 4 3 3 by the formula V ⫽ 3␲R ⫺ 3␲r , where R is the outer radius and r is the inner radius of the shell. Write the right-hand side in completely factored form, then find the volume of a shell where R ⫽ 1.8 cm and r ⫽ 1.5 cm. Answer in exact form and in approximate form rounded to the nearest tenth. 105. Volume of a box: The volume of a rectangular box x inches in height is given by the relationship V ⫽ x3 ⫹ 8x2 ⫹ 15x. Factor the right-hand side to determine: (a) The number of inches that the width exceeds the height, (b) the number of inches the length exceeds the height, and (c) the volume given the height is 2 ft. 106. Shipping textbooks: A publisher ships paperback books stacked x copies high in a box. The total number of books shipped per box is given by the relationship B ⫽ x3 ⫺ 13x2 ⫹ 42x. Factor the

right-hand side to determine (a) how many more or fewer books fit the width of the box (than the height), (b) how many more or fewer books fit the length of the box (than the height), and (c) the number of books shipped per box if they are stacked 10 high in the box. 107. Space-Time relationships: Due to the work of Albert Einstein and other physicists who labored on space-time relationships, it is known that the faster an object moves the shorter it appears to become. This phenomenon is modeled by the v 2 1⫺a b, c B where L0 is the length of the object at rest, L is the relative length when the object is moving at velocity v, and c is the speed of light. Factor the radicand and use the result to determine the relative length of a 12-in. ruler if it is shot past a stationary observer at 0.75 times the speed of light 1v ⫽ 0.75c2 . Lorentz transformation L ⫽ L0

108. Tubular fluid flow: As a fluid flows through a tube, it is flowing faster at the center of the tube than at the sides, where the tube exerts a backward drag. Poiseuille’s law gives the velocity of the flow G 2 1R ⫺ r2 2 , at any point of the cross section: v ⫽ 4␩ where R is the inner radius of the tube, r is the distance from the center of the tube to a point in the flow, G represents what is called the pressure gradient, and ␩ is a constant that depends on the viscosity of the fluid. Factor the right-hand side and find v given R ⫽ 0.5 cm, r ⫽ 0.3 cm, G ⫽ 15, and ␩ ⫽ 0.25.

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Section R.5 Rational Expressions and Equations

53

Solve by factoring.

109. Envelope sizes: Large mailing envelopes often come in standard sizes, with 5- by 7-in. and 9- by 12-in. envelopes being the most common. The next larger size envelope has an area of 143 in2, with a length that is 2 in. longer than the width. What are the dimensions of the larger envelope?

a length that is 6 in. longer than the width. What are the dimensions of the Ledger size paper?

Letter

Legal Ledger

110. Paper sizes: Letter size paper is 8.5 in. by 11 in. Legal size paper is 812 in. by 14 in. The next larger (common) size of paper has an area of 187 in2, with

䊳


111. Factor out a constant that leaves integer coefficients for each term: a. 12x4 ⫹ 18x3 ⫺ 34x2 ⫹ 4 b. 23b5 ⫺ 16b3 ⫹ 49b2 ⫺ 1 112. If x ⫽ 2 is substituted into 2x3 ⫹ hx ⫹ 8, the result is zero. What is the value of h? 113. Factor the expression: 192x3 ⫺ 164x2 ⫺ 270x. 114. As an alternative to evaluating polynomials by direct substitution, nested factoring can be used. The method has the advantage of using only products and sums — no powers. For P ⫽ x3 ⫹ 3x2⫹ 1x ⫹ 5, we begin by grouping all variable terms and factoring x: P ⫽ 3 x3 ⫹ 3x2 ⫹ 1x 4 ⫹ 5 ⫽

R.5

In Section R.5 you will review how to:

A. Write a rational

C. D. E.

Factor each expression completely.

115. x4 ⫺ 81

116. 16n4 ⫺ 1

117. p6 ⫺ 1

118. m6 ⫺ 64

119. q4 ⫺ 28q2 ⫹ 75

120. a4 ⫺ 18a2 ⫹ 32

Rational Expressions and Equations

LEARNING OBJECTIVES

B.

x 3 x2 ⫹ 3x ⫹ 14 ⫹ 5. Then we group the inner terms with x and factor again: P ⫽ x 3 x2 ⫹ 3x ⫹ 14 ⫹ 5 ⫽ x3x1x ⫹ 32 ⫹ 14 ⫹ 5. The expression can now be evaluated using any input and the order of operations. If x ⫽ 2, we quickly find that P ⫽ 27. Use this method to evaluate H ⫽ x3 ⫹ 2x2 ⫹ 5x ⫺ 9 for x ⫽ ⫺3.

expression in simplest form Multiply and divide rational expressions Add and subtract rational expressions Simplify compound fractions Solve rational equations

A rational number is one that can be written as the quotient of two integers. Similarly, a rational expression is one that can be written as the quotient of two polynomials. We can apply the skills developed in a study of fractions (how to reduce, add, subtract, multiply, and divide) to rational expressions, sometimes called algebraic fractions.

A. Writing a Rational Expression in Simplest Form A rational expression is in simplest form when the numerator and denominator have no common factors (other than 1). After factoring the numerator and denominator, we apply the fundamental property of rational expressions. Fundamental Property of Rational Expressions If P, Q, and R are polynomials, with Q, R ⫽ 0, P P P#R P#R ⫽ and 122 ⫽ # 112 # Q R Q Q Q R

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In words, the property says (1) a rational expression can be simplified by canceling common factors in the numerator and denominator, and (2) an equivalent expression can be formed by multiplying numerator and denominator by the same nonzero polynomial. EXAMPLE 1

Simplifying a Rational Expression Write the expression in simplest form:

Solution

1x ⫺ 121x ⫹ 12 x2 ⫺ 1 ⫽ 2 1x ⫺ 121x ⫺ 22 x ⫺ 3x ⫹ 2

1x ⫺ 121x ⫹ 12

x2 ⫺ 1 . x ⫺ 3x ⫹ 2 2

factor numerator and denominator

1

⫽

1x ⫺ 121x ⫺ 22 x⫹1 ⫽ x⫺2

common factors reduce to 1

simplest form


WORTHY OF NOTE If we view a and b as two points on the number line, we note that they are the same distance apart, regardless of the order they are subtracted. This tells us the numerator and denominator will have the same absolute value but be opposite in sign, giving a value of ⫺1 (check using a few test values).

CAUTION

When simplifying rational expressions, we sometimes encounter expressions of a⫺b a⫺b . If we factor ⫺1 from the numerator, we see that ⫽ the form b⫺a b⫺a ⫺11b ⫺ a2 ⫽ ⫺1. b⫺a

When reducing rational numbers or expressions, only common factors can be reduced. It is incorrect to reduce (or divide out) individual terms: x⫹1 1 ⫽ (except for x ⫽ 0) x⫹2 2

Note that after simplifying an expression, we are actually saying the resulting (simpler) expression is equivalent to the original expression for all values where both are defined. The first expression is not defined when x ⫽ 1 or x ⫽ 2, the second when x ⫽ 2 (since the denominators would be zero). The calculator screens shown in Figure R.5 help to illustrate this fact, and it appears that we would very much prefer to be working with the simpler expression!

⫺6 ⫹ 423 ⫽ ⫺3 ⫹ 423, and 2

Figure R.5

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EXAMPLE 2

Simplifying a Rational Expression Write the expression in simplest form:

Solution

55

213 ⫺ x2 6 ⫺ 2x ⫽ 2 1x ⫺ 321x ⫹ 32 x ⫺9 1221⫺12 ⫽ x⫹3 ⫺2 ⫽ x⫹3

6 ⫺ 2x . x2 ⫺ 9

factor numerator and denominator

reduce:

3⫺x ⫽ ⫺1 x⫺3

simplest form

A. You’ve just seen how we can write a rational expression in simplest form


B. Multiplication and Division of Rational Expressions Operations on rational expressions use the factoring skills reviewed earlier, along with much of what we know about rational numbers. Multiplying Rational Expressions Given that P, Q, R, and S are polynomials with Q, S ⫽ 0, P R # ⫽ PR Q S QS 1. Factor all numerators and denominators completely. 2. Reduce common factors. 3. Multiply numerator ⫻ numerator and denominator ⫻ denominator.

EXAMPLE 3

Multiplying Rational Expressions Compute the product:

Solution

2a ⫹ 2 # 3a2 ⫺ a ⫺ 2 . 3a ⫺ 3a2 9a2 ⫺ 4

21a ⫹ 12 13a ⫹ 221a ⫺ 12 2a ⫹ 2 # 3a2 ⫺ a ⫺ 2 # ⫽ 2 2 3a11 ⫺ a2 13a ⫺ 2213a ⫹ 22 3a ⫺ 3a 9a ⫺ 4 ⫽ ⫽

21a ⫹ 12

#

1⫺12

13a ⫹ 22 1a ⫺ 1 2 1

3a11 ⫺ a2 13a ⫺ 2213a ⫹ 22 ⫺21a ⫹ 12 3a13a ⫺ 22

factor

reduce:

a⫺1 ⫽ ⫺1 1⫺a

simplest form


To divide fractions, we multiply the first expression by the reciprocal of the second (we sometimes say, “invert the divisor and multiply”). The quotient of two rational expressions is computed in the same way.

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Dividing Rational Expressions Given that P, Q, R, and S are polynomials with Q, R, S ⫽ 0, P R P S PS ⫼ ⫽ # ⫽ Q S Q R QR Invert the divisor and multiply.

EXAMPLE 4

Dividing Rational Expressions Compute the quotient

Solution

4m3 ⫺ 12m2 ⫹ 9m 10m2 ⫺ 15m ⫼ . m2 ⫺ 49 m2 ⫹ 4m ⫺ 21

10m2 ⫺ 15m 4m3 ⫺ 12m2 ⫹ 9m ⫼ m2 ⫺ 49 m2 ⫹ 4m ⫺ 21 4m3 ⫺ 12m2 ⫹ 9m # m2 ⫹ 4m ⫺ 21 ⫽ m2 ⫺ 49 10m2 ⫺ 15m 2 m14m ⫺ 12m ⫹ 92 1m ⫹ 72 1m ⫺ 32 # ⫽ 1m ⫹ 721m ⫺ 72 5m12m ⫺ 32

m 12m ⫺ 3212m ⫺ 32 1m ⫹ 72 1m ⫺ 32 1

⫽ ⫽

1

invert and multiply

factor

1

1m ⫹ 721m ⫺ 72 12m ⫺ 32 1m ⫺ 32

#

5m12m ⫺ 32

factor and reduce

lowest terms

51m ⫺ 72

Note that we sometimes refer to simplest form as lowest terms. Now try Exercises 21 through 42

CAUTION

1w ⫹ 721w ⫺ 72 w ⫺ 2 # , it is a common mistake to think that all factors 1w ⫺ 721w ⫺ 22 w ⫹ 7 “cancel,” leaving an answer of zero. Actually, all factors reduce to 1, and the result is a value of 1 for all inputs where the product is defined. For products like

1w ⫹ 72 1w ⫺ 72 w ⫺ 2 # ⫽1 1w ⫺ 721w ⫺ 22 w ⫹ 7 1

B. You’ve just seen how we can multiply and divide rational expressions

1

1

C. Addition and Subtraction of Rational Expressions Recall that the addition and subtraction of fractions requires finding the lowest common denominator (LCD) and building equivalent fractions. The sum or difference of the numerators is then placed over this denominator. The procedure for the addition and subtraction of rational expressions is very much the same. Note that the LCD can also be described as the least common multiple (LCM) of all denominators. Addition and Subtraction of Rational Expressions 1. 2. 3. 4.

Find the LCD of all rational expressions. Build equivalent expressions using the LCD. Add or subtract numerators as indicated. Write the result in lowest terms.

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EXAMPLE 5

Adding and Subtracting Rational Expressions Compute as indicated: 7 3 a. ⫹ 10x 25x2

Solution

57

b.

5 10x ⫺ x⫺3 x ⫺9 2

a. The LCM for 10x and 25x2 is 50x2. 3 3 2 7 7 # 5x # ⫹ ⫹ ⫽ 2 10x 10x 5x 25x 25x2 2 35x 6 ⫽ ⫹ 2 50x 50x2 35x ⫹ 6 ⫽ 50x2

find the LCD write equivalent expressions

simplify

add the numerators and write the result over the LCD

The result is in simplest form. b. The LCM for x2 ⫺ 9 and x ⫺ 3 is 1x ⫺ 32 1x ⫹ 32. 5 5 10x 10x #x⫹3 ⫺ ⫺ ⫽ 2 x⫺3 x⫺3 x⫹3 1x ⫺ 321x ⫹ 32 x ⫺9 10x ⫺ 51x ⫹ 32 ⫽ 1x ⫺ 321x ⫹ 32 10x ⫺ 5x ⫺ 15 ⫽ 1x ⫺ 321x ⫹ 32 5x ⫺ 15 ⫽ 1x ⫺ 32 1x ⫹ 32

find the LCD write equivalent expressions

subtract numerators, write the result over the LCD distribute

combine like terms

1

⫽

51x ⫺ 32

1x ⫺ 321x ⫹ 32

⫽

5 x⫹3

factor and reduce


EXAMPLE 6

Adding and Subtracting Rational Expressions Perform the operations indicated: n⫺3 c 5 b2 a. b. ⫺ 2 ⫺ 2 a n⫹2 n ⫺4 4a

Solution

a. The LCM for n ⫹ 2 and n2 ⫺ 4 is 1n ⫹ 22 1n ⫺ 22. 5 n⫺3 n⫺3 5 #n⫺2⫺ ⫽ ⫺ 2 n⫹2 1n ⫹ 22 n ⫺ 2 1n ⫹ 22 1n ⫺ 22 n ⫺4 51n ⫺ 22 ⫺ 1n ⫺ 32 ⫽ 1n ⫹ 221n ⫺ 22 5n ⫺ 10 ⫺ n ⫹ 3 ⫽ 1n ⫹ 221n ⫺ 22 4n ⫺ 7 ⫽ 1n ⫹ 221n ⫺ 22

write equivalent expressions subtract numerators, write the result over the LCD distribute

result

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b. The LCM for a and 4a2 is 4a2:

b2 c c 4a b2 ⫺ ⫺ # ⫽ a a 4a 4a2 4a2 b2 4ac ⫽ 2⫺ 2 4a 4a b2 ⫺ 4ac ⫽ 4a2

write equivalent expressions simplify subtract numerators, write the result over the LCD


CAUTION

When the second term in a subtraction has a binomial numerator as in Example 6a, be sure the subtraction is applied to both terms. It is a common error to write 51n ⫺ 22 n⫺3 5n ⫺ 10 ⫺ n ⫺ 3 X in which the subtraction is applied ⫺ ⫽ 1n ⫹ 221n ⫺ 22 1n ⫹ 221n ⫺ 22 1n ⫹ 221n ⫺ 22 to the first term only. This is incorrect!

C. You’ve just seen how we can add and subtract rational expressions

D. Simplifying Compound Fractions Rational expressions whose numerator or denominator contain a fraction are called 3 2 ⫺ 3m 2 compound fractions. The expression is a compound fraction with a 3 1 ⫺ 4m 3m2 2 3 3 1 numerator of ⫺ and a denominator of ⫺ . The two methods commonly 3m 2 4m 3m2 used to simplify compound fractions are summarized in the following boxes. Simplifying Compound Fractions (Method I) 1. Add/subtract fractions in the numerator, writing them as a single expression. 2. Add/subtract fractions in the denominator, also writing them as a single expression. 3. Multiply the numerator by the reciprocal of the denominator and simplify if possible. Simplifying Compound Fractions (Method II) 1. Find the LCD of all fractions in the numerator and denominator. 2. Multiply the numerator and denominator by this LCD and simplify. 3. Simplify further if possible. Method II is illustrated in Example 7.

EXAMPLE 7

Simplifying a Compound Fraction Simplify the compound fraction: 2 3 ⫺ 3m 2 3 1 ⫺ 4m 3m2

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Solution

The LCD for all fractions is 12m2. 2 3 3 12m2 2 ⫺ ⫺ ba b a 3m 2 3m 2 1 ⫽ 1 1 3 3 12m2 ⫺ ⫺ b a ba 4m 4m 1 3m2 3m2 2 12m2 3 12m2 a ba b ⫺ a ba b 3m 1 2 1 ⫽ 12m2 1 3 12m2 b ⫺ a 2 ba b a ba 4m 1 1 3m 8m ⫺ 18m2 ⫽ 9m ⫺ 4

multiply numerator and denominator by 12m 2 ⫽

59

12m 2 1

distribute

simplify

⫺1

⫽

2m14 ⫺ 9m2 9m ⫺ 4

⫽ ⫺2m

D. You’ve just seen how we can simplify compound fractions

factor and write in lowest terms


E. Solving Rational Equations In Section R.3 we solved linear equations using basic properties of equality. If any equation contained fractional terms, we “cleared the fractions” using the least common denominator (LCD). We can also use this idea to solve rational equations, or equations that contain rational expressions. Solving Rational Equations 1. Identify and exclude any values that cause a zero denominator. 2. Multiply both sides by the LCD and simplify (this will eliminate all denominators). 3. Solve the resulting equation. 4. Check all solutions in the original equation.

EXAMPLE 8

Solving a Rational Equation Solve for m:

Solution

4 1 2 . ⫽ 2 ⫺ m m⫺1 m ⫺m

Since m2 ⫺ m ⫽ m1m ⫺ 12, the LCD is m1m ⫺ 12, where m ⫽ 0 and m ⫽ 1. 4 1 2 d multiply by LCD b ⫽ m1m ⫺ 12 c m1m ⫺ 12a ⫺ m m⫺1 m1m ⫺ 12 m1m ⫺ 12 2 m1m ⫺ 12 m1m ⫺ 12 1 4 a b⫺ a b⫽ a b distribute and simplify m 1 1 m⫺1 1 m1m ⫺ 12 denominators are eliminated 21m ⫺ 12 ⫺ m ⫽ 4 2m ⫺ 2 ⫺ m ⫽ 4 distribute m⫽6 solve for m

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Checking by substitution we have: 2 4 1 ⫽ 2 original equation ⫺ m m⫺1 m ⫺m 1 4 2 ⫺ ⫽ substitute 6 for m 2 162 162 ⫺ 1 162 ⫺ 162 1 4 1 ⫺ ⫽ simplify 3 5 30 3 2 5 ⫺ ⫽ common denominator 15 15 15 2 2 ✓ result ⫽ 15 15 A calculator check is shown in the figure. Now try Exercises 75 through 80 Multiplying both sides of an equation by a variable sometimes introduces a solution that satisfies the resulting equation, but not the original equation—the one we’re trying to solve. Such “solutions” are called extraneous roots and illustrate the need to check all apparent solutions in the original equation. In the case of rational equations, we are particularly aware that any value that causes a zero denominator is outside the domain and cannot be a solution. EXAMPLE 9

Solving a Rational Equation Solve: x ⫹

Solution

4x 12 ⫽1⫹ . x⫺3 x⫺3

The LCD is x ⫺ 3, where x ⫽ 3. multiply both 4x 12 b ⫽ 1x ⫺ 32a1 ⫹ b 1x ⫺ 32ax ⫹ sides by LCD x⫺3 x⫺3 12 x⫺3 4x x⫺3 distribute and ba ba b ⫽ 1x ⫺ 32 112 ⫹ a b simplify 1x ⫺ 32x ⫹ a 1 x⫺3 1 x⫺3 x2 ⫺ 3x ⫹ 12 ⫽ x ⫺ 3 ⫹ 4x denominators are eliminated set equation equal to zero x2 ⫺ 8x ⫹ 15 ⫽ 0 factor 1x ⫺ 321x ⫺ 52 ⫽ 0 zero factor property x ⫽ 3 or x ⫽ 5 Checking shows x ⫽ 3 is an extraneous root, and x ⫽ 5 is the only valid solution.

E. You’ve just seen how we can solve rational equations


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61

R.5 EXERCISES



1. In simplest form, 1a ⫺ b2/1a ⫺ b2 is equal to , while 1a ⫺ b2/1b ⫺ a2 is equal to 3. As with numeric fractions, algebraic fractions require a for addition and subtraction.

.

2. A rational expression is in when the numerator and denominator have no common factors, other than . 4. Since x2 ⫹ 9 is prime, the expression 1x2 ⫹ 92/ 1x ⫹ 32 is already written in .

State T or F and discuss/explain your response.

5.

x x⫹1 1 ⫺ ⫽ x⫹3 x⫹3 x⫹3

6.

1x ⫹ 321x ⫺ 22

1x ⫺ 221x ⫹ 32

⫽0


Reduce to lowest terms.

7. a. 8. a.

a⫺7 ⫺3a ⫹ 21

b.

x⫺4 ⫺7x ⫹ 28

b.

x2 ⫺ 5x ⫺ 14 9. a. 2 x ⫹ 6x ⫺ 7

2x ⫹ 6 4x2 ⫺ 8x 3x ⫺ 18 6x2 ⫺ 12x

a2 ⫹ 3a ⫺ 28 b. a2 ⫺ 49

16. a. c.

x3 ⫹ 4x2 ⫺ 5x x3 ⫺ x 12y2 ⫺ 13y ⫹ 3 27y3 ⫺ 1

b. d.

5p2 ⫺ 14p ⫺ 3 5p2 ⫹ 11p ⫹ 2 ax2 ⫺ 5x2 ⫺ 3a ⫹ 15 ax ⫺ 5x ⫹ 5a ⫺ 25

Compute as indicated. Write final results in lowest terms.

17.

a2 ⫺ 4a ⫹ 4 a2 ⫺ 2a ⫺ 3 # a2 ⫺ 9 a2 ⫺ 4

10. a.

r2 ⫹ 3r ⫺ 10 r2 ⫹ r ⫺ 6

b.

m2 ⫹ 3m ⫺ 4 m2 ⫺ 4m

18.

11. a.

x⫺7 7⫺x

b.

5⫺x x⫺5

b2 ⫹ 5b ⫺ 24 # 2 b 2 b ⫺ 6b ⫹ 9 b ⫺ 64

19.

12. a.

v2 ⫺ 3v ⫺ 28 49 ⫺ v2

b.

u2 ⫺ 10u ⫹ 25 25 ⫺ u2

x2 ⫺ 7x ⫺ 18 # 2x2 ⫹ 7x ⫹ 3 x2 ⫺ 6x ⫺ 27 2x2 ⫹ 5x ⫹ 2

20.

13. a.

⫺12a3b5 4a2b⫺4

b.

7x ⫹ 21 63

6v2 ⫹ 23v ⫹ 21 4v2 ⫺ 25 # 3v ⫹ 7 4v2 ⫺ 4v ⫺ 15

21.

m3n ⫺ m3 d. 4 m ⫺ m4n

22.

a3 ⫺ 4a2 a2 ⫹ 3a ⫺ 28 ⫼ 3 2 a ⫹ 5a ⫺ 14 a ⫺8

y2 ⫺ 9 c. 3⫺y 14. a. c. 15. a. c.

p3 ⫺ 64 p3 ⫺ p2

⫼

p2 ⫹ 4p ⫹ 16 p2 ⫺ 5p ⫹ 4

5m⫺3n5 ⫺10mn2

b.

⫺5v ⫹ 20 25

23.

n2 ⫺ 4 2⫺n

d.

w4 ⫺ w4v w3v ⫺ w3

3⫺x 3x ⫺ 9 ⫼ 4x ⫹ 12 5x ⫹ 15

24.

2n3 ⫹ n2 ⫺ 3n n3 ⫺ n2

b.

6x2 ⫹ x ⫺ 15 4x2 ⫺ 9

2⫺b 5b ⫺ 10 ⫼ 7b ⫺ 28 5b ⫺ 20

25.

x3 ⫹ 8 x2 ⫺ 2x ⫹ 4

d.

mn2 ⫹ n2 ⫺ 4m ⫺ 4 mn ⫹ n ⫹ 2m ⫹ 2

a2 ⫹ a 3a ⫺ 9 # a2 ⫺ 3a 2a ⫹ 2

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26.

R–62

CHAPTER R A Review of Basic Concepts and Skills 2 p2 ⫺ 36 # 2 4p 2p 2p ⫹ 12p

Compute as indicated. Write answers in lowest terms [recall that a ⴚ b ⴝ ⴚ1(b ⴚ a)].

8 # 1a2 ⫺ 2a ⫺ 352 27. 2 a ⫺ 25

43.

3 5 ⫹ 2 2x 8x

44.

15 7 ⫺ 2 16y 2y

m2 ⫺ 5m m2 ⫺ m ⫺ 20

45.

7 1 ⫺ 4x2y3 8xy4

46.

5 3 ⫹ 6a3b 9ab3

28. 1m2 ⫺ 162 29.

#

xy ⫺ 3x ⫹ 2y ⫺ 6 x2 ⫺ 3x ⫺ 10

xy ⫺ 3x xy ⫺ 5y

⫼

47.

4p p ⫺ 36 2

⫺

2 p⫺6

48.

3q q ⫺ 49 2

⫺

3 2q ⫺ 14

30.

ab ⫺ 2a 2a ⫺ ab ⫹ 7b ⫺ 14 ⫼ 2 ab ⫺ 7a b ⫺ 14b ⫹ 49

49.

4 m ⫹ 4⫺m m ⫺ 16

50.

p 2 ⫹ 2 p⫺2 4⫺p

31.

m2 ⫺ 16 m2 ⫹ 2m ⫺ 8 ⫼ m2 ⫺ 2m m2

51.

2 ⫺5 m⫺7

52.

4 ⫺9 x⫺1

32.

2x2 ⫺ 18 18 ⫺ 6x ⫼ 3 2 x ⫺ 25 x ⫺ 2x2 ⫺ 25x ⫹ 50

53.

33.

y⫹3

2

#y

3y2 ⫹ 9y

⫹ 7y ⫹ 12 y2 ⫺ 16

⫼

y2 ⫹ 4y

2

y⫹1 y ⫹ y ⫺ 30 2

⫺

2 y⫹6

54.

3 4n ⫺ 4n ⫺ 20 n ⫺ 5n

y2 ⫺ 4y

2

34.

x2 ⫺ 1 x ⫹ 1 x2 ⫹ 4x ⫺ 5 # ⫼ x2 ⫺ 5x ⫺ 14 x2 ⫺ 4 x ⫹ 5

55.

1 a ⫹ 2 a⫹4 a ⫺ a ⫺ 20

35.

x2 ⫺ x ⫹ 0.21 x2 ⫺ 0.49 ⫼ x2 ⫹ 0.5x ⫺ 0.14 x2 ⫺ 0.09

56.

x⫺5 2x ⫺ 1 ⫺ 2 x ⫹ 3x ⫺ 4 x ⫹ 3x ⫺ 4

36.

x2 ⫺ 0.8x ⫹ 0.15 x2 ⫺ 0.25 ⫼ x2 ⫹ 0.1x ⫺ 0.2 x2 ⫺ 0.16

57.

4 4 n2 ⫹ n ⫹ 3 9 37. ⫼ 13 1 2 n2 ⫺ n ⫹ n2 ⫺ 15 15 25 n2 ⫺

q2 ⫺ 38. q2 ⫺

40. 41.

9 25

1 3 q⫺ 10 10

3 17 q⫹ 20 20 1 q2 ⫺ 16

q2 ⫹ ⫼

2

p3 ⫹ p2 ⫺ 49p ⫺ 49 p2 ⫹ 6p ⫺ 7

42. a

4x ⫺ 25 2x ⫺ x ⫺ 15 # 4x ⫹ 25x ⫺ 21 ⫼ 2 b x ⫺ 11x ⫹ 30 x ⫺ 9x ⫹ 18 12x2 ⫺ 5x ⫺ 3 2

2

y ⫹ 2y ⫹ 1

m⫺5 2 ⫹ 2 m ⫺9 m ⫹ 6m ⫹ 9

60.

m⫹6 m⫹2 ⫺ 2 2 m ⫺ 25 m ⫺ 10m ⫹ 25

2

y⫹2 5y ⫹ 11y ⫹ 2 2

⫹

5 y ⫹y⫺6 2

m m⫺4 ⫹ 2 3m ⫺ 11m ⫹ 6 2m ⫺ m ⫺ 15 2

Write each term as a rational expression. Then compute the sum or difference indicated.

p3 ⫺ 1

4n2 ⫺ 1 6n2 ⫹ 5n ⫹ 1 # 12n2 ⫺ 17n ⫹ 6 # 12n2 ⫺ 5n ⫺ 3 2n2 ⫹ n 6n2 ⫺ 7n ⫹ 2 2

2y ⫺ 5 2

59.

62.

p2 ⫹ p ⫹ 1

y ⫹ 2y ⫹ 1

⫺

⫺2 7 ⫺ 2 3a ⫹ 12 a ⫹ 4a

2

⫼

3y ⫺ 4 2

58.

61.

6a ⫺ 24 3a ⫺ 24a ⫺ 12a ⫹ 96 ⫼ 3 2 a ⫺ 11a ⫹ 24 3a ⫺ 81 3

39.

4 9

2

2

63. a. p⫺2 ⫺ 5p⫺1

64. a. 3a⫺1 ⫹ 12a2 ⫺1

b. x⫺2 ⫹ 2x⫺3

b. 2y⫺1 ⫺ 13y2 ⫺1

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R–63 Simplify each compound fraction. Use either method.

5 1 ⫺ a 4 65. 25 1 ⫺ 2 16 a

1 8 ⫺ 3 27 x 66. 2 1 ⫺ x 3

1 p⫺2 67. 1 1⫹ p⫺2

3 y⫺6 68. 9 y⫹ y⫺6

3 2 ⫹ 3⫺x x⫺3 69. 4 5 ⫹ x x⫺3

2 1 ⫺ y⫺5 5⫺y 70. 3 2 ⫺ y y⫺5

2 y ⫺ y ⫺ 20 71. 3 4 ⫺ y⫹4 y⫺5

2 x ⫺ 3x ⫺ 10 72. 6 4 ⫺ x⫹2 x⫺5

p⫹

2

1⫹

73. a. 74. a.

1 ⫹ 3m⫺1 1 ⫺ 3m⫺1

b.

4 ⫺ 9a⫺2 3a⫺2

b.

Solve each equation. Identify any extraneous roots.

75.

2 5 1 ⫽ 2 ⫹ x x⫹1 x ⫹x

76.

1 5 3 ⫽ ⫺ 2 m m⫹3 m ⫹ 3m

77.

3 21 ⫽ a⫹2 a⫺1

78.

4 7 ⫽ 2y ⫺ 3 3y ⫺ 5

79.

1 1 1 ⫺ ⫽ 2 3y 4y y

81. x ⫹

80.

1 1 3 ⫺ ⫽ 2 5x 2x x

2x 14 ⫽1⫹ x⫺7 x⫺7

82.

2x 10 ⫹x⫽1⫹ x⫺5 x⫺5

83.

5 20 6 ⫽ ⫹ 2 n⫹3 n⫺2 n ⫹n⫺6

84.

1 ⫹ 2x⫺2 1 ⫺ 2x⫺2

2 1 7 ⫽⫺ ⫺ 2 p⫹2 p ⫹ 3 p ⫹ 5p ⫹ 6

85.

3 ⫹ 2n⫺1 5n⫺2

3 2a2 ⫹ 5 a ⫽ ⫺ 2 2a ⫹ 1 a⫺3 2a ⫺ 5a ⫺ 3

86.

⫺18 3n 4n ⫹ ⫽ 2n ⫺ 1 3n ⫹ 1 6n ⫺ n ⫺ 1

2

Rewrite each expression as a compound fraction. Then simplify using either method.

63


2


87. Cost to seize illegal drugs: C ⴝ

450P 100 ⴚ P

The cost C, in millions of 450P P dollars, for a government to find 100 ⴚ P and seize P% 10 ⱕ P 6 1002 of 40 a certain illegal drug is modeled 60 by the rational equation shown. 80 Complete the table (round to the nearest dollar) and answer the 90 following questions. 93 a. What is the cost of seizing 95 40% of the drugs? Estimate 98 the cost at 85%. 100 b. Why does cost increase dramatically the closer you get to 100%? c. Will 100% of the drugs ever be seized?

88. Chemicals in the bloodstream: C ⴝ

200H2 H3 ⴙ 40

Rational equations are often used 200H2 to model chemical concentrations H H3 ⴙ 40 in the bloodstream. The percent 0 concentration C of a certain drug 1 H hours after injection into muscle tissue can be modeled by 2 the equation shown (H ⱖ 0). 3 Complete the table (round to the 4 nearest tenth of a percent) and 5 answer the following questions. 6 a. What is the percent 7 concentration of the drug 3 hr after injection? b. Why is the concentration virtually equal at H ⫽ 4 and H ⫽ 5? c. Why does the concentration begin to decrease? d. How long will it take for the concentration to become less than 10%?

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APPLICATIONS

89. Stock prices: When a hot new stock hits the market, its price will often rise dramatically and then taper 5017d 2 ⫹ 102 off over time. The equation P ⫽ d3 ⫹ 50 models the price of stock XYZ d days after it has “hit the market.” (a) Create a table of values showing the price of the stock for the first 10 days (rounded to the nearest dollar) and comment on what you notice. (b) Find the opening price of the stock. (c) Does the stock ever return to its original price? 90. Population growth: The Department of Wildlife introduces 60 elk into a new game reserve. It is projected that the size of the herd will grow 1016 ⫹ 3t2 , where according to the equation N ⫽ 1 ⫹ 0.05t N is the number of elk and t is the time in years. (a) Approximate the population of elk after 14 yr. (b) If recent counts find 225 elk, approximately how many years have passed? 91. Typing speed: The number of words per minute that a beginner can type is approximated by the 60t ⫺ 120 , where N is the number equation N ⫽ t

䊳

R–64


of words per minute after t weeks, 3 6 t 6 12. Use a table to determine how many weeks it takes for a student to be typing an average of forty-five words per minute. 92. Memory retention: A group of students is asked to memorize 50 Russian words that are unfamiliar to them. The number N of these words that the average student remembers D days later is modeled 5D ⫹ 35 1D ⱖ 12. How many by the equation N ⫽ D words are remembered after (a) 1 day? (b) 5 days? (c) 12 days? (d) 35 days? (e) 100 days? According to this model, is there a certain number of words that the average student never forgets? How many? 93. Pollution removal: For a steel mill, the cost C (in millions of dollars) to remove toxins from the 22P resulting sludge is given by C ⫽ , where 100 ⫺ P P is the percent of the toxins removed. What percent can be removed if the mill spends $88,000,000 on the cleanup? Round to tenths of a percent.


94. One of these expressions is not equal to the others. Identify which and explain why. 20n a. b. 20 # n ⫼ 10 # n 10n 20 n 1 # c. 20n # d. 10n 10 n 95. The average of A and B is x. The average of C, D, and E is y. The average of A, B, C, D, and E is: 3x ⫹ 2y 2x ⫹ 3y a. b. 5 5 21x ⫹ y2 31x ⫹ y2 c. d. 5 5

3 2 and , what is the 5 4 reciprocal of the sum of their reciprocals? Given c a that and are any two numbers—what is the b d reciprocal of the sum of their reciprocals?

96. Given the rational numbers

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R.6

Radicals, Rational Exponents, and Radical Equations


A. Simplify radical

B.

C.

D. E.

F.

expressions of the form n 1a n Rewrite and simplify radical expressions using rational exponents Use properties of radicals to simplify radical expressions Add and subtract radical expressions Multiply and divide radical expressions; write a radical expression in simplest form Solve equations and use formulas involving radicals

Square roots and cube roots come from a much larger family called radical expressions. Expressions containing radicals can be found in virtually every field of mathematical study, and are an invaluable tool for modeling many real-world phenomena. n

A. Simplifying Radical Expressions of the Form 1an In previous coursework, you likely noted that 1a ⫽ b only if b2 ⫽ a. This definition cannot be applied to expressions like 1⫺16, since there is no number b such that b2 ⫽ ⫺16. In other words, the expression 1a represents a real number only if a ⱖ 0 (for a full review of the real numbers and other sets of numbers, see Appendix I at www.mhhe.com/coburn). Of particular interest to us now is an inverse operation for a2. In other words, what operation can be applied to a2 to return a? Consider the following.

EXAMPLE 1

䊳

Evaluating a Radical Expression

Evaluate 2a2 for the values given: a. a ⫽ 3 b. a ⫽ 5 c. a ⫽ ⫺6

Solution

䊳

a. 232 ⫽ 19 ⫽3

b. 252 ⫽ 125 ⫽5

c. 21⫺62 2 ⫽ 136 ⫽6


䊳

The pattern seemed to indicate that 2a2 ⫽ a and that our search for an inverse operation was complete—until Example 1(c), where we found that 21⫺62 2 ⫽ ⫺6. Using the absolute value concept, we can “repair” this apparent discrepancy and state a general rule for simplifying these expressions: 2a2 ⫽ 冟a冟. For expressions like 249x2 and 2y6, the radicands can be rewritten as perfect squares and simplified in the same manner: 249x2 ⫽ 217x2 2 ⫽ 7冟x冟 and 2y6 ⫽ 21y3 2 2 ⫽ 冟y3冟. The Square Root of a2: 2a2 For any real number a, 2a2 円 a円.

EXAMPLE 2

䊳

Simplifying Square Root Expressions Simplify each expression. a. 2169x2

Solution

䊳

b. 2x2 ⫺ 10x ⫹ 25

a. 2169x2 ⫽ 冟13x冟 ⫽ 13冟x冟 b. 2x ⫺ 10x ⫹ 25 ⫽ 21x ⫺ 52 ⫽ 冟x ⫺ 5冟 2

since x could be negative 2

since x 5 could be negative


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䊳

65

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䊳

CAUTION

In Section R.2, we noted that 1A ⫹ B2 2 ⫽ A2 ⫹ B2, indicating that you cannot square the individual terms in a sum (the square of a binomial results in a perfect square trinomial). In a similar way, 2A2 ⫹ B2 ⫽ A ⫹ B, and you cannot take the square root of individual terms. There is a big difference between the expressions 2A2 ⫹ B2 and 21A ⫹ B2 2 ⫽ 冟A ⫹ B冟. Try evaluating each when A ⫽ 3 and B ⫽ 4. 3 3 3 3 To investigate expressions like 2 x , note the radicand in both 1 8 and 1 ⫺64 can be written as a perfect cube. From our earlier definition of cube roots we know 3 3 3 3 1 8⫽ 2 122 3 ⫽ 2, 1 ⫺64 ⫽ 2 1⫺42 3 ⫽ ⫺4, and that every real number has only one real cube root. For this reason, absolute value notation is not used or needed when taking cube roots.

3 3 The Cube Root of a3: 2 a

For any real number a, 3 3 2 a a.

EXAMPLE 3

䊳

Simplifying Cube Root Expressions Simplify each expression. 3 3 a. 2 b. 2 ⫺27x3 ⫺64n6

Solution

䊳

3 3 a. 2 ⫺27x3 ⫽ 2 1⫺3x2 3 ⫽ ⫺3x

3 3 b. 2 ⫺64n6 ⫽ 2 1⫺4n2 2 3 ⫽ ⫺4n2


䊳

We can extend these ideas to fourth roots, fifth roots, and so on. For example, the 5 fifth root of a is b only if b5 ⫽ a. In symbols, 1a ⫽ b implies b5 ⫽ a. Since an odd number of negative factors is always negative: 1⫺22 5 ⫽ ⫺32, and an even number of negative factors is always positive: 1⫺22 4 ⫽ 16, we must take the index into account n when evaluating expressions like 1an. If n is even and the radicand is unknown, absolute value notation must be used. n

WORTHY OF NOTE 2 Just as 1 ⫺16 is not a real number, 4 6 1 ⫺16 and 1 ⫺16 do not represent real numbers. An even number of repeated factors is always positive!

EXAMPLE 4

䊳

The nth Root of an: 2an For any real number a, n 1. 1an ⫽ 冟a冟 when n is even. Simplifying Radical Expressions Simplify each expression. 4 4 a. 1 b. 1 81 ⫺81 4 5 4 e. 216m f. 232p5

Solution

A. You’ve just seen how we can simplify radical n expressions of the form 1a n

䊳

n

2. 1an ⫽ a when n is odd.

4 1 81 ⫽ 3 5 132 ⫽ 2 4 4 2 16m4 ⫽ 2 12m2 4 ⫽ 冟2m冟 or 2冟m冟 6 g. 2 1m ⫹ 52 6 ⫽ 冟m ⫹ 5冟

a. c. e.

5 c. 1 32 6 g. 2 1m ⫹ 52 6

5 d. 1 ⫺32 7 h. 2 1x ⫺ 22 7

4 b. 1 ⫺81 is not a real number 5 d. 1 ⫺32 ⫽ ⫺2 5 5 f. 232p5 ⫽ 2 12p2 5 ⫽ 2p 7 h. 2 1x ⫺ 22 7 ⫽ x ⫺ 2


䊳

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Section R.6 Radicals, Rational Exponents, and Radical Equations

67

B. Radical Expressions and Rational Exponents As an alternative to radical notation, a rational (fractional) exponent can be used, 3 along with the power property of exponents. For 1a3 ⫽ a, notice that an exponent of one-third1 can 3 replace the cube root notation and produce the same result: 3 3 2 a ⫽ 1a3 2 3 ⫽ a3 ⫽ a. In the same way, an exponent of one-half can replace the 1 2 square root notation: 2a2 ⫽ 1a2 2 2 ⫽ a2 ⫽ 冟a冟. In general, we have the following: Rational Exponents If a is a real number and n is an integer greater than 1, 1 n n then 1a ⫽ 2a1 ⫽ an n provided 1a represents a real number.

EXAMPLE 5

䊳

Simplifying Radical Expressions Using Rational Exponents Simplify by rewriting each radicand as a perfect nth power and converting to rational exponent notation. 8w3 3 4 4 a. 2 ⫺125 b. ⫺2 16x20 c. 2 ⫺81 d. 3 B 27

Solution

䊳

a. 2⫺125 ⫽ 2 1⫺52 3 1 ⫽ 3 1⫺52 3 4 3 3 ⫽ 1⫺52 3 ⫽ ⫺5

b. ⫺216x20 ⫽ ⫺ 2 12x5 2 4 1 ⫽ ⫺ 3 12x5 2 4 4 4 ⫽ ⫺ 02x5 0 ⫽ ⫺2冟x冟5

4 c. 2 ⫺81 ⫽ 1⫺812 4 is not a real number

d.

3

3

1

4

4

3 8w3 3 2w ⫽ a b B 27 B 3 1 2w 2w 3 3 ⫽ ca b d ⫽ 3 3 3

Now try Exercises 15 and 16 n

WORTHY OF NOTE Any rational number can be decomposed into the product of a unit fraction and an m 1 integer: ⫽ # m. n n

n

䊳

1

Figure R.6 When a rational exponent is used, as in 1a ⫽ 2a1 ⫽ an, the denominator of the exponent represents the index number, while the m numerator of the exponent represents the original power on a. This an n m is true even when the exponent on a is something other than one! In (兹a ) 4 other words, the radical expression 2163 can be rewritten as 1 3 1 3 1163 2 4 ⫽ 1161 2 4 or 164. This is further illustrated in Figure R.6 where we see the rational exponent has the form, “power over root.” To evaluate this expression without the aid of a calculator, we use the commutative property to rewrite 3 1 1 3 1 3 1161 2 4 as 1164 2 1 and begin with the fourth root of 16: 1164 2 1 ⫽ 23 ⫽ 8. m In general, if m and n have no common factors (other than 1) the expression a n can be interpreted in the following two ways.

Rational Exponents If mn is a rational number expressed in lowest terms with n ⱖ 2, then 112 a n ⫽ 1 2a2 m m

n

n

or

122 a n ⫽ 2am m

n

(compute 1a, then take the mth power), (compute am, then take the nth root), n provided 1a represents a real number.

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Expressions with rational exponents are generally easier to evaluate if we compute the root first, then apply the exponent. Computing the root first also helps us determine whether or not an expression represents a real number.

EXAMPLE 6

䊳

Simplifying Expressions with Rational Exponents Simplify each expression, if possible. b. 1⫺492 2

3

3

a. ⫺492

Solution

3

WORTHY OF NOTE

3 While the expression 1⫺82 3 ⫽ 1 ⫺8 represents the real number ⫺2, the 2 6 2 expression 1⫺82 6 ⫽ 1 1⫺82 is not a 2 1 real number, even though ⫽ . 3 6 Note that the second exponent is not in lowest terms. 1

2

b. 1⫺492 2 ⫽ 3 1⫺492 2 4 3, ⫽ 1 1⫺492 3 not a real number

1

3

c. 1⫺82 3 ⫽ 3 1⫺82 34 2 3 ⫽ 1 1⫺82 2 2

d. ⫺8⫺3

2

a. ⫺492 ⫽ ⫺1492 2 3 ⫽ ⫺1 1492 3 ⫽ ⫺172 3 or ⫺343

䊳

c. 1⫺82 3

1

d.

⫽ 1⫺22 2 or 4

1

⫺8⫺3 ⫽ ⫺183 2 ⫺2 3 ⫽ ⫺1 1 82 ⫺2 2

1

⫽ ⫺2⫺2 or ⫺

B. You’ve just seen how we can rewrite and simplify radical expressions using rational exponents

1 4


䊳

C. Using Properties of Radicals to Simplify Radical Expressions The properties used to simplify radical expressions are closely connected to the properties of exponents. For instance, the1 product to a power1 property holds even when n 1 1 1 1 is a rational number. This means 1xy2 2 ⫽ x2y2 and 14 # 252 2 ⫽ 42 # 252. When the second statement is expressed in radical form, we have 14 # 25 ⫽ 14 # 225, with both forms having a value of 10. This suggests the product property of radicals, which can be extended to include cube roots, fourth roots, and so on. Product Property of Radicals n

n

If 1A and 1B represent real-valued expressions, then n

n

n

1AB ⫽ 1A # 1B

CAUTION

䊳

and

n

n

n

1A # 1B ⫽ 1AB.

Note that this property applies only to a product of two terms, not to a sum or difference. In other words, while 29x2 ⫽ 冟3x冟, 29 ⫹ x2 ⫽ | 3 ⫹ x |!

One application of the product property is to simplify radical expressions. In genn eral, the expression 1a is in simplified form if a has no factors (other than 1) that are perfect nth roots.

EXAMPLE 7

䊳

Simplifying Radical Expressions Write each expression in simplest form using the product property. a. 118

Solution

䊳

3 b. 5 2 125x4

a. 118 ⫽ 19 # 2 ⫽ 1912 ⫽ 3 12

c.

⫺4 ⫹ 220 2

3 3 d. 1.22 16n4 2 4n5

3 3 b. 5 2 125x4 ⫽ 5 # 2 125 # x4 3 3 3 # 3 1 These steps can ⫽ 5 # 2125 # 2 x 2x e 3 be done mentally # # # ⫽ 5 5 x 1x 3 x ⫽ 25x 1

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69


WORTHY OF NOTE

c.

For expressions like those in Example 7(c), students must resist the “temptation” to reduce individual terms as in ⫺4 ⫹ 120 ⫽ ⫺2 ⫹ 120. 2 Remember, only factors can be reduced.

⫺4 ⫹ 14 # 5 ⫺4 ⫹ 120 ⫽ 2 2 ⫺4 ⫹ 215 ⫽ 2 1 21⫺2 ⫹ 152 ⫽ 2 ⫽ ⫺2 ⫹ 15

3 3 3 d. 1.2 2 16n4 2 4n5 ⫽ 1.2 2 64 # n9 3 3 9 ⫽ 1.22642 n 3 3 ⫽ 1.22 642 1n3 2 3 ⫽ 1.2142n3 ⫽ 4.8n3

䊳


When radicals are combined using the product property, the result may contain a perfect nth root, which should be simplified. Note that the index numbers must be the same in order to use this property. The quotient property of radicals can also be established using exponential 100 1100 ⫽ ⫽ 2 suggests the following: properties. The fact that A 25 125 Quotient Property of Radicals n

n

If 1A and 1B represent real-valued expressions with B ⫽ 0, then n n 1A 1A n A n A ⫽ n and n ⫽ . AB 1B 1B A B Many times the product and quotient properties must work together to simplify a radical expression, as shown in Example 8A.

EXAMPLE 8A

䊳

Simplifying Radical Expressions Simplify each expression: 218a5 a. 22a

Solution

䊳

a.

218a5 22a

⫽

b.

18a5 B 2a

b.

⫽ 29a4 ⫽ 3a2

81 A 125x3 3

3 81 1 81 ⫽ 3 3 A 125x 2125x3 3 1 27 # 3 ⫽ 5x 3 31 3 ⫽ 5x 3

Radical expressions can also be simplified using rational exponents.

EXAMPLE 8B

䊳

Using Rational Exponents to Simplify Radical Expressions Simplify using rational exponents: 3 4 a. 236p4q5 b. v 2 v

Solution

C. You’ve just seen how we can use properties of radicals to simplify radical expressions

䊳

a. 236p4q5 ⫽ 136p4q5 2 2 1 4 5 ⫽ 362p2q2 4 1 ⫽ 6p2q12 ⫹ 2 2 1 ⫽ 6p2q2q2 ⫽ 6p2q2 1q 1

3 c. 1 m1m 4

3 4 b. v 2 v ⫽ v1 # v3 3 4 ⫽ v3 # v3 7 ⫽ v3 6 1 ⫽ v3v3 3 ⫽ v2 1 v

1

1

3 c. 1 m1m ⫽ m3m2 1 1 ⫽ m3 ⫹ 2 5 ⫽ m6 6 ⫽ 2 m5


䊳

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D. Addition and Subtraction of Radical Expressions 3 Since 3x and 5x are like terms, we know 3x ⫹ 5x ⫽ 8x. If x ⫽ 1 7, the sum becomes 3 3 3 3 17 ⫹ 5 17 ⫽ 8 17, illustrating how like radical expressions can be combined. Like radicals are those that have the same index and radicand. In some cases, we can identify like radicals only after radical terms have been simplified.

EXAMPLE 9

䊳

Adding and Subtracting Radical Expressions Simplify and combine (if possible). 3 3 a. 145 ⫹ 2 120 b. 2 16x5 ⫺ x 2 54x2

Solution

䊳

a. 145 ⫹ 2 120 ⫽ 3 15 ⫹ 212 152 ⫽ 3 15 ⫹ 4 15 ⫽ 7 15

simplify radicals: 145 ⫽ 19 # 5; 120 ⫽ 14 # 5 like radicals result

3 b. 216x ⫺ x 254x ⫽ 28 # 2 # x # x ⫺ x 2 27 # 2 # x2 3 3 ⫽ 2x 2 2x2 ⫺ 3x 2 2x2 3 2 ⫽ ⫺x 22x 3

5

3

2

3

3

D. You’ve just seen how we can add and subtract radical expressions

2

simplify radicals result


䊳

E. Multiplication and Division of Radical Expressions; Radical Expressions in Simplest Form Multiplying radical expressions is simply an extension of our earlier work. The multiplication can take various forms, from the distributive property to any of the special products reviewed in Section R.2. For instance, 1A ⫾ B2 2 ⫽ A2 ⫾ 2AB ⫹ B2, even if A or B is a radical term.

EXAMPLE 10

䊳

Multiplying Radical Expressions Compute each product and simplify. a. 5 131 16 ⫺ 4 132 b. 12 12 ⫹ 6 132 13 110 ⫹ 1152 c. 1x ⫹ 172 1x ⫺ 172 d. 13 ⫺ 122 2

Solution

䊳

a. 5131 16 ⫺ 4 132 ⫽ 5 118 ⫺ 201 132 2 ⫽ 5132 1 2 ⫺ 1202132 ⫽ 15 12 ⫺ 60

distribute simplify: 1 18 ⫽ 3 1 2, 1 132 2 ⫽ 3 result

b. 1212 ⫹ 613213110 ⫹ 1152 ⫽ 6120 ⫹ 2 130 ⫹ 18130 ⫹ 6145 F-O-I-L ⫽ 12 15 ⫹ 20 130 ⫹ 18 15 extract roots and simplify ⫽ 30 15 ⫹ 20 130 result c. 1x ⫹ 172 1x ⫺ 172 ⫽ x2 ⫺ 1 172 2 ⫽ x2 ⫺ 7

LOOKING AHEAD Notice that the answer for Example 10c contains no radical terms, since the outer and inner products sum to zero. This result will be used to simplify certain radical expressions in this section and later in Chapter 2.

1A ⫹ B2 1A ⫺ B2 ⫽ A 2 ⫺ B 2 result

d. 13 ⫺ 122 2 ⫽ 132 2 ⫺ 2132 1 122 ⫹ 1 122 2 ⫽ 9 ⫺ 6 12 ⫹ 2 ⫽ 11 ⫺ 6 12

1A ⫺ B 2 2 ⫽ A 2 ⫺ 2AB ⫹ B 2 simplify each term result


䊳

One application of products and powers of radical expressions is to evaluate certain quadratic expressions, as illustrated in Example 11.

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EXAMPLE 11

䊳

71

Evaluating a Quadratic Expression Show that when x2 ⫺ 4x ⫹ 1 is evaluated at x ⫽ 2 ⫹ 13, the result is zero.

Solution

䊳

x2 ⫺ 4x ⫹ 1 original expression 12 ⫹ 132 2 ⫺ 412 ⫹ 132 ⫹ 1 substitute 2 ⫹ 13 for x 4 ⫹ 4 13 ⫹ 3 ⫺ 8 ⫺ 4 13 ⫹ 1 square binomial; distribute 14 ⫹ 3 ⫺ 8 ⫹ 12 ⫹ 1413 ⫺ 4 132 commutative and associative properties 0✓ A calculator check is shown in the figure.


䊳

When we applied the quotient property in Example 8A, we obtained a denominator free of radicals. Sometimes the denominator is not automatically free of radicals, and the need to write radical expressions in simplest form comes into play. This process is called rationalizing the denominator. Radical Expressions in Simplest Form A radical expression is in simplest form if: 1. The radicand has no perfect nth root factors. 2. The radicand contains no fractions. 3. No radicals occur in a denominator. As with other types of simplification, the desired form can be achieved in various ways. If the denominator is a single radical term, we multiply the numerator and denominator by the factors required to eliminate the radical in the denominator [see Examples 12(a) and 12(b)]. If the radicand is a rational expression, it is generally easier to build an equivalent fraction within the radical having perfect nth root factors in the denominator [see Example 12(c)].

EXAMPLE 12

䊳

Simplifying Radical Expressions Simplify by rationalizing the denominator. Assume a, x ⫽ 0. 2 ⫺7 a. b. 3 1x 5 23

Solution

䊳

a.

b.

2 2 # 13 ⫽ 5 13 5 13 13 2 13 2 13 ⫽ ⫽ 2 15 51 132 ⫺7 3 1 x

⫽ ⫽

3 3 ⫺71 1 x211 x2 3 3 3 1 x1 1 x211 x2 3 2 ⫺7 2x

3 3 2 x 3 2 ⫺72 x ⫽ x

multiply numerator and denominator by 13

simplify—denominator is now rational

3 multiply using two additional factors of 1 x

product property

3 3 2 x ⫽x


䊳

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In some applications, the denominator may be a sum or difference containing a radical term. In this case, the methods from Example 12 are ineffective, and instead we multiply by a conjugate since 1A ⫹ B21A ⫺ B2 ⫽ A2 ⫺ B2. If either A or B is a square root, the result will be a denominator free of radicals.

EXAMPLE 13

䊳

Simplifying Radical Expressions Using a Conjugate Simplify the expression by rationalizing the denominator. 2 ⫹ 13 Write the answer in exact form and approximate form 16 ⫺ 12 rounded to three decimal places.

Solution

䊳

2 ⫹ 13 2 ⫹ 13 # 16 ⫹ 12 ⫽ 16 ⫺ 12 16 ⫺ 12 16 ⫹ 12 2 16 ⫹ 2 12 ⫹ 118 ⫹ 16 ⫽ 1 162 2 ⫺ 1 122 2 3 16 ⫹ 2 12 ⫹ 3 12 6⫺2 3 16 ⫹ 5 12 ⫽ 4 ⬇ 3.605

⫽

E. You’ve just seen how we can multiply and divide radical expressions and write a radical expression in simplest form

multiply by the conjugate of the denominator FOIL difference of squares simplify

exact form approximate form


䊳

F. Equations and Formulas Involving Radicals A radical equation is any equation that contains terms with a variable in the radicand. To solve a radical equation, we attempt to isolate a radical term on one side, then apply the appropriate nth power to free up the radicand and solve for the unknown. This is an application of the power property of equality. The Power Property of Equality n

n

If 1u and v are real-valued expressions and 1u ⫽ v, then 1 1u2 n ⫽ vn u ⫽ vn for n an integer, n ⱖ 2. n

Raising both sides of an equation to an even power can also introduce a false solution (extraneous root). Note that by inspection, the equation x ⫺ 2 ⫽ 1x has only the solution x ⫽ 4. But the equation 1x ⫺ 22 2 ⫽ x (obtained by squaring both sides) has both x ⫽ 4 and x ⫽ 1 as solutions, yet x ⫽ 1 does not satisfy the original equation. This means we should check all solutions of an equation where an even power is applied.

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EXAMPLE 14

䊳

Solving Radical Equations Solve each radical equation: a. 13x ⫺ 2 ⫹ 12 ⫽ x ⫹ 10

Solution

䊳

73

3 b. 2 1 x⫺5⫹4⫽0

a. 13x ⫺ 2 ⫹ 12 ⫽ x ⫹ 10 13x ⫺ 2 ⫽ x ⫺ 2 1 13x ⫺ 22 2 ⫽ 1x ⫺ 22 2 3x ⫺ 2 ⫽ x2 ⫺ 4x ⫹ 4 0 ⫽ x2 ⫺ 7x ⫹ 6 0 ⫽ 1x ⫺ 62 1x ⫺ 12 x ⫺ 6 ⫽ 0 or x ⫺ 1 ⫽ 0 x ⫽ 6 or x ⫽ 1

Check

䊳

x ⫽ 6:

Check

䊳

x ⫽ 1:

original equation isolate radical term (subtract 12) apply power property, power is even simplify, square binomial set equal to zero factor apply zero product property result, check for extraneous roots

13162 ⫺ 2 ⫹ 12 ⫽ 162 ⫹ 10 116 ⫹ 12 ⫽ 16 16 ⫽ 16 ✓ 13112 ⫺ 2 ⫹ 12 ⫽ 112 ⫹ 10 11 ⫹ 12 ⫽ 11 13 ⫽ 11x

The only solution is x ⫽ 6; x ⫽ 1 is extraneous. A calculator check is shown in the figures. 3 b. 2 1 x⫺5⫹4⫽0 3 1 x ⫺ 5 ⫽ ⫺2 3 1 1x ⫺ 52 3 ⫽ 1⫺22 3 x ⫺ 5 ⫽ ⫺8 x ⫽ ⫺3

original equation isolate radical term (subtract 4, divide by 2) apply power property, power is odd 3 simplify: 1 x ⫺ 52 3 ⫽ x ⫺ 5

solve

Substituting ⫺3 for x in the original equation verifies it is a solution.


䊳

Sometimes squaring both sides of an equation still results in an equation with a radical term, but often there is one fewer than before. In this case, we simply repeat the process, as indicated by the flowchart in Figure R.7.

EXAMPLE 15

䊳

Solving Radical Equations Solve the equation: 1x ⫹ 15 ⫺ 1x ⫹ 3 ⫽ 2.

Solution

䊳

1x ⫹ 15 ⫺ 1x ⫹ 3 ⫽ 2 1x ⫹ 15 ⫽ 1x ⫹ 3 ⫹ 2 1 1x ⫹ 152 2 ⫽ 1 1x ⫹ 3 ⫹ 22 2 x ⫹ 15 ⫽ 1x ⫹ 32 ⫹ 4 1x ⫹ 3 ⫹ 4 x ⫹ 15 ⫽ x ⫹ 4 1x ⫹ 3 ⫹ 7 8 ⫽ 4 1x ⫹ 3 2 ⫽ 1x ⫹ 3 4⫽x⫹3 1⫽x

original equation isolate one radical power property 1A ⫹ B2 2; A ⫽ 1x ⫹ 3, B ⫽ 2 simplify isolate radical divide by four power property possible solution

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Figure R.7

Check

1x ⫹ 15 ⫺ 1x ⫹ 3 ⫽ 2 1112 ⫹ 15 ⫺ 1112 ⫹ 3 ⫽ 2 116 ⫺ 14 ⫽ 2 4⫺2⫽2 2 ⫽ 2✓

䊳

Radical Equations

Isolate radical term

original equation substitute 1 for x simplify solution checks


䊳

Since rational exponents are so closely related to radicals, the solution process for each is very similar. The goal is still to “undo” the radical (rational exponent) and solve for the unknown.

Apply power property

Does the result contain a radical?

Power Property of Equality YES

For real-valued expressions u and v, with positive integers m, n, and mn in lowest terms: If m is odd m and u n ⫽ v,

NO

If m is even m and u n ⫽ v 1v 7 02,

then 1u n 2 m ⫽ vm m n

Solve using properties of equality

then 1u n 2 m ⫽ ⫾v m

n

m n

n

n

n

u ⫽ vm

u ⫽ ⫾vm

The power property of equality basically says that if certain conditions are satisfied, both sides of an equation can be raised to any needed power.

Check results in original equation

EXAMPLE 16

䊳

Solving Equations with Rational Exponents Solve each equation: 3 a. 31x ⫹ 12 4 ⫺ 9 ⫽ 15

Solution

䊳

3

a. 31x ⫹ 12 4 ⫺ 9 ⫽ 15 1x ⫹ 12 ⫽ 8 3 4

3 1x ⫹ 12 4 ⫽ 8 x ⫹ 1 ⫽ 16 x ⫽ 15 3 4 4 3

Check

䊳

4 3

3 4

3115 ⫹ 12 ⫺ 9 ⫽ 15

3116 2 ⫺ 9 ⫽ 15 3122 3 ⫺ 9 ⫽ 15 3182 ⫺ 9 ⫽ 15 15 ⫽ 15 ✓ 1 4

b.

3

1x ⫺ 32 ⫽ 4 2 3

3 1x ⫺ 32 4 ⫽ ⫾ 4 x⫺3⫽ ⫾8 x⫽3⫾8 2 3 3 2

3 2

b. 1x ⫺ 32 3 ⫽ 4 2

original equation; mn ⫽ 34 isolate variable term (add 9, divide by 3) apply power property, note m is odd simplify 383 ⫽ 183 2 4 ⫽ 16 4 4

1

result substitute 15 for x in the original equation simplify, rewrite exponent 4 1 16 ⫽ 2

23 ⫽ 8 solution checks original equation; mn ⫽ 23 apply power property, note m is even simplify 342 ⫽ 142 2 3 ⫽ 8 4 3

1

result

The solutions are 3 ⫹ 8 ⫽ 11 and 3 ⫺ 8 ⫽ ⫺5. Verify by checking both in the original equation. Now try Exercises 55 through 58

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䊳

CAUTION

As you continue solving equations with radicals and rational exponents, be careful not to arbitrarily place the “⫾” sign in front of terms given in radical form. The expression 118 indicates the positive square root of 18, where 118 ⫽ 312. The equation x2 ⫽ 18 becomes x ⫽ ⫾ 118 after applying the power property, with solutions x ⫽ ⫾312 1x ⫽ ⫺312, x ⫽ 3122, since the square of either number produces 18.

In Section R.4, we used a technique called u-substitution to factor expressions in quadratic form. The following equations are also in quadratic form since the de2 1 gree of the leading term is twice the degree of the middle term: x3 ⫺ 3x3 ⫺ 10 ⫽ 0 and 1x2 ⫹ x2 2 ⫺ 81x2 ⫹ x2 ⫹ 12 ⫽ 0. The first equation and its solution appear in Example 17.

EXAMPLE 17

䊳

Solving Equations in Quadratic Form 2

1

Solve using a u-substitution: x3 ⫺ 3x3 ⫺ 10 ⫽ 0.

Solution

䊳

This equation is in quadratic form since it can be rewritten as: 1 1 1x3 2 2 ⫺ 31x3 2 1 ⫺ 10 ⫽ 0, where the2 degree of leading term is twice that of second 1 term. If we let u ⫽ x3, then u2 ⫽ x3 and the equation becomes u2 ⫺ 3u1 ⫺ 10 ⫽ 0, which is factorable. 1u ⫺ 521u ⫹ 22 u⫽5 1 x3 ⫽ 5 1 1x3 2 3 ⫽ 53 x ⫽ 125

⫽0 or u ⫽ ⫺2 1 or x3 ⫽ ⫺2 1 or 1x3 2 3 ⫽ 1⫺22 3 or x ⫽ ⫺8

factor solution in terms of u 1

resubstitute x 3 for u cube both sides: 13 132 ⫽ 1 solve for x

Both solutions check. Now try Exercises 59 and 60 Figure R.8 Hypotenuse

Leg 90⬚

Leg

䊳

A right triangle is one that has a 90° angle (see Figure R.8). The longest side (opposite the right angle) is called the hypotenuse, while the other two sides are simply called “legs.” The Pythagorean theorem is a formula that says if you add the square of each leg, the result will be equal to the square of the hypotenuse. Furthermore, we note the converse of this theorem is also true. Pythagorean Theorem 1. For any right triangle with legs a and b and hypotenuse c, a2 ⫹ b2 ⫽ c2 2. For any triangle with sides a, b, and c, if a2 ⫹ b2 ⫽ c2, then the triangle is a right triangle. A geometric interpretation of the theorem is given in Figure R.9, which shows

Figure R.9

32 ⫹ 42 ⫽ 52.

Area 16 in2

4

5

ea Ar in2 25

3

Area 9 in2

25

13 12

24

⫹ ⫽ 25 ⫹ 144 ⫽ 169 52

122

c

7

5

132

b

⫹ ⫽ 49 ⫹ 576 ⫽ 625 72

242

252

⫹ b2 ⫽ c2 general case

a2

a

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EXAMPLE 18

䊳

Applying the Pythagorean Theorem An extension ladder is placed 9 ft from the base of a building in an effort to reach a third-story window that is 27 ft high. What is the minimum length of the ladder required? Answer in exact form using radicals, and in approximate form by rounding to one decimal place.

Solution

䊳

We can assume the building makes a 90° angle with the ground, and use the Pythagorean theorem to find the required length. Let c represent this length. c2 ⫽ a2 ⫹ b2 c2 ⫽ 192 2 ⫹ 1272 2 c2 ⫽ 81 ⫹ 729 c2 ⫽ 810 c ⫽ 1810 c ⫽ 9 110 c ⬇ 28.5 ft

Pythagorean theorem substitute 9 for a and 27 for b 92 ⫽ 81, 272 ⫽ 729

c

add

27 ft

definition of square root; c 7 0 exact form: 1810 ⫽ 181 # 10 ⫽ 9 110 approximate form

The ladder must be at least 28.5 ft tall.

9 ft

F. You’ve just seen how we can solve equations and use formulas involving radicals


R.6 EXERCISES 䊳


Fill in each blank with the appropriate word or phrase. Carefully reread the section, if necessary. n

1. 1an ⫽ 冟a冟 if n 7 0 is a(n)

integer.

3. By decomposing the rational exponent, we can 3 ? rewrite 16 4 as 116 ? 2 ?. 5. Discuss/Explain what it means when we say an expression like 1A has been written in simplest form. 䊳

2. The conjugate of 2 ⫺ 13 is

.

4. 1x2 2 3 ⫽ x2 3 ⫽ x1 is an example of the property of exponents. 3 2

#

3 2

6. Discuss/Explain why it would be easier x12 to simplify the expression given using 1 rational exponents rather than radicals. x3


Evaluate the expression 2x2 for the values given.

7. a. x ⫽ 9

b. x ⫽ ⫺10

8. a. x ⫽ 7

b. x ⫽ ⫺8

Simplify each expression, assuming that variables can represent any real number.

9. a. 249p2 c. 281m4

b. 21x ⫺ 32 2 d. 2x2 ⫺ 6x ⫹ 9

10. a. 225n2 c. 2v10 3 11. a. 1 64 3 c. 2 216z12 3 12. a. 1 ⫺8 3 c. 2 27q9

b. 21y ⫹ 22 2 d. 24a2 ⫹ 12a ⫹ 9 3 b. 2 ⫺216x3

d.

v3 B ⫺8 3

3 b. 2 ⫺125p3

d.

w3 B ⫺64 3

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6 13. a. 1 64 5 c. 2243x10 5 e. 2 1k ⫺ 32 5

4 14. a. 1 81 5 c. 21024z15 5 e. 2 1q ⫺ 92 5

3 15. a. 1 ⫺125

6 b. 1 ⫺64 5 d. 2⫺243x5 6 f. 2 1h ⫹ 22 6

4 b. 1 ⫺81 5 d. 2⫺1024z20 6 f. 2 1p ⫹ 42 6

4 b. ⫺ 281n12

10

c. 2⫺36

49v B 36 4 b. ⫺ 216m24 d.

16. a. 1 ⫺216 3

c. 2⫺121

25x6 d. B 4

24. a. 28x6 2 3 c. 227a2b6 9 12 ⫺ 248 e. 8 25. a. 2.5 218a22a3 c.

26. a. 5.1 22p232p5 c.

3

16 2 b. a b 25

2 3

17. a. 8

4 ⫺2 c. a b 25 3

d. a

27. a.

⫺27p6 8q3

2 3

b

28. a.

4 b. a b 9

3

16 ⫺4 c. a b 81 3

2

⫺125v9 3 b d. a 27w6 b. a⫺

3

19. a. ⫺1442 c. 1⫺272 ⫺3 2

b. a⫺

3

20. a. ⫺1002 c. 1⫺1252

d. ⫺a

4 b 25

21. a. 12n p 2 22. a. a

1

4x2

b

23y 20 B 4x4

5 29. a. 2 32x10y15

⫺20 ⫹ 232 4 2 b. ⫺ 23b212b2 3 f.

3 3 d. 29v2u23u5v2

4 b. ⫺ 25q220q3 5 d. 25cd2 125cd 3

b.

c. 31 b 4 e. 2b 1 b

27x b 64

3

4 30. a. 2 81a12b16 4

c. 32a 3 4 e. 1 c1 c

3 2

x9 ⫺3 d. ⫺a b 8

3

3 2 108n4

3 2 4n 81 d. 12 3 9 A 8z

b.

3 2 72b5 3 2 3b2

125 d. ⫺9 3 A 27x6 4 5 b. x 2 x

d.

3 1 6

26

5 6 b. a 2 a

d.

3 1 3 4 2 3

4

⫺23

2 ⫺25 5

3

c.

227y7

4

Use properties of exponents to simplify. Answer in exponential form without negative exponents.

24x8

28m5

d. 254m6n8

3 2

3 ⫺43

49 b 36

ab2 25ab4 B 3 B 27

22m 45 c. B 16x2

3 2

18. a. 92

x3y 4x5y B 3 B 12y

3 b. 3 2 128a4b2

2

b. a

3

8y4 3

64y2

b

1 3

b. 12x y 2 ⫺14

3 4

4

Simplify each expression. Assume all variables represent nonnegative real numbers.

23. a. 218m2 3 3 c. 264m3n5 8 ⫺6 ⫹ 228 e. 2

3 b. ⫺2 2 ⫺125p3q7

d. 232p3q6 f.

27 ⫺ 272 6

Simplify and add (if possible).

31. a. b. c. d.

12 272 ⫺ 9 298 8 248 ⫺ 3 2108 7 218m ⫺ 250m 2 228p ⫺ 3 263p

32. a. b. c. d.

⫺3280 ⫹ 2 2125 5 212 ⫹ 2 227 3 212x ⫺ 5 275x 3 240q ⫹ 9 210q

3 3 33. a. 3x 1 54x ⫺ 5 216x4 b. 14 ⫹ 13x ⫺ 112x ⫹ 145 c. 272x3 ⫹ 150 ⫺ 17x ⫹ 127 3 3 34. a. 5 254m3 ⫺ 2m216m3 b. 110b ⫹ 1200b ⫺ 120 ⫹ 140

c. 275r3 ⫹ 132 ⫺ 127r ⫹ 138

77

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Compute each product and simplify the result.

35. a. 17 122 b. 131 15 ⫹ 172 c. 1n ⫹ 1521n ⫺ 152 d. 16 ⫺ 132 2

46. a.

7 17 ⫹ 3

b.

12 1x ⫹ 13

47. a.

110 ⫺ 3 13 ⫹ 12

b.

7 ⫹ 16 3 ⫺ 3 12

37. a. 13 ⫹ 217213 ⫺ 2 172 b. 1 15 ⫺ 1142 1 12 ⫹ 1132 c. 12 12 ⫹ 6 162 13110 ⫹ 172

48. a.

1 ⫹ 12 16 ⫹ 114

b.

1 ⫹ 16 5 ⫹ 2 13

2

36. a. 10.3 152 2 b. 151 16 ⫺ 122 c. 14 ⫹ 13214 ⫺ 132 d. 12 ⫹ 152 2

38. a. 15 ⫹ 4110211 ⫺ 2 1102 b. 1 13 ⫹ 122 1 110 ⫹ 1112 c. 1315 ⫹ 4 122 1 115 ⫹ 162

Use a substitution to verify the solutions to the quadratic equation given. Verify results using a calculator.

39. x2 ⫺ 4x ⫹ 1 ⫽ 0 a. x ⫽ 2 ⫹ 13

b. x ⫽ 2 ⫺ 13

40. x ⫺ 10x ⫹ 18 ⫽ 0 a. x ⫽ 5 ⫺ 17

b. x ⫽ 5 ⫹ 17

41. x2 ⫹ 2x ⫺ 9 ⫽ 0 a. x ⫽ ⫺1 ⫹ 110

b. x ⫽ ⫺1 ⫺ 110

42. x2 ⫺ 14x ⫹ 29 ⫽ 0 a. x ⫽ 7 ⫺ 2 15

b. x ⫽ 7 ⫹ 2 15

2

Rationalize each expression by building perfect nth root factors for each denominator. Assume all variables represent positive quantities.

3 112 27 c. B 50b 5 e. 3 1a

43. a.

20 B 27x3 1 d. 3 A 4p b.

⫺4 125 44. a. b. B 12n3 120 5 3 c. d. 3 B 12x A 2m2 ⫺8 e. 3 3 15 Simplify the following expressions by rationalizing the denominators. Where possible, state results in exact form and approximate form, rounded to hundredths. 45. a.

8 3 ⫹ 111

b.

6 1x ⫺ 12

Solve each equation and check your solutions by substitution. Identify any extraneous roots.

49. a. ⫺313x ⫺ 5 ⫽ ⫺9 b. x ⫽ 13x ⫹ 1 ⫹ 3 50. a. ⫺2 14x ⫺ 1 ⫽ ⫺10 b. ⫺5 ⫽ 15x ⫺ 1 ⫺ x 3 51. a. 2 ⫽ 13m ⫺ 1 3 b. 2 1 7 ⫺ 3x ⫺ 3 ⫽ ⫺7 3 1 2m ⫹ 3 ⫹2⫽3 c. ⫺5 3 3 d. 1 2x ⫺ 9 ⫽ 1 3x ⫹ 7 3 52. a. ⫺3 ⫽ 1 5p ⫹ 2 3 b. 3 1 3 ⫺ 4x ⫺ 7 ⫽ ⫺4 3 1 6x ⫺ 7 ⫺ 5 ⫽ ⫺6 c. 4 3 3 d. 3 1 x ⫹ 3 ⫽ 2 1 2x ⫹ 17

53. a. b. c. d.

1x ⫺ 9 ⫹ 1x ⫽ 9 x ⫽ 3 ⫹ 223 ⫺ x 1x ⫺ 2 ⫺ 12x ⫽ ⫺2 112x ⫹ 9 ⫺ 124x ⫽ ⫺3

54. a. b. c. d.

1x ⫹ 7 ⫺ 1x ⫽ 1 12x ⫹ 31 ⫹ x ⫽ 2 13x ⫽ 1x ⫺ 3 ⫹ 3 13x ⫹ 4 ⫺ 17x ⫽ ⫺2

Write the equation in simplified form, then solve. Check all answers by substitution. 3

55. a. x5 ⫹ 17 ⫽ 9 3 b. ⫺2x4 ⫹ 47 ⫽ ⫺7 5

56. a. 0.3x2 ⫺ 39 ⫽ 42 5 b. 0.5x3 ⫹ 92 ⫽ ⫺43 2

57. a. 21x ⫹ 52 3 ⫺ 11 ⫽ 7 4 b. ⫺31x ⫺ 22 5 ⫹ 29 ⫽ ⫺19 1

3 58. a. 3x3 ⫺ 10 ⫽ 1 x 2 5 2 b. 2 1x ⫺ 4 ⫽ x5 2

1

59. x3 ⫺ 2x3 ⫺ 15 ⫽ 0 3

60. x3 ⫺ 9x2 ⫽ ⫺8

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WORKING WITH FORMULAS 1

61. Fish length to weight relationship: L 1.131W2 3 The length to weight relationship of a female Pacific halibut can be approximated by the formula shown, where W is the weight in pounds and L is the length in feet. A fisherman lands a halibut that weighs 400 lb. Approximate the length of the fish (round to two decimal places). 䊳

79

1s 4 The time it takes an object to fall a certain distance is given by the formula shown, where t is the time in seconds and s is the distance the object has fallen. Approximate the time it takes an object to hit the ground, if it is dropped from the top of a building that is 80 ft in height (round to hundredths).

62. Timing a falling object: t

APPLICATIONS

63. Length of a cable: A radio tower is secured by cables that are anchored in the ground 8 m from its base. If the cables are attached to the tower 24 m above the ground, what is the length of each cable? Answer in (a) exact form using radicals, and (b) approximate form by rounding to one decimal place.

24 m

c

8m 64. Height of a kite: Benjamin Franklin is flying his kite in a storm once again. John Adams has walked to a position directly under the kite and is 75 ft from Ben. If the kite is 50 ft above John Adams’ head, how much string S has Ben let out? Answer in (a) exact form using radicals, and (b) approximate form by rounding to one decimal place.

S

50 ft

75 ft

The time T (in days) required for a planet to make one revolution around3 the sun is modeled by the function T 0.407R2, where R is the maximum radius of the planet’s orbit (in millions of miles). This is known as Kepler’s third law of planetary motion. Use the equation given to approximate the number of days required for one complete orbit of each planet, given its maximum orbital radius.

65. a. Earth: 93 million mi b. Mars: 142 million mi c. Mercury: 36 million mi 66. a. Venus: 67 million mi b. Jupiter: 480 million mi c. Saturn: 890 million mi 67. Accident investigation: After an accident, police officers will try to determine the approximate velocity V that a car was traveling using the formula V ⫽ 2 26L, where L is the length of the skid marks in feet and V is the velocity in miles per hour. (a) If the skid marks were 54 ft long, how fast was the car traveling? (b) Approximate the speed of the car if the skid marks were 90 ft long. 68. Wind-powered energy: If a wind-powered generator is delivering P units of power, the velocity V of the wind (in miles per hour) can be 3 P , where k is a constant determined using V ⫽ Ak that depends on the size and efficiency of the generator. Rationalize the radical expression and use the new version to find the velocity of the wind if k ⫽ 0.004 and the generator is putting out 13.5 units of power.

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69. Surface area: The lateral surface area (surface area excluding the base) h S of a cone is given by the formula 2 2 S ⫽ ␲r 2r ⫹ h , where r is the r radius of the base and h is the height of the cone. Find the lateral surface area of a cone that has a radius of 6 m and a height of 10 m. Answer in simplest form. 70. Surface area: The lateral surface a area S of a frustum (a truncated cone) is given by the formula h S ⫽ ␲1a ⫹ b2 2h2 ⫹ 1b ⫺ a2 2, b where a is the radius of the upper base, b is the radius of the lower base, and h is the height. Find the surface area of a frustum where a ⫽ 6 m, b ⫽ 8 m, and h ⫽ 10 m. Answer in simplest form. 71. Planetary motion: The time T (in days) for a planet to make one revolution (elliptical orbit) 3 around the sun is modeled by T ⫽ 0.407R2, where R is the maximum radius of the planet’s orbit in

䊳

millions of miles (Kepler’s third law of planetary motion). Use the equation to approximate the maximum radius of each orbit, given the number of days it takes for one revolution. (See Exercises 65 and 66.) a. Mercury: 88 days b. Venus: 225 days c. Earth: 365 days d. Mars: 687 days e. Jupiter: 4333 days f. Saturn: 10,759 days 72. Wind-powered energy: If a wind-powered generator is delivering P units of power, the velocity V of the wind (in miles per hour) can be 3 P , where k is a constant determined using V ⫽ Ak that depends on the size and efficiency of the generator. Given k ⫽ 0.004, approximately how many units of power are being delivered if the wind is blowing at 27 miles per hour? (See Exercise 68.)


The expression x2 7 is not factorable using integer values. But the expression can be written in the form x2 1 272 2, enabling us to factor it as a “binomial” and its conjugate: 1x 272 1x 272. Use this idea to factor the following expressions.

73. a. x2 ⫺ 5 b. n2 ⫺ 19 74. a. 4v2 ⫺ 11 b. 9w2 ⫺ 17 75. The following terms form a pattern that continues until the sixth term is found: 23x ⫹ 29x ⫹ 227x ⫹ p (a) Compute the sum of all six terms; (b) develop a system (investigate the pattern further) that will enable you to find the sum of 12 such terms without actually writing out the terms.

76. Find a quick way to simplify the expression without the aid of a calculator.

a a a a a3 b 5 6

3 2

4 5

3 4

2 5

aaaa

80

10 3

1 1 1 1 9 77. If 1x2 ⫹ x⫺2 2 2 ⫽ , find the value of x2 ⫹ x⫺2. 2

78. Rewrite by rationalizing the numerator: 1x ⫹ h ⫺ 1x h Determine the values of x for which each expression represents a real number. 79.

1x ⫺ 1 x2 ⫺ 4

80.

x2 ⫺ 4 1x ⫺ 1

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Overview of Chapter R

OVERVIEW OF CHAPTER R Prerequisite Definitions, Properties, Formulas, and Relationships Notation and Relations concept

• Set notation:

notation {members}

僆 • Is an element of ⭋ or { } • Empty set Is a proper subset of ( • 5x | x p6

• Defining a set

description braces enclose the members of a set

indicates membership in a set a set having no elements indicates the elements of one set are entirely contained in another the set of all x, such that x . . .

Sets of Numbers • Natural: ⺞ ⫽ 51, 2, 3, 4, p6

• Integers: ⺪ ⫽ 5p , ⫺3, ⫺2, ⫺1, 0, 1, 2, 3, p6 • Irrational: ⺘ ⫽ {numbers with a nonterminating,

example set of even whole numbers A ⫽ 50, 2, 4, 6, 8, p6 14 僆 A odd numbers in A S ⫽ 50, 6, 12, 18, 24, p6 S ( A S ⫽ 5x | x ⫽ 6n for n 僆⺧6

• Whole: ⺧ ⫽ 50, 1, 2, 3, p6

• Rational: ⺡ ⫽ e q , where p, q 僆⺪; q ⫽ 0 f • Real: ⺢ ⫽ {all rational and irrational numbers} p

nonrepeating decimal form}

Absolute Value of a Number

Distance between numbers a and b on the number line

• The distance between a number n

冟a ⫺ b冟 or 冟b ⫺ a冟 and zero (always positive) n if n ⱖ 0 0n 0 ⫽ e ⫺n if n 6 0 For a complete review of these ideas, go to www.mhhe.com/coburn.

R.1 Properties of Real Numbers: For real numbers a, b, and c, Commutative Property

• Addition: a ⫹ b ⫽ b ⫹ a • Multiplication: a # b ⫽ b # a Identities

Associative Property

• Addition: 1a ⫹ b2 ⫹ c ⫽ a ⫹ 1b ⫹ c2 • Multiplication: 1a # b2 # c ⫽ a # 1b # c2 Inverses

• Additive: a ⫹ 1⫺a2 ⫽ 0

• Additive: 0 ⫹ a ⫽ a • Multiplicative: 1 # a ⫽ a

p q • Multiplicative: q # p ⫽ 1; p, q ⫽ 0

R.2 Properties of Exponents: For real numbers a and b, and integers m, n, and p • • • •

(excluding 0 raised to a nonpositive power), Product property: b • Power property: 1bm 2 n ⫽ bmn am p amp Product to a power: 1ambn 2 p ⫽ amp # bnp Quotient to a power: b ⫽ np 1b ⫽ 02 a • bm bn b m⫺n Quotient property: n ⫽ b 1b ⫽ 02 0 Zero exponents: ⫽ 1 1b ⫽ 02 b b • ⫺n n 1 a b • Scientific notation: N ⫻ 10k; 1 ⱕ 冟N冟 6 10, k 僆⺪ Negative exponents: b⫺n ⫽ n ; a b ⫽ a b a b b 1a, b ⫽ 02 m

# bn ⫽ bm⫹n

81

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R.2 Polynomials • A polynomial is a sum or difference of monomial terms • Polynomials are classified as a monomial, binomial, trinomial, or polynomial, depending on the number of terms • The degree of a polynomial in one variable is the same as the largest exponent occuring on the variable in any term • A polynomial expression is in standard form when written with the terms in descending order of degree R.3 Solving Linear Equations and Inequalities • Properties of Equality: additive property If A and B are algebraic expressions, where A ⫽ B, then A ⫹ C ⫽ B ⫹ C.

• • • • • • • •

multiplicative property If A and B are algebraic expressions, where A ⫽ B, then A # C ⫽ B # C B A [C can be positive or negative] and ⫽ ;C ⫽ 0 C C A linear equation in one variable is one that can be written in the form ax ⫹ b ⫽ c, where the exponent on the variable is a 1. To solve a linear equation, we attempt to isolate the term containing the variable using the additive property, then solve for the variable using the multiplicative property. An equation can be an identity (always true), a contradiction (never true) or conditional (true or false depending on the input value[s]). If an equation contains fractions, multiplying both sides by the least common denominator of all fractions will “clear the denominators” and reduce the amount of work required to solve. Inequalities are solved using properties similar to those used for solving equations. The one exception is when multiplying or dividing by a negative quantity, as the inequality symbol must then be reversed to maintain the truth of the resulting statement. Solutions to an inequality can be given using a simple inequality, graphed on a number line, stated in set notation, or stated using interval notation. Given two sets A and B: A intersect B 1A ¨ B2 is the set of elements shared by both A and B (elements common to both sets). A union B 1A ´ B2 is the set of elements in either A or B (elements are combined to form a larger set). Compound inequalities are formed using the conjunction “and” or the conjunction “or.” The result can be either a joint inequality as in ⫺3 6 x ⱕ 5, or a disjoint inequality, x 6 ⫺2 or x 7 7.

R.4 Special Factorizations • A2 ⫺ B2 ⫽ 1A ⫹ B21A ⫺ B2 • A2 ⫾ 2AB ⫹ B2 ⫽ 1A ⫾ B2 2 3 3 2 2 • A ⫺ B ⫽ 1A ⫺ B21A ⫹ AB ⫹ B 2 • A3 ⫹ B3 ⫽ 1A ⫹ B21A2 ⫺ AB ⫹ B2 2 • Certain equations of higher degree can be solved using factoring skills and the zero product property. R.5 Rational Expressions: For polynomials P, Q, R, and S with no denominator of zero, P#R P P#R P ⫽ Equivalence: ⫽ • Q#R Q Q Q#R P R P#R PR R P S PS P Multiplication: # ⫽ # ⫽ • Division: ⫼ ⫽ # ⫽ Q S Q S QS Q S Q R QR Q P⫹Q Q P⫺Q P P Addition: ⫹ ⫽ • Subtraction: ⫺ ⫽ R R R R R R Addition/subtraction with unlike denominators: 1. Find the LCD of all rational expressions. 2. Build equivalent expressions using LCD. 3. Add/subtract numerators as indicated. 4. Write the result in lowest terms. To solve rational equations, first clear denominators using the LCD, noting values that must be excluded. Multiplying an equation by a variable quantity sometimes introduces extraneous solutions. Check all results in the original equation.

• Lowest terms: • • •

• •

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Practice Test

R.6 Properties of Radicals • 1a is a real number only for a ⱖ 0 n • 1a ⫽ b, only if bn ⫽ a

83

• 1a ⫽ b, only if b2 ⫽ a n • If n is even, 1a represents a real number only

if a ⱖ 0 n • For any real number a, 1an ⫽ a when n is odd m If a is a real number and n is an integer greater • If n is a rational number written in lowest terms 1 n n m m than 1, then 1 a ⫽ an provided 1 a represents a n n with n ⱖ 2, then a n ⫽ 11a2 m and a n ⫽ 1am real number n provided 2a represents a real number. n n n n If 1A and 1B represent real numbers and B ⫽ 0, If 1A and 1B represent real numbers, • n n n n 1AB ⫽ 1 A # 1B 1A n A ⫽ n BB 1B A radical expression is in simplest form when: 1. the radicand has no factors that are perfect nth roots, 2. the radicand contains no fractions, and 3. no radicals occur in a denominator. To solve radical equations, isolate the radical on one side, then apply the appropriate “nth power” to free up the radicand. Repeat the process if needed. See flowchart on page 74. For equations with a rational exponent mn, isolate the variable term and raise both sides to the mn power. If m is even, there will be two real solutions.

n • For any real number a, 1an ⫽ 冟a冟 when n is even

•

•

•

• •

R.6 Pythagorean Theorem • For any right triangle with legs a and b and hypotenuse c: a2 ⫹ b2 ⫽ c2.

• For any triangle with sides a, b, and c, if

a2 ⫹ b2 ⫽ c2, then the triangle is a right triangle.

PRACTICE TEST 1. State true or false. If false, state why. 7 a. 13 ⫹ 42 2 ⫽ 25 b. ⫽ 0 0 c. x ⫺ 3 ⫽ ⫺3 ⫹ x d. ⫺21x ⫺ 32 ⫽ ⫺2x ⫺ 3 2. State the value of each expression. 3 a. 1 121 b. 1 ⫺125 c. 1⫺36 d. 1400 3. Evaluate each expression: 7 1 5 1 a. ⫺ a⫺ b b. ⫺ ⫺ 8 4 3 6 c. ⫺0.7 ⫹ 1.2 d. 1.3 ⫹ 1⫺5.92 4. Evaluate each expression: 1 a. 1⫺42a⫺2 b b. 1⫺0.621⫺1.52 3 ⫺2.8 c. d. 4.2 ⫼ 1⫺0.62 ⫺0.7 # 0.08 12 10 5. Evaluate using a calculator: 2000 a1 ⫹ b 12 6. State the value of each expression, if possible. a. 0 ⫼ 6 b. 6 ⫼ 0

7. State the number of terms in each expression and identify the coefficient of each. c⫹2 ⫹c a. ⫺2v2 ⫹ 6v ⫹ 5 b. 3 8. Evaluate each expression given x ⫽ ⫺0.5 and y ⫽ ⫺2. Round to hundredths as needed. y a. 2x ⫺ 3y2 b. 12 ⫺ x14 ⫺ x2 2 ⫹ x 9. Translate each phrase into an algebraic expression. a. Nine less than twice a number is subtracted from the number cubed. b. Three times the square of half a number is subtracted from twice the number. 10. Create a mathematical model using descriptive variables. a. The radius of the planet Jupiter is approximately 119 mi less than 11 times the radius of the Earth. Express the radius of Jupiter in terms of the Earth’s radius. b. Last year, Video Venue Inc. earned $1.2 million more than four times what it earned this year. Express last year’s earnings of Video Venue Inc. in terms of this year’s earnings.

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11. Simplify by combining like terms. a. 8v2 ⫹ 4v ⫺ 7 ⫹ v2 ⫺ v b. ⫺413b ⫺ 22 ⫹ 5b c. 4x ⫺ 1x ⫺ 2x2 2 ⫹ x13 ⫺ x2 12. Factor each expression completely. a. 9x2 ⫺ 16 b. 4v3 ⫺ 12v2 ⫹ 9v 3 2 c. x ⫹ 5x ⫺ 9x ⫺ 45 13. Simplify using the properties of exponents. 5 a. ⫺3 b. 1⫺2a3 2 2 1a2b4 2 3 b 5p2q3r4 2 m2 3 c. a b d. a b 2n ⫺2pq2r4 14. Simplify using the properties of exponents. ⫺12a3b5 a. 3a2b4 b. 13.2 ⫻ 10⫺17 2 ⫻ 12.0 ⫻ 1015 2 a⫺3 # b ⫺4 c. a ⫺2 b d. ⫺7x0 ⫹ 1⫺7x2 0 c 15. Compute each product. a. 13x2 ⫹ 5y2 13x2 ⫺ 5y2 b. 12a ⫹ 3b2 2 16. Add or subtract as indicated. a. 1⫺5a3 ⫹ 4a2 ⫺ 32 ⫹ 17a4 ⫹ 4a2 ⫺ 3a ⫺ 152 b. 12x2 ⫹ 4x ⫺ 92 ⫺ 17x4 ⫺ 2x2 ⫺ x ⫺ 92 Simplify or compute as indicated. x⫺5 4 ⫺ n2 b. 2 5⫺x n ⫺ 4n ⫹ 4 3 3x2 ⫺ 13x ⫺ 10 x ⫺ 27 c. 2 d. x ⫹ 3x ⫹ 9 9x2 ⫺ 4 x2 ⫹ x ⫺ 20 x2 ⫺ 25 e. 2 ⫼ 2 3x ⫺ 11x ⫺ 4 x ⫺ 8x ⫹ 16 2 m⫹3 f. 2 ⫺ 51m ⫹ 42 m ⫹ m ⫺ 12

17. a.

18. a. 21x ⫹ 112 2 ⫺3

25 2 c. a b 16 e. 7 140 ⫺ 190 2 g. B 5x

3 ⫺8 A 27v3 ⫺4 ⫹ 132 d. 8 f. 1x ⫹ 152 1x ⫺ 152 8 h. 16 ⫺ 12

b.

19. Maximizing revenue: Due to past experience, the manager of a video store knows that if a popular video game is priced at $30, the store will sell 40 each day. For each decrease of $0.50, one additional sale will be made. The formula for the store’s revenue is then R ⫽ 130 ⫺ 0.5x2 140 ⫹ x2, where x represents the number of times the price is decreased. Multiply the binomials and use a table of values to determine (a) the number of 50¢ decreases that will give the most revenue and (b) the maximum amount of revenue. 20. Diagonal of a rectangular prism: Use the Pythagorean theorem to determine the length of the diagonal of the rectangular prism shown in the figure. (Hint: First find the diagonal of the base.)

42 cm

32 cm

24 cm

21. Solve each linear equation. a. ⫺2b ⫹ 7 ⫽ ⫺5 b. 312n ⫺ 62 ⫹ 1 ⫽ 7 1 2 3 c. 4m ⫺ 5 ⫽ 11m ⫹ 2 d. x ⫹ ⫽ 2 3 4 e. ⫺813p ⫹ 52 ⫺ 9 ⫽ 613 ⫺ 4p2 g 5g 1 f. ⫺ ⫽ 3 ⫺ ⫺ 6 2 12 22. If one-fourth of the sum of a number and twelve is added to three, the result is sixteen. Find an equation model for this statement, then use it to find the number. 23. Solve each polynomial equation by factoring. a. x3 ⫺ 7x2 ⫽ 4x ⫺ 28 b. ⫺7r3 ⫹ 21r2 ⫹ 28r ⫽ 0 c. g4 ⫺ 10g2 ⫹ 9 ⫽ 0 24. Solve each rational equation. 7 1 3 ⫹ ⫽ a. 5x 10 4x 1 3h 7 ⫽ ⫺ 2 b. h⫹3 h h ⫹ 3h 3 4n ⫹ 20 n ⫺ ⫽ 2 c. n⫹2 n⫺4 n ⫺ 2n ⫺ 8 25. Solve each radical equation. 2x2 ⫹ 7 ⫹3⫽5 a. 2 b. 3 1x ⫹ 4 ⫽ x ⫹ 4 c. 13x ⫹ 4 ⫽ 2 ⫺ 1x ⫹ 2

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CHAPTER CONNECTIONS

Relations, Functions, and Graphs CHAPTER OUTLINE 1.1 Rectangular Coordinates; Graphing Circles and Other Relations 86

1.2 Linear Equations and Rates of Change 103 1.3 Functions, Function Notation, and the Graph of a Function 117

1.4 Linear Functions, Special Forms, and

Viewing relations and functions in terms of an equation, a table of values, and the related graph, often brings a clearer understanding of the relationships involved. For instance, while many business are aware that internet use is increasing with time, they are ver y interested in the rate of growth, in order to prepare and develop related goods and ser vices. This application appears as Exercise 109 in Section 1.4. Check out these other real-world connections 䊳䊳䊳䊳

Earthquake Range: Section 1.1, Exercise 96 Garbage Collection: Section 1.2, Exercise 42 Eating Out: Section 1.4, Exercise 112 Credit Card Use: Section 1.6, Exercise 33

More on Rates of Change 134

1.5 Solving Equations and Inequalities Graphically; Formulas and Problem Solving 148

1.6 Linear Function Models and Real Data 163

85

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1.1



A. Express a relation in

In everyday life, we encounter a large variety of relationships. For instance, the time it takes us to get to work is related to our average speed; the monthly cost of heating a home is related to the average outdoor temperature; and in many cases, the amount of our charitable giving is related to changes in the cost of living. In each case we say that a relation exists between the two quantities.

mapping notation and ordered pair form B. Graph a relation C. Graph relations on a calculator D. Develop the equation and graph of a circle using the distance and midpoint formulas

A. Relations, Mapping Notation, and Ordered Pairs

Consumer spending (dollars per year)

Figure 1.1 In the most general sense, a relation is simply a correspondence between two sets. Relations can be repreP B sented in many different ways and may even be very Missy April 12 “unmathematical,” like the one shown in Figure 1.1 Jeff Nov 11 between a set of people and the set of their correspondAngie Sept 10 ing birthdays. If P represents the set of people and B Megan Nov 28 represents the set of birthdays, we say that elements of Mackenzie May 7 Michael P correspond to elements of B, or the birthday relation April 14 Mitchell maps elements of P to elements of B. Using what is called mapping notation, we might simply write P S B. From a purely practical standpoint, we note that while it is possible for two different people to share the same birthday, it is quite impossible for the same person to have two different birthdays. Later, this observation will help us mark the difference between a relation and special kind of relation called a function. Figure 1.2 The bar graph in Figure 1.2 is also an 500 example of a relation. In the graph, each year is related to annual consumer spend($411-est) 400 ($375) ing per person on cable and satellite television. As an alternative to mapping or a bar ($281) 300 graph, this relation could also be repre($234) sented using ordered pairs. For example, ($192) 200 the ordered pair (5, 234) would indicate that in 2005, spending per person on 100 cable and satellite TV in the United States averaged $234. When a relation is 3 11 5 7 9 represented using ordered pairs, we say Year (0 → 2000) Source: 2009 Statistical Abstract of the United States, the relation is pointwise-defined. Table 1089 (some figures are estimates) Over a long period of time, we could collect many ordered pairs of the form (t, s), where consumer spending s depends on the time t. For this reason we often call the second coordinate of an ordered pair (in this case s) the dependent variable, with the first coordinate designated as the independent variable. The set of all first coordinates is called the domain of the relation. The set of all second coordinates is called the range. 1–2

86

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Section 1.1 Rectangular Coordinates; Graphing Circles and Other Relations

EXAMPLE 1

䊳

Expressing a Relation as a Mapping and as a Pointwise-Defined Relation Represent the relation from Figure 1.2 in mapping notation and as a pointwise-defined relation, then state its domain and range.

Solution

䊳

A. You’ve just seen how we can express a relation in mapping notation and ordered pair form

Let t represent the year and s represent consumer spending. The mapping t S s gives the diagram shown. As a pointwisedefined relation we have (3, 192), (5, 234), (7, 281), (9, 375), and (11, 411). The domain is the set {3, 5, 7, 9, 11}; the range is {192, 234, 281, 375, 411}.

t

s

3 5 7 9 11

192 234 281 375 411


䊳

For more on this relation, see Exercise 93.

B. The Graph of a Relation Table 1.1 y ⴝ x ⴚ 1 x

y

4

5

2

3

0

1

2

1

4

3

Table 1.2 x ⴝ 円y円 x

y

2

2

1

1

0

0

1

1

2

2

Relations can also be stated in equation form. The equation y x 1 expresses a relation where each y-value is one less than the corresponding x-value (see Table 1.1). The equation x 冟y冟 expresses a relation where each x-value corresponds to the absolute value of y (see Table 1.2). In each case, the relation is the set of all ordered pairs (x, y) that create a true statement when substituted, and a few ordered pair solutions are shown in the tables for each equation. Relations can be expressed graphically using a recFigure 1.3 tangular coordinate system. It consists of a horizontal y 5 number line (the x-axis) and a vertical number line (the 4 y-axis) intersecting at their zero marks. The point of inter3 QII QI section is called the origin. The x- and y-axes create a 2 flat, two-dimensional surface called the xy-plane and 1 divide the plane into four regions called quadrants. 5 4 3 2 1 1 2 3 4 5 x 1 These are labeled using a capital “Q” (for quadrant) and 2 the Roman numerals I through IV, beginning in the QIII QIV 3 upper right and moving counterclockwise (Figure 1.3). 4 The grid lines shown denote the integer values on each 5 axis and further divide the plane into a coordinate grid, where every point in the plane corresponds to an ordered Figure 1.4 pair. Since a point at the origin has not moved along either y 5 axis, it has coordinates (0, 0). To plot a point (x, y) means we place a dot at its location in the xy-plane. A few of the (4, 3) ordered pairs from y x 1 are plotted in Figure 1.4, where a noticeable pattern emerges—the points seem to (2, 1) lie along a straight line. 5 x If a relation is pointwise-defined, the graph of the 5 (0, 1) relation is simply the plotted points. The graph of a re(2, 3) lation in equation form, such as y x 1, is the set of (4, 5) all ordered pairs (x, y) that are solutions (make the 5 equation true). Solutions to an Equation in Two Variables 1. If substituting x a and y b results in a true equation, the ordered pair (a, b) is a solution and on the graph of the relation. 2. If the ordered pair (a, b) is on the graph of a relation, it is a solution (substituting x a and y b will result in a true equation).

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CHAPTER 1 Relations, Functions, and Graphs

We generally use only a few select points to determine the shape of a graph, then draw a straight line or smooth curve through these points, as indicated by any patterns formed. EXAMPLE 2

䊳

Graphing Relations Graph the relations y x 1 and x 冟y冟 using the ordered pairs given in Tables 1.1 and 1.2.

Solution

䊳

For y x 1, we plot the points then connect them with a straight line (Figure 1.5). For x 冟y冟, the plotted points form a V-shaped graph made up of two half lines (Figure 1.6). Figure 1.5 5

Figure 1.6 y

y yx1

5

x y

(4, 3) (2, 2)

(2, 1) (0, 0) 5

5

x

5

5

(2, 3)

x

(2, 2)

(0, 1)

5

5

(4, 5)

Now try Exercises 13 through 16 WORTHY OF NOTE As the graphs in Example 2 indicate, arrowheads are used where appropriate to indicate the infinite extension of a graph.

䊳

While we used only a few points to graph the relations in Example 2, they are actually made up of an infinite number of ordered pairs that satisfy each equation, including those that might be rational or irrational. This understanding is an important part of reading and interpreting graphs, and is illustrated for you in Figures 1.7 through 1.10. Figure 1.7

Figure 1.8

y x 1: selected integer values

y x 1: selected rational values

y

y 5

5

5

5

x

5

5

5

5

Figure 1.9

Figure 1.10

y x 1: selected real number values

y x 1: all real number values

y

y

5

5

5

5

5

x

x

5

5

5

x

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Since there are an infinite number of ordered pairs forming the graph of y x 1, the domain cannot be given in list form. Here we note x can be any real number and write D: x 僆⺢. Likewise, y can be any real number and for the range we have R: y 僆⺢. All of these points together make these graphs continuous, which for our purposes means you can draw the entire graph without lifting your pencil from the paper. Actually, a majority of graphs cannot be drawn using only a straight line or directed line segments. In these cases, we rely on a “sufficient number” of points to outline the basic shape of the graph, then connect the points with a smooth curve. As your experience with graphing increases, this “sufficient number of points” tends to get smaller as you learn to anticipate what the graph of a given relation should look like. In particular, for the linear graph in Figure 1.5 we notice that both the x- and y-variables have an implied exponent of 1. This is in fact a characteristic of linear equations and graphs. In Example 3 we’ll notice that if the exponent on one of the variables is 2 (either x or y is squared ) while the other exponent is 1, the result is a graph called a parabola. If the x-term is squared (Example 3a) the parabola is oriented vertically, as in Figure 1.11, and its highest or lowest point is called the vertex. If the y-term is squared (Example 3c), the parabola is oriented horizontally, as in Figure 1.13, and the leftmost or rightmost point is the vertex. The graphs and equations of other relations likewise have certain identifying characteristics. See Exercises 85 through 92. EXAMPLE 3

䊳

Graphing Relations Graph the following relations by completing the tables given. Then use the graph to state the domain and range of the relation. a. y x2 2x b. y 29 x2 c. x y2

Solution

䊳

For each relation, we use each x-input in turn to determine the related y-output(s), if they exist. Results can be entered in a table and the ordered pairs used to assist in drawing a complete graph. Figure 1.11 a. y ⴝ x2 2x y

x

y

(x, y) Ordered Pairs

4

24

(4, 24)

3

15

(3, 15)

2

8

(2, 8)

1

3

(1, 3)

0

(0, 0)

1

0 1

(1, 1)

2

0

(2, 0)

3

3

(3, 3)

4

8

(4, 8)

(4, 8)

(2, 8) y x2 2x

5

(1, 3)

(3, 3)

(0, 0)

(2, 0)

5

5 2

x

(1, 1)

The resulting vertical parabola is shown in Figure 1.11. Although (4, 24) and (3, 15) cannot be plotted here, the arrowheads indicate an infinite extension of the graph, which will include these points. This “infinite extension” in the upward direction shows there is no largest y-value (the graph becomes infinitely “tall”). Since the smallest possible y-value is 1 [from the vertex (1, 1)], the range is y 1. However, this extension also continues forever in the outward direction as well (the graph gets wider and wider). This means the x-value of all possible ordered pairs could vary from negative to positive infinity, and the domain is all real numbers. We then have D: x 僆⺢ and R: y 1.

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y ⴝ 29 ⴚ x2

b.

Figure 1.12

x

y


4

not real

—

3

0

(3, 0)

2

15

(2, 15)

1

212

(1, 212)

0

3

(0, 3)

1

212

(1, 212)

2

15

(2, 15)

3

0

(3, 0)

4

not real

—

y 兹9 x2

y 5

(1, 2兹2) (2, 兹5)

(0, 3) (1, 2兹2) (2, 兹5)

(3, 0)

(3, 0)

5

5

x

5

The result is the graph of a semicircle (Figure 1.12). The points with irrational coordinates were graphed by estimating their location. Note that when x 6 3 or x 7 3, the relation y 29 x2 does not represent a real number and no points can be graphed. Also note that no arrowheads are used since the graph terminates at (3, 0) and (3, 0). These observations and the graph itself show that for this relation, D: 3 x 3, and R: 0 x 3. c. Similar to x 冟y冟, the relation x y2 is defined only for x 0 since y2 is always nonnegative (1 y2 has no real solutions). In addition, we reason that each positive x-value will correspond to two y-values. For example, given x 4, (4, 2) and (4, 2) are both solutions to x y2. x ⴝ y2

Figure 1.13 y

x

y


2

not real

—

1

not real

—

0

(0, 0)

0 1

1, 1

(1, 1) and (1, 1)

2

12, 12

(2, 12) and (2, 12)

3

13, 13

(3, 13) and (3, 13)

4

2, 2

(4, 2) and (4, 2)

5

x y2 (4, 2)

(2, 兹2) (0, 0) 5

5

5

x

(2, 兹2) (4, 2)

This relation is a horizontal parabola, with a vertex at (0, 0) (Figure 1.13). The graph begins at x 0 and extends infinitely to the right, showing the domain is x 0. Similar to Example 3a, this “infinite extension” also extends in both the upward and downward directions and the y-value of all possible ordered pairs could vary from negative to positive infinity. We then have D: x 0 and R: y 僆⺢. B. You’ve just seen how we can graph relations


䊳

C. Graphing Relations on a Calculator For relations given in equation form, the TABLE feature of a graphing calculator can be used to compute ordered pairs, and the GRAPH feature to draw the related graph. To use these features, we first solve the equation for the variable y (write y in terms of x), then enter the right-hand expression on the calculator’s Y= (equation editor) screen.

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We can then select either the GRAPH feature, or set-up, create, and use the TABLE feature. We’ll illustrate here using the relation 2x y 3. 1. Solve for y in terms of x. 2x y 3 given equation y 2x 3 add 2x to each side 2. Enter the equation. Press the Y= key to access the equation editor, then enter 2x 3 as Y1 (see Figure 1.14). The calculator automatically highlights the equal sign, showing that equation Y1 is now active. If there are other equations on the screen, you can either them or deactivate them by moving the cursor to overlay the equal sign and pressing . 3. Use the TABLE or GRAPH . To set up the table, we use the keystrokes 2nd (TBLSET). For this exercise, we’ll put the table in the “Indpnt: Auto Ask” mode, which will have the calculator automatically generate the input and output values. In this mode, we can tell the calculator where to start the inputs (we chose TblStart 3), and have the calculator produce the input values using any increment desired (we choose Tbl 1). See Figure 1.15A. Access the table using 2nd GRAPH (TABLE), and the table resulting from this setup is shown in Figure 1.15B. Notice that all ordered pairs satisfy the equation y 2x 3, or “y is twice x increased by 3.”

Figure 1.14

CLEAR

ENTER

Figure 1.15A

WINDOW

Figure 1.15B

Since much of our graphical work is centered at (0, 0) on the coordinate grid, the calculator’s default settings for the standard viewing are [10, 10] for both x and y (Figure 1.16). The Xscl and Yscl values give the scale used on each axis, and indicate here that each “tick mark” will be 1 unit apart. To graph the line in this window, we can use the ZOOM key and select 6:ZStandard (Figure 1.17), which resets the window to these default settings and automatically graphs the line (Figure 1.18). WINDOW

Figure 1.16

Figure 1.18

Figure 1.17

10

10

10

10

In addition to using the calculator’s TABLE feature to find ordered pairs for a given graph, we can also use the calculator’s TRACE feature. As the name implies, this feature allows us to “trace” along the graph by moving a cursor to the left and right using the arrow keys. The calculator displays the coordinates of the cursor’s location each time it moves. After pressing the TRACE key, the marker appears automatically and as you move it to the left or right, the current coordinates are shown at the bottom

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of the screen (Figure 1.19). While not very “pretty,” (0.8510638, 1.2978723) is a point on this line (rounded to seven decimal places) and satisfies its equation. The calculator is displaying these decimal values because the viewing screen is exactly 95 pixels wide, 47 pixels to the left of the y-axis, and 47 pixels to the right. This means that each time you press the left or right arrow, the x-value changes by 1/47th—which is not a nice round number. To have the calculator TRACE through “friendlier” values, we can use the ZOOM 4:ZDecimal feature, which sets Xmin 4.7 and Xmax 4.7, or ZOOM 8:ZInteger, which sets Xmin 47 and Xmax 47. Let’s use the ZOOM 4:ZDecimal option here, noting the calculator automatically regraphs the line. Pressing the TRACE key once again and moving the marker shows that more “friendly” ordered pair solutions are displayed (Figure 1.20). Other methods for finding a friendly window are discussed later in this section. EXAMPLE 4

䊳

Figure 1.19 10

10

10

10

Figure 1.20 3.1

4.7

4.7

3.1

Graphing a Relation Using Technology Use a calculator to graph 2x 3y 6. Then use the TABLE feature to determine the value of y when x 0, and the value of x when y 0. Write each result in ordered pair form.

Solution

䊳

We begin by solving the equation for y, so we can enter it on the given equation 2x 3y 6 3y 2x 6 subtract 2x (isolate the y-term) 2 x 2 divide by 3 y 3

Y=

screen.

2 x 2 on the Y= screen and using ZOOM 6:ZStandard produces 3 the graph shown. Using the TABLE and scrolling as needed, shows that when x 0, y 2, and when y 0, x 3. As ordered pairs we have 10, 22 and 13, 02 . Entering y

10

10

C. You’ve just seen how we can graph a relation using a calculator

10

10


䊳

D. The Equation and Graph of a Circle Using the midpoint and distance formulas, we can develop the equation of another important relation, that of a circle. As the name suggests, the midpoint of a line segment is located halfway between the endpoints. On a standard number line, the midpoint of the line segment with endpoints 1 and 5 is 3, but more important, note that 3 is the

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6 15 3. This observation 2 2 can be extended to find the midpoint between any two points (x1, y1) and (x2, y2) in the xy-plane. We simply find the average distance between the x-coordinates and the average distance between the y-coordinates. average distance (from zero) of 1 unit and 5 units:

The Midpoint Formula

Given any line segment with endpoints P1 1x1, y1 2 and P2 1x2, y2 2 , the midpoint M is given by M: a

x1 x2 y1 y2 , b 2 2

The midpoint formula can be used in many different ways. Here we’ll use it to find the coordinates of the center of a circle. 䊳

EXAMPLE 5

䊳

Solution

Using the Midpoint Formula

The diameter of a circle has endpoints at P1 13, 22 and P2 15, 42 . Use the midpoint formula to find the coordinates of the center, then plot this point. x 1 x2 y 1 y2 , b 2 2 3 5 2 4 , b M: a 2 2 2 2 M: a , b 11, 12 2 2

y 5

Midpoint: a

P2

(1, 1) 5

5

x

P1 5

The center is at (1, 1), which we graph directly on the diameter as shown. Now try Exercises 29 through 38 䊳

The Distance Formula

Figure 1.21 y

c

(x2, y2)

b

x

a

(x1, y1)

(x2, y1)

P2

P1

a 兩 x2 x1兩

b 兩 y2 y1兩

d

In addition to a line segment’s midpoint, we are often interested in the length of the segment. For any two points (x1, y1) and (x2, y2) not lying on a horizontal or vertical line, a right triangle can be formed as in Figure 1.21. Regardless of the triangle’s orientation, the length of side a (the horizontal segment or base of the triangle) will have length 冟x2 x1冟 units, with side b (the vertical segment or height) having length 冟y2 y1冟 units. From the Pythagorean theorem (Section R.6), we see that c2 a2 b2 corresponds to c2 1 冟x2 x1冟 2 2 1 冟y2 y1冟 2 2. By taking the square root of both sides we obtain the length of the hypotenuse, which is identical to the distance between these two points: c 21x2 x1 2 2 1y2 y1 2 2. The result is called the distance formula, although it’s most often written using d for distance, rather than c. Note the absolute value bars are dropped from the formula, since the square of any quantity is always nonnegative. This also means that either point can be used as the initial point in the computation. The Distance Formula

Given any two points P1 1x1, y1 2 and P2 1x2, y2 2, the straight line distance d between them is

d 21x2 x1 2 2 1y2 y1 2 2

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EXAMPLE 6

䊳

Using the Distance Formula Use the distance formula to find the diameter of the circle from Example 5.

Solution

䊳

For 1x1, y1 2 13, 22 and 1x2, y2 2 15, 42, the distance formula gives d

21x2 x1 2 2 1y2 y1 2 2 2 3 5 132 4 2 34 122 4 2 282 62 1100 10

The diameter of the circle is 10 units long. Now try Exercises 39 through 48

䊳

A circle can be defined as the set of all points in a plane that are a fixed distance called the radius, from a fixed point called the center. Since the definition involves distance, we can construct the general equation of a circle using the distance formula. Assume the center has coordinates (h, k), and let (x, y) represent any point on the graph. The distance between these points is equal to the radius r, and the distance formula yields: 21x h2 2 1y k2 2 r. Squaring both sides gives the equation of a circle in standard form: 1x h2 2 1y k2 2 r2. The Equation of a Circle A circle of radius r with center at (h, k) has the equation 1x h2 2 1y k2 2 r2

If h 0 and k 0, the circle is centered at (0, 0) and the graph is a central circle with equation x2 y2 r2. At other values for h or k, the center is at (h, k) with no change in the radius. Note that an open dot is used for the center, as it’s actually a point of reference and not a part of the graph.

y

Circle with center at (h, k) r

k

(x, y)

(h, k)

Central circle

(x h)2 (y k)2 r2 r

(x, y)

(0, 0)

h

x

x2 y2 r2

EXAMPLE 7

䊳

Finding the Equation of a Circle in Standard Form

Solution

䊳

Since the center is at (0, 1) we have h 0, k 1, and r 4. Using the standard form 1x h2 2 1y k2 2 r2 we obtain

Find the equation of a circle with center 10, 1) and radius 4. 1x 02 2 3y 112 4 2 42 x2 1y 12 2 16

substitute 0 for h, 1 for k, and 4 for r simplify

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95

The graph of x2 1y 12 2 16 is shown in the figure. y (0, 3) Circle

r4 (4, 1)

Center: (0, 1) Radius: r 4 Diameter: 2r 8

x (4, 1)

(0, 1)

(0, 5)


䊳

The graph of a circle can be obtained by first identifying the coordinates of the center and the length of the radius from the equation in standard form. After plotting the center point, we count a distance of r units left and right of center in the horizontal direction, and up and down from center in the vertical direction, obtaining four points on the circle. Neatly graph a circle containing these four points. EXAMPLE 8

䊳

Graphing a Circle

Solution

䊳

Comparing the given equation with the standard form, we find the center is at 12, 32 and the radius is r 213 ⬇ 3.5.

Graph the circle represented by 1x 22 2 1y 32 2 12. Clearly label the center and radius.

standard form

S

S

S

1x h2 2 1y k2 2 r2

1x 22 2 1y 32 2 12 h 2 k 3 h2 k 3

given equation

r2 12 r 112 2 13 ⬇ 3.5

radius must be positive

Plot the center 12, 32 and count approximately 3.5 units in the horizontal and vertical directions. Complete the circle by freehand drawing or using a compass. The graph shown is obtained. y Some coordinates are approximate

Circle (2, 0.5) x

r ~ 3.5 (1.5, 3)

(2, 3)

(5.5, 3)

Center: (2, 3) Radius: r 2兹3 Endpoints of horizontal diameter (2 2兹3, 3) and (2 2兹3, 3) Endpoints of vertical diameter (2, 3 2兹3) and (2, 3 2兹3)

(2, 6.5)


䊳

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96

1–12


In Example 8, note the equation is composed of binomial squares in both x and y. By expanding the binomials and collecting like terms, we can write the equation of the circle in general form:

WORTHY OF NOTE After writing the equation in standard form, it is possible to end up with a constant that is zero or negative. In the first case, the graph is a single point. In the second case, no graph is possible since roots of the equation will be complex numbers. These are called degenerate cases. See Exercise 105.

EXAMPLE 9

1x 22 2 1y 32 2 12

x 4x 4 y 6y 9 12 x2 y2 4x 6y 1 0 2

2

standard form expand binomials combine like terms—general form

For future reference, observe the general form contains a sum of second-degree terms in x and y, and that both terms have the same coefficient (in this case, “1”). Since this form of the equation was derived by squaring binomials, it seems reasonable to assume we can go back to the standard form by creating binomial squares in x and y. This is accomplished by completing the square. 䊳

Finding the Center and Radius of a Circle Find the center and radius of the circle with equation x2 y2 2x 4y 4 0. Then sketch its graph and label the center and radius.

Solution

䊳

To find the center and radius, we complete the square in both x and y. x2 y2 2x 4y 4 0 1x2 2x __ 2 1y2 4y __ 2 4 1x2 2x 12 1y2 4y 42 4 1 4 adds 1 to left side

given equation group x-terms and y-terms; add 4 complete each binomial square

add 1 4 to right side

adds 4 to left side

1x 12 2 1y 22 2 9

factor and simplify

The center is at 11, 22 and the radius is r 19 3. (1, 5)

(4, 2)

y

r3 (1, 2)

(2, 2) Circle Center: (1, 2) Radius: r 3

x

(1, 1)


EXAMPLE 10

䊳

䊳

Applying the Equation of a Circle To aid in a study of nocturnal animals, some naturalists install a motion detector near a popular watering hole. The device has a range of 10 m in any direction. Assume the water hole has coordinates (0, 0) and the device is placed at 12, 12 . a. Write the equation of the circle that models the maximum effective range of the device. b. Use the distance formula to determine if the device will detect a badger that is approaching the water and is now at coordinates 111, 52 .

y 5

10

5

x

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Solution

䊳

97

a. Since the device is at (2, 1) and the radius (or reach) of detection is 10 m, any movement in the interior of the circle defined by 1x 22 2 1y 12 2 102 will be detected. b. Using the points (2, 1) and (11, 5) in the distance formula yields: d

21x2 x1 2 2 1y2 y1 2 2 2111 22 2 35 112 4 2 292 142 2 181 16 197 ⬇ 9.85

distance formula substitute given values simplify compute squares result

Since 9.85 6 10, the badger is within range of the device and will be detected. Now try Exercises 95 through 100

䊳

When using a graphing calculator to study circles, it’s important to note that most standard viewing windows have the x- and y-values preset at [10, 10] even though the calculator screen is not square. This tends to compress the y-values and give a skewed image of the graph. If the circle appears oval in shape, use ZOOM 5:ZSquare to obtain the correct perspective. Graphing calculators can produce the graph of a circle in various ways, and the choice of method simply depends on what you’d like to accomplish. To simply view the graph or compare two circular graphs, the DRAW command is used. From the home screen press: 2nd PRGM (DRAW) 9:Circle(. This generates the “Circle(” command, with the left parentheses indicating we need to supply three inputs, separated by commas. These inputs are the x-coordinate of the center, the y-coordinate of the center, and the radius of the circle. For the circle defined by the equation 1x 32 2 1y 22 2 49, we know the center is at 13, 22 and the radius is 7 units. The resulting command and graph are shown in Figures 1.22 and 1.23. While the DRAW command will graph any circle, we are unable to use the TRACE or CALC commands to interact with the graph. To make these features available, we must first solve for x in terms of y, as we did previously (the 1:ClrDraw command is used to clear the graph). Consider the relation x2 y2 25, which we know is the equation of a circle centered at (0, 0) with radius r 5. x2 y2 25

original equation

y 25 x y 225 x2 2

2

isolate y 2 solve for y

Note that we can separate this result into two parts, enabling the calculator to graph Y1 225 x2 (giving the “upper half” of the circle), and Y2 225 x2 (giving the “lower half”). Enter these on the Y= screen (note that Y2 Y1 can be used instead of reentering the entire expression: VARS ). If we graph Y1 and Y2 on the standard screen, the result appears more oval than circular (Figure 1.24). Using the ZOOM 5:ZSquare option, the tick marks become equally spaced on both axes (Figure 1.25). ENTER

Figure 1.22, 1.23 10

15.2

15.2

10

Figure 1.24

Figure 1.25

10

10

10

10

10

15.2

15.2

10

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Although it is a much improved graph, the circle does not appear “closed” as the calculator lacks sufficient pixels to show the proper curvature. A second alternative is to manually set a “friendly” window. Using Xmin 9.4, Xmax 9.4, Ymin 6.2, and Ymax 6.2 will generate a better graph due to the number of pixels available. Note that we can jump between the upper and lower halves of the circle using the up or down arrows. See Exercises 101 and 102.

D. You’ve just seen how we can develop the equation and graph of a circle using the distance and midpoint formulas

1.1 EXERCISES 䊳



䊳

1. If a relation is defined by a set of ordered pairs, the domain is the set of all ________ components, the range is the set of all ________ components.

4. For x2 y2 25, the center of the circle is at ________ and the length of the radius is ________ units. The graph is called a ________ circle.

2. For the equation y x 5 and the ordered pair (x, y), x is referred to as the input or ________ variable, while y is called the ________ or dependent variable.

5. Discuss/Explain how to find the center and radius of the circle defined by the equation x2 y2 6x 7. How would this circle differ from the one defined by x2 y2 6y 7?

3. A circle is defined as the set of all points that are an equal distance, called the ________, from a given point, called the ________.

6. In Example 3(b) we graphed the semicircle defined by y 29 x2. Discuss how you would obtain the equation of the full circle from this equation, and how the two equations are related.


Represent each relation in mapping notation, then state the domain and range.

GPA

7.

9. {(1, 2), (3, 4), (5, 6), (7, 8), (9, 10)}

4.00 3.75 3.50 3.25 3.00 2.75 2.50 2.25 2.00 0

10. {(2, 4), (3, 5), (1, 3), (4, 5), (2, 3)} 11. {(4, 0), (1, 5), (2, 4), (4, 2), (3, 3)} 12. {(1, 1), (0, 4), (2, 5), (3, 4), (2, 3)}

1

2

3

4

Complete each table using the given equation. For Exercises 15, 16, 21, and 22, each input may correspond to two outputs (be sure to find both if they exist). Use these points to graph the relation. For Exercises 17 through 24, also state the domain and range.

5

Year in college

Efficiency rating

8.

95 90 85 80 75 70 65 60 55 0

State the domain and range of each pointwise-defined relation.

2 13. y x 1 3 x

1

2

3

4

Month

5

6

y

5 14. y x 3 4 x

6

8

3

4

0

0

3

4

6

8

8

10

y

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99


15. x 2 冟y冟

16. 冟y 1冟 x

x

x

y

Use a graphing calculator to graph the following relations. Then use the TABLE feature to determine the value of y when x ⴝ 0, and the value(s) of x when y ⴝ 0, and write the results in ordered pair form.

y

2

0

0

1

1

3

3

5

26. x 2y 6

6

6

27. y x2 4x

7

7

28. y x2 2x 3

17. y x2 1 x

25. 2x 5y 10

18. y x2 3

y

x

y

Find the midpoint of each segment with the given endpoints.

3

2

29. (1, 8), (5, 6)

2

1

0

0

30. (5, 6), (6, 8)

2

1

31. (4.5, 9.2), (3.1, 9.8)

3

2

32. (5.2, 7.1), (6.3, 7.1)

4

3

19. y 225 x2 x

20. y 2169 x2

y

x

y

4

12

3

5

0

0

2

3

3

5

4

12

21. x 1 y

2

x

y

35.

x 1

4

0

2

1

1.25

2

1

7

9

2

2

1

1

0

0

1

4

2

7

3

36.

1 2 3 4 5 x

y 5 4 3 2 1 ⴚ5ⴚ4ⴚ3ⴚ2ⴚ1 ⴚ1 ⴚ2 ⴚ3 ⴚ4 ⴚ5

1 2 3 4 5 x

Find the center of each circle with the diameter shown.

37.

24. y 1x 12 3 x

y 5 4 3 2 1 ⴚ5ⴚ4ⴚ3ⴚ2ⴚ1 ⴚ1 ⴚ2 ⴚ3 ⴚ4 ⴚ5

y

5

y

Find the midpoint of each segment.

22. y 2 x 2

x

3 1 3 5 34. a , b, a , b 4 3 8 6

2

10

3 23. y 2x 1

1 3 1 2 33. a , b, a , b 5 3 10 4

y

y 5 4 3 2 1 ⴚ5ⴚ4ⴚ3ⴚ2ⴚ1 ⴚ1 ⴚ2 ⴚ3 ⴚ4 ⴚ5

1 2 3 4 5 x

38.

y 5 4 3 2 1 ⴚ5ⴚ4ⴚ3ⴚ2ⴚ1 ⴚ1 ⴚ2 ⴚ3 ⴚ4 ⴚ5

1 2 3 4 5 x

39. Use the distance formula to find the length of the line segment in Exercise 35. 40. Use the distance formula to find the length of the line segment in Exercise 36. 41. Use the distance formula to find the length of the diameter of the circle in Exercise 37. 42. Use the distance formula to find the length of the diameter of the circle in Exercise 38.

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100

1–16


In Exercises 43 to 48, three points that form the vertices of a triangle are given. Use the distance formula to determine if any of the triangles are right triangles (the three sides satisfy the Pythagorean Theorem a2 ⴙ b2 ⴝ c2).

Write each equation in standard form to find the center and radius of the circle. Then sketch the graph.

73. x2 y2 10x 12y 4 0 74. x2 y2 6x 8y 6 0

43. (3, 7), (2, 2), (5, 5)

75. x2 y2 10x 4y 4 0

44. (7, 0), (1, 0), (7, 4)

76. x2 y2 6x 4y 12 0

45. (4, 3), (7, 1), (3, 2)

77. x2 y2 6y 5 0

46. (5, 2), (0, 3), (4, 4)

78. x2 y2 8x 12 0

47. (3, 2), (1, 5), (6, 4)

79. x2 y2 4x 10y 18 0

48. (0, 0), (5, 2), (2, 5)

80. x2 y2 8x 14y 47 0

Find the equation of a circle satisfying the conditions given, then sketch its graph.

81. x2 y2 14x 12 0 82. x2 y2 22y 5 0

49. center (0, 0), radius 3

83. 2x2 2y2 12x 20y 4 0


84. 3x2 3y2 24x 18y 3 0

51. center (5, 0), radius 13 52. center (0, 4), radius 15 53. center (4, 3), radius 2 54. center (3, 8), radius 9 55. center (7, 4), radius 17 56. center (2, 5), radius 16 57. center (1, 2), diameter 6

In this section we looked at characteristics of equations that generated linear graphs, and graphs of parabolas and circles. Use this information and ordered pairs of your choosing to match the eight graphs given with their corresponding equation (two of the equations given have no matching graph).

a. y x2 6x c. x2 y 9

b. x2 1y 32 2 36 d. 3x 4y 12

3 x4 2

f. 1x 12 2 1y 22 2 49

58. center (2, 3), diameter 10

e. y

59. center (4, 5), diameter 4 13

g. 1x 32 2 y2 16 h. 1x 12 2 1y 22 2 9

60. center (5, 1), diameter 4 15 61. center at (7, 1), graph contains the point (1, 7) 62. center at (8, 3), graph contains the point (3, 15)

i. 4x 3y 12

85.

63. center at (3, 4), graph contains the point (7, 9) 64. center at (5, 2), graph contains the point (1, 3)

ⴚ10ⴚ8ⴚ6ⴚ4ⴚ2 ⴚ2 ⴚ4 ⴚ6 ⴚ8 ⴚ10

65. diameter has endpoints (5, 1) and (5, 7) 66. diameter has endpoints (2, 3) and (8, 3) Identify the center and radius of each circle, then graph. Also state the domain and range of the relation.

67. 1x 22 2 1y 32 2 4 68. 1x 52 2 1y 12 2 9

69. 1x 12 2 1y 22 2 12 70. 1x 72 2 1y 42 2 20 71. 1x 42 2 y2 81 72. x2 1y 32 2 49

y 10 8 6 4 2

87.

86.

y

2 4 6 8 10 x

y 10 8 6 4 2 ⴚ10ⴚ8ⴚ6ⴚ4ⴚ2 ⴚ2 ⴚ4 ⴚ6 ⴚ8 ⴚ10

2 4 6 8 10 x

10 8 6 4 2 ⴚ10ⴚ8ⴚ6ⴚ4ⴚ2 ⴚ2 ⴚ4 ⴚ6 ⴚ8 ⴚ10

j. 6x y x2 9

88.

2 4 6 8 10 x

y 10 8 6 4 2 ⴚ10ⴚ8ⴚ6ⴚ4ⴚ2 ⴚ2 ⴚ4 ⴚ6 ⴚ8 ⴚ10

2 4 6 8 10 x

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89.

y 10 8 6 4 2 ⴚ10ⴚ8ⴚ6ⴚ4ⴚ2 ⴚ2 ⴚ4 ⴚ6 ⴚ8 ⴚ10

䊳

2 4 6 8 10 x

90.

91.

y 10 8 6 4 2 ⴚ10ⴚ8ⴚ6ⴚ4ⴚ2 ⴚ2 ⴚ4 ⴚ6 ⴚ8 ⴚ10

y 10 8 6 4 2 ⴚ10ⴚ8ⴚ6ⴚ4ⴚ2 ⴚ2 ⴚ4 ⴚ6 ⴚ8 ⴚ10

2 4 6 8 10 x

92.

2 4 6 8 10 x

y 10 8 6 4 2 ⴚ10ⴚ8ⴚ6ⴚ4ⴚ2 ⴚ2 ⴚ4 ⴚ6 ⴚ8 ⴚ10

2 4 6 8 10 x


93. Spending on Cable and Satellite TV: s ⴝ 29t ⴙ 96

94. Radius of a circumscribed circle: r ⴝ

The data from Example 1 is closely modeled by the formula shown, where t represents the year (t 0 corresponds to the year 2000) and s represents the average amount spent per person, per year in the United States. (a) List five ordered pairs for this relation using t 3, 5, 7, 9, 11. Does the model give a good approximation of the actual data? (b) According to the model, what will be the average amount spent on cable and satellite TV in the year 2013? (c) According to the model, in what year will annual spending surpass $500? (d) Use the table to graph this relation by hand. 䊳

101


The radius r of a circle circumscribed around a square is found by using the formula given, where A is the area of the square. Solve the formula for A and use the result to find the area of the square shown.

A B2 y

(5, 0) x

APPLICATIONS

95. Radar detection: A luxury liner is located at map coordinates (5, 12) and has a radar system with a range of 25 nautical miles in any direction. (a) Write the equation of the circle that models the range of the ship’s radar, and (b) Use the distance formula to determine if the radar can pick up the liner’s sister ship located at coordinates (15, 36). 97. Inscribed circle: Find the equation for both the red and blue circles, then find the area of the region shaded in blue.

y

(2, 0) x

99. Radio broadcast range: Two radio stations may not use the same frequency if their broadcast areas overlap. Suppose station KXRQ has a broadcast area bounded by x2 y2 8x 6y 0 and WLRT has a broadcast area bounded by x2 y2 10x 4y 0. Graph the circle representing each broadcast area on the same grid to determine if both stations may broadcast on the same frequency.

96. Earthquake range: The epicenter (point of origin) of a large earthquake was located at map coordinates (3, 7), with the quake being felt up to 12 mi away. (a) Write the equation of the circle that models the range of the earthquake’s effect. (b) Use the distance formula to determine if a person living at coordinates (13, 1) would have felt the quake. 98. Inscribed triangle: The area of an equilateral triangle inscribed in a circle is given 313 2 r, by the formula A 4 where r is the radius of the circle. Find the area of the equilateral triangle shown.

y (3, 4)

x

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102

100. Radio broadcast range: The emergency radio broadcast system is designed to alert the population by relaying an emergency signal to all points of the country. A signal is sent from a station whose broadcast area is bounded by x2 y2 2500 (x and y in miles) and the signal is picked up and relayed by a transmitter with range 1x 202 2 1y 302 2 900. Graph the circle representing each broadcast area on the same grid to determine the greatest distance from the original station that this signal can be received. Be sure to scale the axes appropriately.

䊳

101. Graph the circle defined by x2 y2 36 using a friendly window, then use the TRACE feature to find the value of y when x 3.6. Now find the value of y when x 4.8. Explain why the values seem “interchangeable.” 102. Graph the circle defined by 1x 32 2 y2 16 using a friendly window, then use the TRACE feature to find the value of the y-intercepts. Show you get the same intercept by computation.


103. Although we use the word “domain” extensively in mathematics, it is also commonly seen in literature and heard in everyday conversation. Using a collegelevel dictionary, look up and write out the various meanings of the word, noting how closely the definitions given are related to its mathematical use. 104. Consider the following statement, then determine whether it is true or false and discuss why. A graph will exhibit some form of symmetry if, given a point that is h units from the x-axis, k units from the y-axis, and d units from the origin, there is a second point on the graph that is a like distance from the origin and each axis.

䊳

1–18


105. When completing the square to find the center and radius of a circle, we sometimes encounter a value for r2 that is negative or zero. These are called degenerate cases. If r2 6 0, no circle is possible, while if r2 0, the “graph” of the circle is simply the point (h, k). Find the center and radius of the following circles (if possible). a. x2 y2 12x 4y 40 0 b. x2 y2 2x 8y 8 0 c. x2 y2 6x 10y 35 0


106. (R.2) Evaluate/Simplify the following expressions. 2 5

a.

xx x3

107. (R.3) Solve the following equation. x 1 5 3 4 6

b. 33 32 31 30 31

108. (R.4) Solve x2 27 6x by factoring.

c. 1253

109. (R.6) Solve 1 1n 3 n and check solutions by substitution. If a solution is extraneous, so state.

1

2 3

d. 27

e. (2m3n)2

f. 15x2 0 5x0

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Linear Equations and Rates of Change


A. Graph linear equations

B.

C. D. E.

using the intercept method Find the slope of a line and interpret it as a rate of change Graph horizontal and vertical lines Identify parallel and perpendicular lines Apply linear equations in context

In preparation for sketching graphs of other equations, we’ll first look more closely at the characteristics of linear graphs. While linear graphs are fairly simple models, they have many substantive and meaningful applications. For instance, most of us are aware that satellite and cable TV have been increasFigure 1.26 ing in popularity since they were first 500 introduced. A close look at Figure 1.2 from ($411-est) 400 Section 1.1 reveals that spending on these ($375) forms of entertainment increased from ($281) 300 $192 per person per year in 2003 to $281 ($234) in 2007 (Figure 1.26). From an investor’s ($192) 200 or a producer’s point of view, there is a very high interest in the questions, “How 100 fast are sales increasing? Can this relationship be modeled mathematically to help 3 11 5 7 9 predict sales in future years?” Answers to Year (0 → 2000) these and other questions are precisely Source: 2009 Statistical Abstract of the United States, what our study in this section is all about. Table 1089 (some figures are estimates) Consumer spending (dollars per year)

1.2

A. The Graph of a Linear Equation A linear equation can be identified using these three tests: 1. the exponent on any variable is one, 2. no variable occurs in a denominator, and 3. no two variables are multiplied together. The equation 3y 9 is a linear equation in one variable, while 2x 3y 12 and y 23 x 4 are linear equations in two variables. In general, we have the following definition: Linear Equations A linear equation is one that can be written in the form ax by c where a, b, and c are real numbers, with a and b not simultaneously equal to zero. As in Section 1.1, the most basic method for graphing a line is to simply plot a few points, then draw a straight line through the points. EXAMPLE 1

䊳

Graphing a Linear Equation in Two Variables Graph the equation 3x 2y 4 by plotting points.

Solution

WORTHY OF NOTE If you cannot draw a straight line through the plotted points, a computational error has been made. All points satisfying a linear equation lie on a straight line.

1–19

䊳

Selecting x 2, x 0, x 1, and x 4 as inputs, we compute the related outputs and enter the ordered pairs in a table. The result is x input

y output

2

5

0

2

1

1 2

4

4

y

(2, 5)

5

(0, 2)

(x, y) ordered pairs 12, 52

(1, q) 5

5

10, 22 11, 12 2

14, 42

x

(4, 4) 5

Now try Exercises 7 through 12 䊳 103

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Notice that the line in Example 1 crosses the y-axis at (0, 2), and this point is called the y-intercept of the line. In general, y-intercepts have the form (0, y). Although difficult to see graphically, substituting 0 for y and solving for x shows this line crosses the x-axis at ( 43, 0) and this point is called the x-intercept. In general, x-intercepts have the form (x, 0). The x- and y-intercepts are usually easier to calculate than other points (since y 0 or x 0, respectively) and we often graph linear equations using only these two points. This is called the intercept method for graphing linear equations. The Intercept Method 1. Substitute 0 for x and solve for y. This will give the y-intercept (0, y). 2. Substitute 0 for y and solve for x. This will give the x-intercept (x, 0). 3. Plot the intercepts and use them to graph a straight line.

EXAMPLE 2

䊳

Graphing Lines Using the Intercept Method Graph 3x 2y 9 using the intercept method.

Solution

䊳

Substitute 0 for x (y-intercept) 3102 2y 9 2y 9 9 y 2 9 a0, b 2

Substitute 0 for y (x-intercept) 3x 2102 9 3x 9 x3 (3, 0)

5

y 3x 2y 9

冢0, t 冣

(3, 0) 5

5

x

5

A. You’ve just seen how we can graph linear equations using the intercept method


䊳

B. The Slope of a Line and Rates of Change After the x- and y-intercepts, we next consider the slope of a line. We see applications of this concept in many diverse areas, including the grade of a highway (trucking), the pitch of a roof (carpentry), the climb of an airplane (flying), the drainage of a field (landscaping), and the slope of a mountain (parks Figure 1.27 y and recreation). While the general concept is an (x2, y2) intuitive one, we seek to quantify the concept (as- y2 sign it a numeric value) for purposes of comparison and decision making. In each of the preceding y2 y1 examples (grade, pitch, climb, etc.), slope is a rise measure of “steepness,” as defined by the ratio vertical change (x1, y1) . Using a line segment through horizontal change y1 arbitrary points P1 1x1, y1 2 and P2 1x2, y2 2 , x2 x1 run we can create the right triangle shown in Figx ure 1.27 to help us quantify this relationship. The x2 x1 figure illustrates that the vertical change or the

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Section 1.2 Linear Equations and Rates of Change

105

change in y (also called the rise) is simply the difference in y-coordinates: y2 y1. The horizontal change or change in x (also called the run) is the difference in x-coordinates: x2 x1. In algebra, we typically use the letter “m” to represent slope, y y change in y giving m x22 x11 as the change in x. The result is called the slope formula. WORTHY OF NOTE

The Slope Formula

Given two points P1 1x1, y1 2 and P2 1x2, y2 2 , the slope of the nonvertical line through P1 and P2 is

While the original reason that “m” was chosen for slope is uncertain, some have speculated that it was because in French, the verb for “to climb” is monter. Others say it could be due to the “modulus of slope,” the word modulus meaning a numeric measure of a given property, in this case the inclination of a line.

y2 y1 x2 x1 where x2 x1. m

Actually, the slope value does much more than quantify the slope of a line, it expresses a rate of change between the quantities measured along each axis. In ¢y change in y applications of slope, the ratio change in x is symbolized as ¢x . The symbol ¢ is the Greek letter delta and has come to represent a change in some quantity, and the ¢y notation m ¢x is read, “slope is equal to the change in y over the change in x.” Interpreting slope as a rate of change has many significant applications in college algebra and beyond. EXAMPLE 3

䊳

Using the Slope Formula ¢y Find the slope of the line through the given points, then use m to find an ¢x additional point on the line. a. (2, 1) and (8, 4) b. (2, 6) and (4, 2)

Solution

䊳

a. For P1 12, 12 and P2 18, 42 , b. y2 y1 m x2 x1 41 82 3 1 6 2 The slope of this line is 12. ¢y 1 , we note that y Using ¢x 2 increases 1 unit (the y-value is positive), as x increases 2 units. Since (8, 4) is known to be on the line, the point 18 2, 4 12 110, 52 must also be on the line.

For P1 12, 62 and P2 14, 22, y2 y1 m x2 x1 26 4 122 4 2 6 3 The slope of this line is 2 3 . ¢y 2 Using , we note that y ¢x 3 decreases 2 units (the y-value is negative), as x increases 3 units. Since (4, 2) is known to be on the line, the point 14 3, 2 22 17, 02 must also be on the line. Now try Exercises 33 through 40 䊳

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䊳

CAUTION

When using the slope formula, try to avoid these common errors. 1. The order that the x- and y-coordinates are subtracted must be consistent, y y y y since x22 x11 x21 x12. 2. The vertical change (involving the y-values) always occurs in the numerator: y2 y1 x2 x1 x2 x1 y2 y1 . 3. When x1 or y1 is negative, use parentheses when substituting into the formula to prevent confusing the negative sign with the subtraction operation.

EXAMPLE 4

䊳

Interpreting the Slope Formula as a Rate of Change Jimmy works on the assembly line for an auto parts remanufacturing company. By 9:00 A.M. his group has assembled 29 carburetors. By 12:00 noon, they have completed 87 carburetors. Assuming the relationship is linear, find the slope of the line and discuss its meaning in this context.

Solution WORTHY OF NOTE Actually, the assignment of (t1, c1) to (9, 29) and (t2, c2) to (12, 87) was arbitrary. The slope ratio will be the same as long as the order of subtraction is the same. In other words, if we reverse this assignment and use 1t1, c1 2 112, 872 and 1t2, c2 2 19, 292 , we have 87 58 m 29 9 12 3

58 3.

䊳

First write the information as ordered pairs using c to represent the carburetors assembled and t to represent time. This gives 1t1, c1 2 19, 292 and 1t2, c2 2 112, 872. The slope formula then gives: c2 c1 ¢c 87 29 ¢t t2 t1 12 9 58 or 19.3 3 carburetors assembled Here the slope ratio measures , and we see that Jimmy’s group can hours assemble 58 carburetors every 3 hr, or about 1913 carburetors per hour. Now try Exercises 41 through 44 䊳

Positive and Negative Slope If you’ve ever traveled by air, you’ve likely heard the announcement, “Ladies and gentlemen, please return to your seats and fasten your seat belts as we begin our descent.” For a time, the descent of the airplane follows a linear path, but the slope of the line is negative since the altitude of the plane is decreasing. Positive and negative slopes, as well as the rate of change they represent, are important characteristics of linear graphs. In Example 3(a), the slope was a positive number (m 7 0) and the line will slope upward from left to right since the y-values are increasing. If m 6 0 as in Example 3(b), the slope of the line is negative and the line slopes downward as you move left to right since y-values are decreasing.

m 0, positive slope y-values increase from left to right

m 0, negative slope y-values decrease from left to right

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EXAMPLE 5

䊳

Applying Slope as a Rate of Change in Altitude At a horizontal distance of 10 mi after take-off, an airline pilot receives instructions to decrease altitude from their current level of 20,000 ft. A short time later, they are 17.5 mi from the airport at an altitude of 10,000 ft. Find the slope ratio for the descent of the plane and discuss its meaning in this context. Recall that 1 mi 5280 ft.

Solution

䊳

Let a represent the altitude of the plane and d its horizontal distance from the airport. Converting all measures to feet, we have 1d1, a1 2 152,800, 20,0002 and 1d2, a2 2 192,400, 10,0002 , giving 10,000 20,000 a2 a1 ¢a ¢d d2 d1 92,400 52,800 10,000 25 39,600 99

¢altitude Since this slope ratio measures ¢distance , we note the plane is decreasing 25 ft in altitude for every 99 ft it travels horizontally.

B. You’ve just seen how we can find the slope of a line and interpret it as a rate of change


䊳

C. Horizontal Lines and Vertical Lines Horizontal and vertical lines have a number of important applications, from finding the boundaries of a given graph (the domain and range), to performing certain tests on nonlinear graphs. To better understand them, consider that in one dimension, the graph of x 2 is a single point (Figure 1.28), indicating a Figure 1.28 location on the number line 2 units from zero x2 in the positive direction. In two dimensions, the 5 4 3 2 1 0 1 2 3 4 5 equation x 2 represents all points with an x-coordinate of 2. A few of these are graphed in Figure 1.29, but since there are an infinite number, we end up with a solid vertical line whose equation is x 2 (Figure 1.30). Figure 1.29

Figure 1.30

y 5

y (2, 5)

5

x2

(2, 3) (2, 1) 5

(2, 1)

WORTHY OF NOTE If we write the equation x 2 in the form ax by c, the equation becomes x 0y 2, since the original equation has no y-variable. Notice that regardless of the value chosen for y, x will always be 2 and we end up with the set of ordered pairs (2, y), which gives us a vertical line.

5

x

5

5

x

(2, 3) 5

5

The same idea can be applied to horizontal lines. In two dimensions, the equation y 4 represents all points with a y-coordinate of positive 4, and there are an infinite number of these as well. The result is a solid horizontal line whose equation is y 4. See Exercises 49 through 54. Horizontal Lines

Vertical Lines

The equation of a horizontal line is yk where (0, k) is the y-intercept.

The equation of a vertical line is xh where (h, 0) is the x-intercept.

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So far, the slope formula has only been applied to lines that were nonhorizontal or nonvertical. So what is the slope of a horizontal line? On an intuitive level, we expect that a perfectly level highway would have an incline or slope of zero. In general, for any two points on a horizontal line, y2 y1 and y2 y1 0, giving a slope of m x2 0 x1 0. For any two points on a vertical line, x2 x1 and x2 x1 0, making the slope ratio undefined: m

y y 2

Figure 1.31 For any horizontal line, y2 ⴝ y1

1

0

(see Figures 1.31 and 1.32).

Figure 1.32 For any vertical line, x2 ⴝ x1 y

y

(x1, y1)

y y2 y1 x x x 2 1 y1 y1 x x 2 1

(x2, y2)

y y2 y1 x x 2 1 x y2 y1 x x 1 1 y2 y1 x 0 undefined

(x2, y2)

0 x x 2 1

x

0 (x1, y1)

䊳

The Slope of a Vertical Line

The slope of any horizontal line is zero.

The slope of any vertical line is undefined.

Calculating Slopes The federal minimum wage remained constant from 1997 through 2006. However, the buying power (in 1996 dollars) of these wage earners fell each year due to inflation (see Table 1.3). This decrease in buying power is approximated by the red line shown. a. Using the data or graph, find the slope of the line segment representing the minimum wage. b. Select two points on the line representing buying power to approximate the slope of the line segment, and explain what it means in this context. Table 1.3 Time t (years)

5.15

Minimum wage w

Buying power p

1997

5.15

5.03

1998

5.15

4.96

1999

5.15

4.85

2000

5.15

4.69

2001

5.15

4.56

2002

5.15

4.49

4.15

2003

5.15

4.39

4.05

2004

5.15

4.28

2005

5.15

4.14

2006

5.15

4.04

5.05 4.95 4.85 4.75 4.65 4.55 4.45 4.35 4.25

19 97 19 98 19 9 20 9 00 20 01 20 02 20 03 20 04 20 0 20 5 06

EXAMPLE 6

The Slope of a Horizontal Line

Wages/Buying power

108

Time in years

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109


Solution

䊳

WORTHY OF NOTE In the context of lines, try to avoid saying that a horizontal line has “no slope,” since it’s unclear whether a slope of zero or an undefined slope is intended.

a. Since the minimum wage did not increase or decrease from 1997 to 2006, the line segment has slope m 0. b. The points (1997, 5.03) and (2006, 4.04) from the table appear to be on or close to the line drawn. For buying power p and time t, the slope formula yields: ¢p p2 p 1 ¢t t2 t1 4.04 5.03 2006 1997 0.11 0.99 9 1 The buying power of a minimum wage worker decreased by 11¢ per year during this time period.

C. You’ve just seen how we can graph horizontal and vertical lines


䊳

D. Parallel and Perpendicular Lines Two lines in the same plane that never intersect are called parallel lines. When we place these lines on the coordinate grid, we find that “never intersect” is equivalent to saying “the lines have equal slopes but different y-intercepts.” In Figure 1.33, notice the rise ¢y and run of each line is identical, and that by counting ¢x both lines have slope m 34. Figure 1.33

y 5

Generic plane L 1

Run L2

L1 Run

Rise

L2

Rise

5

5

x

5

Coordinate plane

Parallel Lines Given L1 and L2 are distinct, nonvertical lines with slopes of m1 and m2, respectively. 1. If m1 m2, then L1 is parallel to L2. 2. If L1 is parallel to L2, then m1 m2. In symbols, we write L1||L2. Any two vertical lines (undefined slope) are parallel.

EXAMPLE 7A

䊳

Determining Whether Two Lines Are Parallel Teladango Park has been mapped out on a rectangular coordinate system, with a ranger station at (0, 0). Brendan and Kapi are at coordinates 124, 182 and have set a direct course for the pond at (11, 10). Caden and Kymani are at (27, 1) and are heading straight to the lookout tower at (2, 21). Are they hiking on parallel or nonparallel courses?

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Solution

䊳

To respond, we compute the slope of each trek across the park. For Brendan and Kapi: m

y2 y1 x2 x1 10 1182

For Caden and Kymani: m

y2 y1 x2 x1

21 1 2 1272 4 20 25 5

11 1242 28 4 35 5

Since the slopes are equal, the two groups are hiking on parallel courses. Two lines in the same plane that intersect at right angles are called perpendicular lines. Using the coordinate grid, we note that intersect at right angles suggests that their slopes are negative reciprocals. While certainly not a proof, notice in Figure 1.34, the rise 3 4 ratio rise run for L1 is 3 and the ratio run for L2 is 4 . Alternatively, we can say their slopes have a product of ⴚ1, since m1 # m2 1 implies m1 m12. Figure 1.34

Generic plane

y

L1

5

L1

Run

Rise Rise Run 5

5

L2

x

L2 5

Coordinate plane

Perpendicular Lines Given L1 and L2 are distinct, nonvertical lines with slopes of m1 and m2, respectively. 1. If m1 # m2 1, then L1 is perpendicular to L2. 2. If L1 is perpendicular to L2, then m1 # m2 1. In symbols we write L1 ⬜ L2. Any vertical line (undefined slope) is perpendicular to any horizontal line (slope m 0). We can easily find the slope of a line perpendicular to a second line whose slope is known or can be found—just find the reciprocal and make it negative. For a line with slope m1 37, any line perpendicular to it will have a slope of m2 73. For m1 5, the slope of any line perpendicular would be m2 15. EXAMPLE 7B

䊳

Determining Whether Two Lines Are Perpendicular

The three points P1 15, 12, P2 13, 22 , and P3 13, 22 form the vertices of a triangle. Use these points to draw the triangle, then use the slope formula to determine if they form a right triangle.

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111


Solution

䊳

Figure 1.35

For a right triangle to be formed, two of the lines through these points must be perpendicular (forming a right angle). From Figure 1.35, it appears a right triangle is formed, but we must verify that two of the sides are actually perpendicular. Using the slope formula, we have: For P1 and P2

For P1 and P3

2 1 m1 35 3 3 2 2

21 m2 3 5 1 8

y 5

P1

P3 5

x

5

P2

5

For P2 and P3 m3

2 122

3 3 4 2 6 3

Since m1 # m3 1, the triangle has a right angle and must be a right triangle.

D. You’ve just seen how we can identify parallel and perpendicular lines


䊳

E. Applications of Linear Equations The graph of a linear equation can be used to help solve many applied problems. If the numbers you’re working with are either very small or very large, scale the axes appropriately. This can be done by letting each tic mark represent a smaller or larger unit so the data points given will fit on the grid. Also, many applications use only nonnegative values and although points with negative coordinates may be used to graph a line, only ordered pairs in QI can be meaningfully interpreted. EXAMPLE 8

䊳

Applying a Linear Equation Model — Commission Sales Use the information given to create a linear equation model in two variables, then graph the line and answer the question posed: A salesperson gets a daily $20 meal allowance plus $7.50 for every item she sells. How many sales are needed for a daily income of $125? Verify your answer by graphing the line on a calculator and using the

䊳

feature.

Let x represent sales and y represent income. This gives

verbal model: Daily income (y) equals $7.5 per sale 1x2 $20 for meals y equation model: y 7.5x 20

Using x 0 and x 10, we find (0, 20) and (10, 95) are points on this line and these are used to sketch the graph. From the graph, it appears that 14 sales are needed to generate a daily income of $125.00.

substitute 125 for y subtract 20 divide by 7.5

(10, 95)

100

Since daily income is given as $125, we substitute 125 for y and solve for x. 125 7.5x 20 105 7.5x 14 x

y 7.5x 20

150

Income

Algebraic Solution

TRACE

50

(0, 20) 0

2

4

6

8 10 12 14 16

Sales

x

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Graphical Solution

䊳

Begin by entering the equation y 7.5x 20 on the Y= screen, recognizing that in this context, both the input and output values must be positive. Reasoning the 10 sales will net $95 (less than $125) and 20 sales will net $170 (more than $125), we set the viewing as shown in Figure 1.36. We can then GRAPH the equation and use the TRACE feature to estimate the number of sales needed. The result shows that income is close to $125 when x is close to 14 (Figure 1.37). In addition to letting us trace along a graph, the TRACE option enables us to evaluate the equation at specific points. Simply entering the number “14” causes the calculator to accept 14 as the desired input (Figure 1.38), and after pressing , it verifies that (14, 125) is indeed a point on the graph (Figure 1.39).

Figure 1.36

WINDOW

Figure 1.37 200

20

0

ENTER

0

Figure 1.38

E. You’ve just seen how we can apply linear equations in context

Figure 1.39


䊳

1.2 EXERCISES 䊳



1. To find the x-intercept of a line, substitute ______ for y and solve for x. To find the y-intercept, substitute _________ for x and solve for y.

4. The slope of a horizontal line is _______, the slope of a vertical line is _______, and the slopes of two parallel lines are ______.

2. The slope formula is m ______ ______, and indicates a rate of change between the x- and y-variables.

5. Discuss/Explain If m1 2.1 and m2 2.01, will the lines intersect? If m1 23 and m2 23 , are the lines perpendicular?

3. If m 6 0, the slope of the line is ______ and the line slopes _______ from left to right.

6. Discuss/Explain the relationship between the slope formula, the Pythagorean theorem, and the distance formula. Include several illustrations.

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Create a table of values for each equation and sketch the graph.

x

y

3 9. y ⫽ x ⫹ 4 2 x

8. ⫺3x ⫹ 5y ⫽ 10 x

y

36. (⫺3, ⫺1), (0, 7)

37. (1, ⫺8), (⫺3, 7)

38. (⫺5, 5), (0, ⫺5)

39. (⫺3, 6), (4, 2)

40. (⫺2, ⫺4), (⫺3, ⫺1)

41. The graph shown models the relationship between the cost of a new home and the size of the home in square feet. (a) Determine the slope of the line and interpret what the slope ratio means in this context and (b) estimate the cost of a 3000 ft2 home. Exercise 41

5 10. y ⫽ x ⫺ 3 3

y

x

Exercise 42

500

y

1200 960

Volume (m3)

7. 2x ⫹ 3y ⫽ 6

35. (10, 3), (4, ⫺5)

Cost ($1000s)

䊳

113


250

720 480 240

0

1

2

3

4

5

0

ft2 (1000s)

Graph the following equations using the intercept method. Plot a third point as a check.

13. 3x ⫹ y ⫽ 6

14. ⫺2x ⫹ y ⫽ 12

15. 5y ⫺ x ⫽ 5

16. ⫺4y ⫹ x ⫽ 8

17. ⫺5x ⫹ 2y ⫽ 6

18. 3y ⫹ 4x ⫽ 9

19. 2x ⫺ 5y ⫽ 4

20. ⫺6x ⫹ 4y ⫽ 8

21. 2x ⫹ 3y ⫽ ⫺12

22. ⫺3x ⫺ 2y ⫽ 6

1 23. y ⫽ ⫺ x 2

24. y ⫽

25. y ⫺ 25 ⫽ 50x

26. y ⫹ 30 ⫽ 60x

2 27. y ⫽ ⫺ x ⫺ 2 5

3 28. y ⫽ x ⫹ 2 4

29. 2y ⫺ 3x ⫽ 0

30. y ⫹ 3x ⫽ 0

31. 3y ⫹ 4x ⫽ 12

32. ⫺2x ⫹ 5y ⫽ 8

2 x 3

43. The graph shown models the relationship between the distance of an aircraft carrier from its home port and the number of hours since departure. (a) Determine the slope of the line and interpret what the slope ratio means in this context and (b) estimate the distance from port after 8.25 hours. Exercise 43 300

33. (3, 5), (4, 6)

34. (⫺2, 3), (5, 8)

500

150

0

Compute the slope of the line through the given points, ¢y then graph the line and use m ⴝ ¢x to find two additional points on the line. Answers may vary.

Exercise 44

Circuit boards

12. If you completed Exercise 10, verify that 37 (⫺1.5, ⫺5.5) and 1 11 2 , 6 2 also satisfy the equation given. Do these points appear to be on the graph you sketched?

100

42. The graph shown models the relationship between the volume of garbage that is dumped in a landfill and the number of commercial garbage trucks that enter the site. (a) Determine the slope of the line and interpret what the slope ratio means in this context and (b) estimate the number of trucks entering the site daily if 1000 m3 of garbage is dumped per day.

Distance (mi)

11. If you completed Exercise 9, verify that (⫺3, ⫺0.5) and (12, 19 4 ) also satisfy the equation given. Do these points appear to be on the graph you sketched?

50

Trucks

10

Hours

20

250

0

5

10

Hours

44. The graph shown models the relationship between the number of circuit boards that have been assembled at a factory and the number of hours since starting time. (a) Determine the slope of the line and interpret what the slope ratio means in this context and (b) estimate how many hours the factory has been running if 225 circuit boards have been assembled.

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45. Height and weight: While there are many exceptions, numerous studies have shown a close relationship between an average height and average weight. Suppose a person 70 in. tall weighs 165 lb, while a person 64 in. tall weighs 142 lb. Assuming the relationship is linear, (a) find the slope of the line and discuss its meaning in this context and (b) determine how many pounds are added for each inch of height. 46. Rate of climb: Shortly after takeoff, a plane increases altitude at a constant (linear) rate. In 5 min the altitude is 10,000 ft. Fifteen minutes after takeoff, the plane has reached its cruising altitude of 32,000 ft. (a) Find the slope of the line and discuss its meaning in this context and (b) determine how long it takes the plane to climb from 12,200 ft to 25,400 ft. 47. Sewer line slope: Fascinated at how quickly the plumber was working, Ryan watched with great interest as the new sewer line was laid from the house to the main line, a distance of 48 ft. At the edge of the house, the sewer line was 6 in. under ground. If the plumber tied in to the main line at a depth of 18 in., what is the slope of the (sewer) line? What does this slope indicate? 48. Slope (pitch) of a roof: A contractor goes to a lumber yard to purchase some trusses (the triangular frames) for the roof of a house. Many sizes are available, so the contractor takes some measurements to ensure the roof will have the desired slope. In one case, the height of the truss (base to ridge) was 4 ft, with a width of 24 ft (eave to eave). Find the slope of the roof if these trusses are used. What does this slope indicate? Graph each line using two or three ordered pairs that satisfy the equation.

55. Supreme Court justices: The table given shows the total number of justices j sitting on the Supreme Court of the United States for selected time periods t (in decades), along with the number of nonmale, nonwhite justices n for the same years. (a) Use the data to graph the linear relationship between t and j, then determine the slope of the line and discuss its meaning in this context. (b) Use the data to graph the linear relationship between t and n, then determine the slope of the line and discuss its meaning. Exercise 55 Time t (1960 S 0)

Justices j

Nonwhite, nonmale n

0

9

0

10

9

1

20

9

2

30

9

3

40

9

4

50

9

5

56. Boiling temperature: The table shown gives the boiling temperature t of water as related to the altitude h. Use the data to graph the linear relationship between h and t, then determine the slope of the line and discuss its meaning in this context. Exercise 56 Altitude h (ft)

Boiling Temperature t (ⴗF)

0

212.0

1000

210.2

2000

208.4

3000

206.6 204.8

49. x 3

50. y 4

4000

51. x 2

52. y 2

5000

203.0

6000

201.2

Write the equation for each line L1 and L2 shown. Specifically state their point of intersection.

53.

y

54.

L1

L1

L2

4 2 ⴚ4

ⴚ2

2 ⴚ2 ⴚ4

4

x

ⴚ4

ⴚ2

y 5 4 3 2 1 ⴚ1 ⴚ2 ⴚ3 ⴚ4 ⴚ5

L2 2

4

x

Two points on L1 and two points on L2 are given. Use the slope formula to determine if lines L1 and L2 are parallel, perpendicular, or neither.

57. L1: (2, 0) and (0, 6) L2: (1, 8) and (0, 5)

58. L1: (1, 10) and (1, 7) L2: (0, 3) and (1, 5)

59. L1: (3, 4) and (0, 1) 60. L1: (6, 2) and (8, 2) L2: (5, 1) and (3, 0) L2: (0, 0) and (4, 4) 61. L1: (6, 3) and (8, 7) L2: (7, 2) and (6, 0)

62. L1: (5, 1) and (4, 4) L2: (4, 7) and (8, 10)

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115

In Exercises 63 to 68, three points that form the vertices of a triangle are given. Use the points to draw the triangle, then use the slope formula to determine if any of the triangles are right triangles. Also see Exercises 43–48 in Section 1.1.

63. (3, 7), (2, 2), (5, 5)

66. (5, 2), (0, 3), (4, 4)

64. (7, 0), (1, 0), (7, 4)

67. (3, 2), (1, 5), (6, 4)

65. (4, 3), (7, 1), (3,2)

68. (0, 0), (5, 2), (2, 5)

䊳


69. Human life expectancy: L ⴝ 0.15T ⴙ 73.7 In the United States, the average life expectancy has been steadily increasing over the years due to better living conditions and improved medical care. This relationship is modeled by the formula shown, where L is the average life expectancy and T is number of years since 1980. (a) What was the life expectancy in the year 2010? (b) In what year will average life expectancy reach 79 yr?

䊳

70. Interest earnings: 100I ⴝ 35,000T If $5000 dollars is invested in an account paying 7% simple interest, the amount of interest earned is given by the formula shown, where I is the interest and T is the time in years. Begin by solving the formula for I. (a) How much interest is earned in 5 yr? (b) How much is earned in 10 yr? (c) Use the two points (5 yr, interest) and (10 yr, interest) to calculate the slope of this line. What do you notice?

APPLICATIONS

Use the information given to build a linear equation model, then use the equation to respond. For exercises 71 to 74, develop both an algebraic and a graphical solution.

71. Business depreciation: A business purchases a copier for $8500 and anticipates it will depreciate in value $1250 per year. a. What is the copier’s value after 4 yr of use? b. How many years will it take for this copier’s value to decrease to $2250? 72. Baseball card value: After purchasing an autographed baseball card for $85, its value increases by $1.50 per year. a. What is the card’s value 7 yr after purchase? b. How many years will it take for this card’s value to reach $100? 73. Water level: During a long drought, the water level in a local lake decreased at a rate of 3 in. per month. The water level before the drought was 300 in. a. What was the water level after 9 months of drought? b. How many months will it take for the water level to decrease to 20 ft?

74. Gas mileage: When empty, a large dump-truck gets about 15 mi per gallon. It is estimated that for each 3 tons of cargo it hauls, gas mileage decreases by 34 mi per gallon. a. If 10 tons of cargo is being carried, what is the truck’s mileage? b. If the truck’s mileage is down to 10 mi per gallon, how much weight is it carrying? 75. Parallel/nonparallel roads: Aberville is 38 mi north and 12 mi west of Boschertown, with a straight “farm and machinery” road (FM 1960) connecting the two cities. In the next county, Crownsburg is 30 mi north and 9.5 mi west of Dower, and these cities are likewise connected by a straight road (FM 830). If the two roads continued indefinitely in both directions, would they intersect at some point? 76. Perpendicular/nonperpendicular course headings: Two shrimp trawlers depart Charleston Harbor at the same time. One heads for the shrimping grounds located 12 mi north and 3 mi east of the harbor. The other heads for a point 2 mi south and 8 mi east of the harbor. Assuming the harbor is at (0, 0), are the routes of the trawlers perpendicular? If so, how far apart are the boats when they reach their destinations (to the nearest one-tenth mi)?

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77. Cost of college: For the years 2000 to 2008, the cost of tuition and fees per semester (in constant dollars) at a public 4-yr college can be approximated by the equation y 386x 3500, where y represents the cost in dollars and x 0 represents the year 2000. Use the equation to find: (a) the cost of tuition and fees in 2010 and (b) the year this cost will exceed $9000. Source: The College Board

78. Female physicians: In 1960 only about 7% of physicians were female. Soon after, this percentage began to grow dramatically. For the years 1990 to 2000, the percentage of physicians that were female can be approximated by the equation y 0.6x 18.1, where y represents the percentage (as a whole number) and x 0 represents the year 1990. Use the equation to find: (a) the percentage of physicians that were female in 2000 and (b) the projected year this percentage would have exceeded 30%.

79. Decrease in smokers: For the years 1990 to 2000, the percentage of the U.S. adult population who were smokers can be approximated by the equation y 13 25 x 28.7, where y represents the percentage of smokers (as a whole number) and x 0 represents 1990. Use the equation to find: (a) the percentage of adults who smoked in the year 2005 and (b) the year the percentage of smokers is projected to fall below 15%. Source: WebMD

80. Temperature and cricket chirps: Biologists have found a strong relationship between temperature and the number of times a cricket chirps. This is modeled by the equation T 14N 40, where N is the number of times the cricket chirps per minute and T is the temperature in Fahrenheit. Use the equation to find: (a) the outdoor temperature if the cricket is chirping 48 times per minute and (b) the number of times a cricket chirps if the temperature is 70°.

Source: American Journal of Public Health

䊳


81. If the lines 4y 2x 5 and 3y ax 2 are perpendicular, what is the value of a? 82. Let m1, m2, m3, and m4 be the slopes of lines L1, L2, L3, and L4, respectively. Which of the following statements is true? a. m4 6 m1 6 m3 6 m2 y L2 L m 6 m 6 m 6 m b. 3 1 2 4 1 L3 c. m3 6 m4 6 m2 6 m1 L4 x d. m1 6 m3 6 m4 6 m2 e. m1 6 m4 6 m3 6 m2

䊳

83. An arithmetic sequence is a sequence of numbers where each successive term is found by adding a fixed constant, called the common difference d, to the preceding term. For instance 3, 7, 11, 15, . . . is an arithmetic sequence with d 4. The formula for the “nth term” tn of an arithmetic sequence is a linear equation of the form tn t1 1n 12d, where d is the common difference and t1 is the first term of the sequence. Use the equation to find the term specified for each sequence. a. 2, 9, 16, 23, 30, . . . ; 21st term b. 7, 4, 1, 2, 5, . . . ; 31st term c. 5.10, 5.25, 5.40, 5.55, . . . ; 27th term 9 d. 32, 94, 3, 15 4 , 2 , . . . ; 17th term


84. (1.1) Name the center and radius of the circle defined by 1x 32 2 1y 42 2 169 85. (R.6) Compute the sum and product indicated: a. 120 3 145 15 b. 13 152 13 252 86. (R.4) Solve the equation by factoring, then check the result(s) using substitution: 12x2 44x 45 0

87. (R.5) Factor the following polynomials completely: a. x3 3x2 4x 12 b. x2 23x 24 c. x3 125

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1.3

Functions, Function Notation, and the Graph of a Function


A. Distinguish the graph of a function from that of a relation B. Determine the domain and range of a function C. Use function notation and evaluate functions D. Read and interpret information given graphically

In this section we introduce one of the most central ideas in mathematics—the concept of a function. Functions can model the cause-and-effect relationship that is so important to using mathematics as a decision-making tool. In addition, the study will help to unify and expand on many ideas that are already familiar.

A. Functions and Relations There is a special type of relation that merits further attention. A function is a relation where each element of the domain corresponds to exactly one element of the range. In other words, for each first coordinate or input value, there is only one possible second coordinate or output. Functions A function is a relation that pairs each element from the domain with exactly one element from the range. If the relation is defined by a mapping, we need only check that each element of the domain is mapped to exactly one element of the range. This is indeed the case for the mapping P S B from Figure 1.1 (page 2), where we saw that each person corresponded to only one birthday, and that it was impossible for one person to be born on two different days. For the relation x ⫽ 冟y冟 shown in Figure 1.6 (page 4), each element of the domain except zero is paired with more than one element of the range. The relation x ⫽ 冟y冟 is not a function.

EXAMPLE 1

䊳

Determining Whether a Relation is a Function Three different relations are given in mapping notation below. Determine whether each relation is a function. a. b. c.

Solution

䊳

Person

Room

Marie Pesky Bo Johnny Rick Annie Reece

270 268 274 276 272 282

Pet

Weight (lb)

Fido

450 550 2 40 8 3

Bossy Silver Frisky Polly

War

Year

Civil War

1963

World War I

1950

World War II

1939

Korean War

1917

Vietnam War

1861

Relation (a) is a function, since each person corresponds to exactly one room. This relation pairs math professors with their respective office numbers. Notice that while two people can be in one office, it is impossible for one person to physically be in two different offices. Relation (b) is not a function, since we cannot tell whether Polly the Parrot weighs 2 lb or 3 lb (one element of the domain is mapped to two elements of the range). Relation (c) is a function, where each major war is paired with the year it began. Now try Exercises 7 through 10 䊳

1–33

117

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If the relation is pointwise-defined or given as a set of individual and distinct plotted points, we need only check that no two points have the same first coordinate with a different second coordinate. This gives rise to an alternative definition for a function. Functions (Alternate Definition) A function is a set of ordered pairs (x, y), in which each first component is paired with only one second component.

EXAMPLE 2

䊳

Identifying Functions Two relations named f and g are given; f is pointwise-defined (stated as a set of ordered pairs), while g is given as a set of plotted points. Determine whether each is a function. f: 1⫺3, 02, 11, 42, 12, ⫺52, 14, 22, 1⫺3, ⫺22, 13, 62, 10, ⫺12, (4, ⫺5), and (6, 1)

Solution

䊳

The relation f is not a function, since ⫺3 is paired with two different outputs: 1⫺3, 02 and 1⫺3, ⫺22 .

g

5

y (0, 5)

(⫺4, 2)

The relation g shown in the figure is a function. Each input corresponds to exactly one output, otherwise one point would be directly above the other and have the same first coordinate.

(3, 1)

(⫺2, 1) ⫺5

5

x

(4, ⫺1) (⫺1, ⫺3) ⫺5

Now try Exercises 11 through 18 䊳 The graphs of y ⫽ x ⫺ 1 and x ⫽ 冟y冟 from Section 1.1 offer additional insight into the definition of a function. Figure 1.40 shows the line y ⫽ x ⫺ 1 with emphasis on the plotted points (4, 3) and 1⫺3, ⫺42. The vertical movement shown from the x-axis to a point on the graph illustrates the pairing of a given x-value with one related y-value. Note the vertical line shows only one related y-value (x ⫽ 4 is paired with only y ⫽ 3). Figure 1.41 gives the graph of x ⫽ 冟y冟, highlighting the points (4, 4) and (4, ⫺4). The vertical movement shown here branches in two directions, associating one x-value with more than one y-value. This shows the relation y ⫽ x ⫺ 1 is also a function, while the relation x ⫽ 冟y冟 is not. Figure 1.41

Figure 1.40 5

y y⫽x⫺1

y

x ⫽ ⱍyⱍ (4, 4)

5

(4, 3) (2, 2) (0, 0) ⫺5

5

x

⫺5

5

x

(2, ⫺2) (⫺3, ⫺4)

⫺5

⫺5

(4, ⫺4)

This “vertical connection” of a location on the x-axis to a point on the graph can be generalized into a vertical line test for functions.

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119

Section 1.3 Functions, Function Notation, and the Graph of a Function

Vertical Line Test A given graph is the graph of a function, if and only if every vertical line intersects the graph in at most one point. Applying the test to the graph in Figure 1.40 helps to illustrate that the graph of any nonvertical line must be the graph of a function, as is the graph of any pointwisedefined relation where no x-coordinate is repeated. Compare the relations f and g from Example 2. 䊳

EXAMPLE 3

Using the Vertical Line Test Use the vertical line test to determine if any of the relations shown (from Section 1.1) are functions.

䊳

Solution

Visualize a vertical line on each coordinate grid (shown in solid blue), then mentally shift the line to the left and right as shown in Figures 1.42, 1.43, and 1.44 (dashed lines). In Figures 1.42 and 1.43, every vertical line intersects the graph only once, indicating both y ⫽ x2 ⫺ 2x and y ⫽ 29 ⫺ x2 are functions. In Figure 1.44, a vertical line intersects the graph twice for any x 7 0 [for instance, both (4, 2) and 14, ⫺22 are on the graph]. The relation x ⫽ y2 is not a function.

Figure 1.42

Figure 1.43

y (4, 8)

(⫺2, 8) y ⫽ x2 ⫺ 2x

5

Figure 1.44 y

y y ⫽ 兹9 ⫺ x2 (0, 3)

5

x ⫽ y2 (4, 2)

(2, 兹2)

5

(⫺1, 3)

(⫺3, 0)

(3, 3)

(0, 0)

(3, 0)

⫺5

5

x

⫺5

5

(2, 0)

(0, 0) ⫺5

5

(1, ⫺1)

⫺2

x ⫺5

⫺5

x

(2, ⫺兹2) (4, ⫺2)


EXAMPLE 4

䊳

Using the Vertical Line Test Use a table of values to graph the relations defined by a. y ⫽ 冟x冟 b. y ⫽ 1x, then use the vertical line test to determine whether each relation is a function.

Solution

䊳

WORTHY OF NOTE For relations and functions, a good way to view the distinction is to consider a mail carrier. It is possible for the carrier to put more than one letter into the same mailbox (more than one x going to the same y), but quite impossible for the carrier to place the same letter in two different boxes (one x going to two y’s).

a. For y ⫽ 冟x冟, using input values from x ⫽ ⫺4 to x ⫽ 4 produces the following table and graph (Figure 1.45). Note the result is a V-shaped graph that “opens upward.” The point (0, 0) of this absolute value graph is called the vertex. Since any vertical line will intersect the graph in at most one point, this is the graph of a function.

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CHAPTER 1 Relations, Functions, and Graphs y ⴝ 円 x円 x

y ⴝ 円 x円

⫺4

4

⫺3

3

⫺2

2

⫺1

1

0

0

1

1

2

2

3

3

4

4

Figure 1.45 y 5

⫺5

5

x

⫺5

b. For y ⫽ 1x, values less than zero do not produce a real number, so our graph actually begins at (0, 0) (see Figure 1.46). Completing the table for nonnegative values produces the graph shown, which appears to rise to the right and remains in the first quadrant. Since any vertical line will intersect this graph in at most one place, y ⫽ 1x is also a function. Figure 1.46 y

y ⴝ 1x x

5

y ⴝ 1x

0

0

1

1

2

12 ⬇ 1.4

3

13 ⬇ 1.7

4

⫺5

5

x

2 ⫺5

A. You’ve just seen how we can distinguish the graph of a function from that of a relation


B. The Domain and Range of a Function Vertical Boundary Lines and the Domain

WORTHY OF NOTE On a number line, some texts will use an open dot “º” to mark the location of an endpoint that is not included, and a closed dot “•” for an included endpoint.

In addition to its use as a graphical test for functions, a vertical line can help determine the domain of a function from its graph. For the graph of y ⫽ 1x (Figure 1.46), a vertical line will not intersect the graph until x ⫽ 0, and then will intersect the graph for all values x ⱖ 0 (showing the function is defined for these values). These vertical boundary lines indicate the domain is x ⱖ 0. Instead of using a simple inequality to write the domain and range, we will often use (1) a form of set notation, (2) a number line graph, or (3) interval notation. Interval notation is a symbolic way of indicating a selected interval of the real numbers. When a number acts as the boundary point for an interval (also called an endpoint), we use a left bracket “[” or a right bracket “]” to indicate inclusion of the endpoint. If the boundary point is not included, we use a left parenthesis “(” or right parenthesis “).”

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EXAMPLE 5

䊳

121

Using Notation to State the Domain and Range Model the given phrase using the correct inequality symbol. Then state the result in set notation, graphically, and in interval notation: “The set of real numbers greater than or equal to 1.”

Solution

䊳

WORTHY OF NOTE Since infinity is really a concept and not a number, it is never included (using a bracket) as an endpoint for an interval.

Let n represent the number: n ⱖ 1. • Set notation: 5n |n ⱖ 16 [ • Graph: ⫺2 ⫺1

0

1

2

3

4

• Interval notation: n 僆 31, q 2

5

Now try Exercises 35 through 50 䊳 The “僆” symbol says the number n is an element of the set or interval given. The “ q ” symbol represents positive infinity and indicates the interval continues forever to the right. Note that the endpoints of an interval must occur in the same order as on the number line (smaller value on the left; larger value on the right). A short summary of other possibilities is given here for any real number x. Many variations are possible. Conditions (a ⬍ b) x is greater than k x is less than or equal to k x is less than b and greater than a x is less than b and greater than or equal to a x is less than a or x is greater than b

Set Notation 5x| x 7 k6 5x |x ⱕ k6

5x |a 6 x 6 b6 5x| a ⱕ x 6 b6 5x| x 6 a or x 7 b6

Number Line

Interval Notation x 僆 1k, q 2

) k

x 僆 1⫺q, k4

[ k

)

)

a

b

[

)

a

b

x 僆 1a, b2 x 僆 3 a, b2

)

)

a

b

x 僆 1⫺q, a2 ´ 1b, q2

For the graph of y ⫽ 冟x冟 (Figure 1.45), a vertical line will intersect the graph (or its infinite extension) for all values of x, and the domain is x 僆 1⫺q, q2 . Using vertical lines in this way also affirms the domain of y ⫽ x ⫺ 1 (Section 1.1, Figure 1.5) is x 僆 1⫺q, q 2 while the domain of the relation x ⫽ 冟y冟 (Section 1.1, Figure 1.6) is x 僆 30, q 2 .

Range and Horizontal Boundary Lines The range of a relation can be found using a horizontal “boundary line,” since it will associate a value on the y-axis with a point on the graph (if it exists). Simply visualize a horizontal line and move the line up or down until you determine the graph will always intersect the line, or will no longer intersect the line. This will give you the boundaries of the range. Mentally applying this idea to the graph of y ⫽ 1x (Figure 1.46) shows the range is y 僆 3 0, q 2. Although shaped very differently, a horizontal boundary line shows the range of y ⫽ 冟x冟 (Figure 1.45) is also y 僆 30, q2. EXAMPLE 6

䊳

Determining the Domain and Range of a Function Use a table of values to graph the functions defined by 3 a. y ⫽ x2 b. y ⫽ 1 x Then use boundary lines to determine the domain and range of each.

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Solution

䊳

a. For y ⫽ x2, it seems convenient to use inputs from x ⫽ ⫺3 to x ⫽ 3, producing the following table and graph. Note the result is a basic parabola that “opens upward” (both ends point in the positive y direction), with a vertex at (0, 0). Figure 1.47 shows a vertical line will intersect the graph or its extension anywhere it is placed. The domain is x 僆 1⫺q, q 2 . Figure 1.48 shows a horizontal line will intersect the graph only for values of y that are greater than or equal to 0. The range is y 僆 3 0, q 2 . Figure 1.47

Squaring Function x

y ⴝ x2

⫺3

9

⫺2

4

⫺1

1

0

0

1

1

2

4

3

9

5

Figure 1.48

y y ⫽ x2 5

⫺5

5

x

y y ⫽ x2

⫺5

⫺5

5

x

⫺5

b. For y ⫽ 13 x, we select points that are perfect cubes where possible, then a few others to round out the graph. The resulting table and graph are shown. Notice there is a “pivot point” at (0, 0) called a point of inflection, and the ends of the graph point in opposite directions. Figure 1.49 shows a vertical line will intersect the graph or its extension anywhere it is placed. Figure 1.50 shows a horizontal line will likewise always intersect the graph. The domain is x 僆 1⫺q, q 2 , and the range is y 僆 1⫺q, q 2 . Figure 1.49

Cube Root Function x

3 yⴝ 1 x

⫺8

⫺2

⫺4

⬇ ⫺1.6

⫺1

⫺1

0

0

1

1

4

⬇ 1.6

8

2

Figure 1.50 3

5

⫺10

y y ⫽ 兹x

5

10

⫺5

x

⫺10

3 y y ⫽ 兹x

10

x

⫺5


Implied Domains When stated in equation form, the domain of a function is implicitly given by the expression used to define it, since the expression will dictate what input values are allowed. The implied domain is the set of all real numbers for which the function represents a real number. If the function involves a rational expression, the domain will exclude any input that causes a denominator of zero, since division by zero is undefined. If the function involves a square root expression, the domain will exclude inputs that create a negative radicand, since 1A represents a real number only when A ⱖ 0.

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EXAMPLE 7

Solution

䊳

䊳

Figure 1.51

Y1 ⫽ 1X ⫺ 12/1X2 ⫺ 92

123

Determining Implied Domains State the domain of each function using interval notation. 3 a. y ⫽ b. y ⫽ 12x ⫹ 3 x⫹2 x⫺1 c. y ⫽ 2 d. y ⫽ x2 ⫺ 5x ⫹ 7 x ⫺9 a. By inspection, we note an x-value of ⫺2 results in a zero denominator and must be excluded. The domain is x 僆 1⫺q, ⫺22 ´ 1⫺2, q 2. b. Since the radicand must be nonnegative, we solve the inequality 2x ⫹ 3 ⱖ 0, ⫺3 giving x ⱖ ⫺3 2 . The domain is x 僆 3 2 , q 2. c. To prevent division by zero, inputs of ⫺3 and 3 must be excluded (set x2 ⫺ 9 ⫽ 0 and solve by factoring). The domain is x 僆 1⫺q, ⫺32 ´ 1⫺3, 32 ´ 13, q 2 . Note that x ⫽ 1 is in the domain X⫺1 0 ⫽ 0 is defined. See Figure 1.51, where Y1 ⫽ 2 since ⫺8 . X ⫺9 d. Since squaring a number and multiplying a number by a constant are defined for all real numbers, the domain is x 僆 1⫺q, q 2. Now try Exercises 63 through 80 䊳

EXAMPLE 8

䊳

Determining Implied Domains Determine the domain of each function: 2x 14x ⫹ 5 7 7 a. For y ⫽ , we must have ⱖ 0 (for the radicand) and x ⫹ 3 ⫽ 0 Ax ⫹ 3 x⫹3 (for the denominator). Since the numerator is always positive, we need x ⫹ 3 7 0, which gives x 7 ⫺3. The domain is x 僆 1⫺3, q 2 . a. y ⫽

Solution

䊳

7 Ax ⫹ 3

b. y ⫽

2x , we must have 4x ⫹ 5 ⱖ 0 and 14x ⫹ 5 ⫽ 0. This shows 14x ⫹ 5 ⫺5 we need 4x ⫹ 5 7 0, so x 7 ⫺5 4 . The domain is x 僆 1 4 , q 2 .

b. For y ⫽ B. You’ve just seen how we can determine the domain and range of a function


C. Function Notation Figure 1.52 x

Input f Sequence of operations on x as defined by f

Output

y ⫽ f(x)

In our study of functions, you’ve likely noticed that the relationship between input and output values is an important one. To highlight this fact, think of a function as a simple machine, which can process inputs using a stated sequence of operations, then deliver a single output. The inputs are x-values, a program we’ll name f performs the operations on x, and y is the resulting output (see Figure 1.52). Once again we see that “the value of y depends on the value of x,” or simply “y is a function of x.” Notationally, we write “y is a function of x” as y ⫽ f 1x2 using function notation. You are already familiar with letting a variable represent a number. Here we do something quite different, as the letter f is used to represent a sequence of operations to be performed on x. Consider x x the function y ⫽ ⫹ 1, which we’ll now write as f 1x2 ⫽ ⫹ 1 3since y ⫽ f 1x24 . 2 2

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In words the function says, “divide inputs by 2, then add 1.” To evaluate the function at x ⫽ 4 (Figure 1.53) we have: input 4

Figure 1.53

input 4

S

S

f 1x2 ⫽

4

x ⫹1 2 4 f 142 ⫽ ⫹ 1 2 ⫽2⫹1 ⫽3

Input f Divide inputs by 2 then add 1: 4 +1 2

Output

3

Function notation enables us to summarize the three most important aspects of a function using a single expression, as shown in Figure 1.54. Figure 1.54 Output value y = f(x)

f(x) Sequence of operations to perform on the input

Input value

Instead of saying, “. . . when x ⫽ 4, the value of the function is 3,” we simply say “f of 4 is 3,” or write f 142 ⫽ 3. Note that the ordered pair (4, 3) is equivalent to (4, f (4)). 䊳

CAUTION

EXAMPLE 9

Solution

䊳

䊳

Although f(x) is the favored notation for a “function of x,” other letters can also be used. For example, g(x) and h(x) also denote functions of x, where g and h represent different sequences of operations on the x-inputs. It is also important to remember that these represent function values and not the product of two variables: f1x2 ⫽ f # 1x2 .

Evaluating a Function

Given f 1x2 ⫽ ⫺2x2 ⫹ 4x, find 7 a. f 1⫺22 b. f a b 2 a.

f 1x2 ⫽ ⫺2x2 ⫹ 4x f 1⫺22 ⫽ ⫺21⫺22 2 ⫹ 41⫺22 ⫽ ⫺8 ⫹ 1⫺82 ⫽ ⫺16

c. f 1x2 ⫽ ⫺2x2 ⫹ 4x f 12a2 ⫽ ⫺212a2 2 ⫹ 412a2 ⫽ ⫺214a2 2 ⫹ 8a ⫽ ⫺8a2 ⫹ 8a

c. f 12a2 b.

d.

d. f 1a ⫹ 12

f 1x2 ⫽ ⫺2x2 ⫹ 4x 7 7 2 7 f a b ⫽ ⫺2 a b ⫹ 4 a b 2 2 2 ⫺21 ⫺49 ⫹ 14 ⫽ or ⫺10.5 ⫽ 2 2

f 1x2 ⫽ ⫺2x2 ⫹ 4x f 1a ⫹ 12 ⫽ ⫺21a ⫹ 12 2 ⫹ 41a ⫹ 12 ⫽ ⫺21a2 ⫹ 2a ⫹ 12 ⫹ 4a ⫹ 4 ⫽ ⫺2a2 ⫺ 4a ⫺ 2 ⫹ 4a ⫹ 4 ⫽ ⫺2a2 ⫹ 2 Now try Exercises 87 through 102 䊳

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A graphing calculator can evaluate the function Y1 ⫽ ⫺2X2 ⫹ 4X using the TABLE feature, the TRACE feature, or function notation (on the home screen). The first two have been illustrated previously. To use function notation, we access the function names using the VARS key and right arrow to select Y-VARS (Figure 1.55). The 1:Function option is the default, so pressing will enable us to make our choice (Figure 1.56). In this case, we selected 1:Y1, which the calculator then places on the home screen, enabling us to enclose the desired input value in parentheses (function notation). Pressing ENTER completes the evaluation (Figure 1.57), which verifies the result from Example 9(b).

Figure 1.55

ENTER

Figure 1.56

Figure 1.57

C. You’ve just seen how we can use function notation and evaluate functions

D. Reading and Interpreting Information Given Graphically Graphs are an important part of studying functions, and learning to read and interpret them correctly is a high priority. A graph highlights and emphasizes the all-important input/output relationship that defines a function. In this study, we hope to firmly establish that the following statements are synonymous: 1. 2. 3. 4. EXAMPLE 10A

䊳

f 1⫺22 ⫽ 5 1⫺2, f 1⫺22 2 ⫽ 1⫺2, 52 1⫺2, 52 is on the graph of f, and when x ⫽ ⫺2, f 1x2 ⫽ 5

Reading a Graph For the functions f and g whose graphs are shown in Figures 1.58 and 1.59 a. State the domain of the function. b. Evaluate the function at x ⫽ 2. c. Determine the value(s) of x for which y ⫽ 3. d. State the range of the function. Figure 1.58 y

4

3

3

2

2

1

1 1

2

g

5

4

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1

䊳

y

f

5

Solution

Figure 1.59

3

4

5

x

⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1

⫺2

⫺2

⫺3

⫺3

1

2

3

4

5

x

For f, a. The graph is a continuous line segment with endpoints at (⫺4, ⫺3) and (5, 3), so we state the domain in interval notation. Using a vertical boundary line we note the smallest input is ⫺4 and the largest is 5. The domain is x 僆 3⫺4, 54 .

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b. The graph shows an input of x ⫽ 2 corresponds to y ⫽ 1: f (2) ⫽ 1 since (2, 1) is a point on the graph. c. For f (x) ⫽ 3 (or y ⫽ 3) the input value must be x ⫽ 5 since (5, 3) is the point on the graph. d. Using a horizontal boundary line, the smallest output value is ⫺3 and the largest is 3. The range is y 僆 3⫺3, 34 . For g, a. Since g is given as a set of plotted points, we state the domain as the set of first coordinates: D: 5⫺4, ⫺2, 0, 2, 46 . b. An input of x ⫽ 2 corresponds to y ⫽ 2: g(2) ⫽ 2 since (2, 2) is on the graph. c. For g(x) ⫽ 3 (or y ⫽ 3) the input value must be x ⫽ 4, since (4, 3) is a point on the graph. d. The range is the set of all second coordinates: R: 5⫺1, 0, 1, 2, 36.

EXAMPLE 10B

Solution

䊳

䊳

Reading a Graph

Use the graph of f 1x2 given to answer the following questions: a. What is the value of f 1⫺22 ? (⫺2, 4) b. What value(s) of x satisfy f 1x2 ⫽ 1?

y 5

f a. The notation f 1⫺22 says to find the value (0, 1) of the function f when x ⫽ ⫺2. Expressed (⫺3, 1) graphically, we go to x ⫽ ⫺2 and locate the ⫺5 corresponding point on the graph (blue arrows). Here we find that f 1⫺22 ⫽ 4. b. For f 1x2 ⫽ 1, we’re looking for x-inputs that result in an output of y ⫽ 1 3since y ⫽ f 1x2 4 . ⫺5 From the graph, we note there are two points with a y-coordinate of 1, namely, (⫺3, 1) and (0, 1). This shows f 1⫺32 ⫽ 1, f 102 ⫽ 1, and the required x-values are x ⫽ ⫺3 and x ⫽ 0.

5

x

Now try Exercises 103 through 108 䊳 In many applications involving functions, the domain and range can be determined by the context or situation given. EXAMPLE 11

䊳

Determining the Domain and Range from the Context Paul’s 2009 Voyager has a 20-gal tank and gets 18 mpg. The number of miles he can drive (his range) depends on how much gas is in the tank. As a function we have M1g2 ⫽ 18g, where M(g) represents the total distance in miles and g represents the gallons of gas in the tank (see graph). Find the domain and range.

Solution

D. You’ve just seen how we can read and interpret information given graphically

䊳

M 600 480

(20, 360) 360 240

120 Begin evaluating at x ⫽ 0, since the tank cannot (0, 0) hold less than zero gallons. With an empty tank, the 0 10 20 (minimum) range is M102 ⫽ 18102 or 0 miles. On a full tank, the maximum range is M1202 ⫽ 181202 or 360 miles. As shown in the graph, the domain is g 僆 [0, 20] and the corresponding range is M(g) 僆 [0, 360].

g


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1.3 EXERCISES 䊳



䊳

1. If a relation is given in ordered pair form, we state the domain by listing all of the coordinates in a set.

2. A relation is a function if each element of the is paired with element of the range.

3. The set of output values for a function is called the of the function.

4. Write using function notation: The function f evaluated at 3 is negative 5:

5. Discuss/Explain why the relation y ⫽ x2 is a function, while the relation x ⫽ y2 is not. Justify your response using graphs, ordered pairs, and so on.

6. Discuss/Explain the process of finding the domain and range of a function given its graph, using vertical and horizontal boundary lines. Include a few illustrative examples.


Determine whether the mappings shown represent functions or nonfunctions. If a nonfunction, explain how the definition of a function is violated.

7.

8.

9.

Woman

Country

Indira Gandhi Clara Barton Margaret Thatcher Maria Montessori Susan B. Anthony

Britain

Book

Author

Hawaii Roots Shogun 20,000 Leagues Under the Sea Where the Red Fern Grows

Rawls

Basketball star

10.

Country

Language

Canada Japan Brazil Tahiti Ecuador

Japanese Spanish French Portuguese English

U.S. Italy India

Verne Haley Clavell Michener Reported height

Determine whether the relations indicated represent functions or nonfunctions. If the relation is a nonfunction, explain how the definition of a function is violated.

11. (⫺3, 0), (1, 4), (2, ⫺5), (4, 2), (⫺5, 6), (3, 6), (0, ⫺1), (4, ⫺5), and (6, 1) 12. (⫺7, ⫺5), (⫺5, 3), (4, 0), (⫺3, ⫺5), (1, ⫺6), (0, 9), (2, ⫺8), (3, ⫺2), and (⫺5, 7) 13. (9, ⫺10), (⫺7, 6), (6, ⫺10), (4, ⫺1), (2, ⫺2), (1, 8), (0, ⫺2), (⫺2, ⫺7), and (⫺6, 4) 14. (1, ⫺81), (⫺2, 64), (⫺3, 49), (5, ⫺36), (⫺8, 25), (13, ⫺16), (⫺21, 9), (34, ⫺4), and (⫺55, 1) 15.

7'1" 6'6" 6'7" 6'9" 7'2"

y (⫺3, 5)

(2, 4)

(⫺3, 4)

Air Jordan The Mailman The Doctor The Iceman The Shaq

16.

y 5

(⫺1, 1)

5

(3, 4) (1, 3)

(4, 2)

(⫺5, 0) ⫺5

5 x

⫺5

5 x

(0, ⫺2) (5, ⫺3)

(⫺4, ⫺2) ⫺5

(1, ⫺4)

⫺5

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17.

18.

y 5

29.

y 5

(3, 4)

30.

y 5

y 5

(⫺3, 4)

(⫺2, 3)

(3, 3) (1, 2)

(⫺5, 1)

(1, 1)

⫺5

⫺5

5 x

(⫺2, ⫺4)

20.

y 5

⫺5

21.

5 x

22.

y 5

Graph each relation using a table, then use the vertical line test to determine if the relation is a function.

31. y ⫽ x

32. y ⫽ 1 x 3

34. x ⫽ 冟y ⫺ 2冟

Use an inequality to write a mathematical model for each statement, then write the relation in interval notation.

35. To qualify for a secretarial position, a person must type at least 45 words per minute.

⫺5

⫺5

⫺5

33. y ⫽ 1x ⫹ 22 2

y 5

⫺5

5 x

5 x

⫺5

⫺5

Determine whether or not the relations given represent a function. If not, explain how the definition of a function is violated.

19.

⫺5

5 x

(3, ⫺2)

(⫺1, ⫺4)

(4, ⫺5)

⫺5

⫺5

5 x

(⫺5, ⫺2)

36. The balance in a checking account must remain above $1000 or a fee is charged.

y 5

37. To bake properly, a turkey must be kept between the temperatures of 250° and 450°. ⫺5

⫺5

5 x

⫺5

23.

38. To fly effectively, the airliner must cruise at or between altitudes of 30,000 and 35,000 ft.

⫺5

24.

y 5

⫺5

5

⫺5

5 x

Graph each inequality on a number line, then write the relation in interval notation.

y

⫺5

25.

5 x

5 x

⫺5

26.

y 5

39. p 6 3

40. x 7 ⫺2

41. m ⱕ 5

42. n ⱖ ⫺4

43. x ⫽ 1

44. x ⫽ ⫺3

45. 5 7 x 7 2

46. ⫺3 6 p ⱕ 4

Write the domain illustrated on each graph in set notation and interval notation.

y 5

47. ⫺5

⫺5

5 x

5 x

48. 49.

⫺5

⫺5

50. 27.

28.

y 5

⫺5

5 x

⫺5

y

0

1

2

3

)

⫺3 ⫺2 ⫺1

0

[

1

2

3

[

⫺3 ⫺2 ⫺1

0

1

2

3

)

[

⫺3 ⫺2 ⫺1

0

1

2

3

4

Determine whether or not the relations indicated represent functions, then determine the domain and range of each.

5

⫺5

[

⫺3 ⫺2 ⫺1

5 x

51.

52.

y 5

y 5

⫺5 ⫺5

5 x

⫺5

⫺5

5 x

⫺5

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53.

54.

y 5

⫺5

y 5

⫺5

5 x

5 x

⫺5

55.

⫺5

56.

y 5

⫺5

y 5

⫺5

5 x

5 x

⫺5

57.

⫺5

58.

y 5

⫺5

5 x

⫺5

59.

60.

y

⫺5

5 x

62.

y

⫺5

5 x

x x ⫺ 3x ⫺ 10

78. y2 ⫽

x⫺4 x ⫹ 2x ⫺ 15

79. y ⫽

1x ⫺ 2 2x ⫺ 5

80. y ⫽

1x ⫹ 1 3x ⫹ 2

2

2

81. h1x2 ⫽

⫺2 1x ⫹ 4

82.

83. g1x2 ⫽

⫺4 A3 ⫺ x

84. p1x2 ⫽

⫺7 15 ⫺ x

85. r 1x2 ⫽

2x ⫺ 1 13x ⫺ 7

86. s1x2 ⫽

x2 ⫺ 4 111 ⫺ 2x

f 1x2 ⫽

y 5

⫺5

5 x

⫺5

1 87. f 1x2 ⫽ x ⫹ 3 2

2 88. f 1x2 ⫽ x ⫺ 5 3

90. f 1x2 ⫽ 2x2 ⫹ 3x

91. h1x2 ⫽

3 x

93. h1x2 ⫽

5冟x冟 x

2 x2 4冟x冟 94. h1x2 ⫽ x 92. h1x2 ⫽

⫺5

95. g1r2 ⫽ 2␲r

96. g1r2 ⫽ 2␲rh

97. g1r2 ⫽ ␲r

98. g1r2 ⫽ ␲r2h

2

Determine the value of p(5), p1 32 2, p(3a), and p(a ⴚ 1), then simplify.

99. p1x2 ⫽ 12x ⫹ 3

100. p1x2 ⫽ 14x ⫺ 1

3x ⫺ 5 x2 2

Determine the domain of the following functions, and write your response in interval notation.

3 63. f 1x2 ⫽ x⫺5

⫺2 64. g1x2 ⫽ 3⫹x

65. h1a2 ⫽ 13a ⫹ 5 66. p1a2 ⫽ 15a ⫺ 2 67. v1x2 ⫽ 69. u ⫽ 71. y ⫽

x⫹2 x2 ⫺ 25

5 Ax ⫺ 2

Determine the value of g(4), g 1 32 2, g(2c), and g(c ⴙ 3), then simplify.

⫺5

5

77. y1 ⫽

Determine the value of h(3), h1ⴚ23 2 , h(3a), and h(a ⫺ 2), then simplify.

y

⫺5

61.

76. y ⫽ 冟x ⫺ 2冟 ⫹ 3

5

⫺5

5 x

75. y ⫽ 2冟x冟 ⫹ 1

89. f 1x2 ⫽ 3x2 ⫺ 4x

⫺5

5

74. s ⫽ t2 ⫺ 3t ⫺ 10

For Exercises 87 through 102, determine the value of f 1ⴚ62, f 1 32 2, f 12c2, and f 1c ⴙ 12 , then simplify. Verify results using a graphing calculator where possible.

y 5

⫺5

5 x

73. m ⫽ n2 ⫺ 3n ⫺ 10

68. w1x2 ⫽

v⫺5 v2 ⫺ 18

70. p ⫽

17 x ⫹ 123 25

72. y ⫽

x⫺4 x2 ⫺ 49

101. p1x2 ⫽

102. p1x2 ⫽

2x2 ⫹ 3 x2

Use the graph of each function given to (a) state the domain, (b) state the range, (c) evaluate f (2), and (d) find the value(s) x for which f 1x2 ⴝ k (k a constant). Assume all results are integer-valued.

103. k ⫽ 4

104. k ⫽ 3 y

y

5

5

q⫹7 q2 ⫺ 12 11 x ⫺ 89 19

⫺5

5 x

⫺5

⫺5

5 x

⫺5

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105. k ⫽ 1

106. k ⫽ ⫺3

107. k ⫽ 2

y

⫺5

5

5 x

⫺5

⫺5

y

5

5 x

⫺5

⫺5

5

5 x

⫺5

⫺5

5 x

⫺5


9 109. Ideal weight for males: W1H2 ⴝ H ⴚ 151 2 The ideal weight for an adult male can be modeled by the function shown, where W is his weight in pounds and H is his height in inches. (a) Find the ideal weight for a male who is 75 in. tall. (b) If I am 72 in. tall and weigh 210 lb, how much weight should I lose? 5 110. Celsius to Fahrenheit conversions: C ⴝ 1F ⴚ 322 9 The relationship between Fahrenheit degrees and degrees Celsius is modeled by the function shown. (a) What is the Celsius temperature if °F ⫽ 41? (b) Use the formula to solve for F in terms of C, then substitute the result from part (a). What do you notice? 䊳

108. k ⫽ ⫺1 y

y

5

䊳

1–46


1 111. Pick’s theorem: A ⴝ B ⴙ I ⴚ 1 2 Pick’s theorem is an interesting yet little known formula for computing the area of a polygon drawn in the Cartesian coordinate system. The formula can be applied as long as the vertices of the polygon are lattice points (both x and y are integers). If B represents the number of lattice points lying directly on the boundary of the polygon (including the vertices), and I represents the number of points in the interior, the area of the polygon is given by the formula shown. Use some graph paper to carefully draw a triangle with vertices at 1⫺3, 12 , (3, 9), and (7, 6), then use Pick’s theorem to compute the triangle’s area.

APPLICATIONS

112. Gas mileage: John’s old ’87 LeBaron has a 15-gal gas tank and gets 23 mpg. The number of miles he can drive is a function of how much gas is in the tank. (a) Write this relationship in equation form and (b) determine the domain and range of the function in this context. 113. Gas mileage: Jackie has a gas-powered model boat with a 5-oz gas tank. The boat will run for 2.5 min on each ounce. The number of minutes she can operate the boat is a function of how much gas is in the tank. (a) Write this relationship in equation form and (b) determine the domain and range of the function in this context. 114. Volume of a cube: The volume of a cube depends on the length of the sides. In other words, volume is a function of the sides: V1s2 ⫽ s3. (a) In practical terms, what is the domain of this function? (b) Evaluate V(6.25) and (c) evaluate the function for s ⫽ 2x2. 115. Volume of a cylinder: For a fixed radius of 10 cm, the volume of a cylinder depends on its height. In other words, volume is a function of height:

V1h2 ⫽ 100␲h. (a) In practical terms, what is the domain of this function? (b) Evaluate V(7.5) and 8 (c) evaluate the function for h ⫽ . ␲ 116. Rental charges: Temporary Transportation Inc. rents cars (local rentals only) for a flat fee of $19.50 and an hourly charge of $12.50. This means that cost is a function of the hours the car is rented plus the flat fee. (a) Write this relationship in equation form; (b) find the cost if the car is rented for 3.5 hr; (c) determine how long the car was rented if the bill came to $119.75; and (d) determine the domain and range of the function in this context, if your budget limits you to paying a maximum of $150 for the rental. 117. Cost of a service call: Paul’s Plumbing charges a flat fee of $50 per service call plus an hourly rate of $42.50. This means that cost is a function of the hours the job takes to complete plus the flat fee. (a) Write this relationship in equation form; (b) find the cost of a service call that takes 212 hr; (c) find the number of hours the job took if the charge came to $262.50; and (d) determine the

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118. Predicting tides: The graph shown approximates the height of the tides at Fair Haven, New Brunswick, for a 12-hr period. (a) Is this the graph of a function? Why? (b) Approximately what time did high tide occur? (c) How high is the tide at 6 P.M.? (d) What time(s) will the tide be 2.5 m? 5

Meters

4

2 1

5

7

1.0

9

11

1 A.M.

6

8

10

12 2 A.M.

4

Time

3

Time


Distance in meters

120. A father challenges his son to a 400-m race, depicted in the graph shown here. 400 300 200 100 0

10

20

30

40

50

60

70

80

Time in seconds Father:

Son:

a. Who won and what was the approximate winning time? b. Approximately how many meters behind was the second place finisher? c. Estimate the number of seconds the father was in the lead in this race. d. How many times during the race were the father and son tied?

121. Sketch the graph of f 1x2 ⫽ x, then discuss how you could use this graph to obtain the graph of F1x2 ⫽ 冟x冟 without computing additional points. 冟x冟 What would the graph of g1x2 ⫽ look like? x 122. Sketch the graph of f 1x2 ⫽ x2 ⫺ 4, then discuss how you could use this graph to obtain the graph of F1x2 ⫽ 冟x2 ⫺ 4冟 without computing additional points. 冟x2 ⫺ 4冟 Determine what the graph of g1x2 ⫽ 2 would x ⫺4 look like. 123. If the equation of a function is given, the domain is implicitly defined by input values that generate real-valued outputs. But unless the graph is given or can be easily sketched, we must attempt to find the range analytically by solving for x in terms of y. We should note that sometimes this is an easy task, while at other times it is virtually impossible and we must rely on other methods. For the following functions, determine the implicit domain and find the range by solving for x in terms of y. a. y ⫽ xx

䊳

0.5

4 P.M.

3

3 P.M.

119. Predicting tides: The graph shown approximates the height of the tides at Apia, Western Samoa, for a 12-hr period. (a) Is this the graph of a function? Why? (b) Approximately what time did low tide occur? (c) How high is the tide at 2 A.M.? (d) What time(s) will the tide be 0.7 m? Meters

domain and range of the function in this context, if your insurance company has agreed to pay for all charges over $500 for the service call.

䊳

131


⫺ 3 ⫹ 2

b. y ⫽ x2 ⫺ 3


124. (1.1) Find the equation of a circle whose center is 14, ⫺12 with a radius of 5. Then graph the circle.

126. (R.4) Solve the equation by factoring, then check the result(s) using substitution: 3x2 ⫺ 4x ⫽ 7.

125. (R.6) Compute the sum and product indicated: a. 124 ⫹ 6 154 ⫺ 16 b. 12 ⫹ 132 12 ⫺ 132

127. (R.4) Factor the following polynomials completely: a. x3 ⫺ 3x2 ⫺ 25x ⫹ 75 b. 2x2 ⫺ 13x ⫺ 24 c. 8x3 ⫺ 125

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MID-CHAPTER CHECK Exercises 5 and 6 y L1

5

L2

2. Find the slope of the line passing through the given points: 1⫺3, 82 and 14, ⫺102 . 3. In 2009, Data.com lost $2 million. In 2010, they lost $0.5 million. Will the slope of the line through these points be positive or negative? Why? Calculate the slope. Were you correct? Write the slope as a unit rate and explain what it means in this context.

⫺5

5 x

⫺5

Exercises 7 and 8 y 5

h(x)

⫺5

5 x

⫺5

4. To earn some spending money, Sahara takes a job in a ski shop working primarily with her specialty—snowboards. She is paid a monthly salary of $950 pus a commission of $7.50 for each snowboard she sells. (a) Write a function that models her monthly earnings E. (b) Use a graphing calculator to determine her income if she sells 20, 30, or 40 snowboards in one month. (c) Use the results of parts a and b to set an appropriate viewing window and graph the line. (d) Use the TRACE feature to determine the number of snowboards that must be sold for Sahara’s monthly income to top $1300.

6. Write the equation for line L2 shown. Is this the graph of a function? Discuss why or why not. 7. For the graph of function h(x) shown, (a) determine the value of h(2); (b) state the domain; (c) determine the value(s) of x for which h1x2 ⫽ ⫺3; and (d) state the range. 8. Judging from the appearance of the graph alone, compare the rate of change (slope) from x ⫽ 1 to x ⫽ 2 to the rate of change from x ⫽ 4 to x ⫽ 5. Which rate of change is larger? How is that demonstrated graphically? Exercise 9 F 9. Compute the slope of the line F(p) shown, and explain what it means as a rate of change in this context. Then use the slope to predict the fox population when the pheasant population P is 13,000. Pheasant population (1000s) Fox population (in 100s)

1. Sketch the graph of the line 4x ⫺ 3y ⫽ 12. Plot and label at least three points.

10

9 8 7 6 5 4 3 2 1

0

1

2

3

4

5

6

7

8

9 10

10. State the domain and range for each function below. y y a. b. 5

5

⫺5

5 x

⫺5

5 x

⫺5

⫺5

y

c.

5

⫺5

5. Write the equation for line L1 shown. Is this the graph of a function? Discuss why or why not.

5 x

⫺5

REINFORCING BASIC CONCEPTS Finding the Domain and Range of a Relation from Its Graph The concepts of domain and range are an important and fundamental part of working with relations and functions. In this chapter, we learned to determine the domain of any relation from its graph using a “vertical boundary line,” and the range by using a “horizontal boundary line.” These approaches to finding the domain and range can be combined into a single step by envisioning a rectangle drawn around or about the graph. If the entire graph can be “bounded” within the rectangle, the domain and range can be based on the rectangle’s related length and width. If it’s impossible to bound the graph in a particular direction, the related x- or y-values continue infinitely. Consider the graph in Figure 1.60. This is the graph of an ellipse (Section 8.2), and a rectangle that bounds the graph in all directions is shown in Figure 1.61.

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1–49

Reinforcing Basic Concepts

Figure 1.60

Figure 1.61

y

y

10

10

8

8

6

6

4

4

2

2

⫺10⫺8 ⫺6 ⫺4 ⫺2 ⫺2

2

4

6

8 10

x

⫺10⫺8 ⫺6 ⫺4 ⫺2 ⫺2

⫺4

⫺4

⫺6

⫺6

⫺8

⫺8

⫺10

⫺10

2

4

6

8 10

x

The rectangle extends from x ⫽ ⫺3 to x ⫽ 9 in the horizontal direction, and from y ⫽ 1 to y ⫽ 7 in the vertical direction. The domain of this relation is x 僆 3⫺3, 9 4 and the range is y 僆 31, 7 4 . The graph in Figure 1.62 is a parabola, and no matter how large we draw the rectangle, an infinite extension of the graph will extend beyond its boundaries in the left and right directions, and in the upward direction (Figure 1.63). Figure 1.62 Figure 1.63 y

y

10

10

8

8

6

6

4

4

2

2

⫺10⫺8 ⫺6 ⫺4 ⫺2 ⫺2

2

4

6

8 10

x

⫺10⫺8 ⫺6 ⫺4 ⫺2 ⫺2

⫺4

⫺4

⫺6

⫺6

⫺8

⫺8

⫺10

⫺10

2

4

6

8 10

x

The domain of this relation is x 僆 1⫺q, q 2 and the range is y 僆 3⫺6, q 2 . Finally, the graph in Figure 1.64 is the graph of a square root function, and a rectangle can be drawn that bounds the graph below and to the left, but not above or to the right (Figure 1.65). Figure 1.64 Figure 1.65 y

y

10

10

8

8

6

6

4

4

2

2

⫺10⫺8 ⫺6 ⫺4 ⫺2 ⫺2

2

4

6

8 10

x

⫺10⫺8 ⫺6 ⫺4 ⫺2 ⫺2

⫺4

⫺4

⫺6

⫺6

⫺8

⫺8

⫺10

⫺10

2

4

6

8 10

x

The domain of this relation is x 僆 3 ⫺7, q 2 and the range is y 僆 3 ⫺5, q 2 . Use this approach to find the domain and range of the following relations and functions. Exercise 1:

Exercise 2:

y 10 8 6 4 2 ⫺10⫺8⫺6⫺4⫺2 ⫺2 ⫺4 ⫺6 ⫺8 ⫺10

⫺10⫺8⫺6⫺4⫺2 ⫺2 ⫺4 ⫺6 ⫺8 ⫺10

2 4 6 8 10 x

Exercise 4:

y 10 8 6 4 2

10 8 6 4 2 2 4 6 8 10 x

Exercise 3:

y

⫺10⫺8⫺6⫺4⫺2 ⫺2 ⫺4 ⫺6 ⫺8 ⫺10

y 10 8 6 4 2

2 4 6 8 10 x

⫺10⫺8⫺6⫺4⫺2 ⫺2 ⫺4 ⫺6 ⫺8 ⫺10

2 4 6 8 10 x

133

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1.4

Linear Functions, Special Forms, and More on Rates of Change


A. Write a linear equation in slope-intercept form and function form B. Use slope-intercept form to graph linear equations C. Write a linear equation in point-slope form D. Apply the slope-intercept form and point-slope form in context

EXAMPLE 1

䊳

The concept of slope is an important part of mathematics, because it gives us a way to measure and compare change. The value of an automobile changes with time, the circumference of a circle increases as the radius increases, and the tension in a spring grows the more it is stretched. The real world is filled with examples of how one change affects another, and slope helps us understand how these changes are related.

A. Linear Equations, Slope-Intercept Form and Function Form In Section 1.2, we learned that a linear equation is one that can be written in the form ax by c. Solving for y in a linear equation offers distinct advantages to understanding linear graphs and their applications.

Solving for y in a Linear Equation Solve 2y 6x 4 for y, then evaluate at x 4, x 0, and x 13.

Solution

䊳

2y 6x 4 2y 6x 4 y 3x 2

given equation add 6x divide by 2

Since the coefficients are integers, evaluate the function mentally. Inputs are multiplied by 3, then increased by 2, yielding the ordered pairs (4, 14), (0, 2), and 113, 12 . Now try Exercises 7 through 12

䊳

This form of the equation (where y has been written in terms of x) enables us to quickly identify what operations are performed on x in order to obtain y. Once again, for y 3x 2: multiply inputs by 3, then add 2. EXAMPLE 2

䊳

Solving for y in a Linear Equation Solve the linear equation 3y 2x 6 for y, then identify the new coefficient of x and the constant term.

Solution

䊳

3y 2x 6 3y 2x 6 2 y x2 3

given equation add 2x divide by 3

The coefficient of x is 23 and the constant term is 2. Now try Exercises 13 through 18

䊳

WORTHY OF NOTE In Example 2, the final form can be written y 23 x 2 as shown (inputs are multiplied by two-thirds, then increased by 2), or written as y 2x 3 2 (inputs are multiplied by two, the result divided by 3 and this amount increased by 2). The two forms are equivalent.

134

When the coefficient of x is rational, it’s helpful to select inputs that are multiples of the denominator if the context or application requires us to evaluate the equation. This enables us to perform most operations mentally. For y 23x 2, possible inputs might be x 9, 6, 0, 3, 6, and so on. See Exercises 19 through 24. In Section 1.2, linear equations were graphed using the intercept method. When the equation is written with y in terms of x, we notice a powerful connection between the graph and its equation—one that highlights the primary characteristics of a linear graph. 1–50

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Section 1.4 Linear Functions, Special Forms, and More on Rates of Change

EXAMPLE 3

䊳

Noting Relationships between an Equation and Its Graph Find the intercepts of 4x 5y 20 and use them to graph the line. Then, a. Use the intercepts to calculate the slope of the line, then identify the y-intercept. b. Write the equation with y in terms of x and compare the calculated slope and y-intercept to the equation in this form. Comment on what you notice.

Solution

䊳

Substituting 0 for x in 4x 5y 20, we find the y-intercept is 10, 42. Substituting 0 for y gives an x-intercept of 15, 02 . The graph is displayed here. a. The y-intercept is 10, 42 and by calculation or ¢y , the slope is m 4 counting 5 [from the ¢x intercept 15, 02 we count down 4, giving ¢y 4, and right 5, giving ¢x 5, to arrive at the intercept 10, 42 ]. b. Solving for y: given equation 4x 5y 20 5y 4x 20 subtract 4x 4 y x 4 divide by 5 5

y 5 4 3 2

(5, 0)

1

5 4 3 2 1 1

4

1

2

3

4

5

x

2 3

5

4

(0, 4)

5

The slope value seems to be the coefficient of x, while the y-intercept is the constant term. Now try Exercises 25 through 30

䊳

After solving a linear equation for y, an input of x 0 causes the “x-term” to become zero, so the y-intercept automatically involves the constant term. As Example 3 illustrates, we can also identify the slope of the line—it is the coefficient of x. In general, a linear equation of the form y mx b is said to be in slope-intercept form, since the slope of the line is m and the y-intercept is (0, b). Slope-Intercept Form For a nonvertical line whose equation is y mx b, the slope of the line is m and the y-intercept is (0, b). Solving a linear equation for y in terms of x is sometimes called writing the equation in function form, as this form clearly highlights what operations are performed on the input value in order to obtain the output (see Example 1). In other words, this form plainly shows that “y depends on x,” or “y is a function of x,” and that the equations y mx b and f 1x2 mx b are equivalent. Linear Functions A linear function is one of the form

f 1x2 mx b,

where m and b are real numbers. Note that if m 0, the result is a constant function f 1x2 b. If m 1 and b 0, the result is f 1x2 x, called the identity function.

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EXAMPLE 4

䊳

Finding the Function Form of a Linear Equation Write each equation in both slope-intercept form and function form. Then identify the slope and y-intercept of the line. a. 3x 2y 9 b. y x 5 c. 2y x

Solution

䊳

A. You’ve just seen how we can write a linear equation in slope-intercept form and function form

a. 3x 2y 9

b. y x 5

2y 3x 9

y x 5

3 9 y x 2 2 3 9 f 1x2 x 2 2 3 9 m ,b 2 2 9 y-intercept a0, b 2

y 1x 5 f 1x2 1x 5 m 1, b 5

c. 2y x x y 2 1 y x 2 1 f 1x2 x 2 1 m ,b0 2

y-intercept (0, 5)

y-intercept (0, 0)


䊳

Note that we can analytically develop the slope-intercept form of a line using the slope formula. Figure 1.66 shows the graph of a general line through the point (x, y) with a y-intercept of (0, b). Using these points in the slope formula, we have Figure 1.66

y2 y1 m x 2 x1

y 5

(x, y)

5

5

x

(0, b)

yb m x0 yb m x y b mx y mx b

5

slope formula

substitute: (0, b) for (x1, y1), (x, y) for (x2, y2)

simplify multiply by x add b to both sides

This approach confirms the relationship between the graphical characteristics of a line and its slope-intercept form. Specifically, for any linear equation written in the form y mx b, the slope must be m and the y-intercept is (0, b).

B. Slope-Intercept Form and the Graph of a Line If the slope and y-intercept of a linear equation are known or can be found, we can construct its equation by substituting these values directly into the slope-intercept form y mx b. EXAMPLE 5

䊳

y

Finding the Equation of a Line from Its Graph

5

Find the slope-intercept equation of the line shown.

Solution

䊳

Using 13, 22 and 11, 22 in the slope formula, ¢y or by simply counting , the slope is m 42 or 21. ¢x By inspection we see the y-intercept is (0, 4). Substituting 21 for m and 4 for b in the slopeintercept form we obtain the equation y 2x 4.

(1, 2)

5

5

x

(3, 2) 5


䊳

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Actually, if the slope is known and we have any point (x, y) on the line, we can still construct the equation since the given point must satisfy the equation of the line. In this case, we’re treating y mx b as a simple formula, solving for b after substituting known values for m, x, and y. EXAMPLE 6

䊳

Using y ⫽ mx ⫹ b as a Formula Find the slope-intercept equation of a line that has slope m 45 and contains 15, 22. Verify results on a graphing calculator.

Solution

䊳

10

Use y mx b as a “formula,” with m 45, x 5, and y 2. y mx b 2 45 152 b 2 4 b 6b

10

slope-intercept form

10

substitute 45 for m, 5 for x, and 2 for y simplify 10 (5, 2) is on the line

solve for b

The equation of the line is y 6. After entering the equation on the Y= screen of a graphing calculator, we can evaluate x 5 on the home screen, or use the TRACE feature. See the figures provided. 4 5x


䊳

Writing a linear equation in slope-intercept form enables us to draw its graph with a minimum of effort, since we can easily locate the y-intercept and a second point using ¢y ¢y 2 . For instance, the rate of change indicates that counting down 2 and ¢x ¢x 3 right 3 from a known point will locate another point on this line. EXAMPLE 7

䊳

Graphing a Line Using Slope-Intercept Form and the Rate of Change Write 3y 5x 9 in slope-intercept form, then graph the line using the y-intercept and the rate of change (slope).

Solution

WORTHY OF NOTE Noting the fraction 53 is equal to 5 3 , we could also begin at (0, 3) and ¢y 5 count (down 5 and left 3) ¢x 3 to find an additional point on the line: 13, 22 . Also, for any ¢y a , note negative slope ¢x b a a a . b b b

䊳

3y 5x 9 3y 5x 9 y 53x 3

y fx 3 y

Run 3

given equation isolate y term Rise 5

divide by 3

The slope is m 53 and the y-intercept is (0, 3). ¢y 5 (up 5 and Plot the y-intercept, then use ¢x 3 right 3 — shown in blue) to find another point on the line (shown in red). Finish by drawing a line through these points.

(3, 8)

y f x (0, 3)

5

5

x

2


䊳

For a discussion of what graphing method might be most efficient for a given linear equation, see Exercises 103 and 114.

Parallel and Perpendicular Lines From Section 1.2 we know parallel lines have equal slopes: m1 m2, and perpendicular 1 lines have slopes with a product of 1: m1 # m2 1 or m1 . In some applications, m2 we need to find the equation of a second line parallel or perpendicular to a given line, through a given point. Using the slope-intercept form makes this a simple four-step process.

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Finding the Equation of a Line Parallel or Perpendicular to a Given Line 1. Identify the slope m1 of the given line. 2. Find the slope m2 of the new line using the parallel or perpendicular relationship. 3. Use m2 with the point (x, y) in the “formula” y mx b and solve for b. 4. The desired equation will be y m2 x b.

EXAMPLE 8

䊳

Finding the Equation of a Parallel Line

Solution

䊳

Begin by writing the equation in slope-intercept form to identify the slope.

Find the slope-intercept equation of a line that goes through 16, 12 and is parallel to 2x 3y 6. 2x 3y 6 3y 2x 6 y 2 3 x 2

given line isolate y-term result

The original line has slope m1 2 3 and this will also be the slope of any line parallel to it. Using m2 2 3 with 1x, y2 S 16, 12 we have y mx b 2 162 b 1 3 1 4 b 5 b

The equation of the new line is y

2 3 x

slope-intercept form substitute 2 3 for m, 6 for x, and 1 for y simplify solve for b

5. Now try Exercises 63 through 76

䊳

31 Graphing the lines from Example 8 as Y1 and Y2 on a graphing calculator, we note the lines do appear to be parallel (they actually must be since they have identical slopes). Using the 47 ZOOM 8:ZInteger feature of the calculator, we 47 can quickly verify that Y2 indeed contains the point (6, 1). For any nonlinear graph, a straight line 31 drawn through two points on the graph is called a secant line. The slope of a secant line, and lines parallel and perpendicular to this line, play fundamental roles in the further development of the rate-of-change concept.

EXAMPLE 9

䊳

Finding Equations for Parallel and Perpendicular Lines A secant line is drawn using the points (4, 0) and (2, 2) on the graph of the function shown. Find the equation of a line that is a. parallel to the secant line through (1, 4). b. perpendicular to the secant line through (1, 4).

Solution

䊳

¢y Either by using the slope formula or counting , we find the secant line has slope ¢x 2 1 m . 6 3

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WORTHY OF NOTE The word “secant” comes from the Latin word secare, meaning “to cut.” Hence a secant line is one that cuts through a graph, as opposed to a tangent line, which touches the graph at only one point.

a. For the parallel line through (1, 4), m2 y mx b 1 112 b 4 3 12 1 b 3 3 13 b 3

1 . 3

y 5

slope-intercept form substitute 1 3 for m, 1 for x, and 4 for y

5

(1, 4) result

x

5

x

5

13 1 . x 3 3

b. For the perpendicular line through (1, 4), m2 3. y mx b 4 3112 b 4 3 b 1 b

5

simplify 14 12 32

The equation of the parallel line (in blue) is y

y 5

slope-intercept form substitute 3 for m, 1 for x, and 4 for y simplify

5

result

The equation of the perpendicular line (in yellow) is y 3x 1. B. You’ve just seen how we can use the slope-intercept form to graph linear equations

139


(1, 4)

5


䊳

C. Linear Equations in Point-Slope Form As an alternative to using y mx b, we can find the equation of the line using the y2 y1 m, and the fact that the slope of a line is constant. For a given slope formula x2 x1 slope m, we can let (x1, y1) represent a given point on the line and (x, y) represent any y y1 m. Isolating the “y” terms other point on the line, and the formula becomes x x1 on one side gives a new form for the equation of a line, called the point-slope form: y y1 m x x1 1x x1 2 y y1 a b m1x x1 2 x x1 1 y y1 m1x x1 2

slope formula

multiply both sides by (x x1) simplify S point-slope form

The Point-Slope Form of a Linear Equation

For a nonvertical line whose equation is y y1 m1x x1 2 , the slope of the line is m and (x1, y1) is a point on the line.

While using y mx b (as in Example 6) may appear to be easier, both the slope-intercept form and point-slope form have their own advantages and it will help to be familiar with both.

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EXAMPLE 10

䊳

Using y y1 m(x x1) as a Formula Find the equation of the line in point-slope form, if m 23 and (3, 3) is on the line. Then graph the line. y

Solution

䊳

C. You’ve just seen how we can write a linear equation in point-slope form

y y1 m1x x1 2 2 y 132 3 x 132 4 3 2 y 3 1x 32 3

5

point-slope form

y 3 s (x 3)

substitute 23 for m; (3, 3) for (x1, y1) simplify, point-slope form

¢y 2 To graph the line, plot (3, 3) and use ¢x 3 to find additional points on the line.

x3

5

5

x

y2 (3, 3) 5


䊳

D. Applications of Linear Equations As a mathematical tool, linear equations rank among the most common, powerful, and versatile. In all cases, it’s important to remember that slope represents a rate of change. ¢y The notation m literally means the quantity measured along the y-axis, is chang¢x ing with respect to changes in the quantity measured along the x-axis. EXAMPLE 11

䊳

Relating Temperature to Altitude In meteorological studies, atmospheric temperature depends on the altitude according to the formula T1h2 3.5h 58.5, where T(h) represents the approximate Fahrenheit temperature at height h (in thousands of feet, 0 h 36). a. Interpret the meaning of the slope in this context. b. Determine the temperature at an altitude of 12,000 ft. c. If the temperature is 8°F what is the approximate altitude?

Algebraic Solution

䊳

3.5 ¢T , ¢h 1 meaning the temperature drops 3.5°F for every 1000-ft increase in altitude. b. Since height is in thousands, use h 12. a. Notice that h is the input variable and T is the output. This shows

T1h2 3.5h 58.5 T1122 3.51122 58.5 16.5

Technology Solution

original formula substitute 12 for h result

䊳

At a height of 12,000 ft, the temperature is about 16.5°.

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c. Replacing T(h) with 8 and solving gives

Algebraic Solution

䊳

T1h2 3.5h 58.5 8 3.5h 58.5 66.5 3.5h 19 h

original formula substitute 8 for T(h) subtract 58.5 divide by 3.5

The temperature is about 8°F at a height of 19 1000 19,000 ft.

Graphical Solution

䊳

Since we’re given 0 h 36, we can set Xmin 0 and Xmax 40. At ground level 1x 02 , the formula gives a temperature of 58.5°, while at h 36, we have T1362 67.5. This shows appropriate settings for the range would be Ymin 50 and Ymax 50 (see figure). After setting Y1 3.5X 58.5, we press TRACE and move the cursor until we find an output value near 8, which occurs when X is near 19. To check, we input 19 for x and the calculator displays an output of 8, which corresponds with the algebraic result (at 19,000 ft, the temperature is 8°F). 50

0

40

50


䊳

In many applications, outputs that are integer or rational values are rare, making it difficult to use the TRACE feature alone to find an exact solution. In the Section 1.5, we’ll develop additional ways that graphs and technology can be used to solve equations. In some applications, the relationship is known to be linear but only a few points on the line are given. In this case, we can use two of the known data points to calculate the slope, then the point-slope form to find an equation model. One such application is linear depreciation, as when a government allows businesses to depreciate vehicles and equipment over time (the less a piece of equipment is worth, the less you pay in taxes). EXAMPLE 12A

䊳

Using Point-Slope Form to Find a Function Model Five years after purchase, the auditor of a newspaper company estimates the value of their printing press is $60,000. Eight years after its purchase, the value of the press had depreciated to $42,000. Find a linear equation that models this depreciation and discuss the slope and y-intercept in context.

Solution

䊳

Since the value of the press depends on time, the ordered pairs have the form (time, value) or (t, v) where time is the input, and value is the output. This means the ordered pairs are (5, 60,000) and (8, 42,000). v2 v1 t2 t1 42,000 60,000 85 18,000 6000 3 1

m

slope formula 1t1, v1 2 15, 60,0002; 1t2, v2 2 18, 42,0002 simplify and reduce

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6000 ¢value , indicating the printing press loses ¢time 1 $6000 in value with each passing year. The slope of the line is

WORTHY OF NOTE Actually, it doesn’t matter which of the two points are used in Example 12A. Once the point (5, 60,000) is plotted, a constant slope of m 6000 will “drive” the line through (8, 42,000). If we first graph (8, 42,000), the same slope would “drive” the line through (5, 60,000). Convince yourself by reworking the problem using the other point.

v v1 m1t t1 2 v 60,000 60001t 52 v 60,000 6000t 30,000 v 6000t 90,000

point-slope form substitute 6000 for m; (5, 60,000) for (t1, v1) simplify solve for v

The depreciation equation is v1t2 6000t 90,000. The v-intercept (0, 90,000) indicates the original value (cost) of the equipment was $90,000. Once the depreciation equation is found, it represents the (time, value) relationship for all future (and intermediate) ages of the press. In other words, we can now predict the value of the press for any given year. However, note that some equation models are valid for only a set period of time, and each model should be used with care.

EXAMPLE 12B

䊳

Using a Function Model to Gather Information From Example 12A, a. How much will the press be worth after 11 yr? b. How many years until the value of the equipment is $9000? c. Is this function model valid for t 18 yr (why or why not)?

Solution

䊳

a. Find the value v when t 11: v1t2 6000t 90,000 v1112 60001112 90,000 24,000

equation model substitute 11 for t result (11, 24,000)

After 11 yr, the printing press will only be worth $24,000. b. “. . . value is $9000” means v1t2 9000: v1t2 9000 6000t 90,000 9000 6000t 81,000 t 13.5

D. You’ve just seen how we can apply the slopeintercept form and point-slope form in context

value at time t substitute 6000t 90,000 for v (t ) subtract 90,000 divide by 6000

After 13.5 yr, the printing press will be worth $9000. c. Since substituting 18 for t gives a negative quantity, the function model is not valid for t 18. In the current context, the model is only valid while v 0 and solving 6000t 90,000 0 shows the domain of the function in this context is t 30, 15 4 . Now try Exercises 107 through 112

䊳

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1.4 EXERCISES 䊳



7 x 3, the slope is 4 and the y-intercept is .

1. For the equation y

3. Line 1 has a slope of 0.4. The slope of any line perpendicular to line 1 is . 5. Discuss/Explain how to graph a line using only the slope and a point on the line (no equations).

䊳

¢cost indicates the ¢time changing in response to changes in

2. The notation

is .

4. The equation y y1 m1x x1 2 is called the form of a line.

6. Given m 35 and 15, 62 is on the line. Compare and contrast finding the equation of the line using y mx b versus y y1 m1x x1 2.


Solve each equation for y and evaluate the result using x 5, x 2, x 0, x 1, and x 3.

7. 4x 5y 10

8. 3y 2x 9

9. 0.4x 0.2y 1.4 10. 0.2x 0.7y 2.1 11.

1 3x

1 5y

1

12.

1 7y

1 3x

2

For each equation, solve for y and identify the new coefficient of x and new constant term.

13. 6x 3y 9

14. 9y 4x 18

15. 0.5x 0.3y 2.1 16. 0.7x 0.6y 2.4 17. 56x 17y 47

18.

7 12 y

4 15 x 76

Write each equation in slope-intercept form (solve for y) and function form, then identify the slope and y-intercept.

31. 2x 3y 6

32. 4y 3x 12

33. 5x 4y 20

34. y 2x 4

35. x 3y

36. 2x 5y

37. 3x 4y 12 0

38. 5y 3x 20 0

For Exercises 39 to 50, use the slope-intercept form to state the equation of each line. Verify your solutions to Exercises 45 to 47 using a graphing calculator.

39.

Evaluate each equation by selecting three inputs that will result in integer values. Then graph each line.

19. y 43x 5

20. y 54x 1

21. y 32x 2

22. y 25x 3

23. y 16x 4

24. y 13x 3

Find the x- and y-intercepts for each line, then (a) use these two points to calculate the slope of the line, (b) write the equation with y in terms of x (solve for y) and (c) compare the calculated slope and y-intercept to the equation from part (b). Comment on what you notice.

25. 3x 4y 12

26. 3y 2x 6

27. 2x 5y 10

28. 2x 3y 9

29. 4x 5y 15

30. 5y 6x 25

40.

y 5 4 3 2 1 54321 1 2 (3, 1) 3 4 5

41.

(5, 5)

(3, 3) (0, 1) 1 2 3 4 5 x

54321 1 2 3 4 5

y

(1, 0)

5 4 3 (0, 3) 2 1

54321 1 2 (2, 3) 3 4 5

y 5 4 (0, 3) 3 2 1

1 2 3 4 5 x

42. m 2; y-intercept 10, 32 43. m 3; y-intercept (0, 2)

44. m 3 2 ; y-intercept 10, 42

45. m 4; 13, 22 is on the line

(5, 1)

1 2 3 4 5 x

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144

1–60


46. m 2; 15, 32 is on the line 47. m

3 2 ;

48.

y

50.

14, 72 is on the line 49.

10,000

y 1500

8000

1200

6000

900

4000

600

2000

300 12

Write the equations in slope-intercept form and state whether the lines are parallel, perpendicular, or neither.

14

16

18

20 x

26

28

30

32

71. 4y 5x 8 5y 4x 15

72. 3y 2x 6 2x 3y 6

73. 2x 5y 20 4x 3y 18

74. 4x 6y 12 2x 3y 6

75. 3x 4y 12 6x 8y 2

76. 5y 11x 135 11y 5x 77

34 x

A secant line is one that intersects a graph at two or more points. For each graph given, find an equation of the line (a) parallel and (b) perpendicular to the secant line, through the point indicated.

y 2000 1600 1200 800

77.

400

78.

y 5

y 5

(1, 3) 8

10

12

14

16 x

Write each equation in slope-intercept form, then use the rate of change (slope) and y-intercept to graph the line.

51. 3x 5y 20

52. 2y x 4

53. 2x 3y 15

54. 3x 2y 4

5

55. y 23x 3

56. y 52x 1

57. y 1 3 x 2

58. y 4 5 x 2

59. y 2x 5

60. y 3x 4

61. y 12x 3

62. y 3 2 x 2

(2, 4)

5

79.

5 x

5

80.

y 5

Graph each linear equation using the y-intercept and rate of change (slope) determined from each equation.

5

5 x

y 5

(1, 3)

5

5

5 x

5

81.

5 x

5

82.

y 5

(1, 2.5)

y 5

(1, 3)

Find the equation of the line using the information given. Write answers in slope-intercept form.

63. parallel to 2x 5y 10, through the point 15, 22

64. parallel to 6x 9y 27, through the point 13, 52

65. perpendicular to 5y 3x 9, through the point 16, 32 66. perpendicular to x 4y 7, through the point 15, 32 67. parallel to 12x 5y 65, through the point 12, 12

68. parallel to 15y 8x 50, through the point 13, 42 69. parallel to y 3, through the point (2, 5)

70. perpendicular to y 3 through the point (2, 5)

5

5 x

5

5 x

(0, 2) 5

5

Find the equation of the line in point-slope form, then graph the line.

83. m 2; P1 12, 52

84. m 1; P1 12, 32

85. P1 13, 42, P2 111, 12 86. P1 11, 62, P2 15, 12

87. m 0.5; P1 11.8, 3.12

88. m 1.5; P1 10.75, 0.1252

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1–61

Find the equation of the line in point-slope form, and state the meaning of the slope in context — what information is the slope giving us?

89.

90.

y Typewriters in service (in ten thousands)

Income (in thousands)

y 10 9 8 7 6 5 4 3 2 1 0

x

1 2 3 4 5 6 7 8 9

10 9 8 7 6 5 4 3 2 1 0

Student’s final grade (%) (includes extra credit)

x Hours of television per day 0

93.

0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

60 40 20

1

2

3

4

Rainfall per month (in inches)

5

x

y D

x

x y G

x

x y H

x

x

95. While driving today, I got stopped by a state trooper. After she warned me to slow down, I continued on my way.

x Independent investors (1000s) 1 2 3 4 5 6 7 8 9 10

y Eggs per hen per week

Cattle raised per acre

80

y C

y F

y E

x

10 9 8 7 6 5 4 3 2 1 0

94.

y 100

0

x

1 2 3 4 5 6 7 8 9

y Online brokerage houses

92.

y

y B

x

Year (1990 → 0)

100 90 80 70 60 50 40 30 20 10

Using the concept of slope, match each description with the graph that best illustrates it. Assume time is scaled on the horizontal axes, and height, speed, or distance from the origin (as the case may be) is scaled on the vertical axis. y A

Sales (in thousands)

91.

145


96. After hitting the ball, I began trotting around the bases shouting, “Ooh, ooh, ooh!” When I saw it wasn’t a home run, I began sprinting. 97. At first I ran at a steady pace, then I got tired and walked the rest of the way.

10 8 6

98. While on my daily walk, I had to run for a while when I was chased by a stray dog.

4 2 0

60

65

70

75

80

x

Temperature in °F

99. I climbed up a tree, then I jumped out. 100. I steadily swam laps at the pool yesterday. 101. I walked toward the candy machine, stared at it for a while then changed my mind and walked back. 102. For practice, the girls’ track team did a series of 25-m sprints, with a brief rest in between.

䊳


103. General linear equation: ax by c The general equation of a line is shown here, where a, b, and c are real numbers, with a and b not simultaneously zero. Solve the equation for y and note the slope (coefficient of x) and y-intercept (constant term). Use these to find the slope and y-intercept of the following lines, without solving for y or computing points. a. 3x 4y 8 b. 2x 5y 15 c. 5x 6y 12 d. 3y 5x 9

104. Intercept-Intercept form of a linear y x equation: 1 h k The x- and y-intercepts of a line can also be found by writing the equation in the form shown (with the equation set equal to 1). The x-intercept will be (h, 0) and the y-intercept will be (0, k). Find the x- and y-intercepts of the following lines using this method. How is the slope of each line related to the values of h and k? a. 2x 5y 10 b. 3x 4y 12 c. 5x 4y 8

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146 䊳

1–62


APPLICATIONS

105. Speed of sound: The speed of sound as it travels through the air depends on the temperature of the air according to the function V 35T 331, where V represents the velocity of the sound waves in meters per second (m/s), at a temperature of T° Celsius. (a) Interpret the meaning of the slope and y-intercept in this context. (b) Determine the speed of sound at a temperature of 20°C. (c) If the speed of sound is measured at 361 m/s, what is the temperature of the air? 106. Acceleration: A driver going down a straight highway is traveling 60 ft/sec (about 41 mph) on cruise control, when he begins accelerating at a rate of 5.2 ft/sec2. The final velocity of the car is given by V 26 5 t 60, where V is the velocity at time t. (a) Interpret the meaning of the slope and y-intercept in this context. (b) Determine the velocity of the car after 9.4 seconds. (c) If the car is traveling at 100 ft/sec, for how long did it accelerate? 107. Investing in coins: The purchase of a “collector’s item” is often made in hopes the item will increase in value. In 1998, Mark purchased a 1909-S VDB Lincoln Cent (in fair condition) for $150. By the year 2004, its value had grown to $190. (a) Use the relation (time since purchase, value) with t 0 corresponding to 1998 to find a linear equation modeling the value of the coin. (b) Discuss what the slope and y-intercept indicate in this context. (c) How much was the penny worth in 2009? (d) How many years after purchase will the penny’s value exceed $250? (e) If the penny is now worth $170, how many years has Mark owned the penny? 108. Depreciation: Once a piece of equipment is put into service, its value begins to depreciate. A business purchases some computer equipment for $18,500. At the end of a 2-yr period, the value of the equipment has decreased to $11,500. (a) Use the relation (time since purchase, value) to find a linear equation modeling the value of the equipment. (b) Discuss what the slope and y-intercept indicate in this context. (c) What is the equipment’s value after 4 yr? (d) How many years after purchase will the value decrease to $6000? (e) Generally, companies will sell used equipment while it still has value and use the funds to purchase new equipment. According to the function, how many years will it take this equipment to depreciate in value to $1000?

109. Internet connections: The number of households that are hooked up to the Internet (homes that are online) has been increasing steadily in recent years. In 1995, approximately 9 million homes were online. By 2001 this figure had climbed to about 51 million. (a) Use the relation (year, homes online) with t 0 corresponding to 1995 to find an equation model for the number of homes online. (b) Discuss what the slope indicates in this context. (c) According to this model, in what year did the first homes begin to come online? (d) If the rate of change stays constant, how many households were on the Internet in 2006? (e) How many years after 1995 will there be over 100 million households connected? (f) If there are 115 million households connected, what year is it? Source: 2004 Statistical Abstract of the United States, Table 965

110. Prescription drugs: Retail sales of prescription drugs have been increasing steadily in recent years. In 1995, retail sales hit $72 billion. By the year 2000, sales had grown to about $146 billion. (a) Use the relation (year, retail sales of prescription drugs) with t 0 corresponding to 1995 to find a linear equation modeling the growth of retail sales. (b) Discuss what the slope indicates in this context. (c) According to this model, in what year will sales reach $250 billion? (d) According to the model, what was the value of retail prescription drug sales in 2005? (e) How many years after 1995 will retail sales exceed $279 billion? (f) If yearly sales totaled $294 billion, what year is it? Source: 2004 Statistical Abstract of the United States, Table 122

111. Prison population: In 1990, the number of persons sentenced and serving time in state and federal institutions was approximately 740,000. By the year 2000, this figure had grown to nearly 1,320,000. (a) Find a linear function with t 0 corresponding to 1990 that models this data, (b) discuss the slope ratio in context, and (c) use the equation to estimate the prison population in 2010 if this trend continues. Source: Bureau of Justice Statistics at www.ojp.usdoj.gov/bjs

112. Eating out: In 1990, Americans bought an average of 143 meals per year at restaurants. This phenomenon continued to grow in popularity and in the year 2000, the average reached 170 meals per year. (a) Find a linear function with t 0 corresponding to 1990 that models this growth, (b) discuss the slope ratio in context, and (c) use the equation to estimate the average number of times an American will eat at a restaurant in 2010 if the trend continues. Source: The NPD Group, Inc., National Eating Trends, 2002

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113. Locate and read the following article. Then turn in a one-page summary. “Linear Function Saves Carpenter’s Time,” Richard Crouse, Mathematics Teacher, Volume 83, Number 5, May 1990: pp. 400–401. 114. The general form of a linear equation is ax by c, where a and b are not simultaneously zero. (a) Find the x- and y-intercepts using the general form (substitute 0 for x, then 0 for y). Based on what you see, when does the intercept method work most efficiently? (b) Find the slope and y-intercept using the general form (solve for y). Based on what you see, when does the slopeintercept method work most efficiently?

115. Match the correct graph to the conditions stated for m and b. There are more choices than graphs. a. m 6 0, b 6 0 b. m 7 0, b 6 0 c. m 6 0, b 7 0 d. m 7 0, b 7 0 e. m 0, b 7 0 f. m 6 0, b 0 g. m 7 0, b 0 h. m 0, b 6 0 (1)

y

(2)

y

x

(4)

y

y

(3)

x

(5)

x

䊳

147


y

x

y

(6)

x

x


116. (1.3) Determine the domain: a. y 12x 5 5 b. y 2x 5

119. (R.2) Compute the area of the circular sidewalk shown here 1A ␲r2 2 . Use your calculator’s value of ␲ and round the answer (only) to hundredths. 10 yd

117. (R.6) Simply without the use of a calculator. 2 a. 273 b. 281x2 118. (R.3) Three equations follow. One is an identity, another is a contradiction, and a third has a solution. State which is which. 21x 52 13 1 9 7 2x 21x 42 13 1 9 7 2x 21x 52 13 1 9 7 2x

8 yd

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1.5

Solving Equations and Inequalities Graphically; Formulas and Problem Solving


A. Solve equations

B.

C. D.

E.

graphically using the intersection-of-graphs method Solve equations graphically using the x-intercept/zeroes method Solve linear inequalities graphically Solve for a specified variable in a formula or literal equation Use a problem-solving guide to solve various problem types

In this section, we’ll build on many of the ideas developed in Section R.3 (Solving Linear Equations and Inequalities), as we learn to manipulate formulas and employ certain problem-solving strategies. We will also extend our understanding of graphical solutions to a point where they can be applied to virtually any family of functions.

A. Solving Equations Graphically Using the Intersect Method For some background on why a graphical solution is effective, consider the equation 2x 9 31x 12 2. By definition, an equation is a statement that two expressions are equal for some value of the variable (Section R.3). To highlight this fact, the expressions 2x 9 and 31x 12 2 are evaluated independently for selected integers in Tables 1.4 and 1.5. Table 1.4

Table 1.5

x

2x ⫺ 9

x

⫺3(x ⫺ 1) ⫺ 2

3

15

3

10

2

13

2

7

1

11

1

4

0

9

0

1

1

7

1

2

2

5

2

5

3

3

3

8

Note the two expressions are equal (the equation is true) only when the input is x 2. Solving equations graphically is a simple extension of this observation. By treating the expression on the left as the independent function Y1, we have Y1 2X 9 and the related linear graph will contain all ordered pairs shown in Table 1.4 (see Figure 1.67). Doing the same for the right-hand expression yields Y2 31X 12 2, and its related graph will likewise contain all ordered pairs shown in the Table 1.5 (see Figure 1.68).

f

∂

2x 9 31x 12 2 Y1

Y2

The solution is then found where Y1 Y2, or in other words, at the point where these two lines intersect (if it exists). See Figure 1.69. Most graphing calculators have an intersect feature that can quickly find the point(s) where two graphs intersect. On many calculators, we access this ability using the sequence 2nd TRACE (CALC) and selecting option 5:intersect (Figure 1.70).

10

10

10

10

10

148

Figure 1.69

Figure 1.68

Figure 1.67

10

10

10

10

10

10

10

1–64

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Section 1.5 Solving Equations and Inequalities Graphically; Formulas and Problem Solving

Figure 1.71 Figure 1.70

149

Figure 1.72 10

10

10

10

10

10

10

10

Because the calculator can work with up to 10 Figure 1.73 expressions at once, it will ask you to identify 10 each graph you want to work with—even when there are only two. A marker is displayed on each graph in turn, and named in the upper left 10 corner of the window (Figure 1.71). You can 10 select a graph by pressing , or bypass a graph by pressing one of the arrow keys. For situations involving multiple graphs or multiple 10 solutions, the calculator offers a “GUESS?” option that enables you to specify the approximate location of the solution you’re interested in (Figure 1.72). For now, we’ll simply press two times in succession to identify each graph, and a third time to bypass the “GUESS?” option. The calculator then finds and displays the point of intersection (Figure 1.73). Be sure to check the settings on your viewing window before you begin, and if the point of intersection is not visible, try ZOOM 3:Zoom Out or other windowresizing features to help locate it. ENTER

ENTER

EXAMPLE 1A

䊳

Solving an Equation Graphically 1 Solve the equation 21x 32 7 x 2 using 2 a graphing calculator.

Solution

䊳

Begin by entering the left-hand expression as Y1 and the right-hand expression as Y2 (Figure 1.74). To find points of intersection, press 2nd TRACE (CALC) and select option 5:intersect, which automatically places you on the graphing window, and asks you to identify the “First curve?.” As discussed, pressing three times in succession will identify each graph, bypass the “Guess?” option, then find and display the point of intersection (Figure 1.75). Here the point of intersection 10 is (2, 3), showing the solution to this equation is x 2 (for which both expressions equal 3). This can be verified by direct substitution or by using the TABLE feature. ENTER

Figure 1.74

Figure 1.75 10

10

10

This method of solving equations is called the Intersection-of-Graphs method, and can be applied to many different equation types.

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Intersection-of-Graphs Method for Solving Equations For any equation of the form f(x) g(x), 1. Assign f (x) as Y1 and g (x) as Y2. 2. Graph both function and identify any point(s) of intersection, if they exist. The x-coordinate of all such points is a solution to the equation. Recall that in the solution of linear equations, we sometimes encounter equations that are identities (infinitely many solutions) or contradictions (no solutions). These possibilities also have graphical representations, and appear as coincident lines and parallel lines respectively. These possibilities are illustrated in Figure 1.76. Figure 1.76 y

y

y

x

x

One point of intersection (unique solution)

EXAMPLE 1B

䊳

Infinitely many points of intersection (identity)

x

No point of intersection (contradiction)

Solving an Equation Graphically Solve 0.75x 2 0.511 1.5x2 3 using a graphing calculator.

Solution

䊳

With 0.75X 2 as Y1 and 0.511 1.5X2 3 as Y2, we use the 2nd TRACE (CALC) option and select 5:intersect. The graphs appear to be parallel lines (Figure 1.77), and after pressing three times we obtain the error message shown (Figure 1.78), confirming there are no solutions. ENTER

Figure 1.78

Figure 1.77 10

10

A. You’ve just seen how we can solve equations using the Intersection-of-Graphs method

10

10


䊳

B. Solving Equations Graphically Using the x-Intercept/Zeroes Method The intersection-of-graphs method works extremely well when the graphs of f(x) and g(x) (Y1 and Y2) are simple and “well-behaved.” Later in this course, we encounter a number of graphs that are more complex, and it will help to develop alternative methods for solving graphically. Recall that two equations are equivalent if they have the same solution set. For instance, the equations 2x 6 and 2x 6 0 are equivalent (since x 3 is a solution to both), as are 3x 1 x 5 and 2x 6 0 (since

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x 3 is a solution to both). Applying the intersection-of-graphs method to the last two equivalent equations, gives

Y2

2x 6 0 5

Y1

and

f

The intersection-of-graphs method focuses on a point of intersection (a, b), and names x a as the solution. The zeroes method focuses on the input x r, for which the output is 0 3 Y1 1r2 0 4 . All such values r are called the zeroes of the function. Much more will be said about functions and their zeroes in later chapters.

f

WORTHY OF NOTE

f

3x 1 x 5

Y1

Y2

The intersection method will work equally well in both cases, but the equation on the right has only one variable expression, and will produce a single (visible) graph (since Y2 0 is simply the x-axis). Note that here we seek an input value that will result in an output of 0. In other words, all solutions will have the form (x, 0), which is the x-intercept of the graph. For this reason, the method is alternatively called the zeroes method or the x-intercept method. The method employs the approach shown above, in which the equation f 1x2 g1x2 is rewritten as f 1x2 g1x2 0, with f 1x2 g1x2 assigned as Y1. Zeroes/x-Intercept Method for Solving Equations

For any equation of the form f 1x2 g1x2, 1. Rewrite the equation as f 1x2 g1x2 0. 2. Assign f 1x2 g1x2 as Y1. 3. Graph the resulting function and identify any x-intercepts, if they exist. Any x-intercept(s) of the graph will be a solution to the equation. To locate the zero (x-intercept) for 2x 6 0 on a graphing calculator, enter 2X 6 for Y1 and use the 2:zero option found on the same menu as the 5:intercept option (Figure 1.79). Since some equations have more than one zero, the 2:zero option will ask you to “narrow down” the interval it should search, even though there is only one zero here. It does this by asking for a “Left Bound?”, a “Right Bound?”, and a “GUESS?” (the Guess? option can once again be bypassed). You can enter these bounds by tracing along the graph or by inputting a chosen value, then pressing (note how the calculator posts a marker at each Figure 1.79 bound). Figure 1.80 shows we entered x 0 as the left bound and x 4 as the right, and the calculator will search for the x-intercept in this interval (note that in general, the cursor will be either above or below the x-axis for the left bound, but must be on the opposite side of the x-axis for the right bound). Pressing once more bypasses the Guess? option and locates the x-intercept at (3, 0). The solution is x 3 (Figure 1.81). ENTER

ENTER

Figure 1.80

Figure 1.81 10

10

10

10

10

10

10

10

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EXAMPLE 2

䊳

Solving an Equation Using the Zeroes Method Solve the equation 41x 32 6 2x 3 using the zeroes method.

Solution

䊳

Figure 1.82 As given, we have f 1x2 41x 32 6 10 and g1x2 2x 3. Rewriting the equation as f 1x2 g1x2 0 gives 41x 32 6 12x 32 0, where the expression for g(x) is parenthesized to ensure 10 10 the equations remain equivalent. Entering 41X 32 6 12X 32 as Y1 and pressing 2nd TRACE (CALC) 2:zero produces the screen shown in Figure 1.82, with the 10 calculator requesting a left bound. We can input any x-value that is obviously to the left of the x-intercept, or move the cursor to any position left of the x-intercept and press (we input x 4, see Figure 1.83). The calculator then asks for a right bound and as before we can input any x-value obviously to the right, or simply move the cursor to any location on the opposite side of the x-axis and press (we chose x 2, see Figure 1.84). After bypassing the Guess? option (press once again), the calculator locates the x-intercept at (3.5, 0), and the solution to the original equation is x 3.5 (Figure 1.85). ENTER

ENTER

ENTER

Figure 1.83

Figure 1.84

10

10

10

10

B. You’ve just seen how we can solve equations using the x-intercept/zeroes method

Figure 1.85 10

10

10

10

10

10

10

10


䊳

C. Solving Linear Inequalities Graphically The intersection-of-graphs method can also be applied to solve linear inequalities. The point of intersection simply becomes one of the boundary points for the solution interval, and is included or excluded depending on the inequality given. For the inequality f 1x2 7 g1x2 written as Y1 7 Y2, it becomes clear the inequality is true for all inputs x where the outputs for Y1 are greater than the outputs for Y2, meaning the graph of f(x) is above the graph of g(x). A similar statement can be made for f 1x2 6 g1x2 written as Y1 6 Y2. Intersection-of-Graphs Method for Solving Inequalities

For any inequality of the form f 1x2 7 g1x2 , 1. Assign f(x) as Y1 and g(x) as Y2. 2. Graph both functions and identify any point(s) of intersection, if they exist. The solution set is all real numbers x for which the graph of Y1 is above the graph of Y2. For strict inequalities, the boundary of the solution interval is not included. A similar process is used for the inequalities f 1x2 g1x2 , f 1x2 6 g1x2 , and f 1x2 g1x2 . Note that we can actually draw the graphs of Y1 and Y2 differently (one more bold than the

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153

other) to clearly tell them apart in the viewing window. This is done on the Y= screen, by moving the cursor to the far left of the current function and pressing until a bold line appears. From the default setting, pressing one time produces this result. ENTER

ENTER

EXAMPLE 3

䊳

Solving an Inequality Using the Intersection-of-Graphs Method Solve 0.513 x2 5 2x 4 using the intersection-of-graphs method.

Solution

䊳

Figure 1.86

To assist with the clarity of the solution, we set the calculator to graph Y2 using a bolder line than Y1 (Figure 1.86). With 0.513 X2 5 as Y1 and 2X 4 as Y2, we use the 2nd TRACE (CALC) option and select 5:intersect. Pressing three times serves to identify both graphs, bypass the “Guess?” option, and display the point of intersection 13, 22 (Figure 1.87). Since the graph of Y1 is below the graph of Y2 1Y1 Y2 2 for all values of x to the right of 13, 22 , x 3 is the left boundary, with 10 the interval extending to positive infinity. Due to the less than or equal to inequality, we include x 3 and the solution interval is x 僆 33, q 2 . ENTER

C. You’ve just seen how we can solve linear inequalities graphically

Figure 1.87 10

10

10


䊳

D. Solving for a Specified Variable in Literal Equations A formula is an equation that models a known relationship between two or more quantities. A literal equation is simply one that has two or more variables. Formulas are a type of literal equation, but not every literal equation is a formula. For example, the formula A P PRT models the growth of money in an account earning simple interest, where A represents the total amount accumulated, P is the initial deposit, R is the annual interest rate, and T is the number of years the money is left on deposit. To describe A P PRT, we might say the formula has been “solved for A” or that “A is written in terms of P, R, and T.” In some cases, before using a formula it may be convenient to solve for one of the other variables, say P. In this case, P is called the object variable. EXAMPLE 4

䊳

Solving for Specified Variable Given A P PRT, write P in terms of A, R, and T (solve for P).

Solution

䊳

Since the object variable occurs in more than one term, we first apply the distributive property. A P PRT A P11 RT2 P11 RT2 A 1 RT 11 RT2 A P 1 RT

focus on P — the object variable factor out P solve for P [divide by (1 RT )]

result


䊳

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We solve literal equations for a specified variable using the same methods we used for other equations and formulas. Remember that it’s good practice to focus on the object variable to help guide you through the solution process, as again shown in Example 5. EXAMPLE 5

䊳

Solving for a Specified Variable Given 2x 3y 15, write y in terms of x (solve for y).

Solution

䊳

WORTHY OF NOTE In Example 5, notice that in the second step we wrote the subtraction of 2x as 2x 15 instead of 15 2x. For reasons that will become clearer as we continue our study, we generally write variable terms before constant terms.

2x 3y 15 3y 2x 15 1 13y2 13 12x 152 3 y 2 3 x 5

focus on the object variable subtract 2x (isolate y-term) multiply by 13 (solve for y ) simplify and distribute


䊳

Literal Equations and General Solutions Solving literal equations for a specified variable can help us develop the general solution for an entire family of equations. This is demonstrated here for the family of linear equations written in the form ax b c. A side-by-side comparison with a specific linear equation demonstrates that identical ideas are used. Specific Equation 2x 3 15 2x 15 3 x

15 3 2

Literal Equation focus on object variable

ax b c ax c b

subtract constant

x

divide by coefficient

cb a

Of course the solution on the left would be written as x 6 and checked in the original equation. On the right we now have a general formula for all equations of the form ax b c. EXAMPLE 6

䊳

Solving Equations of the Form ax ⫹ b ⫽ c Using a General Formula Solve 6x 1 25 using the formula just developed, and check your solution in the original equation.

Solution

䊳

For this equation, a 6, b 1, and c 25, giving x

cb a 25 112

24 6 4

6

→

Check:

6x 1 25 6142 1 25 24 1 25 25 25 ✓ Now try Exercises 55 through 60

D. You’ve just seen how we can solve for a specified variable in a formula or literal equation

䊳

Developing a general solution for the linear equation ax b c seems to have little practical use. But in Section 3.2 we’ll use this idea to develop a general solution for quadratic equations, a result with much greater significance.

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E. Using a Problem-Solving Guide Becoming a good problem solver is an evolutionary process. Over time and with continued effort, your problem-solving skills grow, as will your ability to solve a wider range of applications. Most good problem solvers develop the following characteristics: • A positive attitude • A mastery of basic facts • Strong mental arithmetic skills

• Good mental-visual skills • Good estimation skills • A willingness to persevere

These characteristics form a solid basis for applying what we call the Problem-Solving Guide, which simply organizes the basic elements of good problem solving. Using this guide will help save you from two common stumbling blocks—indecision and not knowing where to start. Problem-Solving Guide • Gather and organize information. Read the problem several times, forming a mental picture as you read. Highlight key phrases. List given information, including any related formulas. Clearly identify what you are asked to find. • Make the problem visual. Draw and label a diagram or create a table of values, as appropriate. This will help you see how different parts of the problem fit together. • Develop an equation model. Assign a variable to represent what you are asked to find and build any related expressions referred to in the problem. Write an equation model based on the relationships given in the problem. Carefully reread the problem to double-check your equation model. • Use the model and given information to solve the problem. Substitute given values, then simplify and solve. State the answer in sentence form, and check that the answer is reasonable. Include any units of measure indicated.

General Modeling Exercises Translating word phrases into symbols is an important part of building equations from information given in paragraph form. Sometimes the variable occurs more than once in the equation, because two different items in the same exercise are related. If the relationship involves a comparison of size, we often use line segments or bar graphs to model the relative sizes. EXAMPLE 7

䊳

Solving an Application Using the Problem-Solving Guide The largest state in the United States is Alaska (AK), which covers an area that is 230 square miles (mi2) more than 500 times that of the smallest state, Rhode Island (RI). If they have a combined area of 616,460 mi2, how many square miles does each cover?

䊳

Combined area is 616,460 mi2, AK covers 230 more than 500 times the area of RI.

gather and organize information highlight any key phrases

230

…

Solution

make the problem visual

500 times

Rhode Island’s area R

Alaska

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Let R represent the area of Rhode Island. Then 500R 230 represents Alaska’s area.

assign a variable build related expressions

Rhode Island’s area Alaska’s area Total R 1500R 2302 616,460 501R 616,230 R 1230

write the equation model combine like terms, subtract 230 divide by 501

2

Rhode Island covers an area of 1230 mi , while Alaska covers an area of 500112302 230 615,230 mi2. Now try Exercises 63 through 68

䊳

Consecutive Integer Exercises Exercises involving consecutive integers offer excellent practice in assigning variables to unknown quantities, building related expressions, and the problem-solving process in general. We sometimes work with consecutive odd integers or consecutive even integers as well. EXAMPLE 8

䊳

Solving a Problem Involving Consecutive Odd Integers The sum of three consecutive odd integers is 69. What are the integers?

Solution

䊳

The sum of three consecutive odd integers . . . 2

2

4 3 2 1

odd

WORTHY OF NOTE The number line illustration in Example 8 shows that consecutive odd integers are two units apart and the related expressions were built accordingly: n, n 2, n 4, and so on. In particular, we cannot use n, n 1, n 3, . . . because n and n 1 are not two units apart. If we know the exercise involves even integers instead, the same model is used, since even integers are also two units apart. For consecutive integers, the labels are n, n 1, n 2, and so on.

odd

2

2

gather/organize information highlight any key phrases 2 make the problem visual

0

1

odd

2

3

odd

4

n n1 n2 n3 n4

odd

odd

odd

Let n represent the smallest consecutive odd integer, then n 2 represents the second odd integer and 1n 22 2 n 4 represents the third.

In words: first second third odd integer 69 n 1n 22 1n 42 69 3n 6 69 3n 63 n 21


write the equation model equation model combine like terms subtract 6 divide by 3

The odd integers are n 21, n 2 23, and n 4 25. 21 23 25 69 ✓ Now try Exercises 69 through 72

䊳

Uniform Motion (Distance, Rate, Time) Exercises Uniform motion problems have many variations, and it’s important to draw a good diagram when you get started. Recall that if speed is constant, the distance traveled is equal to the rate of speed multiplied by the time in motion: D RT.

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EXAMPLE 9

䊳

157

Solving a Problem Involving Uniform Motion I live 260 mi from a popular mountain retreat. On my way there to do some mountain biking, my car had engine trouble — forcing me to bike the rest of the way. If I drove 2 hr longer than I biked and averaged 60 miles per hour driving and 10 miles per hour biking, how many hours did I spend pedaling to the resort?

Solution

䊳

The sum of the two distances must be 260 mi. The rates are given, and the driving time is 2 hr more than biking time.

Home

gather/organize information highlight any key phrases make the problem visual

Driving

Biking

D1 RT

D2 rt

Resort

D1 D2 Total distance 260 miles

Let t represent the biking time, then T t 2 represents time spent driving. D1 D2 260 RT rt 260 601t 22 10t 260 70t 120 260 70t 140 t2

assign a variable build related expressions write the equation model RT D1, rt D2 substitute t 2 for T, 60 for R, 10 for r distribute and combine like terms subtract 120 divide by 70

I rode my bike for t 2 hr, after driving t 2 4 hr. Now try Exercises 73 through 76

䊳

Exercises Involving Mixtures Mixture problems offer another opportunity to refine our problem-solving skills while using many elements from the problem-solving guide. They also lend themselves to a very useful mental-visual image and have many practical applications. EXAMPLE 10

䊳

Solving an Application Involving Mixtures As a nasal decongestant, doctors sometimes prescribe saline solutions with a concentration between 6% and 20%. In “the old days,” pharmacists had to create different mixtures, but only needed to stock these concentrations, since any percentage in between could be obtained using a mixture. An order comes in for a 15% solution. How many milliliters (mL) of the 20% solution must be mixed with 10 mL of the 6% solution to obtain the desired 15% solution? Provide both an algebraic solution and a graphical solution.

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Algebraic Solution

䊳

Only 6% and 20% concentrations are available; mix some 20% solution with 10 mL of the 6% solution. (See Figure 1.88.)

gather/organize information highlight any key phrases

Figure 1.88 20% solution

6% solution ? mL

10 mL make the problem visual

(10 ?) mL 15% solution

Let x represent the amount of 20% solution, then 10 x represents the total amount of 15% solution. 1st quantity times its concentration

1010.062 0.6

2nd quantity times its concentration

x10.22 0.2x 0.2x 0.05x x


1st2nd quantity times desired concentration

110 x2 10.152 1.5 0.15x 0.9 0.15x 0.9 18

write equation model distribute/simplify subtract 0.6 subtract 0.15x divide by 0.05

To obtain a 15% solution, 18 mL of the 20% solution must be mixed with 10 mL of the 6% solution.

Graphical Solution

䊳

WORTHY OF NOTE For mixture exercises, an estimate assuming equal amounts of each liquid can be helpful. For example, assume we use 10 mL of the 6% solution and 10 mL of the 20% solution. The final concentration would be halfway in between, 6 20 13%. This is too low a 2 concentration (we need a 15% solution), so we know that more than 10 mL of the stronger (20%) solution must be used.

Although both methods work equally well, here we elect to use the intersection-of-graphs method and enter 1010.062 X10.22 as Y1 and 110 X2 10.152 as Y2. Virtually all graphical solutions require a careful study of the context to set the viewing window prior to graphing. If 10 mL of liquid were used from each concentration, we would have 20 mL of a 13% solution (see Worthy of Note), so more of the stronger solution is needed. This shows that an appropriate Xmax might be close to 30. If all 30 mL were used, the output would be 30 10.152 4.5, so an appropriate Ymax might be around 6 (see Figure 1.89). Using 2nd TRACE (CALC) 5:Intersect and pressing three times gives 0 (18, 4.2) as the point of intersection, showing x 18 mL of the stronger solution must be used (Figure 1.90).

Figure 1.89

Figure 1.90 6

ENTER

30

0

E. You’ve just seen how we can use the problemsolving guide to solve various problem types


䊳

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159

1.5 EXERCISES 䊳



1. When using the method, one side of an equation is entered as and the other side as on a graphing calculator. The of the point of is the solution of the equation. 2. To solve a linear inequality using the intersectionof-graphs method, first find the point of . The of this point is a boundary value of the solution interval and if the inequality is not strict, this value is in the solution. 3. A(n) equation is an equation having more unknowns.

䊳

or

4. For the equation S 2␲r2 2␲rh, we can say that S is written in terms of and . 5. Discuss/Explain the similarities and differences between the intersection and zeroes methods for solving equations. How can the zeroes method be applied to solving linear inequalities? Give examples in your discussion. 6. Discuss/Explain each of the four basic parts of the problem-solving guide. Include a solved example in your discussion.


Solve the following equations using a graphing calculator and the intersection-of-graphs method. For Exercises 7 and 8, carefully sketch the graphs you designate as Y1 and Y2 by hand before using your calculator.

7. 3x 7 21x 12 10 8. 2x 1 21x 32 1 9. 0.8x 0.4 0.2512 0.4x2 2.8 10. 0.5x 2.5 0.7513 0.2x2 0.4 11. x 13x 12 0.514x 62 2

12. 3x 14 x2 0.216 10x2 5.2 13. 13x 2 3 x 9 14.

4 5 x

16.

1 3 1x

8 65x

15. 12 1x 42 10 x 12 12x2

62 5 x 16 23x2

Solve the following equations using a graphing calculator and the x-intercept/zeroes method. Compare your results for Exercises 17 and 18 to those of Exercises 7 and 8.

21. 1.51x 42 2.5 3x 3.5 22. 0.813x 12 0.2 2x 3.8 23. 21x 22 1 x 11 x2

24. 312x 12 1 2x 11 4x2

25. 3x 10.7x 1.22 211.1x 0.62 0.1x

26. 3x 210.2x 1.42 410.8x 0.72 0.2x Solve the following inequalities using a graphing calculator and the intersection-of-graphs method. Compare your results for Exercises 27 and 28 to those of Exercises 7 and 8.

27. 3x 7 7 21x 12 10 28. 2x 1 7 21x 32 1

29. 2x 13 x2 215 2x2 7

30. 413x 52 2 312 4x2 24 31. 0.31x 22 1.1 6 0.2x 3 32. 0.2514 x2 1 6 1 0.5x 33. 31x 12 1 x 41x 12

17. 3x 7 21x 12 10

34. 1.112 x2 0.2 7 510.1 0.2x2 0.1x

18. 2x 1 21x 32 1

35. 211.5x 1.12 0.1x 4x 0.313x 42

19. 213 2x2 5 3x 3

36. 41x 12 2x 7 6 21x 1.52

20. 313 x2 4 2x 5

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160

Solve for the specified variable in each formula or literal equation.

37. P ⫽ C ⫹ CM for C (retail) 38. S ⫽ P ⫺ PD for P (retail) 39. C ⫽ 2␲r for r (geometry) 40. V ⫽ LWH for W (geometry) P1V1 P2V2 41. for T2 (science) ⫽ T1 T2 42.

P1 C ⫽ 2 for P2 (communication) P2 d

43. V ⫽ 43␲r2h for h (geometry) 44. V ⫽ 13␲r2h for h (geometry) a1 ⫹ an 45. Sn ⫽ na b for n (sequences) 2 46. A ⫽

h1b1 ⫹ b2 2 2

for h (geometry)

47. S ⫽ B ⫹ 12PS for P (geometry)

䊳

48. s ⫽ 12gt2 ⫹ vt for g (physics) 49. Ax ⫹ By ⫽ C for y 50. 2x ⫹ 3y ⫽ 6 for y 51. 56x ⫹ 38y ⫽ 2 for y 52. 23x ⫺ 79y ⫽ 12 for y

53. y ⫺ 3 ⫽ ⫺4 5 1x ⫹ 102 for y

54. y ⫹ 4 ⫽ ⫺2 15 1x ⫹ 102 for y

The following equations are given in ax ⴙ b ⴝ c form. Solve by identifying the value of a, b, and c, then using cⴚb the formula x ⴝ . a

55. 3x ⫹ 2 ⫽ ⫺19 56. 7x ⫹ 5 ⫽ 47 57. ⫺6x ⫹ 1 ⫽ 33 58. ⫺4x ⫹ 9 ⫽ 43 59. 7x ⫺ 13 ⫽ ⫺27 60. 3x ⫺ 4 ⫽ ⫺25


61. Surface area of a cylinder: SA ⴝ 2␲r2 ⴙ 2␲rh The surface area of a cylinder is given by the formula shown, where h is the height of the cylinder and r is the radius of the base. Find the height of a cylinder that has a radius of 8 cm and a surface area of 1256 cm2. Use ␲ ⬇ 3.14.

䊳

1–76


62. Using the equation-solving process for Exercise 61 as a model, solve the formula SA ⫽ 2␲r2 ⫹ 2␲rh for h.

APPLICATIONS

Solve by building an equation model and using the problem-solving guidelines as needed. Check all answers using a graphing calculator. General Modeling Exercises

63. Two spelunkers (cave explorers) were exploring different branches of an underground cavern. The first was able to descend 198 ft farther than twice the second. If the first spelunker descended 1218 ft, how far was the second spelunker able to descend? 64. The area near the joining of the Tigris and Euphrates Rivers (in modern Iraq) has often been called the Cradle of Civilization, since the area has evidence of many ancient cultures. The length of the Euphrates River exceeds that of the Tigris by 620 mi. If they have a combined length of 2880 mi, how long is each river?

65. U.S. postal regulations require that a package Girth can have a maximum combined length and girth (distance around) L of 108 in. A shipping H carton is constructed so that it has a width of W 14 in., a height of 12 in., and can be cut or folded to various lengths. What is the maximum length that can be used? Source: www.USPS.com

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66. Hi-Tech Home Improvements buys a fleet of identical trucks that cost $32,750 each. The company is allowed to depreciate the value of their trucks for tax purposes by $5250 per year. If company policies dictate that older trucks must be sold once their value declines to $6500, approximately how many years will they keep these trucks? 67. The longest suspension bridge in the world is the Akashi Kaikyo (Japan) with a length of 6532 feet. Japan is also home to the Shimotsui Straight bridge. The Akashi Kaikyo bridge is 364 ft more than twice the length of the Shimotsui bridge. How long is the Shimotsui bridge? Source: www.guinnessworldrecords.com

68. The Mars rover Spirit landed on January 3, 2004. Just over 1 yr later, on January 14, 2005, the Huygens probe landed on Titan (one of Saturn’s moons). At their closest approach, the distance from the Earth to Saturn is 29 million mi more than 21 times the distance from the Earth to Mars. If the distance to Saturn is 743 million mi, what is the distance to Mars?

161

Uniform Motion Exercises 73. At 9:00 A.M., Linda leaves work on a business trip, gets on the interstate, and sets her cruise control at 60 mph. At 9:30 A.M., Bruce notices she’s left her briefcase and cell phone, and immediately starts after her driving 75 mph. At what time will Bruce catch up with Linda? 74. A plane flying at 300 mph has a 3-hr head start on a “chase plane,” which has a speed of 800 mph. How far from the airport will the chase plane overtake the first plane? 75. Jeff had a job interview in a nearby city 72 mi away. On the first leg of the trip he drove an average of 30 mph through a long construction zone, but was able to drive 60 mph after passing through this zone. If driving time for the trip was 112 hr, how long was he driving in the construction zone? 76. At a high-school cross-country meet, Jared jogged 8 mph for the first part of the race, then increased his speed to 12 mph for the second part. If the race was 21 mi long and Jared finished in 2 hr, how far did he jog at the faster pace? Mixture Exercises Give the total amount of the mix that results and the percent concentration or worth of the mix.

77. Two quarts of 100% orange juice are mixed with 2 quarts of water (0% juice). 78. Ten pints of a 40% acid are combined with 10 pints of an 80% acid. Consecutive Integer Exercises 69. Find two consecutive even integers such that the sum of twice the smaller integer plus the larger integer is one hundred forty-six. 70. When the smaller of two consecutive integers is added to three times the larger, the result is fiftyone. Find the smaller integer. 71. Seven times the first of two consecutive odd integers is equal to five times the second. Find each integer. 72. Find three consecutive even integers where the sum of triple the first and twice the second is eight more than four times the third.

79. Eight pounds of premium coffee beans worth $2.50 per pound are mixed with 8 lb of standard beans worth $1.10 per pound. 80. A rancher mixes 50 lb of a custom feed blend costing $1.80 per pound, with 50 lb of cheap cottonseed worth $0.60 per pound. Solve each application of the mixture concept.

81. To help sell more of a lower grade meat, a butcher mixes some premium ground beef worth $3.10/lb, with 8 lb of lower grade ground beef worth $2.05/lb. If the result was an intermediate grade of ground beef worth $2.68/lb, how much premium ground beef was used?

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82. Knowing that the camping/hiking season has arrived, a nutrition outlet is mixing GORP (Good Old Raisins and Peanuts) for the anticipated customers. How many pounds of peanuts worth $1.29/lb, should be mixed with 20 lb of deluxe raisins worth $1.89/lb, to obtain a mix that will sell for $1.49/lb?

䊳

83. How many pounds of walnuts at 84¢/lb should be mixed with 20 lb of pecans at $1.20/lb to give a mixture worth $1.04/lb? 84. How many pounds of cheese worth 81¢/lb must be mixed with 10 lb cheese worth $1.29/lb to make a mixture worth $1.11/lb?


85. Look up and read the following article. Then turn in a one page summary. “Don’t Give Up!,” William H. Kraus, Mathematics Teacher, Volume 86, Number 2, February 1993: pages 110–112. 86. A chemist has four solutions of a very rare and expensive chemical that are 15% acid (cost $120 per ounce), 20% acid (cost $180 per ounce), 35% acid (cost $280 per ounce) and 45% acid (cost $359 per ounce). She requires 200 oz of a 29% acid solution. Find the combination of any two of these concentrations that will minimize the total cost of the mix. 87. P, Q, R, S, T, and U represent numbers. The arrows in the figure show the sum of the two or three numbers added in the indicated direction

䊳

1–78


(Example: Q T 23). Find P Q R S T U. P

Q

26

S

30 40

R

T 19

U 23

34

88. Given a sphere circumscribed by a cylinder, verify the volume of the sphere is 23 that of the cylinder.


1 2 3 2 x x2 x 2x 90. (1.4) Solve for y, then state the slope and y-intercept of the line: 6x 7y 42 89. (R.5) Solve for x:

91. (R.4) Factor each expression: a. 4x2 9 b. x3 27

92. (1.3) Given g1x2 x2 3x 10, evaluate g1 13 2, g122, and g152

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1.6

Linear Function Models and Real Data

LEARNING OBJECTIVES

Collecting and analyzing data is a tremendously important mathematical endeavor, having applications throughout business, industry, science, and government. The link between classroom mathematics and real-world mathematics is called a regression, in which we attempt to find an equation that will act as a model for the raw data. In this section, we focus on situations where the data is best modeled by a linear function.

In Section 1.6 you will see how we can:

A. Draw a scatterplot and

B.

C.

D.

E.

identify positive and negative associations Use a scatterplot to identify linear and nonlinear associations Use a scatterplot to identify strong and weak correlations Find a linear function that models the relationships observed in a set of data Use linear regression to find the line of best fit

A. Scatterplots and Positive/Negative Associations In this section, we continue our study of ordered pairs and functions, but this time using data collected from various sources or from observed real-world relationships. You can hardly pick up a newspaper or magazine without noticing it contains a large volume of data presented in graphs, charts, and tables. In addition, there are many simple experiments or activities that enable you to collect your own data. We begin analyzing the collected data using a scatterplot, which is simply a graph of all of the ordered pairs in a data set. Often, real data (sometimes called raw data) is not very “well behaved” and the points may be somewhat scattered—the reason for the name.

Positive and Negative Associations Earlier we noted that lines with positive slope rise from left to right, while lines with negative slope fall from left to right. We can extend this idea to the data from a scatterplot. The data points in Example 1A seem to rise as you move from left to right, with larger input values generally resulting in larger outputs. In this case, we say there is a positive association between the variables. If the data seems to decrease or fall as you move left to right, we say there is a negative association.

EXAMPLE 1A

䊳

Drawing a Scatterplot and Observing Associations The ratio of the federal debt to the total population is known as the per capita debt. The per capita debt of the United States is shown in the table for the odd-numbered years from 1997 to 2007. Draw a scatterplot of the data and state whether the association is positive, negative, or cannot be determined. Source: Data from the Bureau of Public Debt at www.publicdebt.treas.gov

Per Capita Debt ($1000s)

30

1997

20.0

28

1999

20.7

2001

20.5

2003

23.3

2005

27.6

2007

30.4

Debt ($1000s)

Year

26 24 22 20 1997 1999 2001 2003 2005 2007

Year

Solution

1–79

䊳

Since the amount of debt depends on the year, year is the input x and per capita debt is the output y. Scale the x-axis from 1997 to 2007 and the y-axis from 20 to 30 to comfortably fit the data (the “squiggly lines,” near the 20 and 1997 in the graph are used to show that some initial values have been skipped). The graph indicates a positive association between the variables, meaning the debt is generally increasing as time goes on. 163

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EXAMPLE 1B

䊳

Drawing a Scatterplot and Observing Associations A cup of coffee is placed on a table and allowed to cool. The temperature of the coffee is measured every 10 min and the data are shown in the table. Draw the scatterplot and state whether the association is positive, negative, or cannot be determined. Temperature (ºF)

0

110

10

89

20

76

30

72

40

71

120 110

Temp (°F)

Elapsed Time (minutes)

100 90 80 70 0

Solution

䊳

A. You’ve just seen how we can draw a scatterplot and identify positive and negative associations

5 10 15 20 25 30 35 40

Time (minutes)

Since temperature depends on cooling time, time is the input x and temperature is the output y. Scale the x-axis from 0 to 40 and the y-axis from 70 to 110 to comfortably fit the data. As you see in the figure, there is a negative association between the variables, meaning the temperature decreases over time. Now try Exercises 7 and 8 䊳

B. Scatterplots and Linear/Nonlinear Associations The data in Example 1A had a positive association, while the association in Example 1B was negative. But the data from these examples differ in another important way. In Example 1A, the data seem to cluster about an imaginary line. This indicates a linear equation model might be a good approximation for the data, and we say there is a linear association between the variables. The data in Example 1B could not accurately be modeled using a straight line, and we say the variables time and cooling temperature exhibit a nonlinear association.

䊳

Drawing a Scatterplot and Observing Associations A college professor tracked her annual salary for 2002 to 2009 and the data are shown in the table. Draw the scatterplot and determine if there is a linear or nonlinear association between the variables. Also state whether the association is positive, negative, or cannot be determined.

Year

Salary ($1000s)

2002

30.5

2003

31

2004

32

2005

33.2

2006

35.5

2007

39.5

2008

45.5

2009

52

55 50

Salary ($1000s)

EXAMPLE 2

45

Appears nonlinear

40 35 30

2002

2004

2006

2008

2010

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165

Section 1.6 Linear Function Models and Real Data

Solution

䊳

B. You’ve just seen how we can use a scatterplot to identify linear and nonlinear associations

Since salary earned depends on a given year, year is the input x and salary is the output y. Scale the x-axis from 2002 to 2010, and the y-axis from 30 to 55 to comfortably fit the data. A line doesn’t seem to model the data very well, and the association appears to be nonlinear. The data rises from left to right, indicating a positive association between the variables. This makes good sense, since we expect our salaries to increase over time. Now try Exercises 9 and 10 䊳

C. Identifying Strong and Weak Correlations Using Figures 1.91 and 1.92 shown, we can make one additional observation regarding the data in a scatterplot. While both associations shown appear linear, the data in Figure 1.91 seems to cluster more tightly about an imaginary straight line than the data in Figure 1.92. Figure 1.91

Figure 1.92 y

y 10

10

5

5

0

5

x

10

0

5

10

x

We refer to this “clustering” as the “goodness of fit,” or in statistical terms, the strength of the correlation. To quantify this fit we use a measure called the correlation coefficient r, which tells whether the association is positive or negative: r 7 0 or r 6 0, and quantifies the strength of the association: 0r 0 ⱕ 100% . Actually, the coefficient is given in decimal form, making 0r 0 ⱕ 1. If the data points form a perfectly straight line, we say the strength of the correlation is either ⫺1 or 1, depending on the association. If the data points appear clustered about the line, but are scattered on either side of it, the strength of the correlation falls somewhere between ⫺1 and 1, depending on how tightly/loosely they’re scattered. This is summarized in Figure 1.93. Figure 1.93 Perfect negative correlation Strong negative correlation ⫺1.00

Moderate negative correlation

No correlation Weak negative correlation

Weak positive correlation 0

Moderate positive correlation

Perfect positive correlation Strong positive correlation ⫹1.00

The following scatterplots help to further illustrate this idea. Figure 1.94 shows a linear and negative association between the value of a car and the age of a car, with a strong correlation. Figure 1.95 shows there is no apparent association between family income and the number of children, and Figure 1.96 appears to show a linear and positive association between a man’s height and weight, with a weak correlation.

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Figure 1.94

Figure 1.96

Age of auto

Male weights

Family income

Figure 1.95

Value of auto

Number of children

Male heights

Until we develop a more accurate method of calculating a numerical value for this correlation, the best we can do are these broad generalizations: weak correlation, strong correlation, or no correlation. EXAMPLE 3A

䊳

High School and College GPAs Many colleges use a student’s high school GPA as a possible indication of their future college GPA. Use the data from Table 1.6 (high school/college GPA) to draw a scatterplot. Then a. Sketch a line that seems to approximate the data, meaning it has the same general direction while passing through the observed “center” of the data. b. State whether the association is positive, negative, or cannot be determined. c. Decide whether the correlation is weak or strong. Table 1.6

Solution

䊳

EXAMPLE 3B

䊳

High School GPA

College GPA

1.8

1.8

2.2

2.3

2.8

2.5

3.2

2.9

3.4

3.6

3.8

3.9

4.0 3.5

College GPA

166

3.0 2.5 2.0 1.5

1.5

2.0

2.5

3.0

3.5

4.0

High School GPA

a. A line approximating the data set as a whole is shown in the figure. b. Since the line has positive slope, there is a positive association between a student’s high school GPA and their GPA in college. c. The correlation appears strong.

Natural Gas Consumption The amount of natural gas consumed by homes and offices varies with the season, with the highest consumption occurring in the winter months. Use the data from Table 1.7 (outdoor temperature/gas consumed) to draw a scatterplot. Then a. Sketch an estimated line of best fit. b. State whether the association is positive, negative, or cannot be determined. c. Decide whether the correlation is weak or strong.

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Table 1.7 Gas Consumed (cubic feet)

30

800

40

620

50

570

60

400

70

290

80

220

800

Gas consumption (ft3)

Outdoor Temperature (Fº)

600

400

200

0

30

40

50

60

70

80

Outdoor temperature (⬚F)

Solution

䊳

C. You’ve just seen how we can use a scatterplot to identify strong and weak correlations

a. We again use appropriate scales and sketch a line that seems to model the data (see figure). b. There is a negative association between temperature and the amount of natural gas consumed. c. The correlation appears to be strong. Now try Exercises 11 and 12

䊳

D. Linear Functions That Model Relationships Observed in a Set of Data Finding a linear function model for a set of data involves visually estimating and sketching a line that appears to best “fit” the data. This means answers will vary slightly, but a good, usable model can often be obtained. To find the function, we select two points on this imaginary line and use either the slope-intercept form or the pointslope formula to construct the function. Points on this estimated line but not actually in the data set can still be used to help determine the function.

EXAMPLE 4

䊳

Finding a Linear Function to Model the Relationship Between GPAs Use the scatterplot from Example 3A to find a function model for the line a college might use to project an incoming student’s future GPA.

Solution

䊳

Any two points on or near the estimated best-fit line can be used to help determine the linear function (see the figure in Example 3A). For the slope, it’s best to pick two points that are some distance apart, as this tends to improve the accuracy of the model. It appears (1.8, 1.8) and (3.8, 3.9) are both on the line, giving y2 ⫺ y1 x2 ⫺ x1 3.9 ⫺ 1.8 ⫽ 3.8 ⫺ 1.8 ⫽ 1.05 y ⫺ y1 ⫽ m1x ⫺ x1 2 y ⫺ 1.8 ⫽ 1.051x ⫺ 1.82 y ⫺ 1.8 ⫽ 1.05x ⫺ 1.89 y ⫽ 1.05x ⫺ 0.09 m⫽

slope formula

substitute (x2, y2) for S (3.8, 3.9), (x1, y1) for S (1.8, 1.8) slope point-slope form substitute 1.05 for m, (1.8, 1.8) for (x1, y1) distribute add 1.8 (solve for y )

One possible function model for this data is f 1x2 ⫽ 1.05x ⫺ 0.09. Slightly different functions may be obtained, depending on the points chosen. Now try Exercises 13 through 22

䊳

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WORTHY OF NOTE Sometimes it helps to draw a straight line on an overhead transparency, then lay it over the scatterplot. By shifting the transparency up and down, and rotating it left and right, the line can more accurately be placed so that it’s centered among and through the data.

EXAMPLE 5

䊳

The function from Example 4 predicts that a student with a high school GPA of 3.2 will have a college GPA of almost 3.3: f 13.22 ⫽ 1.0513.22 ⫺ 0.09 ⬇ 3.3, yet the data gives an actual value of only 2.9. When working with data and function models, we should expect some variation when the two are compared, especially if the correlation is weak. Applications of data analysis can be found in virtually all fields of study. In Example 5 we apply these ideas to an Olympic swimming event.

Finding a Linear Function to Model the Relationship (Year, Gold Medal Times) The men’s 400-m freestyle times (gold medal times — to the nearest second) for the 1976 through 2008 Olympics are given in Table 1.8 (1900 S 0). Let the year be the input x, and winning race time be the output y. Based on the data, draw a scatterplot and answer the following questions. a. Does the association appear linear or nonlinear? b. Is the association positive or negative? c. Classify the correlation as weak or strong. d. Find a function model that approximates the data, then use it to predict the winning time for the 2012 Olympics. Table 1.8 Year (x) (1900 S 0)

Time ( y) (sec)

76

232

80

231

84

231

88

227

92

225

96

228

100

221

104

223

108

223

242

234

Time (sec)

168

226

218

210

Solution

䊳

76

84

92

100

108

Year

Begin by choosing appropriate scales for the axes. The x-axis (year) could be scaled from 76 to 112, and the y-axis (swim time) from 210 to 246. This will allow for a “frame” around the data. After plotting the points, we obtain the scatterplot shown in the figure. a. The association appears to be linear. b. The association is negative, showing that finishing times tend to decrease over time. c. There is a moderate to strong correlation. d. The points (76, 232) and (104, 223) appear to be on a line approximating the data, and we select these to develop our equation model. y2 ⫺ y1 x2 ⫺ x1 223 ⫺ 232 ⫽ 104 ⫺ 76 ⬇ ⫺0.32 y ⫺ 232 ⫽ ⫺0.321x ⫺ 762 y ⫺ 232 ⫽ ⫺0.32x ⫹ 24.32 y ⫽ ⫺0.32x ⫹ 256.32 m⫽

slope formula 1x1, y1 2 S 176, 2322 , 1x2, y2 2 S 1104, 2232 slope (rounded to tenths) point-slope form distribute add 232 (solve for y )

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169

One model for this data is y ⫽ ⫺0.32x ⫹ 256.32. Based on this model, the predicted time for the 2012 Olympics would be f 1x2 ⫽ ⫺0.32x ⫹ 256.32 f 11122 ⫽ ⫺0.3211122 ⫹ 256.32 ⫽ 220.48

function model substitute 112 for x (2012) result

In 2012 the winning time is projected to be about 220.5 sec. Now try Exercises 23 and 24

D. You’ve just seen how we can find a linear function that models relationships observed in a set of data

䊳

As a reminder, great care should be taken when using equation models obtained from real data. It would be foolish to assume that in the year 2700, swim times for the 400-m freestyle would be near 0 sec—even though that’s what the model predicts for x ⫽ 800. Most function models are limited by numerous constraining factors, and data collected over a much longer period of time might even be better approximated using a nonlinear model.

E. Linear Regression and the Line of Best Fit There is actually a sophisticated method for calculating the equation of a line that best fits a data set, called the regression line. The method minimizes the vertical distance between all data points and the line itself, making it the unique line of best fit. Most graphing calculators have the ability to perform this calculation quickly. The process involves these steps: (1) clearing old data, (2) entering new data, (3) displaying the data, (4) calculating the regression line, and (5) displaying and using the regression line. We’ll illustrate by finding the regression line for the data shown in Table 1.8 in Example 5, which gives the men’s 400-m freestyle gold medal times (in seconds) for the 1976 through the 2008 Olympics, with 1900S0.

Step 1: Clear Old Data To prepare for the new data, we first clear out any old data. Press the STAT key and select option 4:ClrList. This places the ClrList command on the home screen. We tell the calculator which lists to clear by pressing 2nd 1 to indicate List1 (L1), then enter a comma using the , key, and continue entering other lists we want to clear: 2nd 2nd 2 , 3 will clear List1 (L1), List2 (L2), and List3 (L3). ENTER

Step 2: Enter New Data Press the STAT key and select option 1:Edit. Move the cursor to the first position of List1, and simply enter the data from the first column of Table 1.8 in order: 76 80 84 , and so on. Then use the right arrow to navigate to List2, and enter the data from the second column: 232 231 231 , and so on. When finished, you should obtain the screen shown in Figure 1.97.

Figure 1.97

Step 3: Display the Data

Figure 1.98

ENTER

ENTER

ENTER

ENTER

WORTHY OF NOTE As a rule of thumb, the tick marks for Xscl can be set by mentally 冟Xmax冟 ⫹ 冟Xmin冟 estimating and 10 using a convenient number in the neighborhood of the result (the same goes for Yscl). As an alternative to manually setting the window, the ZOOM 9:ZoomStat feature can be used.

ENTER

ENTER

With the data held in these lists, we can now display the related ordered pairs on the coordinate grid. First press the Y= key and any existing equations. Y= Then press 2nd to access the “STATPLOTS” screen. With the cursor on 1:Plot1, press and be sure the options shown in Figure 1.98 are highlighted. If you need to make any changes, navigate the cursor to CLEAR

ENTER

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the desired option and press . Note the data in L1 ranges from 76 to 108, while the data in L2 ranges from 221 to 232. This means an appropriate viewing window might be [70, 120] for the x-values, and [210, 240] for the y-values. Press the key and set up the window accordingly. After you’re finished, pressing the GRAPH key should produce the graph shown in Figure 1.99.

Figure 1.99

ENTER

WINDOW

240

70

120

Step 4: Calculate the Regression Equation

210

To have the calculator compute the regression equation, press the STAT and keys to move the cursor over to the CALC options (see Figure 1.100). Since it appears the data is best modeled by a linear equation, we choose option 4:LinReg(ax ⴙ b). Pressing the number 4 places this option on the home screen, and pressing computes the values of a and b (the calculator automatically uses the values in L1 and L2 unless instructed otherwise). Rounded to hundredths, the linear regression model is y ⫽ ⫺0.33x ⫹ 257.06 (Figure 1.101).

Figure 1.100

ENTER

Figure 1.101

Step 5: Display and Use the Results Although graphing calculators have the ability to paste the regression equation directly into Y1 on the Y= screen, for now we’ll enter Y1 ⫽ ⫺0.33x ⫹ 257.06 by hand. Afterward, pressing the GRAPH key will plot the data points (if Plot1 is still active) and graph the line. Your display screen should now look like the one in Figure 1.102. The regression line is the best estimator for the set of data as a whole, but there will still be some difference between the values it generates and the values from the set of raw data (the output in Figure 1.102 shows the estimated time for the 2000 Olympics was about 224 sec, when actually it was the year Ian Thorpe of Australia set a world record of 221 sec).

EXAMPLE 6

䊳

Figure 1.102 240

70

䊳

210

Using Regression to Model Employee Performance Riverside Electronics reviews employee performance semiannually, and awards increases in their hourly rate of pay based on the review. The table shows Thomas’ hourly wage for the last 4 yr (eight reviews). Find the regression equation for the data and use it to project his hourly wage for the year 2011, after his fourteenth review.

Solution

120

Following the prescribed sequence produces the equation y ⫽ 0.48x ⫹ 9.09. For x ⫽ 14 we obtain y ⫽ 0.481142 ⫹ 9.09 or a wage of $15.81. According to this model, Thomas will be earning $15.81 per hour in 2011.

Review (x)

Wage ( y)

(2004) 1

$9.58

2

$9.75

(2005) 3

$10.54

4

$11.41

(2006) 5

$11.60

6

$11.91

(2007) 7

$12.11

8

$13.02


䊳

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With each linear regression, the calculator can be set to compute a correlation coefficient that is a measure of how well the equation fits the data (see Subsection C). To 0 display this “r-value” use 2nd (CATALOG) and activate DiagnosticOn. Figure 1.103 shows a scatterplot with perfect negative correlation 1r ⫽ ⫺12 and notice all data points are on the line. Figure 1.104 shows a strong positive correlation 1r ⬇ 0.982 for the data from Example 6. See Exercise 35. Figure 1.103

Figure 1.104

E. You’ve just seen how we can use a linear regression to find the line of best fit

1.6 EXERCISES 䊳



1. When the ordered pairs from a set of data are plotted on a coordinate grid, the result is called a .

2. If the data points seem to form a curved pattern or if no pattern is apparent, the data is said to have a association.

3. If the data points seems to cluster along an imaginary line, the data is said to have a association.

4. If the pattern of data points seems to increase as they are viewed left to right, the data is said to have a association.

5. Compare/Contrast: One scatterplot is linear, with a weak and positive association. Another is linear, with a strong and negative association. Give a written description of each scatterplot.

6. Discuss/Explain how this is possible: Working from the same scatterplot, Demetrius obtained the equation y ⫽ ⫺0.64x ⫹ 44 for his equation model, while Jessie got the equation y ⫽ ⫺0.59x ⫹ 42.

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DEVELOPING YOUR SKILLS 7. For mail with a high priority, x “Express Mail” offers next day 1981 delivery by 12:00 noon to most 1985 destinations, 365 days of the 1988 year. The service was first offered by the U.S. Postal 1991 Service in the early 1980s and 1995 has been growing in use ever 1999 since. The cost of the service 2002 (in cents) for selected years is shown in the table. (a) Draw a 2010 scatterplot of the data, then (b) decide if the association is positive, negative, or cannot be determined.

y 935 1075 1200 1395 1500 1575 1785 1830

Source: 2004 Statistical Abstract of the United States; USPS.com

8. After the Surgeon General’s x y first warning in 1964, 1965 42.4 cigarette consumption began a 1974 37.1 steady decline as advertising was banned from television 1979 33.5 and radio, and public 1985 29.9 awareness of the dangers of 1990 25.3 cigarette smoking grew. The 1995 24.6 percentage of the U.S. adult population who considered 2000 23.1 themselves smokers is shown 2002 22.4 in the table for selected years. 2005 16.9 (a) Draw a scatterplot of the data, then (b) decide if the association is positive, negative, or cannot be determined. Source: 1998 Wall Street Journal Almanac and 2009 Statistical Abstract of the United States, Table 1299

9. Since the 1970s women have x y made tremendous gains in the 1972 32 political arena, with more and 1978 46 more female candidates running 1984 65 for, and winning seats in the U.S. Senate and U.S. Congress. 1992 106 The number of women 1998 121 candidates for the U.S. Congress 2004 141 is shown in the table for selected years. (a) Draw a scatterplot of the data, (b) decide if the association is linear or nonlinear and (c) if the association is positive, negative, or cannot be determined. Source: Center for American Women and Politics at www.cawp.rutgers.edu/Facts3.html

10. The number of shares traded on the New York Stock Exchange experienced dramatic change in the 1990s as more and more individual investors gained access to the stock market via the Internet

and online brokerage houses. The volume is shown in the table for 2002, and the odd numbered years from 1991 to 2001 (in billions of shares). (a) Draw a scatterplot of the data, (b) decide if the association is linear or nonlinear, and (c) if the association is positive, negative, or cannot be determined.

x

y

1991

46

1993

67

1995

88

1997

134

1999

206

2001

311

2002

369

Source: 2000 and 2004 Statistical Abstract of the United States, Table 1202

The data sets in Exercises 11 and 12 are known to be linear.

11. The total value of the goods x and services produced by a (1970 S 0) y nation is called its gross 0 5.1 domestic product or GDP. The 5 7.6 GDP per capita is the ratio of 10 12.3 the GDP for a given year to the 15 17.7 population that year, and is one of many indicators of 20 23.3 economic health. The GDP per 25 27.7 capita (in $1000s) for the 30 35.0 United States is shown in the 33 37.8 table for selected years. (a) Draw a scatterplot using scales that appropriately fit the data, then sketch an estimated line of best fit, (b) decide if the association is positive or negative, then (c) decide whether the correlation is weak or strong. Source: 2004 Statistical Abstract of the United States, Tables 2 and 641

12. Real estate brokers carefully Price Sales track sales of new homes 130’s 126 looking for trends in location, 150’s 95 price, size, and other factors. 170’s 103 The table relates the average selling price within a price 190’s 75 range (homes in the $120,000 210’s 44 to $140,000 range are 230’s 59 represented by the $130,000 250’s 21 figure), to the number of new homes sold by Homestead Realty in 2004. (a) Draw a scatterplot using scales that appropriately fit the data, then sketch an estimated line of best fit, (b) decide if the association is positive or negative, then (c) decide whether the correlation is weak or strong.

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For the scatterplots given: (a) Arrange them in order from the weakest to the strongest correlation, (b) sketch a line that seems to approximate the data, (c) state whether the association is positive, negative, or cannot be determined, and (d) choose two points on (or near) the line and use them to approximate its slope (rounded to one decimal place).

13. A.

y 60 55 50 45 40 35 30 25

B.

y 60 55 50 45 40 35 30 25

0 1 2 3 4 5 6 7 8 9 x

C.

y 60 55 50 45 40 35 30 25

0 1 2 3 4 5 6 7 8 9 x

D.

y 60 55 50 45 40 35 30 25

0 1 2 3 4 5 6 7 8 9 x

14. A.

y 60 55 50 45 40 35 30 25

0 1 2 3 4 5 6 7 8 9 x

B.

y 60 55 50 45 40 35 30 25

0 1 2 3 4 5 6 7 8 9 x

C.

y 60 55 50 45 40 35 30 25 0 1 2 3 4 5 6 7 8 9 x

y 60 55 50 45 40 35 30 25 0 1 2 3 4 5 6 7 8 9 x

For the scatterplots given, (a) determine whether a linear or nonlinear model would seem more appropriate. (b) Determine if the association is positive or negative. (c) Classify the correlation as weak or strong. (d) If linear, sketch a line that seems to approximate the data and choose two points on the line and use them to approximate its slope.

15.

y 60 55 50 45 40 35 30 25

16.

y 60 55 50 45 40 35 30 25

0 1 2 3 4 5 6 7 8 9 x

17.

y 60 55 50 45 40 35 30 25 0 1 2 3 4 5 6 7 8 9 x

0 1 2 3 4 5 6 7 8 9 x

18.

y 60 55 50 45 40 35 30 25 0 1 2 3 4 5 6 7 8 9 x

20.

y 60 55 50 45 40 35 30 25 0 1 2 3 4 5 6 7 8 9 x

21. In most areas of the country, x y law enforcement has become (1990→0) (1000s) a major concern. The number 3 68.8 of law enforcement officers 6 74.5 employed by the federal 8 83.1 government and having the authority to carry firearms 10 88.5 and make arrests is shown in 14 93.4 the table for selected years. (a) Draw a scatterplot using scales that appropriately fit the data and sketch an estimated line of best fit and (b) decide if the association is positive or negative. (c) Choose two points on or near the estimated line of best fit, and use them to find a function model and predict the number of federal law enforcement officers in 1995 and the projected number for 2011. Answers may vary. Source: U.S. Bureau of Justice, Statistics at www.ojp.usdoj.gov/bjs/fedle.htm

0 1 2 3 4 5 6 7 8 9 x

D.

19.

173

y 60 55 50 45 40 35 30 25 0 1 2 3 4 5 6 7 8 9 x

22. Due to atmospheric pressure, x y the temperature at which ⫺1000 213.8 water will boil varies 0 212.0 predictably with the altitude. Using special equipment 1000 210.2 designed to duplicate 2000 208.4 atmospheric pressure, a lab 3000 206.5 experiment is set up to study 4000 204.7 this relationship for altitudes 5000 202.9 up to 8000 ft. The set of data 6000 201.0 collected is shown in the table, with the boiling temperature y 7000 199.2 in degrees Fahrenheit, 8000 197.4 depending on the altitude x in feet. (a) Draw a scatterplot using scales that appropriately fit the data and sketch an estimated line of best fit, (b) decide if the association is positive or negative. (c) Choose two points on or near the estimated line of best fit, and use them to find a function model and predict the boiling point of water on the summit of Mt. Hood in Washington State (11,239 ft height), and along the shore of the Dead Sea (approximately 1312 ft below sea level). Answers may vary. 23. For the data given in Exercise 11 (Gross Domestic Product per Capita), choose two points on or near the line you sketched and use them to find a function model for the data. Based on this model, what is the projected GDP per capita for the year 2010?

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24. For the data given in Exercise 12 (Sales by Real Estate Brokers), choose two points on or near the line you sketched and use them to find a function

䊳

model for the data. Based on this model, how many sales can be expected for homes costing $275,000? $300,000?


25. Circumference of a Circle: C ⴝ 2␲r: The formula for the circumference of a circle can be written as a function of C in terms of r: C1r2 ⫽ 2␲r. (a) Set up a table of values for r ⫽ 1 through 6 and draw a scatterplot of the data. (b) Is the association positive or negative? Why? (c) What can you say about the strength of the correlation? (d) Sketch a line that “approximates” the data. What can you say about the slope of this line? 26. Volume of a Cylinder: V ⴝ ␲r2h: As part of a project, students cut a long piece of PVC pipe with a diameter of 10 cm into sections that are 5, 10, 15, 20, and 25 cm long. The bottom of each is then made watertight and each section is filled to the 䊳

1–90


brim with water. The volume Height Volume is then measured using a flask (cm) (cm3) marked in cm3 and the results 5 380 collected into the table shown. (a) Draw a scatterplot 10 800 of the data. (b) Is the 15 1190 association positive or 20 1550 negative? Why? (c) What can 25 1955 you say about the strength of the correlation? (d) Would the correlation here be stronger or weaker than the correlation in Exercise 25? Why? (e) Run a linear regression to verify your response.

APPLICATIONS

Use the regression capabilities of a graphing calculator to complete Exercises 27 through 34.

27. Height versus wingspan: Leonardo da Vinci’s famous diagram is an illustration of how the human body comes in predictable proportions. One such comparison is a person’s height to their wingspan (the maximum distance from the outstretched tip of one middle finger to the other). Careful measurements were taken on eight students and the set of data is shown here. Using the data, (a) draw the scatterplot; (b) determine whether the association is linear or nonlinear; (c) determine whether the association is positive or negative; and (d) find the regression equation and use it to predict the wingspan of a student with a height of 65 in. Height (x)

Wingspan ( y)

28. Patent applications: Every year the U.S. Patent and Trademark Office (USPTO) receives thousands of applications from scientists and inventors. The table given shows the number of applications received for the odd years from 1993 to 2003 (1990 S 0). Use the data to (a) draw the scatterplot; (b) determine whether the association is linear or nonlinear; (c) determine whether the association is positive or negative; and (d) find the regression equation and use it to predict the number of applications that will be received in 2011. Source: United States Patent and Trademark Office at www.uspto.gov/web

Year (1990 S 0)

Applications (1000s)

61

60.5

3

188.0

61.5

62.5

5

236.7

54.5

54.5

7

237.0

73

71.5

9

278.3

67.5

66

11

344.7

51

50.75

13

355.4

57.5

54

52

51.5

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29. Patents issued: An Year Patents increase in the (1990 S 0) (1000s) number of patent 3 107.3 applications (see Exercise 28), typically 5 114.2 brings an increase in 7 122.9 the number of patents 9 159.2 issued, though many 11 187.8 applications are denied due to 13 189.6 improper filing, lack of scientific support, and other reasons. The table given shows the number of patents issued for the odd years from 1993 to 2003 (1999 S 0). Use the data to (a) draw the scatterplot; (b) determine whether the association is linear or nonlinear; (c) determine whether the association is positive or negative; and (d) find the regression equation and use it to predict the number of applications that will be approved in 2011. Which is increasing faster, the number of patent applications or the number of patents issued? How can you tell for sure? Source: United States Patent and Trademark Office at www.uspto.gov/web

30. High jump records: In the Year Height sport of track and field, the (1900S 0) in. high jumper is an unusual 0 75 athlete. They seem to defy gravity as they launch their 12 76 bodies over the high bar. 24 78 The winning height at the 36 80 summer Olympics (to the 56 84 nearest unit) has steadily increased over time, as 68 88 shown in the table for 80 93 selected years. Using the 88 94 data, (a) draw the 92 92 scatterplot, (b) determine whether the association 96 94 is linear or nonlinear, 100 93 (c) determine whether the 104 association is positive or 108 negative, and (d) find the regression equation using t ⫽ 0 corresponding to 1900 and predict the winning height for the 2004 and 2008 Olympics. How close did the model come to the actual heights? Source: athens2004.com

31. Females/males in the workforce: Over the last 4 decades, the percentage of the female population in the workforce has been increasing at a fairly steady rate. At the same time, the percentage of the male population in the workforce has been declining. The set of data is shown in the tables. Using the data, (a) draw scatterplots for both data sets, (b) determine whether the associations are linear or nonlinear, (c) determine whether the associations are positive or negative, and (d) determine if the percentage of females in the workforce is increasing faster than the percentage of males is decreasing. Discuss/Explain how you can tell for sure. Source: 1998 Wall Street Journal Almanac, p. 316

Exercise 31 (women)

Exercise 31 (men)

Year (x) (1950 S 0)

Percent

Year (x) (1950 S 0)

Percent

5

36

5

85

10

38

10

83

15

39

15

81

20

43

20

80

25

46

25

78

30

52

30

77

35

55

35

76

40

58

40

76

45

59

45

75

50

60

50

73

32. Height versus male shoe Height Shoe Size size: While it seems 66 8 reasonable that taller people should have larger 69 10 feet, there is actually a 72 9 wide variation in the 75 14 relationship between 74 12 height and shoe size. The data in the table show the 73 10.5 height (in inches) 71 10 compared to the shoe size 69.5 11.5 worn for a random sample 66.5 8.5 of 12 male chemistry students. Using the data, 73 11 (a) draw the scatterplot, 75 14 (b) determine whether the 65.5 9 association is linear or nonlinear, (c) determine whether the association is positive or negative, and (d) find the regression equation and use it to predict the shoe size of a man 80 in. tall and another that is 60 in. tall. Note that the heights of these two men fall outside of the range of our data set (see comment after Example 5 on page 168).

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33. Plastic money: The total x amount of business (1990 S 0) y transacted using credit 1 481 cards has been changing 2 539 rapidly over the last 15 to 20 years. The total 4 731 volume (in billions of 7 1080 dollars) is shown in the 8 1157 table for selected years. 9 1291 (a) Use a graphing calculator to draw a 10 1458 scatterplot of the data 12 1638 and decide whether the association is linear or nonlinear. (b) Calculate a regression equation with x ⫽ 0 corresponding to 1990 and display the scatterplot and graph on the same screen. (c) According to the equation model, how many billions of dollars were transacted in 2003? How much will be transacted in the year 2011? Source: Statistical Abstract of the United States, various years

34. Sales of hybrid Year Hybrid Sales cars: Since their (2000 S 0) (in thousands) mass introduction 2 35 near the turn of the century, the 3 48 sales of hybrid 4 88 cars in the United 5 200 States grew 6 250 steadily until late 2007, when 7 352 the price of 8 313 gasoline began 9 292 showing signs of weakening and eventually dipped below $3.00/gal. Estimates for the annual sales of hybrid cars are given in the table for the years 2002 through 2009 12000 S 02 . (a) Use a graphing calculator to draw a scatterplot of the data and decide if the association is linear or nonlinear. (b) If linear, calculate a regression model for the data and display the scatterplot and data on the same screen. (c) Assuming that sales of hybrid cars recover, how many hybrids does the model project will be sold in the year 2012? Source: http://www.hybridcar.com

䊳


35. It can be very misleading to x y rely on the correlation 50 67 coefficient alone when 100 125 selecting a regression model. 150 145 To illustrate, (a) run a linear regression on the data set 200 275 given (without doing a 250 370 scatterplot), and note the 300 550 strength of the correlation 350 600 (the correlation coefficient). (b) Now run a quadratic regression ( STAT CALC 5:QuadReg) and note the strength of the correlation. (c) What do you notice? What factors other than the correlation coefficient must be taken into account when choosing a form of regression?

36. In his book Gulliver’s Travels, Jonathan Swift describes how the Lilliputians were able to measure Gulliver for new clothes, even though he was a giant compared to them. According to the text, “Then they measured my right thumb, and desired no more . . . for by mathematical computation, once around the thumb is twice around the wrist, and so on to the neck and waist.” Is it true that once around the neck is twice around the waist? Find at least 10 willing subjects and take measurements of their necks and waists in millimeters. Arrange the data in ordered pair form (circumference of neck, circumference of waist). Draw the scatterplot for this data. Does the association appear to be linear? Find the equation of the best fit line for this set of data. What is the slope of this line? Is the slope near m ⫽ 2?

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Making Connections


37. (1.3) Is the graph shown here, the graph of a function? Discuss why or why not.

38. (R.2/R.3) Determine the area of the figure shown 1A ⫽ LW, A ⫽ ␲r2 2.

18 cm 24 cm

39. (1.5) Solve for r: A ⫽ P ⫹ Prt 40. (R.3) Solve for w (if possible): ⫺216w2 ⫹ 52 ⫺ 1 ⫽ 7w ⫺ 413w2 ⫹ 12

MAKING CONNECTIONS Making Connections: Graphically, Symbolically, Numerically, and Verbally Eight graphs (a) through (h) are given. Match the characteristics shown in 1 through 16 to one of the eight graphs. y

(a)

⫺5

y

(b)

5

⫺5

5 x

y

5 x

⫺5

5 x

⫺5

1 1. ____ y ⫽ x ⫹ 1 3 2. ____ y ⫽ ⫺x ⫹ 1 3. ____ m 7 0, b 6 0 4. ____ x ⫽ ⫺1

⫺5

5 x

5 x

⫺5

y

(g)

5

⫺5

5

⫺5

y

(f)

5

⫺5

⫺5

5 x

y

(d)

5

⫺5

⫺5

(e)

y

(c)

5

⫺5

y

(h)

5

5 x

5

⫺5

⫺5

5 x

⫺5

9. ____ f 1⫺32 ⫽ 4, f 112 ⫽ 0

10. ____ f 1⫺42 ⫽ 3, f 142 ⫽ 3

11. ____ f 1x2 ⱖ 0 for x 僆 3⫺3, q2 12. ____ x ⫽ 3

5. ____ y ⫽ ⫺2

13. ____ f 1x2 ⱕ 0 for x 僆 31, q 2

6. ____ m 6 0, b 6 0

14. ____ m is zero

7. ____ m ⫽ ⫺2

15. ____ function is increasing, y-intercept is negative

8. ____ m ⫽

2 3

16. ____ function is decreasing, y-intercept is negative

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SUMMARY AND CONCEPT REVIEW SECTION 1.1


KEY CONCEPTS • A relation is a collection of ordered pairs (x, y) and can be stated as a set or in equation form. • As a set of ordered pairs, we say the relation is pointwise-defined. The domain of the relation is the set of all first coordinates, and the range is the set of all corresponding second coordinates. • A relation can be expressed in mapping notation x S y, indicating an element from the domain is mapped to (corresponds to or is associated with) an element from the range. • The graph of a relation in equation form is the set of all ordered pairs (x, y) that satisfy the equation. We plot a sufficient number of points and connect them with a straight line or smooth curve, depending on the pattern formed. • The x- and y-variables of linear equations and their graphs have implied exponents of 1. • With a relation entered on the Y= screen, a graphing calculator can provide a table of ordered pairs and the related graph. x1 ⫹ x2 y1 ⫹ y2 , b. • The midpoint of a line segment with endpoints (x1, y1) and (x2, y2) is a 2 2

• The distance between the points (x1, y1) and (x2, y2) is d ⫽ 21x2 ⫺ x1 2 2 ⫹ 1y2 ⫺ y1 2 2. • The equation of a circle centered at (h, k) with radius r is 1x ⫺ h2 2 ⫹ 1y ⫺ k2 2 ⫽ r2.

EXERCISES 1. Represent the relation in mapping notation, then state the domain and range. 51⫺7, 32, 1⫺4, ⫺22, 15, 12, 1⫺7, 02, 13, ⫺22, 10, 82 6 2. Graph the relation y ⫽ 225 ⫺ x2 by completing the table, then state the domain and range of the relation. x ⫺5 ⫺4 ⫺2 0 2 4 5

y

3. Use a graphing calculator to graph the relation 5x ⫹ 3y ⫽ ⫺15. Then use the TABLE feature to determine the value of y when x ⫽ 0, and the value(s) of x when y ⫽ 0, and write the results in ordered pair form. Mr. Northeast and Mr. Southwest live in Coordinate County and are good friends. Mr. Northeast lives at 19 East and 25 North or (19, 25), while Mr. Southwest lives at 14 West and 31 South or (⫺14, ⫺31). If the streets in Coordinate County are laid out in one mile squares, 4. Use the distance formula to find how far apart they live. 5. If they agree to meet halfway between their homes, what are the coordinates of their meeting place? 6. Sketch the graph of x2 ⫹ y2 ⫽ 16. 7. Sketch the graph of x2 ⫹ y2 ⫹ 6x ⫹ 4y ⫹ 9 ⫽ 0. 8. Find an equation of the circle whose diameter has the endpoints (⫺3, 0) and (0, 4).

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Summary and Concept Review

179

Linear Equations and Rates of Change

SECTION 1.2

KEY CONCEPTS • A linear equation can be written in the form ax ⫹ by ⫽ c, where a and b are not simultaneously equal to 0. y2 ⫺ y1 • The slope of the line through (x1, y1) and (x2, y2) is m ⫽ x ⫺ x , where x1 ⫽ x2. 2 1 ¢y vertical change change in y rise ⫽ ⫽ . • Other designations for slope are m ⫽ run ⫽ change in x ¢x horizontal change • Lines with positive slope 1m 7 02 rise from left to right; lines with negative slope (m 6 0) fall from left to right. • The equation of a horizontal line is y ⫽ k; the slope is m ⫽ 0. • The equation of a vertical line is x ⫽ h; the slope is undefined. • Lines can be graphed using the intercept method. First determine (x, 0) (substitute 0 for y and solve for x), then (0, y) (substitute 0 for x and solve for y). Then draw a straight line through these points. Parallel lines have equal slopes (m1 ⫽ m2); perpendicular lines have slopes that are negative reciprocals • 1 (m1 ⫽ ⫺ or m1 # m2 ⫽ ⫺1). m2 EXERCISES ¢y rise 9. Plot the points and determine the slope, then use the ratio ⫽ to find an additional point on the line: run ¢x a. 1⫺4, 32 and 15, ⫺22 and b. (3, 4) and 1⫺6, 12. 10. Use the slope formula to determine if lines L1 and L2 are parallel, perpendicular, or neither: a. L1: 1⫺2, 02 and (0, 6); L2: (1, 8) and (0, 5) b. L1: (1, 10) and 1⫺1, 72 : L2: 1⫺2, ⫺12 and 11, ⫺32 11. Graph each equation by plotting points: (a) y ⫽ 3x ⫺ 2 (b) y ⫽ ⫺32x ⫹ 1. 12. Find the intercepts for each line and sketch the graph: (a) 2x ⫹ 3y ⫽ 6 (b) y ⫽ 43x ⫺ 2. 13. Identify each line as either horizontal, vertical, or neither, and graph each line. a. x ⫽ 5 b. y ⫽ ⫺4 c. 2y ⫹ x ⫽ 5 14. Determine if the triangle with the vertices given is a right triangle: 1⫺5, ⫺42, (7, 2), (0, 16). 15. Find the slope and y-intercept of the line shown and discuss the slope ratio in this context. Hawk population (100s)

10

y

8 6 4 2 2

4

6

x

8

Rodent population (1000s)

SECTION 1.3

Functions, Function Notation, and the Graph of a Function

KEY CONCEPTS • A function is a relation, rule, or equation that pairs each element from the domain with exactly one element of the range. • The vertical line test says that if every vertical line crosses the graph of a relation in at most one point, the relation is a function. • The domain and range can be stated using set notation, graphed on a number line, or expressed using interval notation.

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1–96


• On a graph, vertical boundary lines can be used to identify the domain, or the set of “allowable inputs” for a • • • •

function. On a graph, horizontal boundary lines can be used to identify the range, or the set of y-values (outputs) generated by the function. When a function is stated as an equation, the implied domain is the set of x-values that yield real number outputs. x-values that cause a denominator of zero or that cause the radicand of a square root expression to be negative must be excluded from the domain. The phrase “y is a function of x,” is written as y ⫽ f 1x2 . This notation enables us to summarize the three most important aspects of a function with a single expression (input, sequence of operations, output).

EXERCISES 16. State the implied domain of each function: a. f 1x2 ⫽ 24x ⫹ 5

b. g1x2 ⫽

17. Determine h1⫺22, h1⫺23 2 , and h(3a) for h1x2 ⫽ 2x2 ⫺ 3x.

x⫺4 x ⫺x⫺6 2

18. Determine if the mapping given represents a function. If not, explain how the definition of a function is violated. Mythological deities

Primary concern

Apollo Jupiter Ares Neptune Mercury Venus Ceres Mars

messenger war craftsman love and beauty music and healing oceans all things agriculture

19. For the graph of each function shown, (a) state the domain and range, (b) find the value of f (2), and (c) determine the value(s) of x for which f 1x2 ⫽ 1. I.

II.

y 5

⫺5

5 x

⫺5

SECTION 1.4

III.

y 5

⫺5

5 x

⫺5

y 5

⫺5

5 x

⫺5

Linear Functions, Special Forms, and More on Rates of Change

KEY CONCEPTS • The equation of a nonvertical line in slope-intercept form is y ⫽ mx ⫹ b or f 1x2 ⫽ mx ⫹ b. The slope of the line is m and the y-intercept is (0, b). ¢y to • To graph a line given its equation in slope-intercept form, plot the y-intercept, then use the slope ratio m ⫽ ¢x find a second point, and draw a line through these points. • If the slope m and a point (x1, y1) on the line are known, the equation of the line can be written in point-slope form: y ⫺ y1 ⫽ m1x ⫺ x1 2 .

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• A secant line is the straight line drawn through two points on a nonlinear graph. ¢y literally means the quantity measured along the y-axis is changing with respect to changes ¢x in the quantity measured along the x-axis.

• The notation m ⫽

27. For the graph given, (a) find the equation of the line in point-slope form, (b) use the equation to predict the x- and y-intercepts, (c) write the equation in slope-intercept form, and (d) find y when x ⫽ 20, and the value of x for which y ⫽ 15.

SECTION 1.5

Exercise 26 W Wolf population (100s)

EXERCISES 20. Write each equation in slope-intercept form, then identify the slope and y-intercept. a. 4x ⫹ 3y ⫺ 12 ⫽ 0 b. 5x ⫺ 3y ⫽ 15 21. Graph each equation using the slope and y-intercept. a. f 1x2 ⫽ ⫺23 x ⫹ 1 b. h1x2 ⫽ 52 x ⫺ 3 22. Graph the line with the given slope through the given point. a. m ⫽ 23; 11, 42 b. m ⫽ ⫺12; 1⫺2, 32 23. What are the equations of the horizontal line and the vertical line passing through 1⫺2, 52? Which line is the point (7, 5) on? 24. Find the equation of the line passing through (1, 2) and 1⫺3, 52. Write your final answer in slope-intercept form. 25. Find the equation for the line that is parallel to 4x ⫺ 3y ⫽ 12 and passes through the point (3, 4). Write your final answer in slope-intercept form. 26. Determine the slope and y-intercept of the line shown. Then write the equation of ¢W the line in slope-intercept form and interpret the slope ratio m ⫽ in the context ¢R of this exercise.

10 8 6 4 2

2

4

6

8

10

R

Rabbit population (100s)

Exercise 27 y 100 80 60 40 20 0

2

4

6

8

10 x

Solving Equations and Inequalities Graphically; Formulas and Problem Solving

KEY CONCEPTS • To use the intersection-of-graphs method for solving equations, assign the left-hand expression as Y1 and the right-hand as Y2. The solution(s) of the original equation are the x-coordinate(s) of the point(s) of intersection of the graphs of Y1 and Y2. • When an equation is written in the form h1x2 ⫽ 0, the solutions can be found using the x-Intercept/Zeroes method. Assign h(x) as Y1, and find the x-intercepts of its graph. • Linear inequalities can be solved by first applying the intersection-of-graphs method to identify the boundary value of the solution interval. Next, the solution is determined by a careful observation of the relative positions of the graphs (is Y1 above or below Y2) and the given inequality. • To solve formulas for a specified variable, focus on the object variable and apply properties of equality to write this variable in terms of all others. • The basic elements of good problem solving include: 1. Gathering and organizing information 2. Making the problem visual 3. Developing an equation model 4. Using the model to solve the application For a complete review, see the problem-solving guide on page 157.

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1–98


EXERCISES 28. Solve the following equation using the intersection-of-graphs method. 31x ⫺ 22 ⫹ 10 ⫽ 16 ⫺ 213 ⫺ 2x2 29. Solve the following equation using the x-intercept/zeroes method. 21x ⫺ 12 ⫹ 32 ⫽ 51 25x ⫹ 15 2 ⫺ 32 30. Solve the following inequality using the intersection-of-graphs method. 31x ⫹ 22 ⫺ 2.2 6 ⫺2 ⫹ 410.2 ⫺ 0.5x2 Solve for the specified variable in each formula or literal equation. 31. V ⫽ ␲r2h for h 32. P ⫽ 2L ⫹ 2W for L 33. ax ⫹ b ⫽ c for x

34. 2x ⫺ 3y ⫽ 6 for y

Use the problem-solving guidelines (page 157) to solve the following applications. 35. At a large family reunion, two kegs of lemonade are available. One is 2% sugar (too sour) and the second is 7% sugar (too sweet). How many gallons of the 2% keg, must be mixed with 12 gallons of the 7% keg to get a 5% mix? 36. A rectangular window with a width of 3 ft and a height of 4 ft is topped by a semi-circular window. Find the total area of the window. 37. Two cyclists start from the same location and ride in opposite directions, one riding at 15 mph and the other at 18 mph. If their radio phones have a range of 22 mi, how many minutes will they be able to communicate?

SECTION 1.6

Linear Function Models and Real Data

KEY CONCEPTS • A scatterplot is the graph of all the ordered pairs in a real data set. • When drawing a scatterplot, be sure to scale the axes to comfortably fit the data. • If larger inputs tend to produce larger output values, we say there is a positive association. • If larger inputs tend to produce smaller output values, we say there is a negative association. • If the data seem to cluster around an imaginary line, we say there is a linear association between the variables. • If the data clearly cannot be approximated by a straight line, we say the variables exhibit a nonlinear association (or sometimes no association). • The correlation coefficient r measures how tightly a set of data points cluster about an imaginary curve. The strength of the correlation is given as a value between ⫺1 and 1. Measures close to ⫺1 or 1 indicate a very strong correlation. Measures close to 0 indicate a very weak correlation. • We can attempt to model linear data sets using an estimated line of best fit. • A regression line minimizes the vertical distance between all data points and the graph itself, making it the unique line of best fit. EXERCISES Exercise 38 38. To determine the value of doing homework, a student in college algebra collects data x y on the time spent by classmates on their homework in preparation for a quiz. Her data (min study) (quiz score) is entered in the table shown. (a) Use a graphing calculator to draw a scatterplot of 45 70 the data. (b) Does the association appear linear or nonlinear? (c) Is the association 30 63 positive or negative? 10 59 39. If the association in Exercise 38 is linear, (a) use a graphing calculator to find a linear 20 67 function that models the relation (study time, grade), then (b) graph the data and the line 60 73 on the same screen. (c) Does the correlation appear weak or strong? 70 85 40. According to the function model from Exercise 39, what grade can I expect if I study 90 82 for 120 minutes? 75

90

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Practice Test

PRACTICE TEST

2. How much water that is 102°F must be mixed with 25 gal of water at 91°F, so that the resulting temperature of the water will be 97°F.

11. My partner and I are at coordinates 1⫺20, 152 on a map. If our destination is at coordinates 135, ⫺122 , (a) what are the coordinates of the rest station located halfway to our destination? (b) How far away is our destination? Assume that each unit is 1 mi. 12. Write the equations for lines L1 and L2 shown. y L1

3. To make the bowling team, Jacques needs a threegame average of 160. If he bowled 141 and 162 for the first two games, what score S must be obtained in the third game so that his average is at least 160? 4. In the 2009 movie Star Trek (Chris Pine, Zachary Quinto, Zoy Zaldana, Eric Bana), Sulu falls off of the drill platform without a parachute, and Kirk dives off the platform to save him. To slow his fall, Zulu uses a spread-eagle tactic, while Kirk keeps his body straight and arms at his side, to maximize his falling speed. If Sulu is falling at a rate of 180 ft/sec, while Kirk is falling at 250 ft/sec, how long would it take Kirk to reach Sulu, if it took Kirk a full 2 sec to react and dive after Sulu? 5. Two relations here are functions and two are not. Identify the nonfunctions (justify your response). a. x ⫽ y2 ⫹ 2y b. y ⫽ 25 ⫺ 2x c. 冟y冟 ⫹ 1 ⫽ x d. y ⫽ x2 ⫹ 2x 6. Determine whether the lines are parallel, perpendicular, or neither: L1: 2x ⫹ 5y ⫽ ⫺15 and L2: y ⫽ 25 x ⫹ 7. 7. Graph the line using the slope and y-intercept: x ⫹ 4y ⫽ 8 8. Find the center and radius of the circle defined by x2 ⫹ y2 ⫺ 4x ⫹ 6y ⫺ 3 ⫽ 0, then sketch its graph. 9. After 2 sec, a car is traveling 20 mph. After 5 sec, its speed is 40 mph. Assuming the relationship is linear, find the velocity equation and use it to determine the speed of the car after 9 sec. 10. Find the equation of the line parallel to 6x ⫹ 5y ⫽ 3, containing the point 12, ⫺22 . Answer in slopeintercept form.

5

L2

⫺5

5 x

⫺5

13. State the domain and range for the relations shown on graphs 13(a) and 13(b). Exercise 13(a)

Exercise 13(b) y

y 5

⫺4

5

6 x

⫺5

⫺4

6 x

⫺5

W(h) 14. For the linear function shown, a. Determine the value of W(24) from the graph. b. What input h will give an output of W1h2 ⫽ 375? c. Find a linear function for Hours worked the graph. d. What does the slope indicate in this context? e. State the domain and range of h. 2 ⫺ x2 15. Given f 1x2 ⫽ , evaluate and simplify: x2 a. f 1 23 2 b. f 1a ⫹ 32 500 400

Wages earned

1. Solve each equation. 2 a. ⫺ x ⫺ 5 ⫽ 7 ⫺ 1x ⫹ 32 3 b. ⫺5.7 ⫹ 3.1x ⫽ 14.5 ⫺ 41x ⫹ 1.52 c. P ⫽ C ⫹ k C; for C d. P ⫽ 2L ⫹ 2W; for W

300 200 100

0

8

16

24

32

40

h

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Exercise 16 16. In 2007, there were 3.3 million Apple iPhones sold worldwide. By 2009, this figure had jumped to approximately 30.3 million [Source: http://brainstormtech.blogs. fortune.cnn.com/2009/03/12/]. Assume that for a time, this growth could be modeled by a linear function. (a) Determine ¢sales the rate of change , and ¢time (b) interpret it in this context. Then use the rate of change to (c) approximate the number of sales in 2008, and what the projected sales would be for 2010 and 2011.

17. Solve the following equations using the x-intercept/zeroes method. 1 a. 2x ⫹ a4 ⫺ xb ⫽ ⫺120 ⫹ x2 3 b. 210.7x ⫺ 1.32 ⫹ 2.6 ⫽ 2x ⫺ 310.2x ⫺ 22 18. Solve the following inequalities using the intersection-of-graphs method. a. 3x ⫺ 15 ⫺ x2 ⱖ 215 ⫺ x2 ⫹ 3 b. 210.75x ⫺ 12 6 0.7 ⫹ 0.513x ⫺ 12

Exercise 19 19. To study how annual rainfall affects the ability Rainfall Cattle to attain certain levels of (in.) per Acre livestock production, a 12 2 local university collects 16 3 data on the average 19 7 annual rainfall for a 23 9 particular area and 28 11 compares this to the average number of 32 22 free-ranging cattle per 37 23 acre for ranchers in that 40 26 area. The data collected are shown in the table. (a) Use a graphing calculator to draw a scatterplot of the data. (b) Does the association appear linear or nonlinear? (c) Is the association positive or negative?

20. If the association in Exercise 19 is linear, (a) use a graphing calculator to find a linear function that models the relation (rainfall, cattle per acre), (b) use the function to find the number of cattle per acre that might be possible for an area receiving 50 in. of rainfall per year, and (c) state whether the correlation is weak or strong.

STRENGTHENING CORE SKILLS The Various Forms of a Linear Equation Learning mathematics is very much like the construction of a skyscraper. The final height of the skyscraper ultimately depends on the strength of the foundation and quality of the frame supporting each new floor as it is built. Our previous work with linear functions and their graphs, while having a number of useful applications, is actually the foundation on which much of our future work is built. For this reason, it’s important you gain a certain fluency with linear functions and relationships — even to a point where things come to you effortlessly and automatically. As noted mathematician Henri Lebesque once said, “An idea reaches its maximum level of usefulness only when you understand it so well that it seems like you have always known it. You then become incapable of seeing the idea as anything but a trivial and immediate result.” These formulas and concepts, while simple, have an endless number of significant and substantial applications. Forms and Formulas

slope formula

point-slope form

slope-intercept form

standard form

y2 ⫺ y1 x2 ⫺ x1 given any two points on the line

y ⫺ y1 ⫽ m1x ⫺ x1 2

y ⫽ mx ⫹ b

Ax ⫹ By ⫽ C

given slope m and any point 1x1, y1 2

given slope m and y-intercept (0, b)

A, B, and C are integers (used in linear systems)

m⫽

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Calculator Exploration and Discovery

185

Characteristics of Lines

y-intercept

x-intercept

increasing

decreasing

(0, y) let x ⫽ 0, solve for y

(x, 0) let y ⫽ 0, solve for x

m 7 0 line slants upward from left to right

m 6 0 line slants downward form left to right

intersecting

parallel

perpendicular

dependent

m1 ⫽ m2

m1 ⫽ m2, b1 ⫽ b2 lines do not intersect

m1m2 ⫽ ⫺1 lines intersect at right angles

m1 ⫽ m2, b1 ⫽ b2 lines intersect at all points

horizontal

vertical

identity

y⫽k horizontal line through k

x⫽h vertical line through h

y⫽x the input value identifies the output

Relationships between Lines

lines intersect at one point Special Lines

Use the formulas and concepts reviewed here to complete the following exercises. For the two points given: (a) compute the slope of the line through the points and state whether the line is increasing or decreasing, (b) find the equation of the line in point-slope form, then write the equation in slope-intercept form, and (c) find the x- and y-intercepts and graph the line. Exercise 1: P1(0, 5)

P2 (6, 7)

Exercise 2: P1(3, 2)

P2 (0, 9)

Exercise 3: P1(3, 2)

P2 (9, 5)

Exercise 4: P1 1⫺5, ⫺42

P2 (3, 2)

Exercise 6: P1 12, ⫺72

P2 1⫺8, ⫺22

Exercise 5: P1 1⫺2, 52

P2 16, ⫺12

CALCULATOR EXPLORATION AND DISCOVERY Evaluating Expressions and Looking for Patterns These “explorations” are designed to explore the full potential of a graphing calculator, as well as to use this potential to investigate patterns and discover connections that might otherwise be overlooked. In this exploration and discovery, we point out the various ways an expression can be evaluated on a graphing calculator. Some ways seem easier, faster, and/or better than others, but each has advantages and disadvantages depending on the task at hand, and it will help to be aware of them all for future use. One way to evaluate an expression is to use the TABLE feature of a graphing calculator, with the expression (TBLSET) screen entered as Y1 on the Y= screen. If you want the calculator to generate inputs, use the 2nd to indicate a starting value (TblStartⴝ) and an increment value (⌬Tbl ⴝ), and set the calculator in Indpnt: AUTO ASK mode (to input specific values, the calculator should be in Indpnt: AUTO ASK mode). After pressing 2nd GRAPH (TABLE), the calculator shows the corresponding Figure 1.105 input and output values. Expressions can also be evaluated on the home screen for a single value or a series of values. Enter the expression ⫺34x ⫹ 5 on the Y= screen (see Figure 1.105) and use 2nd MODE (QUIT) to get back to the home screen. To evaluate this expression, (Y-VARS), and use the first option 1:Function . This access Y1 using VARS brings us to a submenu where any of the equations Y1 through Y0 (actually Y10) can be accessed. Since the default setting is the one we need (1:Y1), simply press and Y1 appears on the home screen. To evaluate a single input, simply enclose it in WINDOW

ENTER

ENTER

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parentheses. To evaluate more than one input, enter the numbers as a set of values with the set enclosed in parentheses. In Figure 1.106, Y1 has been evaluated for x ⫽ ⫺4, then simultaneously for x ⫽ ⫺4, ⫺2, 0, and 2. A third way to evaluate expressions is using a list, with the desired inputs entered in List 1 (L1), then List 2 (L2) defined in terms of L1. For example, L2 ⫽ ⫺34L1 ⫹ 5 will return the same values for inputs of ⫺4, ⫺2, 0, and 2 seen previously on the home screen (remember to clear the lists first). Lists are accessed by pressing STAT 1:Edit. Enter the numbers ⫺4, ⫺2, 0, and 2 in L1, then use the right arrow to move to L2. It is important to note that you next press the up arrow key so that the cursor overlies L2. The bottom of the screen now reads “L2 ⫽ ” and the calculator is waiting for us to define L2. After entering L2 ⫽ ⫺34L1 ⫹ 5 (see Figure 1.107) and pressing we obtain the same outputs as before (see Figure 1.108). The advantage of using the “list” method is that we can further explore or experiment with the output values in a search for patterns.

Figure 1.106

Figure 1.107

ENTER

Exercise 1: Evaluate the expression 0.2L1 ⫹ 3 on the list screen, using consecutive integer inputs from ⫺6 to 6 inclusive. What do you notice about the outputs? Exercise 2: Evaluate the expression 12L1 ⫺ 19.1 on the list screen, using consecutive integer inputs from ⫺6 to 6 inclusive. We suspect there is a pattern to the output values, but this time the pattern is very difficult to see. On the home screen, compute the difference between a few successive outputs from L2 [for example, L2112 ⫺ L212)]. What do you notice?

Figure 1.108

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CHAPTER CONNECTIONS

2.5 Piecewise-Defined Functions 245

Viewing a function in terms of an equation, a table of values, and the related graph, often brings a clearer understanding of the relationships involved. For example, the power generated by a wind turbine is often modeled 8v3 by the function P 1v2 , where P is the 125 power in watts and v is the wind velocity in miles per hour. While the formula enables us to predict the power generated for a given wind speed, the graph offers a visual representation of this relationship, where we note a rapid growth in power output as the wind speed increases. This application appears as Exercise 107 in Section 2.2.

2.6 Variation: The Toolbox Functions in Action 259


More on Functions CHAPTER OUTLINE 2.1 Analyzing the Graph of a Function 188 2.2 The Toolbox Functions and Transformations 202 2.3 Absolute Value Functions, Equations, and Inequalities 218

2.4 Basic Rational Functions and Power Functions; More on the Domain 230

䊳

䊳

䊳

䊳

Analyzing the Path of a Projectile (Section 2.1, Exercise 57) Altitude of the Jet Stream (Section 2.3, Exercise 61) Amusement Arcades (Section 2.5, Exercise 42) Volume of Phone Calls (Section 2.6, Exercise 55)

187

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2.1

Analyzing the Graph of a Function

LEARNING OBJECTIVES

In this section, we’ll consolidate and refine many of the ideas we’ve encountered related to functions. When functions and graphs are applied as real-world models, we create numeric and visual representations that enable an informed response to questions involving maximum efficiency, positive returns, increasing costs, and other relationships that can have a great impact on our lives.

In Section 2.1 you will see how we can

A. Determine whether a

B.

C.

D.

E.

function is even, odd, or neither Determine intervals where a function is positive or negative Determine where a function is increasing or decreasing Identify the maximum and minimum values of a function Locate local maximum and minimum values using a graphing calculator

A. Graphs and Symmetry While the domain and range of a function will remain dominant themes in our study, for the moment we turn our attention to other characteristics of a function’s graph. We begin with the concept of symmetry. Figure 2.1

Symmetry with Respect to the y-Axis

Consider the graph of f 1x2 x 4x shown in Figure 2.1, where the portion of the graph to the left of the y-axis appears to be a mirror image of the portion to the right. A function is symmetric to the y-axis if, given any point (x, y) on the graph, the point 1x, y2 is also on the graph. We note that 11, 32 is on the graph, as is 11, 32, and that 12, 02 is an x-intercept of the graph, as is (2, 0). Functions that are symmetric with respect to the y-axis are also known as even functions and in general we have: 4

2

5

y f(x) x4 4x2 (2.2, ~4)

(2.2, ~4)

(2, 0)

(2, 0)

5

5

x

(1, 3) 5 (1, 3)

Even Functions: y-Axis Symmetry A function f is an even function if and only if, for each point (x, y) on the graph of f, the point (x, y) is also on the graph. In function notation f 1x2 f 1x2

Symmetry can be a great help in graphing new functions, enabling us to plot fewer points and to complete the graph using properties of symmetry. EXAMPLE 1

䊳

Graphing an Even Function Using Symmetry a. The function g(x) in Figure 2.2 (shown in solid blue) is known to be even. Draw the complete graph. 2 Figure 2.2 b. Show that h1x2 x3 is an even function using y the arbitrary value x k [show h1k2 h1k2 ], 5 then sketch the complete graph using h(0), g(x) h(1), h(8), and y-axis symmetry. (1, 2)

Solution

188

䊳

a. To complete the graph of g (see Figure 2.2) use the points (4, 1), (2, 3), (1, 2), and y-axis symmetry to find additional points. The corresponding ordered pairs are (4, 1), (2, 3), and (1, 2), which we use to help draw a “mirror image” of the partial graph given.

(1, 2)

(4, 1)

(4, 1) 5 x

5

(2, 3)

(2, 3) 5

2–2

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Section 2.1 Analyzing the Graph of a Function 2

b. To prove that h1x2 x3 is an even function, we must show h1k2 h1k2 for any constant k. 2 1 After writing x3 as 3x2 4 3 , we have: h1k2 ⱨ h1k2

2

2 1k2 ⱨ 2 1k2 3

WORTHY OF NOTE The proof can also be demon2 1 strated by writing x 3 as 1x 3 2 2, and you are asked to complete this proof in Exercise 69.

2

3

y 5

(8, 4)

first step of proof

3 1k2 4 ⱨ 3 1k2 4 2

1 3

Figure 2.3

1 3

h(x)

(8, 4)

evaluate h (k ) and h(k ) (1, 1)

2

radical form result: 1k2 2 k2

3 2 3 2 2 k 2 k ✓

(1, 1)

10

(0, 0)

Using h102 0, h112 1, and h182 4 with y-axis symmetry produces the graph shown in Figure 2.3.

10

x

5


䊳

Symmetry with Respect to the Origin Another common form of symmetry is known as symmetry to the origin. As the name implies, the graph is somehow “centered” at (0, 0). This form of symmetry is easy to see for closed figures with their center at (0, 0), like certain polygons, circles, and ellipses (these will exhibit both y-axis symmetry and symmetry with respect to the origin). Note the relation graphed in Figure 2.4 contains the points (3, 3) and (3, 3), along with (1, 4) and (1, 4). But the function f (x) in Figure 2.5 also contains these points and is, in the same sense, symmetric to the origin (the paired points are on opposite sides of the x- and y-axes, and a like distance from the origin). Figure 2.4

Figure 2.5

y

y

5

5

(1, 4)

(1, 4)

(3, 3)

(3, 3)

5

5

x

f(x)

5

5

(3, 3) (1, 4)

(3, 3) (1, 4)

5

x

5

Functions symmetric to the origin are known as odd functions and in general we have: Odd Functions: Symmetry About the Origin A function f is an odd function if and only if, for each point (x, y) on the graph of f, the point (x, y) is also on the graph. In function notation f 1x2 f 1x2

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CHAPTER 2 More on Functions

EXAMPLE 2

䊳

Graphing an Odd Function Using Symmetry a. In Figure 2.6, the function g(x) given (shown in solid blue) is known to be odd. Draw the complete graph. b. Show that h1x2 x3 4x is an odd function using the arbitrary value x k [show h1x2 h1x2 ], then sketch the graph using h122 , h112 , h(0), and odd symmetry.

Solution

䊳

a. To complete the graph of g, use the points (6, 3), (4, 0), and (2, 2) and odd symmetry to find additional points. The corresponding ordered pairs are (6, 3), (4, 0), and (2, 2), which we use to help draw a “mirror image” of the partial graph given (see Figure 2.6). Figure 2.6

Figure 2.7

y

y

10

5

(1, 3)

g(x) (6, 3)

(2, 2) (4, 0)

10

h(x)

(2, 0) x (6, 3) 10

(4, 0)

(2, 2)

5

(2, 0) (0, 0)

5

x

(1, 3) 10

5

b. To prove that h1x2 x3 4x is an odd function, we must show that h1k2 h1k2. h1k2 ⱨ h1k2 1k2 41k2 ⱨ 3 k3 4k 4 k3 4k k3 4k ✓ 3

Using h122 0, h112 3, and h102 0 with symmetry about the origin produces the graph shown in Figure 2.7. Now try Exercises 13 through 24 A. You’ve just seen how we can determine whether a function is even, odd, or neither

䊳

Finally, some relations also exhibit a third form of symmetry, that of symmetry to the x-axis. If the graph of a circle is centered at the origin, the graph has both odd and even symmetry, and is also symmetric to the x-axis. Note that if a graph exhibits x-axis symmetry, it cannot be the graph of a function.

B. Intervals Where a Function Is Positive or Negative

Consider the graph of f 1x2 x2 4 shown in Figure 2.8, which has x-intercepts at (2, 0) and (2, 0). As in Section 1.5, the x-intercepts have the form (x, 0) and are called the zeroes of the function (the x-input causes an output of 0). Just as zero on the number line separates negative numbers from positive numbers, the zeroes of a function that crosses the x-axis separate x-intervals where a function is negative from x-intervals where the function is positive. Noting that outputs (y-values) are positive in Quadrants I and II, f 1x2 7 0 in intervals where its graph is above the x-axis. Conversely, f 1x2 6 0

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Section 2.1 Analyzing the Graph of a Function

in x-intervals where its graph is below the x-axis. To illustrate, compare the graph of f in Figure 2.8, with that of g in Figure 2.9. Figure 2.8 5

(2, 0)

Figure 2.9

y f(x) x2 4

5

y g(x) (x 4)2

(2, 0)

5

5

x

3

(4, 0)

x

5

(0, 4) 5

The graph of f is a parabola, with x-intercepts of (2, 0) and (2, 0). Using our previous observations, we note f 1x2 0 for x 1q, 24 ´ 32, q 2 since the graph is above the x-axis, and f 1x2 6 0 for x 12, 22 . The graph of g is also a parabola, but is entirely above or on the x-axis, showing g1x2 0 for x ⺢. The difference is that zeroes coming from factors of the form (x r) (with degree 1) allow the graph to cross the x-axis. The zeroes of f came from 1x 22 1x 22 0. Zeroes that come from factors of the form 1x r2 2 (with degree 2) cause the graph to “bounce” off the x-axis (intersect without crossing) since all outputs must be nonnegative. The zero of g came from 1x 42 2 0.

WORTHY OF NOTE These observations form the basis for studying polynomials of higher degree in Chapter 4, where we extend the idea to factors of the form 1x r2 n in a study of roots of multiplicity.

EXAMPLE 3

5

䊳

Solving an Inequality Using a Graph Use the graph of g1x2 x3 2x2 4x 8 given to solve the inequalities a. g1x2 0 b. g1x2 6 0

Solution

䊳

From the graph, the zeroes of g (x-intercepts) occur at (2, 0) and (2, 0). a. For g1x2 0, the graph must be on or above the x-axis, meaning the solution is x 32, q 2 . b. For g1x2 6 0, the graph must be below the x-axis, and the solution is x 1q, 22 . As we might have anticipated from the graph, factoring by grouping gives g1x2 1x 221x 22 2, with the graph crossing the x-axis at 2, and bouncing off the x-axis (intersects without crossing) at x 2.

y (0, 8) g(x) 5

5

5

x

2


䊳

Even if the function is not a polynomial, the zeroes can still be used to find x-intervals where the function is positive or negative.

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EXAMPLE 4

Solution

䊳

䊳

Solving an Inequality Using a Graph

y

For the graph of r 1x2 1x 1 2 shown, solve a. r 1x2 0 b. r 1x2 7 0 a. The only zero of r is at (3, 0). The graph is on or below the x-axis for x 31, 3 4 , so r 1x2 0 in this interval. b. The graph is above the x-axis for x 13, q 2 , and r 1x2 7 0 in this interval.

10

r(x) 10

10

10

Now try Exercises 29 through 32 B. You’ve just seen how we can determine intervals where a function is positive or negative

x

䊳

This study of inequalities shows how the graphical solutions studied in Section 1.5 are easily extended to the graph of a general function. It also strengthens the foundation for the graphical solutions studied throughout this text.

C. Intervals Where a Function Is Increasing or Decreasing In our study of linear graphs, we said a graph was increasing if it “rose” when viewed from left to right. More generally, we say the graph of a function is increasing on a given interval if larger and larger x-values produce larger and larger y-values. This suggests the following tests for intervals where a function is increasing or decreasing. Increasing and Decreasing Functions Given an interval I that is a subset of the domain, with x1 and x2 in I and x2 7 x1, 1. A function is increasing on I if f 1x2 2 7 f 1x1 2 for all x1 and x2 in I (larger inputs produce larger outputs). 2. A function is decreasing on I if f 1x2 2 6 f 1x1 2 for all x1 and x2 in I (larger inputs produce smaller outputs).

3. A function is constant on I if f 1x2 2 f 1x1 2 for all x1 and x2 in I (larger inputs produce identical outputs).

f(x)

f (x) is increasing on I

f(x2)

f (x)

f (x) is decreasing on I

f(x) is constant on I

f (x)

f (x1)

f(x1)

f (x2)

f (x2)

f (x1)

f (x1)

f (x1) x1 Interval I

x2

x2 x1 and f (x2) f (x1) for all x I graph rises when viewed from left to right

x

x1 Interval I

x2

x2 x1 and f (x2) f (x1) for all x I graph falls when viewed from left to right

f(x2)

f(x1)

f (x2) x

x1 Interval I

x2

x

x2 x1 and f(x2) f(x1) for all x I graph is level when viewed from left to right

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Consider the graph of f 1x2 x2 4x 5 given in Figure 2.10. Since the parabola opens downward with the vertex at (2, 9), the function must increase until it reaches this peak at x 2, and decrease thereafter. Notationally we’ll write this as f 1x2c for x 1q, 22 and f 1x2T for x 12, q 2. Using the interval 13, 22 shown below the figure, we see that any larger input value from the interval will indeed produce a larger output value, and f 1x2c on the interval. For instance, 1 7 2

x2 7 x1

and

and

f 112 7 f 122 8 7 7

Figure 2.10 10

y f(x) x2 4x 5 (2, 9) (0, 5)

(1, 0)

(5, 0)

5

5

x

10

x (3, 2)

f 1x2 2 7 f 1x1 2

A calculator check is shown in the figure. Note the outputs are increasing until x 2, then they begin decreasing. EXAMPLE 5

䊳

Finding Intervals Where a Function Is Increasing or Decreasing

y 5

Use the graph of v(x) given to name the interval(s) where v is increasing, decreasing, or constant.

Solution

䊳

From left to right, the graph of v increases until leveling off at (2, 2), then it remains constant until reaching (1, 2). The graph then increases once again until reaching a peak at (3, 5) and decreases thereafter. The result is v1x2c for x 1q, 22 ´ 11, 32, v1x2T for x 13, q 2, and v(x) is constant for x 12, 12 .

v(x)

5

5

x

5


䊳

WORTHY OF NOTE Questions about the behavior of a function are asked with respect to the y outputs: is the function positive, is the function increasing, etc. Due to the input/ output, cause/effect nature of functions, the response is given in terms of x, that is, what is causing outputs to be positive, or to be increasing.

EXAMPLE 6

Notice the graph of f in Figure 2.10 and the graph of v in Example 5 have something in common. It appears that both the far left and far right branches of each graph point downward (in the negative y-direction). We say that the end-behavior of both graphs is identical, which is the term used to describe what happens to a graph as 冟x冟 becomes very large. For x 7 0, we say a graph is, “up on the right” or “down on the right,” depending on the direction the “end” is pointing. For x 6 0, we say the graph is “up on the left” or “down on the left,” as the case may be. 䊳

Describing the End-Behavior of a Graph

The graph of f 1x2 x 3x is shown. Use the graph to name intervals where f is increasing or decreasing, and comment on the end-behavior of the graph.

y 5

3

Solution

C. You’ve just seen how we can determine where a function is increasing or decreasing

䊳

From the graph we observe that f 1x2c for x 1q, 12 ´ 11, q 2 , and f 1x2T for x 11, 12 . The end-behavior of the graph is down on the left, and up on the right (down/up).

f(x) x2 3x

5

5

x

5


䊳

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D. Maximum and Minimum Values The y-coordinate of the vertex of a parabola that opens downward, and the y-coordinate of “peaks” from other graphs are called maximum values. A global maximum (also called an absolute maximum) names the largest y-value over the entire domain. A local maximum (also called a relative maximum) gives the largest range value in a specified interval; and an endpoint maximum can occur at an endpoint of the domain. The same can be said for any corresponding minimum values. We will soon develop the ability to locate maximum and minimum values for quadratic and other functions. In future courses, methods are developed to help locate maximum and minimum values for almost any function. For now, our work will rely chiefly on a function’s graph. EXAMPLE 7

䊳

Analyzing Characteristics of a Graph Analyze the graph of function f shown in Figure 2.11. Include specific mention of a. domain and range, b. intervals where f is increasing or decreasing, c. maximum (max) and minimum (min) values, d. intervals where f 1x2 0 and f 1x2 6 0, and e. whether the function is even, odd, or neither.

Solution

D. You’ve just seen how we can identify the maximum and minimum values of a function

䊳

a. Using vertical and horizontal boundary lines show the domain is x ⺢, with a range of: y 1q, 7 4 . b. f 1x2c for x 1q, 32 ´ 11, 52 shown in blue in Figure 2.12, and f 1x2T for x 13, 12 ´ 15, q 2 as shown in red. c. From Part (b) we find that y 5 at (3, 5) and y 7 at (5, 7) are local maximums, with a local minimum of y 1 at (1, 1). The point (5, 7) is also a global maximum (there is no global minimum). d. f 1x2 0 for x 36, 84 ; f 1x2 6 0 for x 1q, 62 ´ 18, q 2 e. The function is neither even nor odd.

Figure 2.11 y 10

(5, 7) f(x)

(3, 5)

(1, 1) 10

10

x

10

Figure 2.12 y 10

(5, 7) (3, 5) (6, 0)

(1, 1)

10

(8, 0) 10 x

10


䊳

E. Locating Maximum and Minimum Values Using Technology In Section 1.5, we used the 2nd TRACE (CALC) 2:zero option of a graphing calculator to locate the zeroes/x-intercepts of a function. The maximum or minimum values of a function are located in much the same way. To illustrate, enter the function y x3 3x 2 as Y1 on the Y= screen, then Figure 2.13 graph it in the window shown, where x 3 4, 44 5 and y 3 5, 54 . As seen in Figure 2.13, it appears a local maximum occurs at x 1 and a local minimum at x 1. To actually find the local maximum, we access the 2nd TRACE 4 4 (CALC) 4:maximum option, which returns you to the graph and asks for a Left Bound?, a Right Bound?, and a Guess? as before. Here, we entered a left bound of “3,” a right bound 5

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Figure 2.14 of “0” and bypassed the guess option by pressing a third time (the calculator again sets 5 the “䉴” and “䉳” markers to show the bounds chosen). The cursor will then be located at the local maximum in your selected interval, with 4 the coordinates displayed at the bottom of the 4 screen (Figure 2.14). Due to the algorithm the calculator uses to find these values, a decimal number very close to the expected value is 5 sometimes displayed, even if the actual value is an integer (in Figure 2.14, 0.9999997 is displayed instead of 1). To check, we evaluate f 112 and find the local maximum is indeed 0. ENTER

EXAMPLE 8

䊳

Locating Local Maximum and Minimum Values on a Graphing Calculator Find the maximum and minimum values of f 1x2

Solution

䊳

1 4 1x 8x2 72 . 2

1 4 1X 8X2 72 as Y1 on the Y= screen, and graph the 2 function in the ZOOM 6:ZStandard window. To locate the leftmost minimum value, we access the 2nd TRACE (CALC) 3:minimum option, and enter a left bound of “4,” and a right bound of “1” (Figure 2.15). After pressing once more, the cursor is located at the minimum in the interval we selected, and we find that a local minimum of 4.5 occurs at x 2 (Figure 2.16). Repeating these steps using the appropriate options shows a local maximum of y 3.5 occurs at x 0, and a second local minimum of y 4.5 occurs at x 2. Note that y 4.5 is also a global minimum.

Begin by entering

ENTER

Figure 2.15

Figure 2.16 10

10

10

E. You’ve just seen how we can locate local maximum and minimum values using a graphing calculator

10

10

10

10

10


䊳

The ideas presented here can be applied to functions of all kinds, including rational functions, piecewise-defined functions, step functions, and so on. There is a wide variety of applications in Exercises 57 through 64.

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2.1 EXERCISES 䊳



1. The graph of a polynomial will cross through the x-axis at zeroes of factors of degree 1, and off the x-axis at the zeroes from linear factors of degree 2.

3. If f 1x2 2 7 f 1x1 2 for x1 6 x2 for all x in a given interval, the function is in the interval.

5. Discuss/Explain the following statement and give an example of the conclusion it makes. “If a function f is decreasing to the left of (c, f (c)) and increasing to the right of (c, f (c)), then f (c) is either a local or a global minimum.” 䊳

2. If f 1x2 f 1x2 for all x in the domain, we say that f is an function and symmetric to the axis. If f 1x2 f 1x2 , the function is and symmetric to the . 4. If f 1c2 f 1x2 for all x in a specified interval, we say that f (c) is a local for this interval.

6. Without referring to notes or textbook, list as many features/attributes as you can that are related to analyzing the graph of a function. Include details on how to locate or determine each attribute.


The following functions are known to be even. Complete each graph using symmetry.

7.

8.

y 5

5

5 x

3 15. f 1x2 4 1 xx

1 16. g1x2 x3 6x 2 1 17. p1x2 3x3 5x2 1 18. q1x2 x x

y 10

10

10 x

Determine whether the following functions are even, odd, or neither.

10

5

Determine whether the following functions are even: f 1k2 f 1k2 .

9. f 1x2 7冟 x 冟 3x2 5 10. p1x2 2x4 6x 1

1 11. g1x2 x4 5x2 1 3

1 冟x冟 x2 The following functions are known to be odd. Complete each graph using symmetry. 13.

12. q1x2

14.

y 10

Determine whether the following functions are odd: f 1k2 f 1k2 .

19. w1x2 x3 x2

3 20. q1x2 x2 3冟x冟 4

1 3 21. p1x2 2 1 x x3 4

22. g1x2 x3 7x

23. v1x2 x3 3冟x冟

Use the graphs given to solve the inequalities indicated. Write all answers in interval notation.

25. f 1x2 x3 3x2 x 3; f 1x2 0

y 10

y

5

10

10 x

10

10 x 5

10

24. f 1x2 x4 7x2 30

5 x

10 5

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26. f 1x2 x3 2x2 4x 8; f 1x2 7 0

32. g1x2 1x 12 3 1; g1x2 6 0 y

y

5

5 5

5 x

g(x) 5

5 x

1

27. f 1x2 x4 2x2 1; f 1x2 7 0 y

5

5

5 x

5

Name the interval(s) where the following functions are increasing, decreasing, or constant. Write answers using interval notation. Assume all endpoints have integer values.

33. y V1x2

34. y H1x2 y

y 5

10

5

28. f 1x2 x3 2x2 4x 8; f 1x2 0 y

1

10

10 x

5

5 x

H(x)

5

5 x 5

10

35. y f 1x2

5

36. y g1x2 y

y

10 10

3 29. p1x2 1 x 1 1; p1x2 0 y

f(x)

8

g(x)

6

10

5

10 x 4 2

10 5

2

4

6

8

x

10

5 x

p(x)

For Exercises 37 through 40, determine (a) interval(s) where the function is increasing, decreasing or constant, and (b) comment on the end-behavior.

5

30. q1x2 1x 1 2; q1x2 7 0 y

37. p1x2 0.51x 22 3

3 38. q1x2 1 x1

y

5

y

5

5

(0, 4) q(x) 5

5 x

(2, 0)

5

31. f 1x2 1x 12 1; f 1x2 0 3

y

5

(1, 0)

5

5 x

5

5

39. y f 1x2

5 x

(0, 1)

5

40. y g1x2 y

y 10 5

5

f(x)

5 x

10

5 5

10 x

5 x 3

10

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For Exercises 41 through 48, determine the following (answer in interval notation as appropriate): (a) domain and range of the function; (b) zeroes of the function; (c) interval(s) where the function is greater than or equal to zero, or less than or equal to zero; (d) interval(s) where the function is increasing, decreasing, or constant; and (e) location of any local max or min value(s).

42. y f 1x2

41. y H1x2 5

y (2, 5)

45. y Y1

46. y Y2 y

y

5

5

5

5

5 x

5 x

5

5

47. p1x2 1x 32 3 1 48. q1x2 冟x 5冟 3

y 5

y

y

10 10

(1, 0)

(3.5, 0)

(3, 0)

5

5 x

5

8

5 x

6

10 5 (0, 5)

10 x

4

5 2

43. y g1x2

44. y h1x2

10

y

y

5 5 x

g(x) 2

5

x

2

4

6

8

10

x

Use a graphing calculator to find the maximum and minimum values of the following functions. Round answers to nearest hundredth when necessary.

5

5

2

5

3 3 6 1x 5x2 6x2 50. y 1x3 4x2 3x2 4 5 51. y 0.0016x5 0.12x3 2x 49. y

52. y 0.01x5 0.03x4 0.25x3 0.75x2 54. y x2 2x 3 2

53. y x 24 x 䊳


55. Conic sections—hyperbola: y 13 24x2 36 y While the conic sections are 5 not covered in detail until f(x) later in the course, we’ve already developed a number 5 5 x of tools that will help us understand these relations and their graphs. The 5 equation here gives the “upper branches” of a hyperbola, as shown in the figure. Find the following by analyzing the equation: (a) the domain and range; (b) the zeroes of the relation; (c) interval(s) where y is increasing or decreasing; (d) whether the relation is even, odd, or neither, and (e) solve for x in terms of y.

56. Trigonometric graphs: y sin1x2 and y cos1x2 The trigonometric functions are also studied at some future time, but we can apply the same tools to analyze the graphs of these functions as well. The graphs of y sin x and y cos x are given, graphed over the interval x 3360°, 360°4 . Use them to find (a) the range of the functions; (b) the zeroes of the functions; (c) interval(s) where y is increasing/decreasing; (d) location of minimum/maximum values; and (e) whether each relation is even, odd, or neither. y

y

(90, 1)

1

1

y cos x

y sin x

(90, 0) 360 270 180

90

90

1

180

270

360 x

360 270 180

90

90

1

180

270

360 x

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199


APPLICATIONS c. d. e. f. g. h.

Height (feet)

57. Catapults and projectiles: Catapults have a long and interesting history that dates back to ancient times, when they were used to launch javelins, rocks, and other projectiles. The diagram given illustrates the path of the projectile after release, which follows a parabolic arc. Use the graph to determine the following: 80 70 60 50 40 30

20

60

100

140

180

220

260

Distance (feet)

a. State the domain and range of the projectile. b. What is the maximum height of the projectile? c. How far from the catapult did the projectile reach its maximum height? d. Did the projectile clear the castle wall, which was 40 ft high and 210 ft away? e. On what interval was the height of the projectile increasing? f. On what interval was the height of the projectile decreasing? P (millions of dollars)

58. Profit and loss: The profit of DeBartolo Construction Inc. is illustrated by the graph shown. Use the graph to t (years since 1990) estimate the point(s) or the interval(s) for which the profit P was: a. increasing b. decreasing 16 12 8 4 0 4 8

1 2 3 4 5 6 7 8 9 10

constant a maximum a minimum positive negative zero

59. Functions and rational exponents: The graph of 2 f 1x2 x3 1 is shown. Use the graph to find: a. domain and range of the function b. zeroes of the function c. interval(s) where f 1x2 0 or f 1x2 6 0 d. interval(s) where f (x) is increasing, decreasing, or constant e. location of any max or min value(s) Exercise 59

Exercise 60

y

y

5

5

(1, 0) (1, 0) 5

(0, 1)

5

(3, 0) 5 x

(3, 0) (0, 1)

5

5 x

5

60. Analyzing a graph: Given h1x2 冟x2 4冟 5, whose graph is shown, use the graph to find: a. domain and range of the function b. zeroes of the function c. interval(s) where h1x2 0 or h1x2 6 0 d. interval(s) where f(x) is increasing, decreasing, or constant e. location of any max or min value(s)

61. Analyzing interest rates: The graph shown approximates the average annual interest rates I on 30-yr fixed mortgages, rounded to the nearest 14 % . Use the graph to estimate the following (write all answers in interval notation). a. domain and range b. interval(s) where I(t) is increasing, decreasing, or constant c. location of any global maximum or d. the one-year period with the greatest rate of increase and minimum values the one-year period with the greatest rate of decrease Source: 2009 Statistical Abstract of the United States, Table 1157 16

Mortgage rate

14 12 10 8 6 4 2 0

t

83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 00 01 02 03 04 05 06 07 08 09

Year (1983 → 83)

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2–14


62. Analyzing the surplus S: The following graph approximates the federal surplus S of the United States. Use the graph to estimate the following. Write answers in interval notation and estimate all surplus values to the nearest $10 billion. a. the domain and range b. interval(s) where S(t) is increasing, decreasing, or constant c. the location of any global maximum and minimum values d. the one-year period with the greatest rate of increase, and the one-year period with the greatest rate of decrease S(t): Federal Surplus (in billions)

Source: 2009 Statistical Abstract of the United States, Table 451 400 200 0 ⫺200 ⫺400 ⫺600 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108

t

Year (1980 → 80)

64. Constructing a graph: Draw a continuous function g that has the following characteristics, then state the zeroes and the location of all maximum and minimum values. [Hint: Write them as (c, g(c)).] a. Domain: x 僆 1⫺q, 8 4 Range: y 僆 3⫺6, q2 b. g102 ⫽ 4.5; g162 ⫽ 0 c. g1x2c for x 僆 1⫺6, 32 ´ 16, 82 g1x2T for x 僆 1⫺q, ⫺62 ´ 13, 62 d. g1x2 ⱖ 0 for x 僆 1⫺q, ⫺9 4 ´ 3⫺3, 8 4 g1x2 6 0 for x 僆 1⫺9, ⫺32

63. Constructing a graph: Draw a continuous function f that has the following characteristics, then state the zeroes and the location of all maximum and minimum values. [Hint: Write them as (c, f (c)).] a. Domain: x 僆 1⫺10, q2 Range: y 僆 1⫺6, q2 b. f 102 ⫽ 0; f 142 ⫽ 0 c. f 1x2c for x 僆 1⫺10, ⫺62 ´ 1⫺2, 22 ´ 14, q2 f 1x2T for x 僆 1⫺6, ⫺22 ´ 12, 42 d. f 1x2 ⱖ 0 for x 僆 3⫺8, ⫺4 4 ´ 30, q 2 f 1x2 6 0 for x 僆 1⫺q, ⫺82 ´ 1⫺4, 02 䊳

EXTENDING THE CONCEPT Exercise 65

65. Does the function shown have a maximum value? Does it have a minimum value? Discuss/explain/justify why or why not.

y 5

Distance (meters)

66. The graph drawn here depicts a 400-m race between a mother and her daughter. Analyze the graph to answer questions (a) through (f). a. Who wins the race, the mother or daughter? b. By approximately how many meters? c. By approximately how many seconds? Exercise 66 Mother Daughter d. Who was leading at t ⫽ 40 seconds? 400 e. During the race, how many seconds was 300 the daughter in the lead? f. During the race, how many seconds was 200 the mother in the lead?

⫺5

5 x

⫺5

100

10

20

30

40

50

Time (seconds)

60

70

80

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201

67. The graph drawn here depicts the last 75 sec of the competition between Ian Thorpe (Australia) and Massimiliano Rosolino (Italy) in the men’s 400-m freestyle at the 2000 Olympics, where a new Olympic record was set. a. Who was in the lead at 180 sec? 210 sec? b. In the last 50 m, how many times were they tied, and when did the ties occur? c. About how many seconds did Rosolino have the lead? d. Which swimmer won the race? e. By approximately how many seconds? f. Use the graph to approximate the new Olympic record set in the year 2000. Thorpe

Rosolino

Distance (meters)

400

350

300

250

150

155

160

165

170

175

180

185

190

195

200

205

210

215

220

225

Time (seconds)

68. Draw the graph of a general function f (x) that has a local maximum at (a, f (a)) and a local minimum at (b, f (b)) but with f 1a2 6 f 1b2 . 䊳

2

69. Verify that h1x2 ⫽ x3 is an even function, by first rewriting h as h1x2 ⫽ 1x3 2 2. 1


70. (R.4) Solve the given quadratic equation by factoring: x2 ⫺ 8x ⫺ 20 ⫽ 0.

71. (R.5) Find the (a) sum and (b) product of the 3 3 rational expressions and . x⫹2 2⫺x

72. (1.4) Write the equation of the line shown, in the form y ⫽ mx ⫹ b.

73. (R.2) Find the surface area and volume of the cylinder shown 1SA ⫽ 2␲r 2 ⫹ ␲r 2h, V ⫽ ␲r 2h2 .

y

36 cm

5

12 cm ⫺5

5 x

⫺5

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2.2

The Toolbox Functions and Transformations


A. Identify basic

B. C.

D.

E.

characteristics of the toolbox functions Apply vertical/horizontal shifts of a basic graph Apply vertical/horizontal reflections of a basic graph Apply vertical stretches and compressions of a basic graph Apply transformations on a general function f (x )

Many applications of mathematics require that we select a function known to fit the context, or build a function model from the information supplied. So far we’ve looked at linear functions. Here we’ll introduce the absolute value, squaring, square root, cubing, and cube root functions. Together these are the six toolbox functions, so called because they give us a variety of “tools” to model the real world (see Section 2.6). In the same way a study of arithmetic depends heavily on the multiplication table, a study of algebra and mathematical modeling depends (in large part) on a solid working knowledge of these functions. More will be said about each function in later sections.

A. The Toolbox Functions While we can accurately graph a line using only two points, most functions require more points to show all of the graph’s important features. However, our work is greatly simplified in that each function belongs to a function family, in which all graphs from a given family share the characteristics of one basic graph, called the parent function. This means the number of points required for graphing will quickly decrease as we start anticipating what the graph of a given function should look like. The parent functions and their identifying characteristics are summarized here.

The Toolbox Functions Identity function

Absolute value function y

y

5

x

f(x) x

3

3

2

2

1

1

0

0

0

1

1

1

1

2

2

2

2

3

3

3

3

x

f (x) x

3

3

2

2

1

1

0

f(x) x 5

5

x

5

5

Square root function y

y

5

f (x) 1x

f(x) x2

x

3

9

2

2

4

1

1

1

0

0

0

0

1

1

1

1

2

1.41

2

4

3

1.73

9

4

2

x

3

202

x

Domain: x (q, q), Range: y [0, q) Symmetry: even Decreasing: x (q, 0); Increasing: x (0, q ) End-behavior: up on the left/up on the right Vertex at (0, 0)

Domain: x (q, q), Range: y (q, q) Symmetry: odd Increasing: x (q, q) End-behavior: down on the left/up on the right

Squaring function

5

5

x

Domain: x (q, q), Range: y [0, q) Symmetry: even Decreasing: x (q, 0); Increasing: x (0, q) End-behavior: up on the left/up on the right Vertex at (0, 0)

5

5

x

Domain: x [0, q), Range: y [0, q) Symmetry: neither even nor odd Increasing: x (0, q) End-behavior: up on the right Initial point at (0, 0)

2–16

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Section 2.2 The Toolbox Functions and Transformations

Cubing function

Cube root function y

y

10

x

f (x) 1 x

27

27

3

2

8

8

2

1

1

1

1

0

0

0

0

1

1

1

1

2

8

8

2

3

27

27

3

x

f (x) x

3

3

5

x

5

3

f(x) 3 x

10

10

x

5

Domain: x (q, q), Range: y (q, q) Symmetry: odd Increasing: x (q, q) End-behavior: down on the left/up on the right Point of inflection at (0, 0)

Domain: x (q, q), Range: y (q, q) Symmetry: odd Increasing: x (q, q) End-behavior: down on the left/up on the right Point of inflection at (0, 0)

In applications of the toolbox functions, the parent graph may be “morphed” and/or shifted from its original position, yet the graph will still retain its basic shape and features. The result is called a transformation of the parent graph. EXAMPLE 1

Solution

䊳

䊳

Identifying the Characteristics of a Transformed Graph The graph of f 1x2 x2 2x 3 is given. Use the graph to identify each of the features or characteristics indicated. a. function family b. domain and range c. vertex d. max or min value(s) e. intervals where f is increasing or decreasing f. end-behavior g. x- and y-intercept(s) a. b. c. d. e. f. g.

y 5

5

5

x

5

The graph is a parabola, from the squaring function family. domain: x 1q, q2 ; range: y 3 4, q 2 vertex: (1, 4) minimum value y 4 at (1, 4) decreasing: x 1q, 12, increasing: x 11, q 2 end-behavior: up/up y-intercept: (0, 3); x-intercepts: (1, 0) and (3, 0) Now try Exercises 7 through 34

A. You’ve just seen how we can identify basic characteristics of the toolbox functions

䊳

Note that for Example 1(f), we can algebraically verify the x-intercepts by substituting 0 for f(x) and solving the equation by factoring. This gives 0 1x 121x 32 , with solutions x 1 and x 3. It’s also worth noting that while the parabola is no longer symmetric to the y-axis, it is symmetric to the vertical line x 1. This line is called the axis of symmetry for the parabola, and for a vertical parabola, it will always be a vertical line that goes through the vertex.

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B. Vertical and Horizontal Shifts As we study specific transformations of a graph, try to develop a global view as the transformations can be applied to any function. When these are applied to the toolbox functions, we rely on characteristic features of the parent function to assist in completing the transformed graph.

Vertical Translations We’ll first investigate vertical translations or vertical shifts of the toolbox functions, using the absolute value function to illustrate. EXAMPLE 2

䊳

Graphing Vertical Translations

Solution

䊳

A table of values for all three functions is given, with the corresponding graphs shown in the figure.

Construct a table of values for f 1x2 x, g1x2 x 1, and h1x2 x 3 and graph the functions on the same coordinate grid. Then discuss what you observe.

x

f (x) x

g(x) x 1

h(x) x 3

3

3

4

0

2

2

3

1

1

1

2

2

0

0

1

3

1

1

2

2

2

2

3

1

3

3

4

0

(3, 4)5

y g(x) x 1

(3, 3) (3, 0)

1

f(x) x

5

5

x

h(x) x 3 5

Note that outputs of g(x) are one more than the outputs of f (x), and that each point on the graph of f has been shifted upward 1 unit to form the graph of g. Similarly, each point on the graph of f has been shifted downward 3 units to form the graph of h, since h1x2 f 1x2 3. Now try Exercises 35 through 42

䊳

We describe the transformations in Example 2 as a vertical shift or vertical translation of a basic graph. The graph of g is the graph of f shifted up 1 unit, and the graph of h, is the graph of f shifted down 3 units. In general, we have the following: Vertical Translations of a Basic Graph

Given k 7 0 and any function whose graph is determined by y f 1x2 , 1. The graph of y f 1x2 k is the graph of f(x) shifted upward k units. 2. The graph of y f 1x2 k is the graph of f(x) shifted downward k units. Graphing calculators are wonderful tools for exploring graphical transformations. To emphasize that a given graph is being shifted vertically as in 3 Example 2, try entering 1 X as Y1 on the Y= screen, then Y2 Y1 2 and Y3 Y1 3 (Figure 2.17 — recall the Y-variables are accessed using VARS (Y-VARS) ). Using the Y-variables in this way enables us to study identical transformations on a variety of graphs, simply by changing the function in Y1. ENTER

Figure 2.17

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205


Using a window size of x 35, 54 and y 3 5, 54 for the cube root function, produces the graphs shown in Figure 2.18, which demonstrate the cube root graph has been shifted upward 2 units (Y2), and downward 3 units (Y3). Try this exploration again using Y1 1X.

Figure 2.18 5 Y2

5

5 Y3

5

Horizontal Translations

The graph of a parent function can also be shifted left or right. This happens when we alter the inputs to the basic function, as opposed to adding or subtracting something to the function itself. For Y1 x2 2 note that we first square inputs, then add 2, which results in a vertical shift. For Y2 1x 22 2, we add 2 to x prior to squaring and since the input values are affected, we might anticipate the graph will shift along the x-axis—horizontally. EXAMPLE 3

䊳

Graphing Horizontal Translations

Solution

䊳

Both f and g belong to the quadratic family and their graphs are parabolas. A table of values is shown along with the corresponding graphs.

Construct a table of values for f 1x2 x2 and g1x2 1x 22 2, then graph the functions on the same grid and discuss what you observe.

x

f (x) x2

y

g(x) (x 2)2

3

9

1

2

4

0

1

1

1

0

0

4

1

1

9

2

4

16

3

9

25

9 8

(3, 9)

(1, 9)

7

f(x) x2

6 5

(0, 4)

4

(2, 4)

3

g(x) (x 2)2

2 1

5 4 3 2 1 1

1

2

3

4

5

x

It is apparent the graphs of g and f are identical, but the graph of g has been shifted horizontally 2 units left. Now try Exercises 43 through 46

䊳

We describe the transformation in Example 3 as a horizontal shift or horizontal translation of a basic graph. The graph of g is the graph of f, shifted 2 units to the left. Once again it seems reasonable that since input values were altered, the shift must be horizontal rather than vertical. From this example, we also learn the direction of the shift is opposite the sign: y 1x 22 2 is 2 units to the left of y x2. Although it may seem counterintuitive, the shift opposite the sign can be “seen” by locating the new x-intercept, which in this case is also the vertex. Substituting 0 for y gives 0 1x 22 2 with x 2, as shown in the graph. In general, we have Horizontal Translations of a Basic Graph

Given h 7 0 and any function whose graph is determined by y f 1x2 , 1. The graph of y f 1x h2 is the graph of f(x) shifted to the left h units. 2. The graph of y f 1x h2 is the graph of f(x) shifted to the right h units.

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Figure 2.19

To explore horizontal translations on a graphing calculator, we input a basic function in Y1 and indicate how we want the inputs altered in Y2 and Y3. Here we’ll enter X3 as Y1 on the Y= screen, then Y2 Y1 1X 52 and Y3 Y1 1X 72 (Figure 2.19). Note how this 10 duplicates the definition and notation for horizontal shifts in the orange box. Based on what we saw in Example 3, we expect the graph of y x3 will first be shifted 5 units left (Y2), then 7 units right (Y3). This in confirmed in Figure 2.20. Try this exploration again using Y1 abs1X2.

Figure 2.20 Y3

10

10

10

EXAMPLE 4

䊳

Graphing Horizontal Translations Sketch the graphs of g1x2 x 2 and h1x2 1x 3 using a horizontal shift of the parent function and a few characteristic points (not a table of values).

Solution

䊳

The graph of g1x2 x 2 (Figure 2.21) is the absolute value function shifted 2 units to the right (shift the vertex and two other points from y x 2 . The graph of h1x2 1x 3 (Figure 2.22) is a square root function, shifted 3 units to the left (shift the initial point and one or two points from y 1x). Figure 2.21 5

Figure 2.22

y g(x) x 2

y h(x) x 3

(1, 3)

5

(6, 3)

(5, 3) (1, 2) 5

Vertex

(2, 0)

5

x 4

B. You’ve just seen how we can perform vertical/ horizontal shifts of a basic graph

(3, 0)

5

x


䊳

C. Vertical and Horizontal Reflections The next transformation we investigate is called a vertical reflection, in which we compare the function Y1 f 1x2 with the negative of the function: Y2 f 1x2 .

Vertical Reflections EXAMPLE 5

䊳

Graphing Vertical Reflections Construct a table of values for Y1 x2 and Y2 x2, then graph the functions on the same grid and discuss what you observe.

Solution

䊳

A table of values is given for both functions, along with the corresponding graphs.

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207

Section 2.2 The Toolbox Functions and Transformations y 5

x

Y1 x 2

Y2 x2

2

4

4

1

1

1

0

0

0

1

1

1

2

4

4

Y1 x2

(2, 4)

5 4 3 2 1

Y2 x2

1

2

3

4

5

x

(2, 4) 5

As you might have anticipated, the outputs for f and g differ only in sign. Each output is a reflection of the other, being an equal distance from the x-axis but on opposite sides. Now try Exercises 51 and 52

䊳

The vertical reflection in Example 5 is called a reflection across the x-axis. In general, Vertical Reflections of a Basic Graph

For any function y f 1x2 , the graph of y f 1x2 is the graph of f(x) reflected across the x-axis.

To view vertical reflections on a graphing calculator, we simply define Y2 Y1, as seen here 3 using 1 X as Y1 (Figure 2.23). As in Section 1.5, we can have the calculator graph Y2 using a bolder line, to easily distinguish between the original graph and its reflection (Figure 2.24). To aid in the viewing, we have set a window size of x 3 5, 54 and y 33, 3 4. Try this exploration again using Y1 X2 4.

Figure 2.23

Figure 2.24 3

5

5

3

Horizontal Reflections It’s also possible for a graph to be reflected horizontally across the y-axis. Just as we noted that f (x) versus f 1x2 resulted in a vertical reflection, f(x) versus f 1x2 results in a horizontal reflection. EXAMPLE 6

䊳

Graphing a Horizontal Reflection

Solution

䊳

A table of values is given here, along with the corresponding graphs.

Construct a table of values for f 1x2 1x and g1x2 1x, then graph the functions on the same coordinate grid and discuss what you observe.

x

f(x) 1x

g(x) 1x

4

not real

2

2

not real

12 1.41

1

not real

1

0

0

0

1

1

not real

2

12 1.41

not real

4

2

not real

y (4, 2)

(4, 2) 2

g(x) x

f(x) x

1

5 4 3 2 1 1 2

1

2

3

4

5

x

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2–22


The graph of g is the same as the graph of f, but it has been reflected across the y-axis. A study of the domain shows why — f represents a real number only for nonnegative inputs, so its graph occurs to the right of the y-axis, while g represents a real number for nonpositive inputs, so its graph occurs to the left. Now try Exercises 53 and 54

䊳

The transformation in Example 6 is called a horizontal reflection of a basic graph. In general, Horizontal Reflections of a Basic Graph

For any function y f 1x2 , the graph of y f 1x2 is the graph of f(x) reflected across the y-axis.

C. You’ve just seen how we can apply vertical/horizontal reflections of a basic graph

D. Vertically Stretching/Compressing a Basic Graph As the words “stretching” and “compressing” imply, the graph of a basic function can also become elongated or flattened after certain transformations are applied. However, even these transformations preserve the key characteristics of the graph. EXAMPLE 7

䊳

Stretching and Compressing a Basic Graph

Solution

䊳

A table of values is given for all three functions, along with the corresponding graphs.

Construct a table of values for f 1x2 x2, g1x2 3x2, and h1x2 13x2, then graph the functions on the same grid and discuss what you observe.

x

f (x) x2

g(x) 3x2

h(x) 13 x2

3

9

27

3

2

4

12

4 3

1

1

3

1 3

0

0

0

0

1

1

3

1 3

2

4

12

4 3

3

9

27

3

y g(x) 3x2

(2, 12) (2, 4)

f(x) x2

10

h(x) ax2 (2, d) 5 4 3 2 1

1

2

3

4

5

x

4

The outputs of g are triple those of f, making these outputs farther from the x-axis and stretching g upward (making the graph more narrow). The outputs of h are one-third those of f, and the graph of h is compressed downward, with its outputs closer to the x-axis (making the graph wider). Now try Exercises 55 through 62

䊳

WORTHY OF NOTE In a study of trigonometry, you’ll find that a basic graph can also be stretched or compressed horizontally, a phenomenon known as frequency variations.

The transformations in Example 7 are called vertical stretches or compressions of a basic graph. Notice that while the outputs are increased or decreased by a constant factor (making the graph appear more narrow or more wide), the domain of the function remains unchanged. In general,

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209


Stretches and Compressions of a Basic Graph

For any function y f 1x2 , the graph of y af 1x2 is 1. the graph of f(x) stretched vertically if a 7 1, 2. the graph of f(x) compressed vertically if 0 6 a 6 1. Figure 2.25

Figure 2.26

To use a graphing calculator in a study of stretches and compressions, we simply define Y2 and Y3 as constant multiples of Y1 (Figure 2.25). As seen in Example 7, if a 7 1 the graph will be stretched vertically, if 0 6 a 6 1, the graph will be vertically compressed. This is further illus- 0 trated here using Y1 1X, with Y2 2Y1 and Y3 0.5Y1. Since the domain of y 1x is restricted to nonnegative values, a window size of x 30, 10 4 and y 3 1, 7 4 was used (Figure 2.26). Try this exploration again using Y1 abs1X2 4.

D. You’ve just seen how we can apply vertical stretches and compressions of a basic graph

7

10

1

E. Transformations of a General Function If more than one transformation is applied to a basic graph, it’s helpful to use the following sequence for graphing the new function. General Transformations of a Basic Graph

Given a function y f 1x2 , the graph of y af 1x h2 k can be obtained by applying the following sequence of transformations: 1. horizontal shifts 2. reflections 3. stretches/compressions 4. vertical shifts We generally use a few characteristic points to track the transformations involved, then draw the transformed graph through the new location of these points. EXAMPLE 8

䊳

Graphing Functions Using Transformations Use transformations of a parent function to sketch the graphs of 3 a. g1x2 1x 22 2 3 b. h1x2 2 1 x21

Solution

a. The graph of g is a parabola, shifted left 2 units, reflected across the x-axis, and shifted up 3 units. This sequence of transformations is shown in Figures 2.27 through 2.29. Note that since the graph has been shifted 2 units left and 3 units up, the vertex of the parabola has likewise shifted from (0, 0) to 12, 32 .

䊳

Figure 2.27 y (x

Figure 2.28

y

2)2

(4, 4)

5

y x2

5

Figure 2.29

y y (x 2)2

5

y g(x) ⫽ ⫺(x ⫹ 2)2 ⫹ 3

(⫺2, 3)

(0, 4)

(2, 0) 5

(2, 0) Vertex

5

Shifted left 2 units

5

x

5

5

x

⫺5

(⫺4, ⫺1) (4, 4)

5

(0, 4)

Reflected across the x-axis

(0, ⫺1)

⫺5

Shifted up 3 units

5

x

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2–24


b. The graph of h is a cube root function, shifted right 2, stretched by a factor of 2, then shifted down 1. This sequence is shown in Figures 2.30 through 2.32 and illustrate how the inflection point has shifted from (0, 0) to 12, 12 . Figure 2.30 y 5

Figure 2.31

3

y x 2

5

Figure 2.32

3 y y 2x 2

5

3 y h(x) 2x 21

(3, 2) (3, 1) (2, 0) 6 x Inflection (1, 1)

4

(2, 0) 6

x

4

(2, 1)

(1, 2)

6

x

(1, 3) 5

5

5

Shifted right 2

(3, 1)

4

Shifted down 1

Stretched by a factor of 2


䊳

It’s important to note that the transformations can actually be applied to any function, even those that are new and unfamiliar. Consider the following pattern: Parent Function

Transformation of Parent Function y 21x 32 2 1

quadratic: y x2

absolute value: y 0 x 0

y 2 0 x 3 0 1

3 y 2 1 x31

cube root: y 1x 3

general: y f 1x2

y 2f 1x 32 1

In each case, the transformation involves a horizontal shift 3 units right, a vertical reflection, a vertical stretch, and a vertical shift up 1. Since the shifts are the same regardless of the initial function, we can generalize the results to any function f(x).

WORTHY OF NOTE Since the shape of the initial graph does not change when translations or reflections are applied, these are called rigid transformations. Stretches and compressions of a basic graph are called nonrigid transformations, as the graph is distended in some way.

vertical reflections, vertical stretches and compressions

S

y af 1x h2 k S

y f 1x2

Transformed Function S

General Function

horizontal shift h units, opposite direction of sign

vertical shift k units, same direction as sign

Also bear in mind that the graph will be reflected across the y-axis (horizontally) if x is replaced with x. This process is illustrated in Example 9 for selected transformations. Remember — if the graph of a function is shifted, the individual points on the graph are likewise shifted.

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211


EXAMPLE 9

䊳

Graphing Transformations of a General Function

Solution

䊳

For g, the graph of f is (1) shifted horizontally 1 unit left (Figure 2.34), (2) reflected across the x-axis (Figure 2.35), and (3) shifted vertically 2 units down (Figure 2.36). The final result is that in Figure 2.36.

Given the graph of f(x) shown in Figure 2.33, graph g1x2 f 1x 12 2.

Figure 2.34

Figure 2.33 y

y

5

5

(2, 3)

(3, 3)

f (x)

(0, 0) 5

5

x

5

(1, 0)

(2, 3)

5

x

5

x

(1, 3)

5

5

Figure 2.36

Figure 2.35

y

y 5

5

(1, 3) (1, 1) g (x)

(1, 0) 5

5

x

5

(3, 2) (1, 2)

(5, 2) (3, 3) 5

(3, 5)

5


䊳

As noted in Example 9, these shifts and transformation are often combined— particularly when the toolbox functions are used as real-world models (Section 2.6). On a graphing calculator we again define Y1 as needed, then define Y2 as any desired combination of shifts, stretches, and/or reflections. For Y1 X2, we’ll define Y2 as 2 Y1 1X 52 3 (Figure 2.37), and expect that the graph of Y2 will be that of Y1 shifted left 5 units, reflected across the x-axis, stretched vertically, and shifted up three units. This shows the new vertex should be at 15, 32 , which is confirmed in Figure 2.38 along with the other transformations. Figure 2.38 Figure 2.37

10

10

10

10

Try this exploration again using Y1 abs1X2 .

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Using the general equation y af 1x h2 k, we can identify the vertex, initial point, or inflection point of any toolbox function and sketch its graph. Given the graph of a toolbox function, we can likewise identify these points and reconstruct its equation. We first identify the function family and the location (h, k) of any characteristic point. By selecting one other point (x, y) on the graph, we then use the general equation as a formula (substituting h, k, and the x- and y-values of the second point) to solve for a and complete the equation. EXAMPLE 10

䊳

Writing the Equation of a Function Given Its Graph Find the equation of the function f(x) shown in the figure.

Solution

䊳

The function f belongs to the absolute value family. The vertex (h, k) is at (1, 2). For an additional point, choose the x-intercept (3, 0) and work as follows: y ax h k 0 a 132 1 2

E. You’ve just seen how we can apply transformations on a general function f(x)

0 4a 2 2 4a 1 a 2

general equation (function is shifted right and up) substitute 1 for h and 2 for k, substitute 3 for x and 0 for y simplify

y 5

f(x) 5

5

x

subtract 2 5

solve for a

The equation for f is y 12 0 x 1 0 2. Now try Exercises 97 through 102

䊳

2.2 EXERCISES 䊳



1. After a vertical , points on the graph are farther from the x-axis. After a vertical , points on the graph are closer to the x-axis. 3. The vertex of h1x2 31x 52 2 9 is at and the graph opens . 5. Given the graph of a general function f (x), discuss/ explain how the graph of F1x2 2f 1x 12 3 can be obtained. If (0, 5), (6, 7), and 19, 42 are on the graph of f, where do they end up on the graph of F?

2. Transformations that change only the location of a graph and not its shape or form, include and .

4. The inflection point of f 1x2 21x 42 3 11 is at and the end-behavior is , .

6. Discuss/Explain why the shift of f 1x2 x2 3 is a vertical shift of 3 units in the positive direction, while the shift of g1x2 1x 32 2 is a horizontal shift 3 units in the negative direction. Include several examples along with a table of values for each.

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2–27 䊳

213



By carefully inspecting each graph given, (a) identify the function family; (b) describe or identify the end-behavior, vertex, intervals where the function is increasing or decreasing, maximum or minimum value(s) and x- and y-intercepts; and (c) determine the domain and range. Assume required features have integer values.

7. f 1x2 x2 4x

15. r 1x2 314 x 3 16. f 1x2 21x 1 4 y

y

5

5

⫺5

5 x

8. g1x2 x2 2x

y

5 x

f(x)

⫺5

⫺5

y

5

⫺5

r(x)

5

17. g1x2 2 14 x

18. h1x2 21x 1 4

y ⫺5

5 x

⫺5

y 5

5

5 x

g(x) h(x)

⫺5

⫺5

9. p1x2 x2 2x 3

⫺5

5 x

⫺5

5 x

10. q1x2 x2 2x 8

y

⫺5

⫺5

y

5

10

⫺5

5 x

⫺10

10 x

⫺5

⫺10

11. f 1x2 x2 4x 5

12. g1x2 x2 6x 5

y

For each graph given, (a) identify the function family; (b) describe or identify the end-behavior, vertex, intervals where the function is increasing or decreasing, maximum or minimum value(s) and x- and y-intercepts; and (c) determine the domain and range. Assume required features have integer values.

19. p1x2 2x 1 4

10

20. q1x2 3x 2 3

y

y

y

5

10

5

q(x)

⫺10

10 x

⫺10

10 x

⫺10

⫺5

5 x

⫺5

For each graph given, (a) identify the function family; (b) describe or identify the end-behavior, initial point, intervals where the function is increasing or decreasing, and x- and y-intercepts; and (c) determine the domain and range. Assume required features have integer values.

5 x

⫺5

⫺5

⫺10

13. p1x2 2 1x 4 2

p(x)

21. r 1x2 2x 1 6 22. f 1x2 3x 2 6 y

y 4

6

r(x) ⫺5 ⫺5

5 x

5 x

f(x)

14. q1x2 2 1x 4 2 ⫺6

⫺4

y

y 5

5

23. g1x2 3x 6

p(x)

24. h1x2 2x 1

y

y 6

6 ⫺5

5 x

⫺5

5 x

q(x)

g(x) ⫺5

h(x)

⫺5 ⫺5

5 x

⫺4

⫺5

5 x

⫺4

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2–28


For each graph given, (a) identify the function family; (b) describe or identify the end-behavior, inflection point, and x- and y-intercepts; and (c) determine the domain and range. Assume required features have integer values. Be sure to note the scaling of each axis.

25. f 1x2 ⫽ ⫺1x ⫺ 12 3

26. g1x2 ⫽ 1x ⫹ 12 3

y 5

f(x)

⫺5

5 x

5 x

⫺5

3 28. p1x2 ⫽ ⫺ 2x ⫹ 1

y

⫺5

5 x

5 x

⫺5

⫺5

29. q1x2 ⫽ 2x ⫺ 1 ⫺ 1 3

30. r 1x2 ⫽ ⫺ 2x ⫹ 1 ⫺1 3

y

y

⫺5

⫺5

5 x

q(x)

⫺5

5 x

r(x)

y 5

32.

f(x)

42. t1x2 ⫽ 0 x 0 ⫺ 3

45. Y1 ⫽ 0 x 0 , Y2 ⫽ 0 x ⫺ 4 0

46. h1x2 ⫽ x3, H1x2 ⫽ 1x ⫺ 42 3 Sketch each graph by hand using transformations of a parent function (without a table of values).

47. p1x2 ⫽ 1x ⫺ 32 2

48. q1x2 ⫽ 1x ⫺ 1

51. g1x2 ⫽ ⫺ 0 x 0

52. j1x2 ⫽ ⫺ 1x

3 53. f 1x2 ⫽ 2 ⫺x

3 50. f 1x2 ⫽ 1 x⫹2

54. g1x2 ⫽ 1⫺x2 3

Use a graphing calculator to graph the functions given in the same window. Comment on what you observe.

⫺5

For Exercises 31–34, identify and state the characteristic features of each graph, including (as applicable) the function family, end-behavior, vertex, axis of symmetry, point of inflection, initial point, maximum and minimum value(s), x- and y-intercepts, and the domain and range.

31.

40. g1x2 ⫽ 1x ⫺ 4


49. h1x2 ⫽ 冟x ⫹ 3冟

5

5

39. f 1x2 ⫽ x3 ⫺ 2

44. f 1x2 ⫽ 1x, g1x2 ⫽ 1x ⫹ 4

p(x) h(x) ⫺5

37. p1x2 ⫽ 冟x冟, q1x2 ⫽ 冟x冟 ⫺ 5, r 1x2 ⫽ 冟x冟 ⫹ 2

43. p1x2 ⫽ x2, q1x2 ⫽ 1x ⫹ 52 2

y 5

5

h1x2 ⫽ 1x ⫺ 3

3 3 g1x2 ⫽ 2 x ⫺ 3, h1x2 ⫽ 2 x⫹4

41. h1x2 ⫽ x2 ⫹ 3

⫺5

27. h1x2 ⫽ x3 ⫹ 1

3 36. f 1x2 ⫽ 2 x,

g1x2 ⫽ 1x ⫹ 2,

Sketch each graph by hand using transformations of a parent function (without a table of values).

g(x)

⫺5

35. f 1x2 ⫽ 1x,

38. p1x2 ⫽ x2, q1x2 ⫽ x2 ⫺ 7, r 1x2 ⫽ x2 ⫹ 3

y

5


y 5

55. p1x2 ⫽ x2, q1x2 ⫽ 3x2, r 1x2 ⫽ 15x2 56. f 1x2 ⫽ 1⫺x, g1x2 ⫽ 41⫺x,

h1x2 ⫽ 14 1⫺x

57. Y1 ⫽ 0 x 0 , Y2 ⫽ 3 0 x 0 , Y3 ⫽ 13 0 x 0

58. u1x2 ⫽ x3, v1x2 ⫽ 8x3, w1x2 ⫽ 15x3

g(x)

Sketch each graph by hand using transformations of a parent function (without a table of values). ⫺5

⫺5

5 x

5 x

59. f 1x2 ⫽ 4 2x 3

61. p1x2 ⫽ 13x3 y 5

⫺5

34.

f(x)

5 x

⫺5

62. q1x2 ⫽ 34 1x

⫺5

⫺5

33.

60. g1x2 ⫽ ⫺2 0x 0

y 5

⫺5

Use the characteristics of each function family to match a given function to its corresponding graph. The graphs are not scaled — make your selection based on a careful comparison.

g(x)

5 x

⫺5

63. f 1x2 ⫽ 12x3

64. f 1x2 ⫽ ⫺2 3 x ⫹ 2

3 65. f 1x2 ⫽ ⫺1x ⫺ 32 2 ⫹ 2 66. f 1x2 ⫽ ⫺ 1 x⫺1⫺1

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215


67. f 1x2 x 4 1

68. f 1x2 1x 6

71. f 1x2 1x 42 2 3

72. f 1x2 x 2 5

Graph each function using shifts of a parent function and a few characteristic points. Clearly state and indicate the transformations used and identify the location of all vertices, initial points, and/or inflection points.

69. f 1x2 1x 6 1 70. f 1x2 x 1 73. f 1x2 1x 3 1 y a.

74. f 1x2 1x 32 2 5 y b.

x

x

75. f 1x2 1x 2 1

76. g1x2 1x 3 2

79. p1x2 1x 32 3 1

80. q1x2 1x 22 3 1

83. f 1x2 x 3 2

84. g1x2 x 4 2

77. h1x2 1x 32 2 2 78. H1x2 1x 22 2 5 3 81. s1x2 1 x12

3 82. t1x2 1 x31

85. h1x2 21x 12 2 3 86. H1x2 12x 2 3 c.

d.

y

3 87. p1x2 13 1x 22 3 1 88. q1x2 4 1 x12

y

89. u1x2 2 1x 1 3 90. v1x2 3 1x 2 1 x

x

91. h1x2 15 1x 32 2 1

92. H1x2 2x 3 4

Apply the transformations indicated for the graph of the general functions given.

e.

f.

y

93.

y

y 5

94.

f(x)

y 5

g(x)

(⫺1, 4) (⫺4, 4)

(3, 2)

(⫺1, 2)

x

x

⫺5

⫺5

5 x

5 x

(⫺4, ⫺2) ⫺5

⫺5

g.

h.

y

y

a. f 1x 22 b. f 1x2 3 c. 12 f 1x 12 d. f 1x2 1

x

x

95. i.

y

j.

y 5

(2, ⫺2)

a. b. c. d. 96.

h(x)

g1x2 2 g1x2 3 2g1x 12 1 2 g1x 12 2 y 5

y

(⫺1, 3)

(2, 0)

(⫺1, 0) ⫺5

x

5 x

⫺5

y

l.

x

⫺5

y

x

a. b. c. d.

(1, ⫺3)

(2, ⫺4)

h1x2 3 h1x 22 h1x 22 1 1 4 h1x2 5

5 x

(⫺2, 0)

x (⫺4, ⫺4)

k.

H(x)

⫺5

a. b. c. d.

H1x 32 H1x2 1 2H1x 32 1 3 H1x 22 1

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Use the graph given and the points indicated to determine the equation of the function shown using the general form y ⴝ af(x ⴞ h) ⴞ k.

97.

98.

y 5

99.

y (⫺5, 6)

y

5

(6, 4.5)

5

p(x) g(x)

(2, 0) ⫺5

5 x

f(x)

⫺5

5 x

⫺3(⫺3, 0)

(0, ⫺4)

⫺5

100.

⫺4

⫺3

(0, ⫺4)

101.

y (⫺4, 5) 5

x

5

102.

y 5

y (3, 7)

7

(1, 4) f(x)

h(x)

r(x) ⫺4

5

x

⫺8

(5, ⫺1)

⫺5

䊳

⫺3 ⫺5

7 x ⫺3

(0, ⫺2)


103. Volume of a sphere: V(r) ⴝ 43␲r3 The volume of a sphere is given by the function shown, where V(r) is the volume in cubic units and r is the radius. Note this function belongs to the cubic family of functions. (a) Approximate the value of 43␲ to one decimal place, then graph the function on the interval [0, 3]. (b) From your graph, estimate the volume of a sphere with radius 2.5 in., then compute the actual volume. Are the results close? (c) For V ⫽ 43 ␲r3, solve for r in terms of V.

䊳

2 x

(⫺4, 0)

104. Fluid motion: V(h) ⴝ ⴚ4 1h ⴙ 20 Suppose the velocity of a fluid flowing from an open tank (no top) through an opening in its side is given by the function shown, where V(h) is the velocity of the fluid (in feet per second) at water height h (in feet). Note this function belongs to the square root family of functions. An open tank is 25 ft deep and filled to the brim with fluid. (a) Use a table of values to graph the 25 ft function on the interval [0, 25]. (b) From your graph, estimate the velocity of the fluid when the water level is 7 ft, then find the actual velocity. Are the answers close? (c) If the fluid velocity is 5 ft/sec, how high is the water in the tank?

APPLICATIONS

105. Gravity, distance, time: After being released, the time it takes an object to fall x ft is given by the function T1x2 ⫽ 14 1x, where T(x) is in seconds. (a) Describe the transformation applied to obtain the graph of T from the graph of y ⫽ 1x, then sketch the graph of T for x 僆 30, 100 4 . (b) How long would it take an object to hit the ground if it were dropped from a height of 81 ft? 106. Stopping distance: In certain weather conditions, accident investigators will use the function v1x2 ⫽ 4.9 1x to estimate the speed of a car (in miles per hour) that has been involved in an accident, based on the length of the skid marks x (in feet). (a) Describe the transformation applied to

obtain the graph of v from the graph of y ⫽ 1x, then sketch the graph of v for x 僆 30, 4004 . (b) If the skid marks were 225 ft long, how fast was the car traveling? Is this point on your graph? 107. Wind power: The power P generated by a certain 8 3 v wind turbine is given by the function P1v2 ⫽ 125 where P(v) is the power in watts at wind velocity v (in miles per hour). (a) Describe the transformation applied to obtain the graph of P from the graph of y ⫽ v3, then sketch the graph of P for v 僆 30, 254 (scale the axes appropriately). (b) How much power is being generated when the wind is blowing at 15 mph?

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108. Wind power: If the power P (in watts) being generated by a wind turbine is known, the velocity of the wind can be determined using the function 3 v1P2 ⫽ 52 2 P. (a) Describe the transformation applied to obtain the graph of v from the graph of 3 y⫽ 2 P, then sketch the graph of v for P 僆 3 0, 512 4 (scale the axes appropriately). (b) How fast is the wind blowing if 343W of power is being generated? Is this point on your graph? 109. Distance rolled due to gravity: The distance a ball rolls down an inclined plane is given by the function d1t2 ⫽ 2t2, where d(t) represents the distance in feet after t sec. (a) Describe the transformation applied to obtain the graph of d from the graph

䊳

of y ⫽ t2, then sketch the graph of d for t 僆 3 0, 3 4 . (b) How far has the ball rolled after 2.5 sec? 110. Acceleration due to gravity: The velocity of a steel ball bearing as it rolls down an inclined plane is given by the function v1t2 ⫽ 4t, where v(t) represents the velocity in feet per second after t sec. (a) Describe the transformation applied to obtain the graph of v from the graph of y ⫽ t, then sketch the graph of v for t 僆 30, 3 4 . (b) What is the velocity of the ball bearing after 2.5 sec? Is this point on your graph?


111. Carefully graph the functions f 1x2 ⫽ 冟x冟 and g1x2 ⫽ 2 1x on the same coordinate grid. From the graph, in what interval is the graph of g(x) above the graph of f (x)? Pick a number (call it h) from this interval and substitute it in both functions. Is g1h2 7 f 1h2? In what interval is the graph of g(x) below the graph of f(x)? Pick a number from this interval (call it k) and substitute it in both functions. Is g1k2 6 f 1k2?

䊳

217

112. Sketch the graph of f 1x2 ⫽ ⫺2冟x ⫺ 3冟 ⫹ 8 using transformations of the parent function, then determine the area of the region in quadrant I that is beneath the graph and bounded by the vertical lines x ⫽ 0 and x ⫽ 6.

113. Sketch the graph of f 1x2 ⫽ x2 ⫺ 4, then sketch the graph of F1x2 ⫽ 冟x2 ⫺ 4冟 using your intuition and the meaning of absolute value (not a table of values). What happens to the graph?


114. (1.1) Find the distance between the points 1⫺13, 92 and 17, ⫺122, and the slope of the line containing these points. 115. (R.2) Find the perimeter of the figure shown. 5x ⫹ 2

2x2 ⫹3x

5x 2x2 ⫹3x ⫹ 5

1 1 7 2 116. (1.5) Solve for x: x ⫹ ⫽ x ⫺ . 3 4 2 12 117. (2.1) Without graphing, state intervals where f 1x2c and f 1x2T for f 1x2 ⫽ 1x ⫺ 42 2 ⫹ 3.

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2.3

Absolute Value Functions, Equations, and Inequalities While the equations x 1 5 and 冟x 1冟 5 are similar in many respects, note the first has only the solution x 4, while either x 4 or x 6 will satisfy the second. The fact there are two solutions shouldn’t surprise us, as it’s a natural result of how absolute value is defined.


A. Solve absolute value equations

A. Solving Absolute Value Equations

B. Solve “less than” absolute value inequalities C. Solve “greater than” absolute value inequalities D. Solve absolute value equations and inequalities graphically E. Solve applications involving absolute value

The absolute value of a number x can be thought of as its distance from zero on the number line, regardless of direction. This means 冟x冟 4 will have two solutions, since there are two numbers that are four units from zero: x 4 and x 4 (see Figure 2.39). Exactly 4 units from zero

Figure 2.39

⫺5 ⫺4

Exactly 4 units from zero ⫺3 ⫺2 ⫺1

0

1

2

3

4

5

This basic idea can be extended to include situations where the quantity within absolute value bars is an algebraic expression, and suggests the following property. Property of Absolute Value Equations If X represents an algebraic expression and k is a positive real number,

WORTHY OF NOTE Note if k 6 0, the equation 冟X冟 k has no solutions since the absolute value of any quantity is always positive or zero. On a related note, we can verify that if k 0, the equation 冟X冟 0 has only the solution X 0.

then 冟X冟 k implies X k or X k As the statement of this property suggests, it can only be applied after the absolute value expression has been isolated on one side.

EXAMPLE 1

䊳

Solving an Absolute Value Equation Solve: 5冟x 7冟 2 13.

Solution

䊳

Begin by isolating the absolute value expression. 5冟x 7冟 2 13 original equation 5冟x 7冟 15 subtract 2 冟x 7冟 3 divide by 5 (simplified form) Now consider x 7 as the variable expression “X” in the property of absolute value equations, giving or x 7 3 x 7 3 apply the property of absolute value equations x4 or x 10 add 7 Substituting into the original equation verifies the solution set is {4, 10}. Now try Exercises 7 through 18

CAUTION

218

䊳

䊳

For equations like those in Example 1, be careful not to treat the absolute value bars as simple grouping symbols. The equation 51x 72 2 13 has only the solution x 10, and “misses” the second solution since it yields x 7 3 in simplified form. The equation 5冟x 7冟 2 13 simplifies to 冟x 7冟 3 and there are actually two solutions. Also note that 5冟x 7冟冟5x 35冟!

2–32

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Section 2.3 Absolute Value Functions, Equations, and Inequalities

219

If an equation has more than one solution as in Example 1, they cannot be simultaneously stored using the, X,T,␪,n key to perform a calculator check (in function or “Func” mode, this is the variable X). While there are other ways to “get around” this (using Y 1 on the home screen, using a TABLE in ASK mode, enclosing the solutions in braces as in {4, 10}, etc.), we can also store solutions using the ALPHA keys. To illustrate, we’ll place the solution x 4 in storage location A, using 4 STO ALPHA MATH (A). Using this “ STO ALPHA ” sequence we’ll next place the solution x 10 in storage location B (Figure 2.40). We can then check both solutions in turn. Note that after we check the first solution, we can recall the expression using 2nd and simply change the A to B (Figure 2.41). ENTER

Figure 2.40

Figure 2.41

Absolute value equations come in many different forms. Always begin by isolating the absolute value expression, then apply the property of absolute value equations to solve.

EXAMPLE 2

䊳

Solving an Absolute Value Equation Solve:

Solution

䊳

2 ` 5 x ` 9 8. 3 2 `5 x ` 9 8 3 2 ` 5 x ` 17 3 2 5 x 17 3 2 x 22 3 x 33

Check WORTHY OF NOTE As illustrated in both Examples 1 and 2, the property we use to solve absolute value equations can only be applied after the absolute value term has been isolated. As you will see, the same is true for the properties used to solve absolute value inequalities.

䊳

2 For x 33: ` 5 1332 ` 3 |5 21112 | 05 22 0 0 17 0 17

original equation

add 9

or or or

98 98 98 98 98 8 8✓

2 5 x 17 3 2 x 12 3 x 18

apply the property of absolute value equations subtract 5 multiply by 32

2 1182 ` 9 8 3 | 5 2162 | 9 8 0 5 12 0 9 8 0 17 0 9 8 17 9 8 8 8✓

For x 18: ` 5

Both solutions check. The solution set is 518, 336.


䊳

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For some equations, it’s helpful to apply the multiplicative property of absolute value: Multiplicative Property of Absolute Value If A and B represent algebraic expressions, then 冟AB冟冟A冟冟B冟. Note that if A 1 the property says 冟1 # B冟冟1冟冟B冟冟B冟. More generally the property is applied where A is any constant.

EXAMPLE 3

䊳

Solution

䊳

Solving Equations Using the Multiplicative Property of Absolute Value Solve: 冟2x冟 5 13. 冟2x冟 5 13 冟2x冟 8 冟2冟冟x冟 8 2冟x冟 8 冟x冟 4 x 4 or x 4

original equation subtract 5 apply multiplicative property of absolute value simplify divide by 2 apply property of absolute value equations

Both solutions check. The solution set is 54, 46. Now try Exercises 23 and 24

䊳

In some instances, we have one absolute value quantity equal to another, as in 冟A冟冟B冟. From this equation, four possible solutions are immediately apparent: (1) A B

(2) A B

(3) A B

(4) A B

However, basic properties of equality show that equations (1) and (4) are equivalent, as are equations (2) and (3), meaning all solutions can be found using only equations (1) and (2).

EXAMPLE 4

䊳

Solving Absolute Value Equations with Two Absolute Value Expressions Solve the equation 冟2x 7冟冟x 1冟.

Solution

䊳

This equation has the form 冟A冟冟B冟, where A 2x 7 and B x 1. From our previous discussion, all solutions can be found using A B and A B. AB 2x 7 x 1 2x x 8 x 8

solution template substitute subtract 7 subtract x

A B 2x 7 1x 12 2x 7 x 1 3x 6 x 2

solution template substitute distribute add x, subtract 7 divide by 3

The solutions are x 8 and x 2. Verify the solutions by substituting them into the original equation. A. You’ve just seen how we can solve absolute value equations


䊳

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221

B. Solving “Less Than” Absolute Value Inequalities Absolute value inequalities can be solved using the basic concept underlying the property of absolute value equalities. Whereas the equation 冟x冟 4 asks for all numbers x whose distance from zero is equal to 4, the inequality 冟x冟 6 4 asks for all numbers x whose distance from zero is less than 4. Distance from zero is less than 4

Figure 2.42

)

⫺5 ⫺4

⫺3 ⫺2 ⫺1

) 0

1

2

3

4

5

As Figure 2.42 illustrates, the solutions are x 7 4 and x 6 4, which can be written as the joint inequality 4 6 x 6 4. This idea can likewise be extended to include the absolute value of an algebraic expression X as follows. Property I: Absolute Value Inequalities (Less Than) If X represents an algebraic expression and k is a positive real number, then 冟X冟 6 k implies k 6 X 6 k Property I can also be applied when the “” symbol is used. Also notice that if k 6 0, the solution is the empty set since the absolute value of any quantity is always positive or zero.

EXAMPLE 5

Solution

䊳

䊳

Solving “Less Than” Absolute Value Inequalities Solve the inequalities: 冟3x 2冟 a. 1 4 冟3x 2冟 a. 1 4 冟3x 2冟 4 4 3x 2 4 6 3x 2 2 2 x 3

b. 冟2x 7冟 6 5 original inequality multiply by 4 apply Property I subtract 2 from all three parts divide all three parts by 3

The solution interval is 32, 23 4.

b. 冟2x 7冟 6 5

original inequality

Since the absolute value of any quantity is always positive or zero, the solution for this inequality is the empty set: { }. Now try Exercises 27 through 38

䊳

As with the inequalities from Section 1.5, solutions to absolute value inequalities can be checked using a test value. For Example 5(a), substituting x 0 from the solution interval yields: 1 1✓ 2

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In addition to checking absolute value inequalities using a test value, the TABLE feature of a graphing calculator can be used, alone or in conjunction with a relational test. Relational tests have the calculator return a “1” if a given statement is true, and a “0” otherwise. To illustrate, consider the inequality 2冟x 3冟 1 5. Enter the expression on the left as Y1, recalling the “abs(” notation is accessed in the MATH menu: MATH (NUM) “1:abs(” (note this option gives only the left parenthesis, you must supply the right). We can then simply inspect the Y1 column of the TABLE to find outputs that are less than or equal to 5. To use a relational test, we enter Y1 5 as Y2 (Figure 2.43), with the “less than or equal to” symbol accessed using 2nd MATH 6:ⱕ. Now the calculator will automatically check the truth of the statement for any value of x (but note we are only checking integer values), and display the result in the Y2 column of the TABLE (Figure 2.44). After scrolling through the table, both approaches show that 2冟x 3冟 1 5 for x 僆 [1, 5]. ENTER

Figure 2.43

Figure 2.44

B. You’ve just seen how we can solve “less than” absolute value inequalities

C. Solving “Greater Than” Absolute Value Inequalities For “greater than” inequalities, consider 冟x冟 7 4. Now we’re asked to find all numbers x whose distance from zero is greater than 4. As Figure 2.45 shows, solutions are found in the interval to the left of 4, or to the right of 4. The fact the intervals are disjoint (disconnected) is reflected in this graph, in the inequalities x 6 4 or x 7 4, as well as the interval notation x 僆 1q, 42 ´ 14, q 2. Distance from zero is greater than 4

Figure 2.45

)

⫺7 ⫺6 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1

Distance from zero is greater than 4

) 0

1

2

3

4

5

6

7

As before, we can extend this idea to include algebraic expressions, as follows: Property II: Absolute Value Inequalities (Greater Than) If X represents an algebraic expression and k is a positive real number, then 冟X冟 7 k implies X 6 k or X 7 k

EXAMPLE 6

䊳

Solving “Greater Than” Absolute Value Inequalities Solve the inequalities: 1 x a. ` 3 ` 6 2 3 2

b. 冟5x 2冟

3 2

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Solution

䊳

223

a. Note the exercise is given as a less than inequality, but as we multiply both sides by 3, we must reverse the inequality symbol. x 1 ` 3 ` 6 2 3 2 x `3 ` 7 6 2 x x or 3 7 6 3 6 6 2 2 x x 6 9 or 7 3 2 2 or x 7 6 x 6 18

original inequality multiply by ⴚ3, reverse the symbol

apply Property II

subtract 3 multiply by 2

Property II yields the disjoint intervals x 僆 1q, 182 ´ 16, q 2 as the solution. )

⫺30 ⫺24 ⫺18 ⫺12 ⫺6

) 0

6

12

18

24

30

3 original inequality 2 Since the absolute value of any quantity is always positive or zero, the solution for this inequality is all real numbers: x 僆⺢.

b. 冟5x 2冟


䊳

A calculator check is shown for part (a) in Figures 2.46 through 2.48. Figure 2.46

Figure 2.47

Figure 2.48

This helps to verify the solution interval is x 僆 1q, 182 ´ 16, q 2 . Due to the nature of absolute value functions, there are times when an absolute value relation cannot be satisfied. For instance the equation 冟x 4冟 2 has no solutions, as the left-hand expression will always represent a non-negative value. The inequality 冟2x 3冟 6 1 has no solutions for the same reason. On the other hand, the inequality 冟9 x冟 0 is true for all real numbers, since any value substituted for x will result in a nonnegative value. We can generalize many of these special cases as follows.

C. You’ve just seen how we can solve “greater than” absolute value inequalities

Absolute Value Functions — Special Cases Given k is a positive real number and A represents an algebraic expression, 冟A冟 k 冟A冟 6 k 冟A冟 7 k has no solutions

has no solutions

is true for all real numbers

See Exercises 51 through 54. CAUTION

䊳

Be sure you note the difference between the individual solutions of an absolute value equation, and the solution intervals that often result from solving absolute value inequalities. The solution 52, 56 indicates that both x 2 and x 5 are solutions, while the solution 32, 52 indicates that all numbers between 2 and 5, including 2, are solutions.

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D. Solving Absolute Value Equations and Inequalities Graphically The concepts studied in Section 1.5 (solving linear equations and inequalities graphically) are easily extended to other kinds of relations. Essentially, we treat each expression forming the equation or inequality as a separate function, then graph both functions to find points of intersection (equations) or where one graph is above or Figure 2.49

Figure 2.50

3.1

4.7

3.1

4.7

4.7

3.1

4.7

3.1

below the other (inequalities). For 2冟x 1冟 3 6 2, enter the expression 2冟X 1冟 3 as Y1 on the Y= screen, and 2 as Y2. Using ZOOM 4:ZDecimal produces the graph shown in Figure 2.49. Using 2nd TRACE (CALC) 5:intersect, we find the graphs intersect at x 1.5 and x 3.5 (Figure 2.50), and the graph of Y1 is above the graph of Y2 in this interval. Since this is a “less than” inequality, the solutions are outside of this interval, which gives x 僆 1q, 1.52 ´ 13.5, q2 as the solution interval. Note that the zeroes/x-intercept method could also have been used. EXAMPLE 7

䊳

Solving Absolute Equations and Inequalities Graphically 1 For f 1x2 2.5冟x 2冟 8 and g1x2 x 3, solve 2 a. f 1x2 g1x2 b. f 1x2 g1x2 c. f 1x2 7 g1x2

Solution

䊳

a. With f 1x2 2.5冟x 2冟 8 as Y1 and 1 g1x2 x 3 as Y2 (set to graph in bold), 2 using 2nd TRACE (CALC) 5:intersect 10 shows the graphs intersect 1Y1 Y2 2 at x 0 and x 5 (see figure). These are 1 the solutions to 2.5冟x 2冟 8 x 3. 2

10

10

10

b. The graph of Y1 is below the graph of Y2 1Y1 6 Y2 2 between these points of 1 intersection, so the solution interval for 2.5冟x 2冟 8 x 3 is x 僆 [0, 5]. 2 c. The graph of Y1 is above the graph of Y2 1Y1 7 Y2 2 outside this interval, 1 giving a solution of x 僆 1q, 02 ´ 15, q 2 for 2.5冟x 2冟 8 7 x 3. 2 D. You’ve just seen how we can solve absolute value equations and inequalities graphically


䊳

E. Applications Involving Absolute Value Applications of absolute value often involve finding a range of values for which a given statement is true. Many times, the equation or inequality used must be modeled after a given description or from given information, as in Example 8.

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EXAMPLE 8

䊳

225

Solving Applications Involving Absolute Value Inequalities For new cars, the number of miles per gallon (mpg) a car will get is heavily dependent on whether it is used mainly for short trips and city driving, or primarily on the highway for longer trips. For a certain car, the number of miles per gallon that a driver can expect varies by no more than 6.5 mpg above or below its field tested average of 28.4 mpg. What range of mileage values can a driver expect for this car?

Solution

䊳

Field tested average: 28.4 mpg mileage varies by no more than 6.5 mpg ⫺6.5

gather information highlight key phrases

⫹6.5

28.4

make the problem visual

Let m represent the miles per gallon a driver can expect. Then the difference between m and 28.4 can be no more than 6.5, or 冟m 28.4冟 6.5. 冟m 28.4冟 6.5 6.5 m 28.4 6.5 21.9 m 34.9

assign a variable write an equation model equation model apply Property I add 28.4 to all three parts

The mileage that a driver can expect ranges from a low of 21.9 mpg to a high of 34.9 mpg. E. You’ve just seen how we can solve applications involving absolute value


䊳

2.3 EXERCISES 䊳


Fill in the blank with the appropriate word or phrase. Carefully reread the section if needed.

1. When multiplying or dividing by a negative quantity, we the inequality symbol to maintain a true statement.

2. To write an absolute value equation or inequality in simplified form, we the absolute value expression on one side.

3. The absolute value equation 冟2x 3冟 7 is true when 2x 3 or when 2x 3 .

4. The absolute value inequality 冟3x 6冟 6 12 is true when 3x 6 7 and 3x 6 6

Describe the solution set for each inequality (assume k > 0). Justify your answer.

5. 冟ax b冟 6 k

6. 冟ax b冟 7 k

.

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226 䊳

2–40



Solve each absolute value equation. Write the solution in set notation. For Exercises 7 to 18, verify solutions by substituting into the original equation. For Exercises 19–26 verify solutions using a calculator.

7. 2冟m 1冟 7 3 8. 3冟n 5冟 14 2 9. 3冟x 5冟 6 15 10. 2冟y 3冟 4 14 11. 2冟4v 5冟 6.5 10.3 12. 7冟2w 5冟 6.3 11.2 13. 冟7p 3冟 6 5 14. 冟3q 4冟 3 5 15. 2冟b冟 3 4 16. 3冟c冟 5 6 17. 2冟3x冟 17 5 18. 5冟2y冟 14 6 19. 3 `

w 4 ` 1 4 2

20. 2 ` 3

v ` 1 5 3

21. 8.7冟p 7.5冟 26.6 8.2

35.

冟5v 1冟 8 6 9 4

36.

冟3w 2冟 6 6 8 2

37. `

1 7 4x 5 ` 3 2 6

38. `

2y 3 3 15 ` 4 8 16

39. 冟n 3冟 7 7 40. 冟m 1冟 7 5 41. 2冟w冟 5 11 42. 5冟v冟 3 23 43. 44.

冟q冟 2 冟p冟 5

5 1 6 3

3 9 2 4

45. 3冟5 7d冟 9 15 46. 5冟2c 7冟 1 11 47. 2 6 ` 3m

1 4 ` 5 5

3 5 2n ` 4 4

22. 5.3冟q 9.2冟 6.7 43.8

48. 4 `

23. 8.7冟2.5x冟 26.6 8.2

49. 4冟5 2h冟 9 7 11

24. 5.3冟1.25n冟 6.7 43.8

50. 3冟7 2k冟 11 7 10

25. 冟x 2冟冟3x 4冟

51. 3.9冟4q 5冟 8.7 22.5

26. 冟2x 1冟冟x 3冟

52. 0.9冟2p 7冟 16.11 10.89

Solve each absolute value inequality. Write solutions in interval notation. Check solutions by back substitution, or using a calculator.

27. 3冟p 4冟 5 6 8 28. 5冟q 2冟 7 8 29. 3 冟m冟 2 7 4 30. 2冟n冟 3 7 7 31. 冟3b 11冟 6 9 32. 冟2c 3冟 5 6 1 33. 冟4 3z冟 12 6 7 34. 冟2 3u冟 5 4

53. 冟4z 9冟 6 4 54. 冟5u 3冟 8 7 6 Use the intersect command on a graphing calculator and the given functions to solve (a) f 1x2 g1x2 , (b) f 1x2 g1x2 , and (c) f 1x2 6 g1x2 .

55. f 1x2 冟x 3冟 2, g1x2 12x 2

56. f 1x2 冟x 2冟 1, g1x2 32x 9

57. f 1x2 0.5冟x 3冟 1, g1x2 2冟x 1冟 5 58. f 1x2 2冟x 3冟 2, g1x2 冟x 4冟 6

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59. Spring Oscillation: 冟d x冟 L A weight attached to a spring hangs at rest a distance of x in. off the ground. If the weight is pulled down (stretched) a distance of L inches and released, the weight begins to bounce and its distance d off the ground must satisfy the indicated formula. (a) If x equals 4 ft and the spring is stretched 3 in. and released, solve the inequality to find what distances from the ground the weight will oscillate between. (b) Solve for x in terms of L and d. 䊳

227

60. A “Fair” Coin: `

h 50 ` 6 1.645 5

If we flipped a coin 100 times, we expect “heads” to come up about 50 times if the coin is “fair.” In a study of probability, it can be shown that the number of heads h that appears in such an experiment should satisfy the given inequality to be considered “fair.” (a) Solve this inequality for h. (b) If you flipped a coin 100 times and obtained 40 heads, is the coin “fair”?

APPLICATIONS

Solve each application of absolute value.

61. Altitude of jet stream: To take advantage of the jet stream, an airplane must fly at a height h (in feet) that satisfies the inequality 冟h 35,050冟 2550. Solve the inequality and determine if an altitude of 34,000 ft will place the plane in the jet stream. 62. Quality control tests: In order to satisfy quality control, the marble columns a company produces must earn a stress test score S that satisfies the inequality 冟S 17,750冟 275. Solve the inequality and determine if a score of 17,500 is in the passing range. 63. Submarine depth: The sonar operator on a submarine detects an old World War II submarine net and must decide to detour over or under the net. The computer gives him a depth model 冟d 394冟 20 7 164, where d is the depth in feet that represents safe passage. At what depth should the submarine travel to go under or over the net? Answer using simple inequalities. 64. Optimal fishing depth: When deep-sea fishing, the optimal depths d (in feet) for catching a certain type of fish satisfy the inequality 28冟d 350冟 1400 6 0. Find the range of depths that offer the best fishing. Answer using simple inequalities. For Exercises 65 through 68, (a) develop a model that uses an absolute value inequality, and (b) solve.

65. Stock value: My stock in MMM Corporation fluctuated a great deal in 2009, but never by more than $3.35 from its current value. If the stock is worth $37.58 today, what was its range in 2009?

66. Traffic studies: On a given day, the volume of traffic at a busy intersection averages 726 cars per hour (cph). During rush hour the volume is much higher, during “off hours” much lower. Find the range of this volume if it never varies by more than 235 cph from the average. 67. Physical training for recruits: For all recruits in the 3rd Armored Battalion, the average number of sit-ups is 125. For an individual recruit, the amount varies by no more than 23 sit-ups from the battalion average. Find the range of sit-ups for this battalion. 68. Computer consultant salaries: The national average salary for a computer consultant is $53,336. For a large computer firm, the salaries offered to their employees vary by no more than $11,994 from this national average. Find the range of salaries offered by this company. 69. Tolerances for sport balls: According to the official rules for golf, baseball, pool, and bowling, (a) golf balls must be within 0.03 mm of d 42.7 mm, (b) baseballs must be within 1.01 mm of d 73.78 mm, (c) billiard balls must be within 0.127 mm of d 57.150 mm, and (d) bowling balls must be within 12.05 mm of d 2171.05 mm. Write each statement using an absolute value inequality, then (e) determine which sport gives the least width of interval b for the diameter tolerance t at average value of the ball.

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70. Automated packaging: The machines that fill boxes of breakfast cereal are programmed to fill each box within a certain tolerance. If the box is overfilled, the company loses money. If it is underfilled, it is considered unsuitable for sale. 䊳

Suppose that boxes marked “14 ounces” of cereal must be filled to within 0.1 oz. Find the acceptable range of weights for this cereal.


71. Determine the value or values (if any) that will make the equation or inequality true. x a. 冟x冟 x 8 b. 冟x 2冟 2 c. x 冟x冟 x 冟x冟 d. 冟x 3冟 6x e. 冟2x 1冟 x 3 72. The equation 冟5 2x冟冟3 2x冟 has only one solution. Find it and explain why there is only one. 73. In many cases, it can be helpful to view the solutions to absolute value equations and inequalities as follows. For any algebraic expression X and positive

䊳

2–42


constant k, the equation 冟X冟 k has solutions X k and X k, since the absolute value of either quantity on the left will indeed yield the positive constant k. Likewise, 冟X冟 6 k has solutions X 6 k and X 6 k. Note the inequality symbol has not been reversed as yet, but will naturally be reversed as part of the solution process. Solve the following equations or inequalities using this idea. a. 冟x 3冟 5 b. 冟x 7冟 7 4 c. 3冟x 2冟 12 d. 3冟x 4冟 7 11


74. (R.4) Factor the expression completely: 18x3 21x2 60x. 76. (R.7) Simplify

1

by rationalizing the 3 23 denominator. State the result in exact form and approximate form (to hundredths).

75. (1.5) Solve V2

2W for ␳ (physics). C␳A

77. (R.3) Solve the inequality, then write the solution set in interval notation: 312x 52 7 21x 12 7.

MID-CHAPTER CHECK 1. Determine whether the following function is even, 冟x冟 odd, or neither. f 1x2 x2 4x 2. Use a graphing calculator to find the maximum and minimum values of f 1x2 1.91x4 2.3x3 2.2x 5.12 . Round to the nearest hundredth. 3. Use interval notation to identify the interval(s) where the function from Exercise 2 is increasing, decreasing, or constant. Round to the nearest hundredth.

4. Write the equation of the function that has the same graph of f 1x2 2x, shifted left 4 units and up 2 units. 5. For the graph given, (a) identify the function family, (b) describe or identify the end-behavior, inflection point, and x- and y-intercepts, (c) determine the domain and range, and (d) determine the value of k if f 1k2 2.5. Assume required features have integer values.

Exercise 5 y 5

f(x)

⫺5

5 x

⫺5

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6. Use a graphing calculator to graph the given functions in the same window and comment on what you observe. p1x2 1x 32 2

r1x2 12 1x 32 2

229

9. Solve the following absolute value inequalities. Write solutions in interval notation. a. 3.1冟d 2冟 1.1 7.3 冟1 y冟 11 2 7 b. 3 2 c. 5冟k 2冟 3 6 4

q1x2 1x 32 2

7. Solve the following absolute value equations. Write the solution in set notation. 2 11 a. 冟d 5冟 1 7 b. 5 冟s 3冟 3 2

10. Kiteboarding: With the correct sized kite, a person can kiteboard when the wind is blowing at a speed w (in mph) that satisfies the inequality 冟w 17冟 9. Solve the inequality and determine if a person can kiteboard with a windspeed of (a) 5 mph? (b) 12 mph?

8. Solve the following absolute value inequalities. Write solutions in interval notation. x a. 3冟q 4冟 2 6 10 b. ` 2 ` 5 5 3

REINFORCING BASIC CONCEPTS Using Distance to Understand Absolute Value Equations and Inequalities For any two numbers a and b on the number line, the distance between a and b can be written 冟a b冟 or 冟b a冟. In exactly the same way, the equation 冟x 3冟 4 can be read, “the distance between 3 and an unknown number is equal to 4.” The advantage of reading it in this way (instead of “the absolute value of x minus 3 is 4”), is that a much clearer visualization is formed, giving a constant reminder there are two solutions. In diagram form we have Figure 2.51. Distance between 3 and x is 4. ⫺5 ⫺4 ⫺3 ⫺2

Figure 2.51

4 units ⫺1

0

1

4 units 2

3

4

5

Distance between 3 and x is 4. 6

7

8

9

From this we note the solutions are x 1 and x 7. In the case of an inequality such as 冟x 2冟 3, we rewrite the inequality as 冟x 122 冟 3 and read it, “the distance between 2 and an unknown number is less than or equal to 3.” With some practice, visualizing this relationship mentally enables a quick statement of the solution: x 僆 35, 14 . In diagram form we have Figure 2.52. Distance between 2 and x is less than or equal to 3. 8 7 6

Figure 2.52

3 units

Distance between 2 and x is less than or equal to 3.

3 units

5 4 3 2 1

0

1

2

3

4

5

6

Equations and inequalities where the coefficient of x is not 1 still lend themselves to this form of conceptual understanding. For 冟2x 1冟 3 we read, “the distance between 1 and twice an unknown number is greater than or equal to 3.” On the number line (Figure 2.53), the number 3 units to the right of 1 is 4, and the number 3 units to the left of 1 is 2. Distance between 1 and 2x is greater than or equal to 3.

Figure 2.53

3 units

6 5 4 3

ⴚ2 1

0

3 units 1

2

3

Distance between 1 and 2x is greater than or equal to 3 4

5

6

7

8

For 2x 2, x 1, and for 2x 4, x 2, and the solution set is x 僆 1q, 1 4 ´ 32, q 2. Attempt to solve the following equations and inequalities by visualizing a number line. Check all results algebraically. Exercise 1: 冟x 2冟 5

Exercise 2: 冟x 1冟 4

Exercise 3: 冟2x 3冟 5

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2.4

Basic Rational Functions and Power Functions; More on the Domain


A. Graph basic rational functions, identify vertical and horizontal asymptotes, and describe end-behavior B. Use transformations to graph basic rational functions and write the equation for a given graph C. Graph basic power functions and state their domains D. Solve applications involving basic rational and power functions

In this section, we introduce two new kinds of relations, rational functions and power functions. While we’ve already studied a variety of functions, we still lack the ability to model a large number of important situations. For example, functions that model the amount of medication remaining in the bloodstream over time, the relationship between altitude and weightlessness, and the equations modeling planetary motion come from these two families.

A. Rational Functions and Asymptotes Just as a rational number is the ratio of two integers, a rational function is the ratio of two polynomials. In general, Rational Functions A rational function V(x) is one of the form V1x2

p1x2 d1x2

,

where p and d are polynomials and d1x2 0. The domain of V(x) is all real numbers, except the zeroes of d. The simplest rational functions are the reciprocal function y 1x and the reciprocal square function y x12, as both have a constant numerator and a single term in the denominator. Since division by zero is undefined, the domain of both excludes x 0. A preliminary study of these two functions will provide a strong foundation for our study of general rational functions in Chapter 4.

The Reciprocal Function: y ⴝ

1 x

The reciprocal function takes any input (other than zero) and gives its reciprocal as the output. This means large inputs produce small outputs and vice versa. A table of values (Table 2.1) and the resulting graph (Figure 2.54) are shown. Table 2.1

230

Figure 2.54

x

y

x

y

1000

1/1000

1/1000

1000

5

1/5

1/3

3

4

1/4

1/2

2

3

1/3

1

1

2

1/2

2

1/2

1

1

3

1/3

1/2

2

4

1/4

1/3

3

5

1/5

1/1000

1000

1000

1/1000

0

undefined

y 3

冢a, 3冣

y

1 x

2

(1, 1)

冢3, a冣

1

5

冢3, a冣 5

冢5, Q冣 (1, 1)

冢 a, 3冣

冢5, Q冣 x

1 2 3

2–44

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231

Section 2.4 Basic Rational Functions and Power Functions; More on the Domain

WORTHY OF NOTE The notation used for graphical behavior always begins by describing what is happening to the x-values, and the resulting effect on the y-values. Using Figure 2.55, visualize that for a point (x, y) on the graph of y 1x , as x gets larger, y must become smaller, particularly since their product must always be 1 1y 1x 1 xy 12 .

Table 2.1 and Figure 2.54 reveal some interesting features. First, the graph passes the vertical line test, verifying y 1x is indeed a function. Second, since division by zero is undefined, there can be no corresponding point on the graph, creating a break at x 0. In line with our definition of rational functions, the domain is x 僆 1q, 02 ´ 10, q 2 . Third, this is an odd function, with a “branch” of the graph in the first quadrant and one in the third quadrant, as the reciprocal of any input maintains its sign. Finally, we note in QI that as x becomes an infinitely large positive number, y becomes infinitely small and closer to zero. It seems convenient to symbolize this endbehavior using the following notation:

Figure 2.55

as x S q,

y

yS0

as x becomes an infinitely large positive number

Graphically, the curve becomes very close to, or approaches the x-axis. We also note that as x approaches zero from the right, y becomes an infinitely large positive number: as x S 0 , y S q . Note a superscript or sign is used to indicate the direction of the approach, meaning from the positive side (right) or from the negative side (left).

y x

x

EXAMPLE 1

y approaches 0

䊳

Describing the End-Behavior of Rational Functions For y 1x in QIII (Figure 2.54), a. Describe the end-behavior of the graph. b. Describe what happens as x approaches zero.

Solution

䊳

Similar to the graph’s behavior in QI, we have a. In words: As x becomes an infinitely large negative number, y approaches zero. In notation: As x S q , y S 0. b. In words: As x approaches zero from the left, y becomes an infinitely large negative number. In notation: As x S 0 , y S q . Now try Exercises 7 and 8

The Reciprocal Square Function: y ⴝ

䊳

1 x2

From our previous work, we anticipate this graph will also have a break at x 0. But since the square of any negative number is positive, the branches of the reciprocal square function are both above the x-axis. Note the result is the graph of an even function. See Table 2.2 and Figure 2.56. Table 2.2

Figure 2.56

x

y

x

y

1000

1/1,000,000

1/1000

1,000,000

5

1/25

1/3

9

4

1/16

1/2

4

3

1/9

1

1

2

1/4

2

1/4

1

1

3

1/9

1/2

4

4

1/16

1/3

9

5

1/25

1/1000

1,000,000

1000

1/1,000,000

0

undefined

y x12

y 3

(1, 1)

冢5,

1

1 25 冣 5

2

冢3, 19冣

(1, 1)

冢3, 19 冣

冢5, 5

1 2 3

1 25 冣

x

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Similar to y 1x , large positive inputs generate small, positive outputs: as x S q, y S 0. This is one indication of asymptotic behavior in the horizontal direction, and we say the line y 0 (the x-axis) is a horizontal asymptote for the reciprocal and reciprocal square functions. In general, Horizontal Asymptotes Given a constant k, the line y k is a horizontal asymptote for V if, as x increases or decreases without bound, V(x) approaches k: as x S q, V1x2 S k

or

as x S q, V1x2 S k

As shown in Figures 2.57 and 2.58, asymptotes are represented graphically as dashed lines that seem to “guide” the branches of the graph. Figure 2.57 shows a horizontal asymptote at y 1, which suggests the graph of f (x) is the graph of y 1x shifted up 1 unit. Figure 2.58 shows a horizontal asymptote at y 2, which suggests the graph of g(x) is the graph of y x12 shifted down 2 units. Figure 2.57 y 3

f(x)

1 x

Figure 2.58 1

y 3

2

2

y1

1

5

EXAMPLE 2

䊳

g(x) x12 2

5

x

1

5

5

1

1

2

2

3

3

x

y 2

Describing the End-Behavior of Rational Functions For the graph in Figure 2.58, use mathematical notation to a. Describe the end-behavior of the graph and name the horizontal asymptote. b. Describe what happens as x approaches zero.

Solution

䊳

a. as x S q, g1x2 S 2, as x S q, g1x2 S 2,

b. as x S 0 , g1x2 S q , as x S 0 , g1x2 S q

y 2 is a horizontal asymptote Now try Exercises 9 and 10

䊳

While the graphical view of Example 2(a) (Figure 2.58) makes these concepts believable, a numerical view of this end-behavior can be even more compelling. Try entering x12 2 as Y1 on the Y= screen, then go to the TABLE feature 1TblStart 3, ¢Tbl 1; Figure 2.59). Scrolling in either direction shows that as 冟x冟 becomes very large, Y1 becomes closer and closer to 2, but will never be equal to 2 (Figure 2.60). Figure 2.59

Figure 2.60

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233

From Example 2(b), we note that as x becomes smaller and close to 0, g becomes very large and increases without bound. This is one indication of asymptotic behavior in the vertical direction, and we say the line x 0 (the y-axis) is a vertical asymptote for g (x 0 is also a vertical asymptote for f in Figure 2.57). In general, Vertical Asymptotes Given a constant h, the vertical line x h is a vertical asymptote for a function V if, as x approaches h, V(x) increases or decreases without bound: as x S h , V1x2 S q

or

as x S h , V1x2 S q

Here is a brief summary: Reciprocal Function f 1x2

A. You’ve just seen how we can graph basic rational functions, identify vertical and horizontal asymptotes, and describe end-behavior

1 x Domain: x 僆 1q, 02 ´ 10, q 2 Range: y 僆 1q, 02 ´ 10, q2 Horizontal asymptote: y 0 Vertical asymptote: x 0

Reciprocal Quadratic Function 1 x2 Domain: x 僆 1q, 02 ´ 10, q2 Range: y 僆 10, q 2 Horizontal asymptote: y 0 Vertical asymptote: x 0 g1x2

B. Using Asymptotes to Graph Basic Rational Functions Identifying these asymptotes is useful because the graphs of y 1x and y x12 can be transformed in exactly the same way as the toolbox functions. When their graphs shift — the vertical and horizontal asymptotes shift with them and can be used as guides to redraw the graph. In shifted form, a k for the reciprocal function, and f 1x2 xh a k for the reciprocal square function. g1x2 1x h2 2 When horizontal and/or vertical shifts are applied to simple rational functions, we first apply them to the asymptotes, then calculate the x- and y-intercepts as before. An additional point or two can be computed as needed to round out the graph. EXAMPLE 3

䊳

Graphing Transformations of the Reciprocal Function 1 1 using transformations of the parent function. x2 1 The graph of g is the same as that of y , but shifted 2 units right and 1 unit upward. x This means the vertical asymptote is also shifted 2 units right, and the horizontal 1 asymptote is shifted 1 unit up. The y-intercept is g102 . For the x-intercept: 2 1 1 substitute 0 for g (x ) 0 x2 1 1 subtract 1 x2 11x 22 1 multiply by 1x 22 x1 solve Sketch the graph of g1x2

Solution

y 5

䊳

x2

4 3

y1 5

(0, 0.5) (1, 0)

2 1 1 2 3 4 5

5

x

The x-intercept is (1, 0). Knowing the graph is from the reciprocal function family and shifting the asymptotes and intercepts yields the graph shown. Now try Exercises 11 through 26

䊳

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These ideas can be “used in reverse” to determine the equation of a basic rational function from its given graph, as in Example 4. EXAMPLE 4

䊳

Writing the Equation of a Basic Rational Function, Given Its Graph Identify the function family for the graph given, then use the graph to write the equation of the function in “shifted form.” Assume 冟a冟 1.

Solution

䊳

The graph appears to be from the reciprocal square family, and has been shifted 2 units right (the vertical asymptote is at x 2), and 1 unit down (the horizontal asymptote is at y 1). From y x12, we obtain f 1x2 1x 1 22 2 1 as the shifted form.

y 6

6

6 x

6

Now try Exercises 27 through 38 B. You’ve just seen how we can use asymptotes and transformations to graph basic rational functions and write the equation for a given graph

䊳

Using the definition of negative exponents, the basic reciprocal and reciprocal square functions can be written as y x1 and y x2, respectively. In this form, we note that these functions also belong to a family of functions known as the power functions (see Exercise 80).

C. Graphs of Basic Power Functions Italian physicist and astronomer Galileo Galilei (1564–1642) made numerous contributions to astronomy, physics, and other fields. But perhaps he is best known for his experiments with gravity, in which he dropped objects of different weights from the Leaning Tower of Pisa. Due in large part to his work, we know that the velocity of an object after it has fallen a certain distance is v 12gs, where g is the acceleration due to gravity (32 ft/sec2), s is the distance in feet the object has fallen, and v is the velocity of the object in feet per second (see Exercise 71). As you will see, this is an example of a formula that uses a power function. From previous coursework or a review of radicals and rational exponents (Sec1 1 3 2 3 x 1 x x tion R.6), we know that 1x can be written as , and as , enabling us to write 1 1 these functions in exponential form: f 1x2 x2 and g1x2 x3. In this form, we see that these actually belong to a larger family of functions, where x is raised to some power, called the power functions. Power Functions and Root Functions For any constant real number p and variable x, functions of the form f 1x2 x p

are called power functions in x. If p is of the form

1 for integers n 2, the functions n

f 1x2 xn 3 f 1x2 1 x 1

are called root functions in x.

n

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1

3

5 The functions y x2, y x4, y x3, y 1 x, and y x2 are all power functions, 1 5 4 but only y x and y 1 x are also root functions. Initially we will focus on power functions where p 7 0.

EXAMPLE 5

䊳

Comparing the Graphs of Power Functions

Use a graphing calculator to graph the power functions f 1x2 x4, g1x2 x3, 3 1 2 h1x2 x , p1x2 x , and q1x2 x2 in the standard viewing window. Make an observation in QI regarding the effect of the 7exponent on each function, then 1 discuss what the graphs of y x6 and y x2 would look like. 1

Solution

䊳

First we enter the functions in sequence as Y1 through Y5 on the Y= screen (Figure 2.61). Using ZOOM 6:ZStandard produces the graphs shown in Figure 2.62. Narrowing the window to focus on QI (Figure 2.63: x 僆 3 4, 104, y 僆 34, 10 4 ), we quickly see that for x 1, larger values of p cause the graph of y x p to increase at a faster rate, and smaller values at a slower rate. In other words 1 1 (for x 1), since 6 , the graph of 6 4 1 y x6 would increase slower and appear 1 to be “under” the graph of Y1 X4. 7 7 Since 7 2, the graph of y x2 would 2 increase faster and appear to be “more narrow” than the graph of Y5 X2 (verify this).

2

Figure 2.61, 2.62

10

10

10

10

Figure 2.63 10 Y5

Y4

Y3 Y2 Y1

4

10

4


䊳

The Domain of a Power Function In addition to the observations made in Example 5, we can make other important notes, particularly regarding the domains of power functions. When the exponent on a power m 7 0 in simplest form, it appears the domain is all real function is a rational number n 2 1 numbers if n 2 is odd, as seen in the graphs of g1x2 x3 , h1x2 x1 x1 , and 2 q1x2 x x1. If n is an even1 number, the domain is all nonnegative real numbers as 3 seen in the graphs of f 1x2 x4 and p1x2 x2. Further exploration will show that if p is irrational, as in y x␲, the domain is also all nonnegative real numbers and we have the following:

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The Domain of a Power Function

Given a power function f 1x2 ⫽ x p with p 7 0. m 1. If p ⫽ is a rational number in simplest form, n a. the domain of f is all real numbers if n is odd: x 僆 1⫺q, q 2 , b. the domain of f is all nonnegative real numbers if n is even: x 僆 3 0, q 2 . 2. If p is an irrational number, the domain of f is all nonnegative real numbers: x 僆 30, q 2 . Further confirmation of statement 1 can be found by recalling the graphs of 1 1 3 y ⫽ 1x ⫽ x2 and y ⫽ 1 x ⫽ x3 from Section 2.2 (Figures 2.64 and 2.65). Figure 2.64 y

Figure 2.65

f(x) ⫽ 兹x

5

5

(note n is even) (9, 3) (4, 2) (6, 2.4)

(0, 0) ⫺1

(note n is odd)

(8, 2)

(1, 1)

(0, 0) 4

8

3 y g(x) ⫽ 兹x

x

⫺8

⫺4

(1, 1) 4

8

x

(⫺1, ⫺1) (⫺8, ⫺2) ⫺5

⫺5

Domain: x 僆 30, q 2 Range: y 僆 3 0, q 2

EXAMPLE 6

䊳

Domain: x 僆 3⫺q, q2 Range: y 僆 3⫺q, q2

Determining the Domains of Power Functions State the domain of the following power functions, and identity whether each is also a root function. 4 1 2 8 a. f 1x2 ⫽ x5 b. g1x2 ⫽ x10 c. h1x2 ⫽ 1 x d. q1x2 ⫽ x3 e. r 1x2 ⫽ x1 5

Solution

䊳

a. Since n is odd, the domain of f is all real numbers; f is not a root function. b. Since n is even, the domain of g is x 僆冤0, q 2 ; g is a root function. 1 c. In exponential form h1x2 ⫽ x8. Since n is even, the domain of h is x 僆冤0, q 2 ; h is a root function. d. Since n is odd, the domain of q is all real numbers; q is not a root function e. Since p is irrational, the domain of r is x 僆冤0, q 2 ; r is not a root function Now try Exercises 49 through 58

䊳

Transformations of Power and Root Functions As we saw in Section 2.2 (Toolbox Functions and Transformations), the graphs of the 3 root functions y ⫽ 1x and y ⫽ 1 x can be transformed using shifts, stretches, reflections, and so on. In Example 8(b) (Section 2.2) we noted the graph of 3 3 h1x2 ⫽ 2 1 x ⫺ 2 ⫺ 1 was the graph of y ⫽ 1 x shifted 2 units right, stretched by a factor of 2, and shifted 1 unit down. Graphs of other power functions can be transformed in exactly the same way.

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237


EXAMPLE 7

䊳

Graphing Transformations of Power Functions Based on our previous observations, 2 3 a. Determine the domain of f 1x2 x3 and g1x2 x2 , then verify by graphing them on a graphing calculator. 2 3 b. Next, discuss what the graphs of F1x2 1x 22 3 3 and G1x2 x2 2 will look like, then graph each on a graphing calculator to verify.

Solution

䊳

m

a. Both f and g are power functions of the form y x n . For f, n is odd so its domain is all real numbers. For g, n is even and the domain is x 僆 30, q2 . Their graphs support this conclusion (Figures 2.66 and 2.67). Figure 2.66

Figure 2.67

10

10

10

10

10

10

10

10

b. The graph of F will be the same as the graph of f, but shifted two units right and three units down, moving the vertex to 12, 32 . The graph of G will be the same as the graph of g, but reflected across the x-axis, and shifted 2 units up (Figures 2.68 and 2.69). Figure 2.69

Figure 2.68 10

10

C. You’ve just seen how we can graph basic power functions and state their domains

10

10

10

10

10

10


䊳

D. Applications of Rational and Power Functions These new functions have a variety of interesting and significant applications in the real world. Examples 8 through 10 provide a small sample, and there are a number of additional applications in the Exercise Set. In many applications, the coefficients may be rather large, and the axes should be scaled accordingly. EXAMPLE 8

䊳

Modeling the Cost to Remove Waste For a large urban-centered county, the cost to remove chemical waste and other 18,000 180, pollutants from a local river is given by the function C1p2 p 100 where C( p) represents the cost (in thousands of dollars) to remove p percent of the pollutants.

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a. Find the cost to remove 25%, 50%, and 75% of the pollutants and comment on the results. b. Graph the function using an appropriate scale. c. Use mathematical notation to state what happens as the county attempts to remove 100% of the pollutants.

Solution

䊳

a. We evaluate the function as indicated, finding that C1252 60, C1502 180, and C1752 540. The cost is escalating rapidly. The change from 25% to 50% brought a $120,000 increase, but the change from 50% to 75% brought a $360,000 increase! C(p) b. From the context, we need only graph the x 100 1200 portion from 0 p 6 100. For the C-intercept we substitute p 0 and find C102 0, which 900 seems reasonable as 0% would be removed (75, 540) 600 if $0 were spent. We also note there must be a vertical asymptote at x 100, since this 300 x-value causes a denominator of 0. Using (25, 60) (50, 180) p this information and the points from part (a) 100 50 75 25 produces the graph shown. c. As the percentage of pollutants removed y 180 approaches 100%, the cost of the cleanup skyrockets. Using notation: as p S 100 , C S q . Now try Exercises 65 through 70

䊳

While not obvious at first, the function C(p) in Example 8 is from the family of 1 reciprocal functions y . A closer inspection shows it has the form x 18,000 1 a k S 180, showing the graph of y is shifted right y x xh x 100 100 units, reflected across the x-axis, stretched by a factor of 18,000 and shifted 180 units down (the horizontal asymptote is y 180). As sometimes occurs in real-world applications, portions of the graph were ignored due to the context. To see the full graph, we reason that the second branch occurs on the opposite side of the vertical and horizontal asymptotes, and set the window as shown in Figure 2.70. After entering C(p) as Y1 on the Y= screen and pressing GRAPH , the full graph appears as shown in Figure 2.71 (for effect, the vertical and horizontal asymptotes were drawn separately using the 2nd PRGM (DRAW) options). Figure 2.71 Figure 2.70

2000

200

0

2000

Next, we’ll use a root function to model the distance to the horizon from a given height.

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239


EXAMPLE 9

䊳

The Distance to the Horizon On a clear day, the distance a person can see from a certain height (the distance to the horizon) is closely approximated by the root function d1h2 ⫽ 3.57 1h, where d(h) represents the viewing distance (in kilometers) from a height of h meters above sea level. a. To the nearest kilometer, how far can a person see when standing on the observation level of the John Hancock building in Chicago, Illinois, about 335 m high? b. To the nearest meter, how high is the observer’s eyes, if the viewing distance is 130 km?

Solution

䊳

a. Substituting 335 for h we have d1h2 ⫽ 3.57 1h d13352 ⫽ 3.57 1335 ⬇ 65.34

original function substitute 335 for h result

On a clear day, a person can see about 65 kilometers. b. We substitute 130 for d(h): d1h2 ⫽ 3.57 1h 130 ⫽ 3.57 1h 36.415 ⬇ 1h 1326.052 ⬇ h

original function substitute 130 for d(h) divide by 3.57 square both sides

If the distance to the horizon is 130 km, the observer’s eyes are at a height of approximately 1326 m. Check the answer to part (b) by solving graphically. Now try Exercises 71 through 74

䊳

One area where power functions and modeling with regression are used extensively is allometric studies. This area of inquiry studies the relative growth of a part of an animal in relation to the growth of the whole, like the wingspan of a bird compared to its weight, or the daily food intake of a mammal or bird compared to its size.

EXAMPLE 10

䊳

Modeling the Food Requirements of Certain Bird Species To study the relationship between the weight of a nonpasserine bird and its daily food intake, the data shown in the table was collected (nonpasserine: nonsinging, nonperching birds). a. On a graphing calculator, enter the data in L1 and L2, then set an appropriate window to view a scatterplot of the data. Does a power regression STAT CALC, A:PwrReg seem appropriate?

Average weight (g)

Daily food intake (g)

Common pigeon

350

25

Ring-necked duck

725

50

Ring-necked pheasant

1400

70

Canadian goose

4525

165

White swan

9075

240

Bird

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b. Use a graphing calculator to find an equation model using a power regression on the data, and enter the equation in Y1 (round values to three decimal places). c. Use the equation to estimate the daily food intake required by a barn owl (470 g), and a gray-headed albatross (6800 g). d. Use the intersection of graphs method to find the weight of a Great-Spotted Kiwi, given the daily food requirement is 130 g.

Solution

䊳

a. After entering the weights in L1 and Figure 2.72 food intake in L2, we set a window that 300 will comfortably fit the data. Using x 僆 30, 10,000 4 and y 僆 330, 300 4 produces the scatterplot shown (Figure 2.72). The data does not appear 0 10,000 linear, and based on our work in Example 5, a power function seems appropriate. 30 b. To access the power regression option, use STAT (CALC) A:PwrReg. To Figure 2.73 three decimal places the equation for Y1 would be 0.493 X0.685 (Figure 2.73). c. For the barn owl, x 470 and we find the estimated food requirement is about 33.4 g per day (Figure 2.74). For the grayheaded albatross x 6800 and the model estimates about 208.0 g of food daily is required. d. Here we’re given the food intake of the Great-Spotted Kiwi (the output value), and want to know what input value (weight) was used. Entering Y2 130, we’ll attempt to find where the graphs of Y1 and Y2 intersect (it will help to deactivate Plot1 on the Y= screen, so that only the graphs of Y1 and Y2 appear). Using 2nd TRACE (CALC) 5:Intersect shows the graphs intersect at about (3423.3, 130) (Figure 2.75), indicating the average weight of a Great-Spotted Kiwi is near 3423.3 g (about 7.5 lb). Figure 2.75 Figure 2.74

300

10,000

0

D. You’ve just seen how we can solve applications involving basic rational and power functions

30


䊳

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2.4 EXERCISES 䊳



1. Write the following in notational form. As x becomes an infinitely large negative number, y approaches 2. 1 ⫹ 2, 1x ⫺ 32 2 a asymptote occurs at x ⫽ 3 and a horizontal asymptote at .

䊳

2. For any constant k, the notation “as 冟 x 冟 S ⫹q, y S k ” is an indication of a asymptote, while “x S k, 冟y冟 S ⫹q ” indicates a asymptote. 1 has branches in Quadrants I x 1 and III. The graph of Y2 ⫽ ⫺ has branches in x Quadrants and .

3. Given the function g1x2 ⫽

4. The graph of Y1 ⫽

5. Discuss/explain how and why the range of the reciprocal function differs from the range of the reciprocal quadratic function. In the reciprocal quadratic function, all range values are positive.

6. If the graphs of Y1 ⫽

1 1 and Y2 ⫽ 2 were drawn x x on the same grid, where would they intersect? In what interval(s) is Y1 7 Y2?


For each graph given, (a) use mathematical notation to describe the end-behavior of each graph and (b) describe what happens as x approaches 1.

1 ⫹2 x⫺1

7. V1x2 ⫽

8. v1x2 ⫽

1 ⫺2 x⫺1

y

y

6 5 4 3 2 1 ⫺4⫺3⫺2⫺1 ⫺1 ⫺2 ⫺3 ⫺4

3 2 1

1 2 3 4 5 6 x

1 x ⫺5⫺4⫺3⫺2⫺1 ⫺1 1 2 3 4 5 ⫺2 ⫺3 ⫺4 ⫺5 ⫺6 ⫺7

For each graph given, (a) use mathematical notation to describe the end-behavior of each graph, (b) name the horizontal asymptote, and (c) describe what happens as x approaches ⴚ2.

9. Q1x2 ⫽

1 ⫺1 ⫹ 1 10. q1x2 ⫽ ⫹2 2 1x ⫹ 22 1x ⫹ 22 2 y

6 5 4 3 2 1 ⫺7⫺6⫺5⫺4⫺3⫺2⫺1 ⫺1 ⫺2 ⫺3 ⫺4

y

5 4 3 2 1 1 2 3 x

⫺7⫺6⫺5⫺4⫺3⫺2⫺1 ⫺1 ⫺2 ⫺3 ⫺4 ⫺5

Sketch the graph of each function using transformations of the parent function (not by plotting points). Clearly state the transformations used, and label the horizontal and vertical asymptotes as well as the x- and y-intercepts (if they exist). Also state the domain and range of each function.

11. f 1x2 ⫽

12. g1x2 ⫽

13.

14.

15. 17. 19. 21. 23. 25.

1 2 3 x

1 ⫺1 x 1 h1x2 ⫽ x⫹2 ⫺1 g1x2 ⫽ x⫺2 1 f 1x2 ⫽ ⫺1 x⫹2 1 h1x2 ⫽ 1x ⫺ 12 2 ⫺1 g1x2 ⫽ 1x ⫹ 22 2 1 f 1x2 ⫽ 2 ⫺ 2 x 1 h1x2 ⫽ 1 ⫹ 1x ⫹ 22 2

16. 18. 20. 22. 24. 26.

1 ⫹2 x 1 f 1x2 ⫽ x⫺3 ⫺1 h1x2 ⫽ ⫺2 x 1 g1x2 ⫽ ⫹2 x⫺3 1 f 1x2 ⫽ 1x ⫹ 52 2 ⫺1 h1x2 ⫽ 2 ⫺ 2 x 1 g1x2 ⫽ 2 ⫹ 3 x 1 g1x2 ⫽ ⫺2 ⫹ 1x ⫺ 12 2

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Identify the parent function for each graph given, then use the graph to construct the equation of the function in shifted form. Assume |a| ⴝ 1. y

27. S(x)

654321 1 2 3 4 5 6

y

28.

4 3 2 1

5 4 3 2 1

1 2 3 4 x

76 54321 1

1 2 3 x

2 3 4 5

s(x)

For each pair of functions given, state which function increases faster for x ⬎ 1, then use the INTERSECT command of a graphing calculator to find where (a) f(x) ⫽ g(x), (b) f(x) ⬎ g(x), and (c) f(x) ⬍ g(x).

39. f 1x2 x2, g1x2 x3

40. f 1x2 x4, g1x2 x5

43. f 1x2 x , g1x2 x

44. f 1x2 x 4, g1x2 x2

41. f 1x2 x4, g1x2 x2 2 3

42. f 1x2 x3, g1x2 x5

4 5

7

6 3 5 4 45. f 1x2 1 x, g1x2 1 x 46. f 1x2 1 x, g1x2 1 x

3 2 4 3 47. f 1x2 2 x , g1x2 x4 48. f 1x2 x2, g1x2 2 x 5

29.

3 2 1

y 4 3 2 1

Q(x)

654321 1 2 3 4 5 6 7

31.

30.

y

1 2 3 4 x

4321 1 2 3 4 5 6

y 2 1 7654321 1 2 v(x) 3 4 5 6 7 8

1 2 3 x

q(x)

y 2 1 7654321 1 2 w(x) 3 4 5 6 7 8

1 2 3 x

Use the graph shown to Exercises 33 through 38 y complete each statement using 10 the direction/approach notation.

33. As x S q, y ______. 34. As x S q, y ______.

10

10 x

35. As x S 1, y ______.

49. f 1x2 x8 7

50. g1x2 x7

51. h1x2 x5

6

52. q1x2 x6

7 53. r1x2 1 x

54. s1x2 x6

6

5

1

Using the functions from Exercises 49–54, identify which of the following are defined and which are not. Do not use a calculator or evaluate.

55. a. f 122

b. f(2)

56. a. h(0.3)

b. h10.32 c. q(0.3)

d. q(0.3)

57. a. h11.22

b. r172

d. s(0)

7 58. a. f a b 8

8 b. ga b c. q11.92 7

c. g122

36. As x S 1 , y ______.

10

37. The line x 1 is a vertical asymptote, since: as x S ____, y S ____. 38. The line y 2 is a horizontal asymptote, since: as x S ____, y S ____.

c. s1␲2

d. g122

d. q(0)

Compare and discuss the graphs of the following functions. Verify your answer by graphing both on a graphing calculator.

59. f 1x2 x8; F1x2 1x 12 8 2 7

䊳

3

State the domain of the following functions.

1 2 3 4 5 6 x

32.

3

7

60. g1x2 x7; G1x2 1x 32 7 2 8

8

6

6

61. p1x2 x5; P1x2 1x 22 5 5

5

62. q1x2 x6; Q1x2 2x6 5


63. Gravitational attraction: F ⴝ

km1m2 2

d The gravitational force F between two objects with masses m1 and m2 depends on the distance d between them and some constant k. (a) If the masses of the two objects are constant while the distance between them gets larger and larger, what happens to F? (b) Let m1 and m2 equal 1 mass unit with k 1 as well, and investigate using a table of values. What family does this function belong to? (c) Solve for m2 in terms of k, m1, d and F.

mⴙM 22gh m For centuries, the velocity v of a bullet of mass m has been found using a device called a ballistic pendulum. In one such device, a bullet is fired into a stationary block of wood of mass M, suspended from the end of a pendulum. The height h the pendulum swings after impact is measured, and the approximate velocity of the bullet can then be calculated using g 9.8 m/sec2 (acceleration due to gravity). When a .22-caliber bullet of mass 2.6 g is fired into a wood block of mass 400 g, their combined mass swings to a height of 0.23 m. To the nearest meter per second, find the velocity of the bullet the moment it struck the wood.

64. Velocity of a bullet: v ⴝ

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243

APPLICATIONS

65. Deer and predators: By banding deer over a period of 10 yr, a capture-and-release project determines the number of deer per square mile in the Mark Twain National Forest can be modeled by 75 the function D1p2 , where p is the number of p predators present and D is the number of deer. Use this model to answer the following. a. As the number of predators increases, what will happen to the population of deer? Evaluate the function at D(1), D(3), and D(5) to verify. b. What happens to the deer population if the number of predators becomes very large? c. Graph the function using an appropriate scale. Judging from the graph, use mathematical notation to describe what happens to the deer population if the number of predators becomes very small (less than 1 per square mile). 66. Balance of nature: A marine biology research group finds that in a certain reef area, the number of fish present depends on the number of sharks in the area. The relationship can be modeled by the 20,000 , where F(s) is the fish function F1s2 s population when s sharks are present. a. As the number of sharks increases, what will happen to the population of fish? Evaluate the function at F(10), F(50), and F(200) to verify. b. What happens to the fish population if the number of sharks becomes very large? c. Graph the function using an appropriate scale. Judging from the graph, use mathematical notation to describe what happens to the fish population if the number of sharks becomes very small. 67. Intensity of light: The intensity I of a light source depends on the distance of the observer from the source. If the intensity is 100 W/m2 at a distance of 5 m, the relationship can be modeled by the 2500 function I1d2 2 . Use the model to answer the d following. a. As the distance from the lightbulb increases, what happens to the intensity of the light? Evaluate the function at I(5), I(10), and I(15) to verify. b. If the intensity is increasing, is the observer moving away or toward the light source?

c. Graph the function using an appropriate scale. Judging from the graph, use mathematical notation to describe what happens to the intensity if the distance from the lightbulb becomes very small. 68. Electrical resistance: The resistance R (in ohms) to the flow of electricity is related to the length of the wire and its gauge (diameter in fractions of an inch). For a certain wire with fixed length, this relationship can be modeled by the function 0.2 R1d2 2 , where R(d) represents the resistance in d a wire with diameter d. a. As the diameter of the wire increases, what happens to the resistance? Evaluate the function at R(0.05), R(0.25), and R(0.5) to verify. b. If the resistance is increasing, is the diameter of the wire getting larger or smaller? c. Graph the function using an appropriate scale. Judging from the graph, use mathematical notation to describe what happens to the resistance in the wire as the diameter gets larger and larger. 69. Pollutant removal: For a certain coal-burning power plant, the cost to remove pollutants from plant emissions can be modeled by 8000 80, where C(p) represents the C1p2 p 100 cost (in thousands of dollars) to remove p percent of the pollutants. (a) Find the cost to remove 20%, 50%, and 80% of the pollutants, then comment on the results; (b) graph the function using an appropriate scale; and (c) use mathematical notation to state what happens if the power company attempts to remove 100% of the pollutants. 70. City-wide recycling: A large city has initiated a new recycling effort, and wants to distribute recycling bins for use in separating various recyclable materials. City planners anticipate the cost of the program can be modeled by the 22,000 220, where C(p) function C1p2 p 100 represents the cost (in $10,000) to distribute the bins to p percent of the population. (a) Find the cost to distribute bins to 25%, 50%, and 75% of the population, then comment on the results; (b) graph the function using an appropriate scale; and (c) use mathematical notation to state what happens if the city attempts to give recycling bins to 100% of the population.

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71. Hot air ballooning: If air resistance is neglected, the velocity (in ft/s) of a falling object can be closely approximated by the function V1s2 8 1s, where s is the distance the object has fallen (in feet). A balloonist suddenly finds it necessary to release some ballast in order to quickly gain altitude. (a) If she were flying at an altitude of 1000 ft, with what velocity will the ballast strike the ground? (b) If the ballast strikes the ground with a velocity of 225 ft/sec, what was the altitude of the balloon? 72. River velocities: The ability of a river or stream to move sand, dirt, or other particles depends on the size of the particle and the velocity of the river. This relationship can be used to approximate the velocity (in mph) of the river using the function V1d2 1.77 1d, where d is the diameter (in inches) of the particle being moved. (a) If a creek can move a particle of diameter 0.095 in., how fast is it moving? (b) What is the largest particle that can be moved by a stream flowing 1.1 mph? 73. Shoe sizes: Although there may be some notable exceptions, the size of shoe worn by the average man is related to his height. This relationship is 3 modeled by the function S1h2 0.75h2, where h is the person’s height in feet and S is the U.S. shoe size. (a) Approximate Denzel Washington’s shoe size given he is 6 ft, 0 in. tall. (b) Approximate Dustin Hoffman’s height given his shoe size is 9.5. 74. Whale weight: For a certain species of whale, the relationship between the length of the whale and the weight of the 27whale can be modeled by the function W1l2 0.03l 11, where l is the length of the whale in meters and W is the weight of the whale in metric tons (1 metric ton ⬇ 2205 pounds). (a) Estimate the weight of a newborn calf that is 6 m long. (b) At 81 metric tons, how long is an average adult? 75. Gestation periods: The data shown in the table can be used to study the relationship between the weight of mammal and its length of pregnancy. Use a graphing calculator to (a) graph a scatterplot of the data and (b) find an equation model using a power regression (round to three decimal places). Use the equation to estimate (c) the length of pregnancy of a racoon (15.5 kg) and (d) the weight of a fox, given the length of pregnancy is 52 days. Average Weight (kg)

Gestation (days)

Rat

0.4

24

Rabbit

3.5

50

Armadillo

6.0

51

Coyote

13.1

62

Dog

24.0

64

Mammal

76. Bird wingspans: The data in the table explores the relationship between a bird’s weight and its wingspan. Use a graphing calculator to (a) graph a scatterplot of the data and (b) find an equation model using a power regression (round to three decimal places). Use the equation to estimate (c) the wingspan of a Bald Eagle (16 lb) and (d) the weight of a Bobwhite Quail with a wingspan of 0.9 ft. Weight Wingspan (lb) (ft)

Bird Golden Eagle

10.5

6.5

Horned Owl

3.1

2.6

Peregrine Falcon

3.3

4.0

Whooping Crane

17.0

7.5

1.5

2.0

Raven

77. Species-area relationship: To study the relationship between the number of species of birds on islands in the Caribbean, the data shown in the table was collected. Use a graphing calculator to (a) graph a scatterplot of the data and (b) find an equation model using a power regression (round to three decimal places). Use the equation to estimate (c) the number of species of birds on Andros (2300 mi2) and (d) the area of Cuba, given there are 98 such species. Island

Area (mi2)

Great Inagua

Species

600

16

Trinidad

2000

41

Puerto Rico

3400

47

Jamaica Hispaniola

4500

38

30,000

82

78. Planetary orbits: The table shown gives the time required for the first five planets to make one complete revolution around the Sun (in years), along with the average orbital radius of the planet in astronomical units (1 AU 92.96 million miles). Use a graphing calculator to (a) graph a scatterplot of the data and (b) find an equation model using a power regression (round to four decimal places). Use the equation to estimate (c) the average orbital radius of Saturn, given it orbits the Sun every 29.46 yr, and (d) estimate how many years it takes Uranus to orbit the Sun, given it has an average orbital radius of 19.2 AU. Planet

Years

Radius

Mercury

0.24

0.39

Venus

0.62

0.72

Earth

1.00

1.00

Mars Jupiter

1.88

1.52

11.86

5.20

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Section 2.5 Piecewise-Defined Functions


79. Consider the graph of f 1x2

1 once again, and the x x by f(x) rectangles mentioned in the Worthy of Note on page 231. Calculate the area of each rectangle formed for x 僆 51, 2, 3, 4, 5, 66 . What do 1 you notice? Repeat the exercise for g1x2 2 and x the x by g(x) rectangles. Can you detect the pattern formed here?

80. All of the power functions presented in this section had positive exponents, but the definition of these types of functions does allow for negative exponents as well. In addition to the reciprocal and reciprocal square functions (y x1 and y x2), these types

䊳

245

of power functions have Depth Temp significant applications. For (meters) (°C) example, the temperature of 125 13.0 ocean water depends on several 250 9.0 factors, including salinity, 500 6.0 latitude, depth, and density. 750 5.0 However, between depths of 125 m and 2000 m, ocean 1000 4.4 temperatures are relatively 1250 3.8 predictable, as indicated by the 1500 3.1 data shown for tropical oceans 1750 2.8 in the table. Use a graphing 2000 2.5 calculator to find the power regression model and use it to estimate the water temperature at a depth of 2850 m.


81. (1.4) Solve the equation for y, then sketch its graph using the slope/intercept method: 2x 3y 15. 82. (1.3) Using a scale from 1 (lousy) to 10 (great), Charlie gave the following ratings: {(The Beatles, 9.5), (The Stones, 9.6), (The Who, 9.5), (Queen, 9.2), (The Monkees, 6.1), (CCR, 9.5), (Aerosmith, 9.2), (Lynyrd Skynyrd, 9.0), (The Eagles, 9.3), (Led

2.5

Zeppelin, 9.4), (The Stones, 9.8)}. Is the relation (group, rating) as given, also a function? State why or why not. 83. (1.5) Solve for c: E mc2. 84. (2.3) Use a graphing calculator to solve 冟x 2冟 1 2冟x 1冟 3.

Piecewise-Defined Functions


A. State the equation, domain, and range of a piecewise-defined function from its graph B. Graph functions that are piecewise-defined C. Solve applications involving piecewisedefined functions

Most of the functions we’ve studied thus far have been smooth and continuous. Although “smooth” and “continuous” are defined more formally in advanced courses, for our purposes smooth simply means the graph has no sharp turns or jagged edges, and continuous means you can draw the entire graph without lifting your pencil. In this section, we study a special class of functions, called piecewisedefined functions, whose graphs may be various combinations of smooth/not smooth and continuous/not continuous. The absolute value function is one example (see Exercise 31). Such functions have a tremendous number of applications in the real world.

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A. The Domain of a Piecewise-Defined Function For the years 1990 to 2000, the American bald eagle remained on the nation’s endangered species list, although the number of breeding pairs was growing slowly. After 2000, the population of eagles grew at a much faster rate, and they were removed from the list soon afterward. From Table 2.3 and plotted points modeling this growth (see Figure 2.76), we observe that a linear model would fit the period from 1992 to 2000 very well, but a line with greater slope would be needed for the years 2000 to 2006 and (perhaps) beyond. Figure 2.76 10,000

Bald eagle breeding pairs

9,000

Table 2.3 Year (1990 S 0)

Bald Eagle Breeding Pairs

Year (1990 S 0)

Bald Eagle Breeding Pairs

2

3700

10

6500

4

4400

12

7600

6

5100

14

8700

8

5700

16

9800

8,000 7,000 6,000 5,000 4,000 3,000

Source: www.fws.gov/midwest/eagle/population 0

2

4

6

8

10

12

14

16

18

t (years since 1990)

WORTHY OF NOTE For the years 1992 to 2000, we can estimate the growth in breeding pairs ¢pairs ¢time using the points (2, 3700) and (10, 6500) in the slope formula. The result is 350 1 , or 350 pairs per year. For 2000 to 2006, using (10, 6500) and (16, 9800) shows the rate of growth is significantly larger: ¢pairs 550 ¢years 1 or 550 pairs per year.

The combination of these two lines would be a single function that modeled the population of breeding pairs from 1990 to 2006, but it would be defined in two pieces. This is an example of a piecewise-defined function. The notation for these functions is a large “left brace” indicating the equations it groups are part of a single function. Using selected data points and techniques from Section 1.4, we find equations that could represent each piece are p1t2 350t 3000 for 0 t 10 and p1t2 550t 1000 for t 7 10, where p(t) is the number of breeding pairs in year t. The complete function is then written: function name

function pieces

350t 3000, p1t2 e 550t 1000,

domain of each piece

2 t 10 t 7 10

In Figure 2.76, note that we indicated the exclusion of t 10 from the second piece of the function using an open half-circle.

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EXAMPLE 1

䊳

Writing the Equation and Domain of a Piecewise-Defined Function The linear piece of the function shown has an equation of y ⫽ ⫺2x ⫹ 10. The equation of the quadratic piece is y ⫽ ⫺x2 ⫹ 9x ⫺ 14. 10 a. Use the correct notation to write them as a 8 single piecewise-defined function and state the domain of each piece by inspecting the graph. 6 b. State the range of the function.

Solution

䊳

y

f(x)

4

a. From the graph we note the linear portion is defined between 0 and 3, with these endpoints 2 included as indicated by the closed dots. The domain here is 0 ⱕ x ⱕ 3. The quadratic 0 portion begins at x ⫽ 3 but does not include 3, as indicated by the half-circle notation. The equation is function name

f 1x2 ⫽ e

function pieces

⫺2x ⫹ 10, ⫺x2 ⫹ 9x ⫺ 14,

(3, 4)

2

4

6

8

x

10

domain

0ⱕxⱕ3 3 6 xⱕ7

b. The largest y-value is 10 and the smallest is zero. The range is y 僆 3 0, 104 .

A. You’ve just seen how we can state the equation, domain, and range of a piecewise-defined function from its graph


䊳

Piecewise-defined functions can be composed of more than two pieces, and can involve functions of many kinds.

B. Graphing Piecewise-Defined Functions As with other functions, piecewise-defined functions can be graphed by simply plotting points. Careful attention must be paid to the domain of each piece, both to evaluate the function correctly and to consider the inclusion/exclusion of endpoints. In addition, try to keep the transformations of a basic function in mind, as this will often help graph the function more efficiently.

EXAMPLE 2

䊳

Graphing a Piecewise-Defined Function Evaluate the piecewise-defined function by noting the effective domain of each piece, then graph by plotting these points and using your knowledge of basic functions. h1x2 ⫽ e

Solution

䊳

⫺x ⫺ 2, 2 1x ⫹ 1 ⫺ 1,

⫺5 ⱕ x 6 ⫺1 x ⱖ ⫺1

The first piece of h is a line with negative slope, while the second is a transformed square root function. Using the endpoints of each domain specified and a few additional points, we obtain the following: For h1x2 ⫽ ⫺x ⫺ 2, ⫺5 ⱕ x 6 ⫺1, x

h(x)

For h1x2 ⫽ 2 1x ⫹ 1 ⫺ 1, x ⱖ ⫺1, x

h(x)

⫺5

3

⫺1

⫺1

⫺3

1

0

1

⫺1

(⫺1)

3

3

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After plotting the points from the first piece, we connect them with a line segment noting the left endpoint is included, while the right endpoint is not (indicated using a semicircle around the point). Then we plot the points from the second piece and draw a square root graph, noting the left endpoint here is included, and the graph rises to the right. From the graph we note the complete domain of h is x 僆 35, q 2 , and the range is y 僆 31, q 2 .

h(x) 5

h(x) x 2 h(x) 2 x 1 1 5

5

x

5


䊳

Most graphing calculators are able to graph piecewise-defined functions. Consider Example 3.

EXAMPLE 3

Solution

䊳

䊳

Graphing a Piecewise-Defined Function Using Technology x 5, 5 x 6 2 Graph the function f 1x2 e on a graphing calculator 2 1x 42 3, x 2 and evaluate f (2). Figure 2.77 10 Both “pieces” are well known—the first is a line with slope m 1 and y-intercept (0, 5). The second is a parabola that opens upward, shifted 4 units to the right and 3 units up. If we attempt to graph 10 f(x) using Y1 X 5 and Y2 1X 42 2 3 10 as they stand, the resulting graph may be difficult to analyze because the pieces overlap and intersect (Figure 2.77). To graph the functions 10 we must indicate the domain for each piece, separated by a slash and enclosed in parentheses. Figure 2.78 For instance, for the first piece we enter Y1 X 5/1X 5 and X 6 22 , and for the second, Y2 1X 42 2 3 1X 22 (Figure 2.78). The slash looks like (is) the division symbol, but in this context, the calculator interprets it as a means of separating the function from the domain. The inequality symbols are accessed using the 2nd MATH (TEST) keys. As shown for Y , compound 1 inequalities must be entered in two parts, using the logical connector “and”: 2nd MATH (LOGIC) 1:and. The graph is shown in Figure 2.79, where we see the function is linear for x 僆 [5, 2) and quadratic for x 僆 [2, q ). Using the 2nd GRAPH (TABLE) feature reveals the calculator will give an ERR: (ERROR) message for inputs outside the domains of Y1 and Y2, and we see that f is defined for x 2 only for Y2: f 122 7 (Figure 2.80). Figure 2.79 Figure 2.80 10

10

10

10


䊳

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As an alternative to plotting points, we can graph each piece of the function using transformations of a basic graph, then erase those parts that are outside of the corresponding domain. Repeat this procedure for each piece of the function. One interesting and highly instructive aspect of these functions is the opportunity to investigate restrictions on their domain and the ranges that result.

Piecewise and Continuous Functions EXAMPLE 4

䊳

Graphing a Piecewise-Defined Function Graph the function and state its domain and range: f 1x2 e

Solution

䊳

1x 32 2 12, 3,

0 6 x6 x 7 6

The first piece of f is a basic parabola, shifted three units right, reflected across the x-axis (opening downward), and shifted 12 units up. The vertex is at (3, 12) and the axis of symmetry is x 3, producing the following graphs. 1. Graph first piece of f (Figure 2.81)

2. Erase portion outside domain. of 0 6 x 6 (Figure 2.82).

Figure 2.81

Figure 2.82 y

y 12

y (x 3)2 12

12

10

10

8

8

6

6

4

4

2

2

1

1 2 3 4 5 6 7 8 9 10

x

y (x 3)2 12

1

1 2 3 4 5 6 7 8 9 10

x

The second function is simply a horizontal line through (0, 3). 3. Graph second piece of f (Figure 2.83).

4. Erase portion outside domain of x 7 6 (Figure 2.84).

Figure 2.83

Figure 2.84

y 12

y y (x 3)2 12

12

10

10

8

8

6

6

4

y3

4

2 1

f (x)

2

1 2 3 4 5 6 7 8 9 10

x

1

1 2 3 4 5 6 7 8 9 10

x

The domain of f is x 僆 10, q 2, and the corresponding range is y 僆 3 3, 124. Now try Exercises 15 through 18

䊳

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Piecewise and Discontinuous Functions Notice that although the function in Example 4 was piecewise-defined, the graph was actually continuous—we could draw the entire graph without lifting our pencil. Piecewise graphs also come in the discontinuous variety, which makes the domain and range issues all the more important.

EXAMPLE 5

䊳

Graphing a Discontinuous Piecewise-Defined Function Graph g(x) and state the domain and range: g1x2 ⫽ e

Solution

䊳

⫺12x ⫹ 6, ⫺x ⫺ 6 ⫹ 10,

0ⱕxⱕ4 4 6 xⱕ9

The first piece of g is a line, with y-intercept (0, 6) and slope 1. Graph first piece of g (Figure 2.85)

⫽ ⫺12.

¢y ¢x

2. Erase portion outside domain. of 0 ⱕ x ⱕ 4 (Figure 2.86).

Figure 2.85

Figure 2.86

y

y

10

10

8

8

6

6

y ⫽ ⫺qx ⫹ 6

4

4

2

2

1

2

3

4

5

6

7

8

9 10

y ⫽ ⫺qx ⫹ 6

x

1

2

3

4

5

6

7

8

9 10

x

The second is an absolute value function, shifted right 6 units, reflected across the x-axis, then shifted up 10 units. WORTHY OF NOTE As you graph piecewise-defined functions, keep in mind that they are functions and the end result must pass the vertical line test. This is especially important when we are drawing each piece as a complete graph, then erasing portions outside the effective domain.

3. Graph second piece of g (Figure 2.87).

4. Erase portion outside domain of 4 6 x ⱕ 9 (Figure 2.88).

Figure 2.87

Figure 2.88

y ⫽ ⫺x ⫺ 6 ⫹ 10

y

y

10

10

8

8

6

6

4

4

2

2

1

2

3

4

5

6

7

8

9 10

x

g(x)

1

2

3

4

5

6

7

8

9 10

x

Note that the left endpoint of the absolute value portion is not included (this piece is not defined at x ⫽ 4), signified by the open dot. The result is a discontinuous graph, as there is no way to draw the graph other than by “jumping” the pencil from where one piece ends to where the next begins. Using a vertical boundary line, we note the domain of g includes all values between 0 and 9 inclusive: x 僆 30, 9 4. Using a horizontal boundary line shows the smallest y-value is 4 and the largest is 10, but no range values exist between 6 and 7. The range is y 僆 34, 6 4 ´ 3 7, 10 4. Now try Exercises 19 and 20

䊳

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EXAMPLE 6

䊳

Graphing a Discontinuous Function The given piecewise-defined function is not continuous. Graph h(x) to see why, then comment on what could be done to make it continuous. x2 4 , h1x2 • x 2 1,

Solution

䊳

x2 x2

The first piece of h is unfamiliar to us, so we elect to graph it by plotting points, noting x 2 is outside the domain. This produces the table shown. After connecting the points, the graph turns out to be a straight line, but with no corresponding y-value for x 2. This leaves a “hole” in the graph at (2, 4), as designated by the open dot (see Figure 2.89). Figure 2.89

Figure 2.90 y

y

WORTHY OF NOTE The discontinuity illustrated here is called a removable discontinuity, as the discontinuity can be removed by redefining a single point on the function. Note that after factoring the first piece, the denominator is a factor of the numerator, and writing the result in lowest terms gives h1x2 1x x221x2 22 x 2, x 2. This is precisely the equation of the line in Figure 2.89 3y x 2 4 .

x

h(x)

4

2

2

0

0

2

2

—

4

6

5

5

5

5

x

5

5

x

5

5

The second piece is pointwise-defined, and its graph is simply the point (2, 1) shown in Figure 2.90. It’s interesting to note that while the domain of h is all real numbers (h is defined at all points), the range is y 僆 1q, 42 ´ 14, q2 as the function never takes on the value y 4. In order for h to be continuous, we would need to redefine the second piece as y 4 when x 2. Now try Exercises 21 through 26

䊳

To develop these concepts more fully, it will help to practice finding the equation of a piecewise-defined function given its graph, a process similar to that of Example 10 in Section 2.2.

EXAMPLE 7

䊳

Determining the Equation of a Piecewise-Defined Function y

Determine the equation of the piecewise-defined function shown, including the domain for each piece.

Solution

䊳

By counting ¢y ¢x from (2, 5) to (1, 1), we find the linear portion has slope m 2, and the y-intercept must be (0, 1). The equation of the line is y 2x 1. The second piece appears to be a parabola with vertex (h, k) at (3, 5). Using this vertex with the point (1, 1) in the general form y a1x h2 2 k gives y a1x h2 k 1 a11 32 2 5 4 a122 2 4 4a 1 a 2

5

4

6

5

general form, parabola is shifted right and up substitute 1 for x, 1 for y, 3 for h, 5 for k simplify; subtract 5 122 2 4 divide by 4

x

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The equation of the parabola is y 1x 32 2 5. Considering the domains shown in the figure, the equation of this piecewise-defined function must be p1x2 e

2x 1, 1x 32 2 5,

B. You’ve just seen how we can graph functions that are piecewise-defined

2 x 6 1 x1


䊳

C. Applications of Piecewise-Defined Functions The number of applications for piecewise-defined functions is practically limitless. It is actually fairly rare for a single function to accurately model a situation over a long period of time. Laws change, spending habits change, and technology can bring abrupt alterations in many areas of our lives. To accurately model these changes often requires a piecewise-defined function.

EXAMPLE 8

䊳

Modeling with a Piecewise-Defined Function For the first half of the twentieth century, per capita spending on police protection can be modeled by S1t2 0.54t 12, where S(t) represents per capita spending on police protection in year t (1900 corresponds to year 0). After 1950, perhaps due to the growth of American cities, this spending greatly increased: S1t2 3.65t 144. Write these as a piecewise-defined function S(t), state the domain for each piece, then graph the function. According to this model, how much was spent (per capita) on police protection in 2000 and 2010? How much will be spent in 2014? Source: Data taken from the Statistical Abstract of the United States for various years.

Solution

䊳

function name

S1t2 e

function pieces

effective domain

0.54t 12, 3.65t 144,

0 t 50 t 7 50

Since both pieces are linear, we can graph each part using two points. For the first function, S102 12 and S1502 39. For the second function S1502 39 and S1802 148. The graph for each piece is shown in the figure. Evaluating S at t 100: S1t2 3.65t 144 S11002 3.6511002 144 365 144 221

S(t) 240 200 160

(80, 148)

120 80 40 0

(50, 39) 10 20 30 40 50 60 70 80 90 100 110 t (1900 → 0)

About $221 per capita was spent on police protection in the year 2000. For 2010, the model indicates that $257.50 per capita was spent: S11102 257.5. By 2014, this function projects the amount spent will grow to S11142 272.1 or $272.10 per capita. Now try Exercises 33 through 44

䊳

Step Functions The last group of piecewise-defined functions we’ll explore are the step functions, so called because the pieces of the function form a series of horizontal steps. These functions find frequent application in the way consumers are charged for services, and have several applications in number theory. Perhaps the most common is called the greatest integer function, though recently its alternative name, floor function, has gained popularity (see Figure 2.91). This is in large part due to an improvement in notation

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and as a better contrast to ceiling functions. The floor function of a real number x, denoted f 1x2 :x ; or Œ x œ (we will use the first), is the largest integer less than or equal to x. For instance, :5.9 ; 5, : 7; 7, and :3.4 ; 4. In contrast, the ceiling function C1x2 <x = is the smallest integer greater than or equal to x, meaning < 5.9 = 6, b

Minor axis

• If a2 7 b2, the major axis is horizontal (parallel to the x-axis) with length 2a, and the minor axis is vertical with length 2b (see Example 3). • If a2 6 b2 the major axis is vertical (parallel to the y-axis) with length 2b, and the minor axis is horizontal with length 2a (see Example 4). Generalizing this observation we obtain the equation of an ellipse in standard form. The Equation of an Ellipse in Standard Form 1x h2 2

1y k2 2

1. a2 b2 If a b the equation represents the graph of an ellipse with center at (h, k). • 冟a冟 gives the horizontal distance from center to graph. • 冟b冟 gives the vertical distance from center to graph. Given

Finally, note the line segment from center to vertex is called the semimajor axis, with the perpendicular line segment from center to graph called the semiminor axis. EXAMPLE 3

䊳

Graphing a Horizontal Ellipse Sketch the graph of the ellipse defined by 1y 12 2 1x 22 2 1. 25 9

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8–12

CHAPTER 8 Analytic Geometry and the Conic Sections

Solution

䊳

Noting a b, we have an ellipse with center 1h, k2 12, 12. The horizontal distance from the center to the graph is a 5, and the vertical distance from the center to the graph is b 3. After plotting the corresponding points and connecting them with a smooth curve, we obtain the graph shown.

y Ellipse (2, 2)

(3, 1)

b3 a5 (2, 1)

x

(7, 1)

(2, 4)


As with the circle, the equation of an ellipse can be given in polynomial form, and here our knowledge of circles is helpful. For the equation 25x2 4y2 100, we know the graph cannot be a circle since the coefficients are unequal, and the center of the graph must be at the origin since h k 0. To actually draw the graph, we convert the equation to standard form. Note that a circle whose center is at (0, 0) is called a central circle, and an ellipse with center at (0, 0) is called a central ellipse.

WORTHY OF NOTE In general, for the equation Ax2 By2 F (A, B, F 7 0), the equation represents a circle if A B, and an ellipse if A B.

EXAMPLE 4

䊳

䊳

Graphing a Vertical Ellipse For 25x2 4y2 100, a. Write the equation in standard form and identify the center and the values of a and b. b. Identify the major and minor axes and name the vertices. c. Sketch the graph. d. Graph the relation on a graphing calculator using a “friendly” window, then use the TRACE feature to find four additional points on the graph whose coordinates are rational.

Solution

䊳

The coefficients of x2 and y2 are unequal, and 25, 4, and 100 have like signs. The equation represents an ellipse with center at (0, 0). To obtain standard form: a. 25x2 4y2 100 given equation 4y2 25x2 1 divide by 100 100 100 y2 x2 1 standard form 4 25 y2 x2 1 write denominators in squared form; a 2, b 5 22 52 b. The result shows a 2 and b 5, indicating the major axis will be vertical and the minor axis will be horizontal. With the center at the origin, the x-intercepts will Figure 8.14 be 12, 02 and (2, 0), y with the vertices (and Vertical ellipse (0, 5) y-intercepts) at Center at (0, 0) 10, 52 and (0, 5). b5 Endpoints of major axis (vertices) c. Plotting these (0, 5) and (0, 5) intercepts and (2, 0) (2, 0) Endpoints of minor axis x sketching the ellipse (2, 0) and (2, 0) a 2 results in the graph Length of major axis 2b: 2(5) 10 shown in Length of minor axis 2a: 2(2) 4 Figure 8.14. (0, 5)

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Section 8.2 The Circle and the Ellipse

d. As with the circle, we begin by solving for y. 25x2 4y2 100 4y2 100 25x2 100 25x2 y2 4 100 25x2 y B 4 Y1

719

original equation isolate term containing y divide by 4

take square roots

100 25X2 100 25X2 , Y2 B B 4 4 Figure 8.15 6.2

The graph is shown in Figure 8.15, where we note that (1.6, 3) is a point on the graph. Due to the symmetry of the ellipse, 11.6, 32 , 11.6, 32 , and 11.6, 32 are also on the graph.

9.4

9.4

6.2

Now try Exercises 25 through 36 WORTHY OF NOTE After writing the equation in standard form, it is possible to end up with a constant that is zero or negative. In the first case, the graph is a single point. In the second case, no graph is possible since roots of the equation will be complex numbers. These are called degenerate cases. See Exercise 84.

EXAMPLE 5

䊳

If the center of the ellipse is not at the origin, the polynomial form has additional linear terms and we must first complete the square in x and y, then write the equation in standard form to sketch the graph (see the Reinforcing Basic Concepts feature for more on completing the square). Figure 8.16 illustrates how the central ellipse and the shifted ellipse are related.

䊳

Figure 8.16 y Ellipse with center at (h, k) k

(h, k)

Central ellipse (0, b) (a, 0) (a, 0) (0, 0) (0, b)

All points shift h units horizontally, k units vertically, opposite the sign (x h)2 (y k)2 1 a2 b2 a2 b2

x h x2 y2 2 1 a2 b ab

Completing the Square to Graph an Ellipse Sketch the graph of 25x2 4y2 150x 16y 141 0, then state the domain and range of the relation.

Solution

䊳

The coefficients of x2 and y2 are unequal and have like signs, and we assume the equation represents an ellipse but wait until we have the factored form to be certain (it could be a degenerate ellipse). 25x2 4y2 150x 16y 141 0 25x2 150x 4y2 16y 141 2 251x 6x __ 2 41y2 4y __ 2 141 251x2 6x 92 41y2 4y 42 141 225 16 c

c

adds 25192 225

c

c adds 4142 16

add 225 16 to right

given equation (polynomial form) group like terms; subtract 141 factor out leading coefficient from each group complete the square

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251x 32 2 41y 22 2 100 41y 22 2 251x 32 2 100 100 100 100 1y 22 2 1x 32 2 1 4 25 1x 32 2 1y 22 2 1 22 52 The result is a vertical ellipse with center at 13, 22, with a 2 and b 5. The vertices are a vertical distance of 5 units from center, and the endpoints of the minor axis are a horizontal distance of 2 units from center. Note this is the same ellipse as in Example 4, but shifted 3 units left and 2 up. The domain of this relation is x 僆 3 5, 1 4 , and the range is y 僆 3 3, 7 4.

factor divide both sides by 100

simplify (standard form)

write denominators in squared form

(3, 7)

y

Vertical ellipse Center at (3, 2)

(5, 2)

B. You’ve just seen how we can use the equation of an ellipse to graph central and noncentral ellipses

(3, 2)

Endpoints of major axis (vertices) (3, 3) and (3, 7) (1, 2)

Endpoints of minor axis (5, 2) and (1, 2) x Length of major axis 2b: 2(5) 10 Length of minor axis 2a: 2(2) 4

(3, 3)


䊳

C. The Foci of an Ellipse In Section 8.1, we noted that an ellipse could also be defined in terms of two special points called the foci. The Museum of Science and Industry in Chicago, Illinois (http://www.msichicago.org), has a permanent exhibit called the Whispering Gallery. The construction of the room is based on some of the reflective properties of an ellipse. If two people stand at designated points in the room and one of them whispers very softly, the other person can hear the whisper quite clearly—even though they are over 40 ft apart! The point where each person stands is a focus of an ellipse. This reflective property also applies to light and radiation, giving the ellipse some powerful applications in science, medicine, acoustics, and other areas. To understand and appreciate these applications, we introduce the analytic definition of an ellipse. WORTHY OF NOTE You can easily draw an ellipse that satisfies the definition. Press two pushpins (these form the foci of the ellipse) halfway down into a piece of heavy cardboard about 6 in. apart. Take an 8-in. piece of string and loop each end around the pins. Use a pencil to draw the string taut and keep it taut as you move the pencil in a circular motion—and the result is an ellipse! A different length of string or a different distance between the foci will produce a different ellipse.

Definition of an Ellipse Given two fixed points f1 and f2 in a plane, an ellipse is the set of all points (x, y) where the distance from f1 to (x, y) added to the distance from f2 to (x, y) remains constant.

y P(x, y) d1

d1 d2 k The fixed points f1 and f2 are called the foci of the ellipse, and the points P(x, y) are on the graph of the ellipse.

f1

d2 f2

x

d1 d2 k

6 in. 3 in.

5 in.

To find the equation of an ellipse in terms of a and b we combine the definition just given with the distance formula. Consider the ellipse shown in Figure 8.17 (for

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Figure 8.17 y (0, b) P(x, y) (a, 0) x

(a, 0) (c, 0)

(c, 0)

calculating ease we use a central ellipse). Note the vertices have coordinates 1a, 02 and (a, 0), and the endpoints of the minor axis have coordinates 10, b2 and (0, b) as before. It is customary to assign foci the coordinates f1 S 1c, 02 and f2 S 1c, 02. We can calculate the distance between (c, 0) and any point P(x, y) on the ellipse using the distance formula: 21x c2 2 1y 02 2

Likewise the distance between 1c, 02 and any point (x, y) is 21x c2 2 1y 02 2

(0, b)

According to the definition, the sum must be constant: 21x c2 2 y2 21x c2 2 y2 k EXAMPLE 6

䊳

Finding the Value of k from the Definition of an Ellipse Use the definition of an ellipse and the diagram given to determine the constant k used for this ellipse (also see the following Worthy of Note). Note that a 5, b 3, and c 4. y (0, 3)

P(3, 2.4)

(5, 0) (4, 0)

(4, 0)

(5, 0) x

(0, 3)

Solution

䊳

21x c2 2 1y 02 2 21x c2 2 1y 02 2 k

213 42 12.4 02 213 42 12.4 02 k 2

WORTHY OF NOTE Note that if the foci are coincident (both at the origin) the “ellipse” will k actually be a circle with radius ; 2 2x2 y2 2x2 y2 k leads to k2 x2 y2 . In Example 6 we 4 10 5, and if found k 10, giving 2 we used the “string” to draw the circle, the pencil would be 5 units from the center, creating a circle of radius 5.

2

2

2

2112 2.4 27 2.4 k 16.76 154.76 k 2.6 7.4 k 10 k The constant value for this ellipse is 10 units. 2

2

2

2

given substitute add simplify radicals compute square roots result

Now try Exercises 45 through 48 In Example 6, the sum of the distances could also be found by moving the point (x, y) to the location of a vertex (a, 0), then using the symmetry of the ellipse. The sum is identical to the length of the major axis, since the overlapping part of the string from (c, 0) to (a, 0) is the same length as from (a, 0) to (c, 0) (see Figure 8.18). This shows the constant k is equal to 2a regardless of the distance between foci. As we noted, the result is

䊳

Figure 8.18 y d1 d2 2a d1

d2

(a, 0) (c, 0)

21x c2 2 y2 21x c2 2 y2 2a

(c, 0)

(a, 0) x

These two segments are equal substitute 2a for k

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The details for simplifying this expression are given in Appendix V, and the result is very close to the standard form seen previously: y2 x2 1 a2 a2 c2 y2 y2 x2 x2 1 1, we might with a2 b2 a2 a2 c 2 suspect that b2 a2 c2, and this is indeed the case. Note from Example 6 the relationship yields By comparing the standard form

b2 a2 c2 3 2 52 42 9 25 16 Additionally, when we consider that (0, b) is Figure 8.19 a point on the ellipse, the distance from (0, b) to y (c, 0) must be equal to a due to symmetry (the (0, b) “constant distance” used to form the ellipse is always 2a). We then see in Figure 8.19, that a a b2 c2 a2 (Pythagorean Theorem), yielding b (a, 0) (a, 0) 2 2 2 b a c as above. x (c, 0) (c, 0) With this development, we now have the ability to locate the foci of any ellipse —an important step toward using the ellipse in practical (0, b) applications. Because we’re often asked to find the location of the foci, it’s best to rewrite the relationship in terms of c2, using absolute value bars to allow for a major axis that is vertical: c2 冟a2 b2冟. EXAMPLE 7

䊳

Completing the Square to Graph an Ellipse and Locate the Foci For the ellipse defined by 25x2 9y2 100x 54y 44 0, find the coordinates of the center, vertices, foci, and endpoints of the minor axis. Then sketch the graph.

Solution

䊳

25x2 9y2 100x 54y 44 0 25x2 100x 9y2 54y 44 2 251x 4x __ 2 91y2 6y __ 2 44 251x2 4x 42 91y2 6y 92 44 100 81 c

c adds 25142 100

c adds 9192 81

c

given group terms; add 44 factor out lead coefficients

add 100 81 to right-hand side

251x 22 2 91y 32 2 225 91y 32 2 251x 22 2 225 225 225 225 1x 22 2 1y 32 2 1 9 25 2 2 1x 22 1y 32 1 32 52

factored form divide by 225

simplify (standard form) write denominators in squared form

The result shows a vertical ellipse with a 3 and b 5. The center of the ellipse is at (2, 3). The vertices are a vertical distance of b 5 units from center at (2, 8) and (2, 2). The endpoints of the minor axis are a horizontal distance of a 3 units from center at (1, 3) and (5, 3). To locate the foci, we use the foci formula

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723

for an ellipse: c2 冟a2 b2冟, giving c2 冟32 52冟 16. This shows the foci “” are located a vertical distance of 4 units from center at (2, 7) and (2, 1). y (2, 8)

Vertical ellipse Center at (2, 3)

(2, 7)

(1, 3)

(2, 3)

(2, 1) (2, 2)

Endpoints of major axis (vertices) (2, 8) and (2, 2) (5, 3)

x

Endpoints of minor axis (1, 3) and (5, 3) Location of foci (2, 7) and (2, 1) Length of major axis: 2b 2(5) 10 Length of minor axis: 2a 2(3) 6


䊳

For an ellipse, a focal chord is a line segment perpendicular to the major axis, through a focus and with endpoints on the ellipse. In the Exercise Set, you are asked 2m2 to verify that the focal chord of an ellipse has length L , where m is the length n of the semiminor axis and n is the length of the semimajor axis. This means the m2 distance from the foci to the graph (along a focal chord) is , a fact can often be used n to help graph an ellipse. For Example 7, m 3 and n 5, so the horizontal distance Figure 8.20 from focus to graph (in either direction) is 9.2 9 32 . From the upper focus (2, 7), we can 5 5 now graph the additional points 12 1.8, 72 11.4 10.2, 72 and (2 1.8, 7) 13.8, 72, and 7.4 from the lower focus 12, 12 we obtain 10.2, 12 and 13.8, 12 without having to evaluate the original equation. Graphical ver3.2 ification is provided in Figure 8.20. Also see Exercises 83 and 85. For future reference, remember the foci of an ellipse always occur on the major axis, with a 7 c and a2 7 c2 for a horizontal ellipse, with b 7 c and b2 7 c2 for a vertical ellipse. This makes it easier to remember the foci formula for ellipses: c2 冟a2 b2冟. If any two of the values for a, b, and c are known, the relationship between them can be used to construct the equation of the ellipse.

EXAMPLE 8

䊳

Finding the Equation of an Ellipse Find the equation of the ellipse (in standard form) that has foci at (0, 2) and (0, 2), with a minor axis 6 units in length. Then graph the ellipse a. By hand. b. On a graphing calculator. m2 c. Find the distance from foci to graph along a focal chord ausing b, and use n the result to verify that the endpoints of both focal chords are all on the graph.

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Solution

䊳

LOOKING AHEAD

Since the foci are on the y-axis and an equal distance from (0, 0), we know this is a vertical and central ellipse with c 2 and c2 4. The minor axis has a length of 2a 6 units, meaning a 3 and a2 9. To find b2, use the foci equation and solve. Figure 8.21 foci equation (ellipse) c2 冟a2 b2冟 4 冟9 b2冟 4 9 b2 4 9 b2 b2 13 b2 5

For the hyperbola, we’ll find that c 7 a, and the formula for the foci of a hyperbola will be c2 a2 b2.

y

substitute

(0, √13)

solve the absolute value equation result

2

Since we know b must be greater than a2 (the major axis is always longer), b2 5 can be discarded. The

(0, 2) (3, 0)

2

y x2 1. 2 3 1 2132 2 a. The graph is shown in Figure 8.21. b. For a calculator generated graph, begin by solving for y.

(3, 0) (0, 2)

x

standard form is

(0, √13)

y2 x2 1 original equation 9 13 13x2 9y2 117 clear denominators isolate y-term 9y2 117 13x2 2 117 13x y2 divide by 9 9.4 9 117 13x2 y take square roots B 9 117 13X2 117 13X2 Y1 , Y2 B 9 B 9 The graph is shown in Figure 8.22. c. From the discussion prior to Example 8, the horizontal distance from foci to graph m2 9 must be . Using the TRACE feature and entering x 9/ 213 verifies n 213

Figure 8.22 6.2

9.4

6.2

Figure 8.23 6.2

9.4

9.4

that a

9

, 2b is a point on the graph (Figure 8.23), and that a

9

, 2b, 213 213 9 9 a , 2b, and a , 2b must also be on the graph due to symmetry. 213 213

6.2

C. You’ve just seen how we can locate the foci of an ellipse and use the foci and other features to write the equation


䊳

D. Applications Involving Foci Applications involving the foci of a conic section can take various forms. In many cases, only partial information about the conic section is available and the ideas from Example 8 must be used to “fill in the gaps.” In other applications, we must rewrite a known or given equation to find information related to the values of a, b, and c.

EXAMPLE 9

䊳

Solving Applications Using the Characteristics of an Ellipse In Washington, D.C., there is a park called the Ellipse located between the White House and the Washington Monument. The park is surrounded by a path that forms an ellipse with the length of the major axis being about 1502 ft and the minor axis having a length of 1280 ft. Suppose the park manager wants to install water fountains at the location of the foci. Find the distance between the fountains rounded to the nearest foot.

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Solution

䊳

725

Since the major axis has length 2a 1502, we know a 751 and a2 564,001. The minor axis has length 2b 1280, meaning b 640 and b2 409,600. To find c, use the foci equation: c2 a2 b2 564,001 409,600 154,401 c ⬇ 393 and c ⬇ 393

since we know a 7 b substitute subtract square root property

The distance between the water fountains would be 213932 786 ft.

D. You’ve just seen how we can solve applications involving the foci


䊳

8.2 EXERCISES 䊳



1. For an ellipse, the relationship between a, b, and c is given by the foci equation , since c 6 a or c 6 b.

2. The greatest distance across an ellipse is called the and the endpoints are called .

3. For a vertical ellipse, the length of the minor axis is and the length of the major axis is .

4. To write the equation 2x2 y2 6x 7 in standard form, the in x.

5. Explain/Discuss how the relations a 7 b, a b and a 6 b affect the graph of a conic section with

6. Suppose foci are located at (3, 2) and (5, 2). Discuss/Explain the conditions necessary for the graph to be an ellipse.

equation 䊳

1x h2 2 a2

1y k2 2 b2

1.


Find an equation of the circle satisfying the conditions given, then graph the result on a graphing calculator and locate two additional points on the graph.

15. x2 y2 4x 10y 4 0 16. x2 y2 4x 6y 3 0


17. x2 y2 6x 5 0


18. x2 y2 8y 5 0


Sketch the graph of each ellipse.

19.

1x 12 2

20.

1x 32 2

21.

1x 22 2

22.

1x 52 2

10. center (0, 4), radius 15 11. diameter has endpoints (4, 9) and (2, 1) 12. diameter has endpoints (2, 32 , and (3, 9) Write each equation in standard form to identify the center and radius of the circle, then sketch its graph.

13. x2 y2 12x 10y 52 0 14. x2 y2 8x 6y 11 0

9 4

25 1

1y 22 2

1y 12 2

1y 32 2

1y 22 2

16 25 4

16

1 1 1 1

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726

8–20


23.

1x 12 2

24.

1x 12

16

1y 22 2

1y 32

2

36

47.

1

9

48.

(0, b) y (0, 8)

(0, b) y

(4.8, 6) (0, 28)

2

1

9

(6, 0) x

(6, 0)

(96, 0) x

(96, 0) (0, 28 )

For each exercise, (a) write the equation in standard form, then identify the center and the values of a and b, (b) state the coordinates of the vertices and the coordinates of the endpoints of the minor axis, (c) sketch the graph, and (d) for 25–28 (only) graph the relations on a graphing calculator and identify four additional points on the graph whose coordinates are rational.

(0, 8)

(76.8, 60)

(0, b)

(0, b)

Find the coordinates of the (a) center, (b) vertices, (c) foci, and (d) endpoints of the minor axis. Then (e) sketch the graph.

49. 4x2 25y2 16x 50y 59 0

25. x2 4y2 16

26. 9x2 y2 36

50. 9x2 16y2 54x 64y 1 0

27. 16x2 9y2 144

28. 25x2 9y2 225

51. 25x2 16y2 200x 96y 144 0

29. 2x2 5y2 10

30. 3x2 7y2 21

52. 49x2 4y2 196x 40y 100 0 53. 6x2 24x 9y2 36y 6 0

Identify each equation as that of an ellipse or circle, then sketch its graph.

54. 5x2 50x 2y2 12y 93 0

31. 1x 12 41y 22 16 2

2

32. 91x 22 2 1y 32 2 36

Find the equation of the ellipse (in standard form) that satisfies the following conditions. Then (a) graph the ellipse by hand, (b) confirm your graph by graphing the ellipse on a graphing calculator, and (c) find the length of the focal chords and verify the endpoints of the chords are on the graph.

33. 21x 22 2 21y 42 2 18 34. 1x 62 2 y2 49

35. 41x 12 2 91y 42 2 36 36. 251x 32 2 41y 22 2 100 Complete the square in both x and y to write each equation in standard form. Then draw a complete graph of the relation and identify all important features, including the domain and range.

37. 4x2 y2 6y 5 0

55. vertices at (6, 0) and (6, 0); foci at (4, 0) and (4, 0) 56. vertices at (8, 0) and (8, 0); foci at (5, 0) and (5, 0) 57. foci at (3, 6) and (3, 2); length of minor axis: 6 units 58. foci at (4, 3) and (8, 3); length of minor axis: 8 units

38. x2 3y2 8x 7 0 39. x2 4y2 8y 4x 8 0

Use the characteristics of an ellipse and the graph given to write the related equation and find the location of the foci.

40. 3x2 y2 8y 12x 8 0 41. 5x2 2y2 20y 30x 75 0

59.

60.

y

y

42. 4x 9y 16x 18y 11 0 2

2

43. 2x2 5y2 12x 20y 12 0 44. 6x2 3y2 24x 18y 3 0

x

x

Use the definition of an ellipse to find the constant k for each ellipse (figures are not drawn to scale).

45.

46.

y (0, 8) (6, 6.4) (a, 0) (6, 0)

(0, 8)

61.

y

62.

y

y

(0, 12)

(6, 0)

(9, 9.6)

(a, 0) (a, 0) x

(9, 0)

(0, 12)

(9, 0)

(a, 0) x

x

x

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8–21 䊳



63. Area of an Ellipse: A ⴝ ␲ab The area of an ellipse is given by the formula shown, where a is the distance from the center to the graph in the horizontal direction and b is the distance from center to graph in the vertical direction. Find the area of the ellipse defined by 16x2 9y2 144. 䊳

727

a2 ⴙ b2 B 2 The perimeter of an ellipse can be approximated by the formula shown, where a represents the length of the semimajor axis and b represents the length of the semiminor axis. Find the perimeter of the y2 x2 ellipse defined by the equation 1. 49 4

64. The Perimeter of an Ellipse: P ⴝ 2␲

APPLICATIONS

65. Decorative fireplaces: A bricklayer intends to build an elliptical fireplace 3 ft high and 8 ft wide, with two glass doors that open at the middle. The hinges to these doors are to be screwed onto a spine that is perpendicular to the hearth and goes through the foci of the ellipse. How far from center will the spines be located? How tall will each spine be? 8 ft

68. Medical procedures: The medical procedure called lithotripsy is a noninvasive medical procedure that is used to break up kidney and bladder stones in the body. A machine called a lithotripter uses its three-dimensional semielliptical shape and the foci properties of an ellipse to concentrate shock waves generated at one focus, on a kidney stone located at the other focus (see diagram—not drawn to scale). If the lithotripter has a length (semimajor axis) of 16 cm and a radius (semiminor axis) of 10 cm, how far from the vertex should a kidney stone be located for the best result? Round to the nearest hundredth.

3 ft

Exercise 68 Vertex Spines

Focus Lithotripter

66. Decorative gardens: A retired math teacher decides to present her husband with a beautiful elliptical garden to help celebrate their 50th anniversary. The ellipse is to be 8 m long and 5 m across, with decorative fountains located at the foci. How far from the center of the ellipse should the fountains be located (round to the nearest 100th of a meter)? How far apart are the fountains? 67. Attracting attention to art: As part of an art show, a gallery owner asks a student from the local university to design a unique exhibit that will highlight one of the more significant pieces in the collection, an ancient sculpture. The student decides to create an elliptical showroom with reflective walls, with a rotating laser light on a stand at one focus, and the sculpture placed at the other focus on a stand of equal height. The laser light then points continually at the sculpture as it rotates. If the elliptical room is 24 ft long and 16 ft wide, how far from the center of the ellipse should the stands be located (round to the nearest 10th of a foot)? How far apart are the stands?

Exercise 69 69. Elliptical arches: In some situations, bridges are built using uniform elliptical 8 ft archways as shown in the 60 ft figure given. Find the equation of the ellipse forming each arch if it has a total width of 30 ft and a maximum center height (above level ground) of 8 ft. What is the height of a point 9 ft to the right of the center of each arch?

70. Elliptical arches: An elliptical arch bridge is built across a one-lane highway. The arch is 20 ft across and has a maximum center height of 12 ft. Will a farm truck hauling a load 10 ft wide with a clearance height of 11 ft be able to go under the bridge without damage? (Hint: See Exercise 69.)

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728


8–22

71. Plumbing: By allowing the free flow of air, a properly vented home enables water to run freely throughout its plumbing system, while helping to prevent sewage gases from entering the home. Find the equation of the elliptical hole cut in a roof in order to allow a 3-in. vent pipe to exit, if the roof 4 has a slope of . 3 72. Light projection: Standing a short distance from a wall, Kymani’s flashlight projects a circle of radius 30 cm. When holding the flashlight at an angle, a vertical ellipse 50 cm long is formed, with the focus 10 cm from the vertex (see Worthy of Note, page 717). Find the equation of the circle and ellipse, and the area of the wall that each illuminates.

75. Planetary orbits: Except for small variations, a planet’s orbit around the Sun is elliptical with the Sun at one focus. The aphelion (maximum distance from the Sun) of the planet Mars is approximately 156 million miles, while the perihelion (minimum distance from the Sun) of Mars is about 128 million miles. Use this information to find the lengths of the semimajor and semiminor axes, rounded to the nearest million. If Mars has an orbital velocity of 54,000 miles per hour (1.296 million miles per day), how many days does it take Mars to orbit the Sun? (Hint: Use the formula from Exercise 64.)

As a planet orbits around the Sun, it traces out an ellipse. If the center of the ellipse were placed at (0, 0) on a coordinate grid, the Sun would be actually offcentered (located at the focus of the ellipse). Use this information and the graphs provided to complete Exercises 73 through 78.

79. Area of a race track: Suppose the Toronado 500 is a car race that is run on an elliptical track. The track is bounded by two ellipses with equations of 4x2 9y2 900 and 9x2 25y2 900, where x and y are in hundreds of yards. Use the formula given in Exercise 63 to find the area of the race track.

y

Sun x

70.5 million miles

Mercury

72 million miles

Exercise 74 y Pluto

Sun x

3650 million miles

3540 million miles

74. Orbit of Pluto: The approximate orbit of the Kuiper object formerly known as Pluto is shown in the figure given. Find an equation that models this orbit.

77. Orbital velocity of Earth: The planet Earth has a perihelion (minimum distance from the Sun) of about 91 million mi, an aphelion (maximum distance from the Sun) of close to 95 million mi, and completes one orbit in about 365 days. Use this information and the formula from Exercise 64 to find Earth’s orbital speed around the Sun in miles per hour. 78. Orbital velocity of Jupiter: The planet Jupiter has a perihelion of 460 million mi, an aphelion of 508 million mi, and completes one orbit in about 4329 days. Use this information and the formula from Exercise 64 to find Jupiter’s orbital speed around the Sun in miles per hour.

Exercise 73

73. Orbit of Mercury: The approximate orbit of the planet Mercury is shown in the figure given. Find an equation that models this orbit.

76. Planetary orbits: The aphelion (maximum distance from the Sun) of the planet Saturn is approximately 940 million miles, while the perihelion (minimum distance from the Sun) of Saturn is about 840 million miles. Use this information to find the lengths of the semimajor and semiminor axes, rounded to the nearest million. If Saturn has an orbital velocity of 21,650 miles per hour (about 0.52 million miles per day), how many days does it take Saturn to orbit the Sun? How many years?

Exercise 80 80. Area of a border: The tablecloth for a large oval table is elliptical in shape. It is designed with two concentric ellipses (one within the other) as shown in the figure. The equation of the outer ellipse is 9x2 25y2 225, and the equation of the inner ellipse is 4x2 16y2 64 with x and y in feet. Use the formula given in Exercise 63 to find the area of the border of the tablecloth.

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8–23

Mid-Chapter Check

729

81. Whispering galleries: Due to their unique properties, ellipses are used in the construction of whispering galleries like those in St. Paul’s Cathedral (London) and Statuary Hall in the U.S. Capitol. Regarding the latter, it is known that John Quincy Adams (1767–1848), while a member of the House of Representatives, situated his desk at a focal point of the elliptical ceiling, easily eavesdropping on the private conversations of other House members located near the other focal point. Suppose a whispering gallery was built using the equation y2 x2 1, with the dimensions in feet. (a) How tall is the ceiling at its highest point? (b) How wide is 2809 2025 the gallery vertex to vertex? (c) How far from the base of the doors at either end, should a young couple stand so that one can clearly hear the other whispering, “I love you.”? 82. While an elliptical billiard table has little practical value, it offers an excellent illustration of elliptical properties. A ball placed at one focus and hit with the cue stick from any angle, will hit the cushion and immediately rebound to the other focus and continue through each focus until coming to rest. Suppose one such table was y2 x2 1 as a model, with the dimensions in feet. (a) How far apart are the constructed using the equation 9 4 vertices? (b) How far apart are the foci? As a side note, Lewis Carroll (1832–1898) did invent a game of circular billiards, complete with rules. 䊳


83. For 6x2 36x 3y2 24y 74 28, does the equation appear to be that of a circle, ellipse, or parabola? Write the equation in factored form. What do you notice? What can you say about the graph of this equation? 䊳

84. Algebraically verify that for the ellipse y2 x2 1 with b 7 a, the length of the focal a2 b2 2a2 chord is still . b


85. (5.4) Evaluate the expression using the change-ofbase formula: log320. z1 86. (3.1) Compute the product z1z2 and quotient of: z2 z1 213 2i13; z2 5 13 5i 87. (2.3) Solve the absolute value inequality (a) graphically and (b) analytically: 2冟x 3冟 10 7 4.

88. (2.6) The resistance R to current flow in an electrical wire varies directly as the length L of the wire and inversely as the square of its diameter d. (a) Write the equation of variation; (b) find the constant of variation if a wire 2 m long with diameter d 0.005 m has a resistance of 240 ohms ( ); and (c) find the resistance in a similar wire 3 m long and 0.006 m in diameter.

MID-CHAPTER CHECK Sketch the graph of each conic section.

1. 1x 42 2 1y 32 2 9

2. x2 y2 10x 4y 4 0 3.

1x 22 2 16

1y 32 2 1

1

4. 9x2 4y2 18x 24y 9 0

5.

1x 32 2 9

1y 42 2 4

1

6. 9x2 16y2 36x 96y 36 0 7. Find the equation for all points located an equal distance from the point (0, 3) and the line y 3.

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730

8–24


9. Find the equation of the ellipse having foci at (0, 13) and 10, 132 , with a minor axis of length 10 units.

8. Find the equation of each relation and state its domain and range. a.

(3, 5)

(5, 1)

b.

y

y 10 8 6 4 (1, 2) 2

5 4 3 2 (1, 1) 1

54321 1 2 3 (3, 3)4 5

108642 2 4 6 8 10

1 2 3 4 5 x

(3, 6) (7, 2) 2 4 6 8 10 x

10. Find the equation of the ellipse (in standard form) if the vertices are (4, 0) and (4, 0) and the distance between the foci is 4 13 units.

(3, 2)

REINFORCING BASIC CONCEPTS More on Completing the Square From our work so far in Chapter 8, we realize the process of completing the square has much greater use than simply as a tool for working with quadratic equations. It is a valuable tool in the application of the conic sections, as well as other areas. The purpose of this Reinforcing Basic Concepts is to strengthen the ability and confidence needed to apply the process correctly. This is important because in some cases the values of a and b are rational or irrational numbers. No matter what the context, 1. The process begins with a coefficient of 1. For 20x2 120x 27y2 54y 192 0, we recognize the equation of an ellipse, since the coefficients of the squared terms are positive and unequal. To study or graph this ellipse, we’ll use the standard form to identify the values of a, b, and c. Grouping the like-variable terms gives 120x2 120x

2 127y2 54y

2 192 0

and to complete the square, we factor out the lead coefficient of each group (to get a coefficient of 1): 201x2 6x

2 271y2 2y

2 192 0

Subtracting 192 from both sides brings us to the fundamental step for completing the square. 2 1 2. The quantity a # linear cofficientb will complete a trinomial square. For this example we obtain 2 2 2 1 1 a # 6b 9 for x, and a # 2b 1 for y, with these numbers inserted in the appropriate group: 2 2 201x2 6x 92 271y2 2y 12 192

complete the square

Due to the distributive property, we have in effect added 20 # 9 180 and 27 # 1 27 (for a total of 207) to the left side of the equation: 201x2 6x 92 271y2 2y 12 192 adds 20 # 9 180 adds 27 # 1 27 to left side

to left side

This brings us to the final step. 3. Keep the equation in balance. Since the left side was increased by 207, we also increase the right side by 207. 201x2 6x 92 271y2 2y 12 192 207 adds 20 # 9 180 adds 27 # 1 27 add 180 27 207 to left side

to left side

to right side

The quantities in parentheses factor, giving 201x 32 271y 12 15. We then divide by 15 and simplify, obtaining 2

41x 32 2

91y 12 2

1. Note the coefficient of each binomial square is not 1, even after setting 3 5 the equation equal to 1. In the Strengthening Core Skills feature of this chapter, we’ll look at how to write equations of this type in standard form to obtain the values of a and b. For now, practice completing the square using these exercises. the standard form

2

Exercise 1: 100x2 400x 18y2 108y 554 0 Exercise 2: 28x2 56x 48y2 192y 195 0

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8.3

The Hyperbola


A. Use the equation of a hyperbola to graph central and noncentral hyperbolas B. Distinguish between the equations of circles, ellipses, and hyperbolas C. Locate the foci of a hyperbola and use the foci and other features to write its equation D. Solve applications involving foci

EXAMPLE 1

䊳

As seen in Section 8.1 (see Figure 8.24), a hyperbola is a conic section formed by a plane that cuts both nappes of a right circular cone. A hyperbola has two Axis symmetric parts called branches, which open in opposite directions. Although the branches appear to resemble parabolas, we will soon discover they are actually a very different curve.

Figure 8.24

Hyperbola

A. The Equation of a Hyperbola In Section 8.2, we noted that for the equation Ax2 By2 F, if A B, the equation is that of a circle, if A B, the equation represents an ellipse. Both cases contain a sum of second-degree terms. Perhaps driven by curiosity, we might wonder what happens if the equation has a difference of second-degree terms. Consider the equation 9x2 16y2 144. It appears the graph will be centered at (0, 0) since no shifts are applied (h and k are both zero). Using the intercept method to graph this equation reveals an entirely new curve, called a hyperbola. Graphing a Central Hyperbola Graph the equation 9x2 16y2 144 using intercepts and additional points as needed.

Solution

9x2 16y2 144 9102 2 16y2 144 16y2 144 y2 9

䊳

given substitute 0 for x simplify divide by 16

Since y2 can never be negative, we conclude that the graph has no y-intercepts. Substituting y 0 to find the x-intercepts gives 9x2 16y2 144 9x2 16102 2 144 9x2 144 x2 16 x 116 and x 116 x 4 and x 4 (4, 0) and 14, 02

given substitute 0 for y simplify divide by 9 square root property simplify

x-intercepts

Knowing the graph has no y-intercepts, we select inputs greater than 4 and less than 4 to help sketch the graph. Using x 5 and x 5 yields 9x2 16y2 144 9152 2 16y2 144 91252 16y2 144 225 16y2 144 16y2 81 81 y2 16 9 9 y y 4 4 y 2.25 y 2.25 15, 2.252 15, 2.252 8–25

given substitute for x 5 152 2 25 2

simplify subtract 225

9x2 16y2 9152 2 16y2 91252 16y2 225 16y2 16y2

divide by 16

square root property decimal form ordered pairs

9 4 y 2.25 15, 2.252 y

144 144 144 144 81 81 y2 16 9 y 4 y 2.25 15, 2.252 731

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732

8–26


Plotting these points and connecting them with a smooth curve, while knowing there are no y-intercepts, produces the graph in the figure. The point at the origin (in blue) is not a part of the graph, and is given only to indicate the “center” of the hyperbola. The points 14, 02 and (4, 0) are called vertices, and the center of the hyperbola is always the point halfway between them.

y Hyperbola (5, 2.25) (5, 2.25) (4, 0)

(4, 0) (0, 0)

x

(5, 2.25) (5, 2.25)


䊳

As with the circle and ellipse, the hyperbola fails the vertical line test and we must graph the relation on a calculator by writing its equation in two parts, each of which is a function. For the hyperbola in Example 1, this gives 9x2 16y2 144 original equation 2 2 isolate y-term 16y 144 9x multiply by 1 16y2 9x2 144 9x2 144 y2 divide by 16 16 9x2 144 y take square roots B 16 9.4 2 2 9X 144 9X 144 , Y2 Y1 B 16 B 16

Figure 8.25 6.2

9.4

The graph is shown in Figure 8.25. 6.2 Since the hyperbola in Example 1 crosses a horizontal line of symmetry, it is referred to as a horizontal hyperbola. If the center is at the origin, we have a central hyperbola. The line passing through the center and both vertices is called the transverse axis (vertices are always on the transverse axis), and the line passing through the center and perpendicular to this axis is called the conjugate axis (see Figure 8.26). In Example 1, the coefficient of x2 was positive and we were subtracting 16y2: 2 9x 16y2 144. The result was a horizontal hyperbola. If the y2-term is positive and we subtract the term containing x2, the result is a vertical hyperbola (Figure 8.27). Figure 8.26

Figure 8.27

y Conjugate axis

Center Vertex

Transverse axis

Transverse axis Vertex

Horizontal hyperbola

x

y

Vertex Vertex Center

Vertical hyperbola

Conjugate axis x

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8–27

733

Section 8.3 The Hyperbola

EXAMPLE 2

䊳

Identifying the Axes, Vertices, and Center of a Hyperbola from Its Graph For the hyperbola shown, state the location of the vertices and the equation of the transverse axis. Then identify the location of the center and the equation of the conjugate axis.

Solution

䊳

By inspection we locate the vertices at (0, 0) and (0, 4). The equation of the transverse axis is x 0. The center is halfway between the vertices at (0, 2), meaning the equation of the conjugate axis is y 2.

y

5

5

5

x

4


䊳

Standard Form As with the ellipse, the polynomial form of the equation is helpful for identifying hyperbolas, but not very helpful when it comes to graphing a hyperbola (since we still must go through the laborious process of finding additional points). For graphing, standard form is once again preferred. Consider the hyperbola 9x2 16y2 144 from Example 1. To write the equation in standard form, we divide by 144 and obtain y2 x2 1. By comparing the standard form to the graph, we note a 4 represents 42 32 the distance from center to vertices, similar to the way we used a previously. But since the graph has no y-intercepts, what could b 3 represent? The answer lies in the fact that branches of a hyperbola are asymptotic, meaning they will approach and become very close to imaginary lines that can be used to sketch the graph. For b a central hyperbola, the slopes of the asymptotic lines are given by the ratios and a b b b , with the related equations being y x and y x. The graph from Example 1 a a a is repeated in Figure 8.28, with the asymptotes drawn. For a clearer understanding of how the equations for the asymptotes were determined, see Exercise 87. A second method of drawing the asymptotes involves drawing a central rectangle with dimensions 2a by 2b, as shown in Figure 8.29. The asymptotes will be the extended diagonals of this rectangle. This brings us to the equation of a hyperbola in standard form. Figure 8.28

Figure 8.29

y Slope m

y

34

rise b3 (4, 0)

Slope m

3 4

run a4 (4, 0) (0, 0)

(4, 0)

2b

x

2a

Slope method

Central rectangle method

x

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734

8–28


The Equation of a Hyperbola in Standard Form The equation 1x h2 2 a2

1y k2 2 b2

The equation 1y k2 2

1

b2

represents a horizontal hyperbola with center (h, k) • transverse axis y k • conjugate axis x h • 冟a冟 gives the distance from center to vertices.

1x h2 2 a2

1

represents a vertical hyperbola with center (h, k) • transverse axis x h • conjugate axis y k • 冟b冟 gives the distance from center to vertices.

b • Asymptotes can be drawn by starting at (h, k) and using slopes m . a

EXAMPLE 3

䊳

Graphing a Hyperbola Using Its Equation in Standard Form Sketch the graph of 161x 22 2 91y 12 2 144, and label the center, vertices, and asymptotes.

Solution

䊳

Begin by noting a difference of the second-degree terms, with the x2-term occurring first. This means we’ll be graphing a horizontal hyperbola whose center is at (2, 1). Continue by writing the equation in standard form. 161x 22 2 91y 12 2 144 91y 12 2 161x 22 2 144 144 144 144 1y 12 2 1x 22 2 1 9 16 1x 22 2 1y 12 2 1 32 42

given equation divide by 144

simplify


Since a 3 the vertices are a horizontal distance of 3 units from the center (2, 1), giving 12 3, 12 S 15, 12 and 12 3, 12 S 11, 12 . After plotting the center and b 4 vertices, we can begin at the center and count off slopes of m , or a 3 draw a rectangle centered at (2, 1) with dimensions 2132 6 (horizontal dimension) by 2142 8 (vertical dimension) to sketch the asymptotes. The complete graph is shown here. Horizontal hyperbola

y

Center at (2, 1)

md

Vertices at (1, 1) and (5, 1) Transverse axis: y 1 Conjugate axis: x 2

(2, 1) (1, 1)

(5, 1) x

m d

冢

Width of rectangle horizontal dimension and distance between vertices 2a 2(3) 6

冣

Length of rectangle (vertical dimension) 2b 2(4) 8


䊳

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735


If the hyperbola in Example 3 were a central hyperbola, the equations of the 4 4 asymptotes would be y x and y x. But the center of this graph has been 3 3 shifted 2 units right and 1 unit up. Using our knowledge of shifts and translations, the equations for the asymptotes of the shifted hyperbola must be 4 4 5 1. 1y 12 1x 22 , or y x in simplified form, and 3 3 3 4 11 4 2. 1y 12 1x 22 or y x . 3 3 3 Using Y1

161X 22 2 144

161X 22 2 144

1, and Y2 1 B 9 B 9 (obtained by solving for y in the original equation), a calculator generated graph of the hyperbola and its asymptotes is shown here (Figures 8.30 and 8.31). Figure 8.31 Figure 8.30

10.3

7.4

11.4

8.3

Polynomial Form If the equation is given as a polynomial in expanded form, complete the square in x and y, then write the equation in standard form.

EXAMPLE 4

䊳

Graphing a Hyperbola by Completing the Square Graph the equation 9y2 x2 54y 4x 68 0 by completing the square. Label the center and vertices and sketch the asymptotes. Then graph the hyperbola on a graphing calculator and use the TRACE feature with a “friendly” window to locate four additional points whose coordinates are rational.

Solution

䊳

Since the y2-term occurs first, we assume the equation represents a vertical hyperbola, but wait for the factored form to be sure (see Exercise 91). 9y2 x2 54y 4x 68 0 9y2 54y x2 4x 68 2 91y 6y ___ 2 11x2 4x ___ 2 68 91y2 6y 92 11x2 4x 42 68 81 142 c

c

adds 9 19 2 81

c

c

adds 1 14 2 4

91y 32 2 11x 22 2 9 1x 22 2 1y 32 2 1 1 9 1y 32 2 1x 22 2 1 12 32

add 81 14 2 to right

given collect like-variable terms; subtract 68 factor out 9 from y-terms and 1 from x-terms complete the square

factor S vertical hyperbola divide by 9 (standard form)


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The center of the hyperbola is 12, 32 with a 3, b 1, and a transverse axis of x 2. The vertices are at 12, 3 12 and 12, 3 12 S 12, 22 and 12, 42 . After plotting the center and vertices, we draw a rectangle centered at 12, 32 with a horizontal “width” of 2132 6 and a vertical “length” of 2112 2 to sketch the asymptotes. The completed graph is given in Figure 8.32. Figure 8.32 Vertical hyperbola

y

Center at (2, 3) Vertices at (2, 2) and (2, 4)

m 3 1

(2, 2)

Transverse axis: x 2 Conjugate axis: y 3

x 1 m3 (2, 3) center

Width of rectangle (horizontal dimension) 2a 2(3) 6

(2, 4)

冢

Length of rectangle vertical dimension and distance between vertices 2b 2(1) 2

冣

To graph the hyperbola on a calculator, we again solve for y.

91y 32 2 1x 22 2 9

factored form

91y 32 9 1x 22 isolate term containing y 2 9 1x 22 1y 32 2 divide by 9 9 9 1x 22 2 y3 take square roots B 9 9 1x 22 2 y subtract 3 3 B 9 9 1X 22 2 9 1X 22 2 Y1 3 3, Y2 Figure 8.33 B B 9 9 2

2

Using the friendly window shown, we can use the arrow keys to TRACE though x-values and find the points 15.2, 0.42 and 19.2, 0.42 on the upper branch, with 15.2, 5.62 and 19.2, 5.62 on the lower branch. Note how these points show that a hyperbola is symmetric to its center, as well as the horizontal line and vertical line through its center. The graph is shown in Figure 8.33. A. You’ve just seen how we can use the equation of a hyperbola to graph central and noncentral hyperbolas

3.2

7.4

11.4

9.2


䊳

B. Distinguishing between the Equations of Circles, Ellipses, and Hyperbolas So far we’ve explored numerous graphs of circles, ellipses, and hyperbolas. In Example 5 we’ll attempt to identify a given conic section from its equation alone (without graphing the equation). As you’ve seen, the corresponding equations have unique characteristics that can help distinguish one from the other.

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737


EXAMPLE 5

䊳

Identifying a Conic Section from Its Equation Identify each equation as that of a circle, ellipse, or hyperbola. Justify your choice and name the center, but do not draw the graphs. a. y2 36 9x2 b. 4x2 16 4y2 c. x2 225 25y2 d. 25x2 100 4y2 2 2 e. 31x 22 41y 32 12 f. 41x 52 2 36 91y 42 2

Solution

B. You’ve just seen how we can distinguish between the equations of circles, ellipses, and hyperbolas

䊳

a. Writing the equation as y2 9x2 36 shows h 0 and k 0. Since the equation contains a difference of second-degree terms, it is the equation of a (vertical) hyperbola (A and B have opposite signs). The center is at (0, 0). b. Rewriting the equation as 4x2 4y2 16 and dividing by 4 gives x2 y2 4. The equation represents a circle of radius 2 1A B2 , with the center at (0, 0). c. Writing the equation as x2 25y2 225 we note a sum of second-degree terms with unequal coefficients. The equation is that of an ellipse 1A B2 , with the center at (0, 0). d. Rewriting the equation as 25x2 4y2 100 we note the equation contains a difference of second-degree terms. The equation represents a central (horizontal) hyperbola (A and B have opposite signs), whose center is at (0, 0). e. The equation is in factored form and contains a sum of second-degree terms with unequal coefficients. This is the equation of an ellipse 1A B2 with the center at 12, 32 . f. Rewriting the equation as 41x 52 2 91y 42 2 36 we note a difference of second-degree terms. The equation represents a horizontal hyperbola (A and B have opposite signs) with center 15, 42. Now try Exercises 49 through 60

䊳

C. The Foci of a Hyperbola Like the ellipse, the foci of a hyperbola play an important part in their application. A long distance radio navigation system (called LORAN for short), can be used to determine the location of ships and airplanes and is based on the characteristics of a hyperbola (see Exercises 85 and 86). Hyperbolic mirrors are also used in some telescopes, and have the property that a beam of light directed at one focus will be reflected to the second focus. To understand and appreciate these applications, we use the analytic definition of a hyperbola: Definition of a Hyperbola Given two fixed points f1 and f2 in a plane, a hyperbola is the set of all points (x, y) such that the distance d1 from f1 to (x, y) and the distance d2 from f2 to (x, y), satisfy the equation 冟d1 d2冟 k. In other words, the difference of these two distances is a positive constant. The fixed points f1 and f2 are called the foci of the hyperbola, and all such points (x, y) are on the graph of the hyperbola.

y d1

(x, y) d2

f1

f2

|d1 d2| k k>0

x

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Figure 8.34 y

(x, y)

(c, 0) x

(⫺c, 0) (⫺a, 0)

(a, 0)

EXAMPLE 6

䊳

As with the analytic definition of the ellipse, it can be shown that the constant k is again equal to 2a (for horizontal hyperbolas). To find the equation of a hyperbola in terms of a and b, we use an approach similar to that of the ellipse (see Appendix V), y2 x2 and the result is identical to that seen earlier: 2 ⫺ 2 ⫽ 1 where b2 ⫽ c2 ⫺ a2 a b (see Figure 8.34). We now have the ability to find the foci of any hyperbola —and can use this information in many significant applications. Since the location of the foci play such an important role, it is best to remember the relationship as c2 ⫽ a2 ⫹ b2 (called the foci formula for hyperbolas), noting that for a hyperbola, c 7 a and c2 7 a2 (also c 7 b and c2 7 b2). Be sure to note that for ellipses, the foci formula is c2 ⫽ 冟a2 ⫺ b2冟 since a 7 c (horizontal ellipses) or b 7 c (vertical ellipses). Graphing a Hyperbola and Identifying Its Foci by Completing the Square For the hyperbola defined by 7x2 ⫺ 9y2 ⫺ 14x ⫹ 72y ⫺ 200 ⫽ 0, find the coordinates of the center, vertices, foci, and the dimensions of the central rectangle. Then sketch the graph, including the asymptotes.

Solution

䊳

given 7x2 ⫺ 9y2 ⫺ 14x ⫹ 72y ⫺ 200 ⫽ 0 2 2 group terms; add 200 7x ⫺ 14x ⫺ 9y ⫹ 72y ⫽ 200 factor out leading coefficients 71x2 ⫺ 2x ⫹ ____ 2 ⫺ 91y2 ⫺ 8y ⫹ ____ 2 ⫽ 200 2 2 71x ⫺ 2x ⫹ 12 ⫺ 91y ⫺ 8y ⫹ 162 ⫽ 200 ⫹ 7 ⫹ 1⫺1442 complete the square c

c

c

S add 7 ⫹ 1⫺1442 to right-hand side

c

adds ⫺9 116 2 ⫽ ⫺144

adds 7112 ⫽ 7

71x ⫺ 12 2 ⫺ 91y ⫺ 42 2 ⫽ 63 1y ⫺ 42 2 1x ⫺ 12 2 ⫺ ⫽1 9 7 1x ⫺ 12 2 1y ⫺ 42 2 ⫺ ⫽1 32 1 172 2

factored form divide by 63 and simplify


This is a horizontal hyperbola with a ⫽ 3 1a2 ⫽ 92 and b ⫽ 17 1b2 ⫽ 72. The center is at (1, 4), with vertices 1⫺2, 42 and (4, 4). Using the foci formula c2 ⫽ a2 ⫹ b2 yields c2 ⫽ 9 ⫹ 7 ⫽ 16, showing the foci are 1⫺3, 42 and (5, 4) (4 units from center). The central rectangle is 2132 ⫽ 6 by 2 17 ⬇ 5.29. Drawing the rectangle and sketching the asymptotes results in the graph shown. Horizontal hyperbola

y 10

Center at (1, 4) Vertices at (⫺2, 4) and (4, 4) (1, 4)

(⫺3, 4) ⫺10

(⫺2, 4)

⫺10

(4, 4)

Transverse axis: y ⫽ 4 Conjugate axis: x ⫽ 1 Location of foci: (⫺3, 4) and (5, 4)

(5, 4) 10

x

Width of rectangle horizontal dimension and distance between vertices 2a ⫽ 2(3) ⫽ 6

冢

冣

Length of rectangle (vertical dimension) 2b ⫽ 2(√7) ≈ 5.29


䊳

The focal chord for a horizontal hyperbola is a vertical line segment through the focus with endpoints on the hyperbola. Similar to the focal chord of an ellipse, we can use its length to find additional points on the graph of the hyperbola. The total length 2b2 is once again L ⫽ (for a horizontal hyperbola), meaning the distance from the foci a

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739

Figure 8.35 b2 . 10.2 a 2 For Example 6, a ⫽ 3 and b ⫽ 7, so the vertical distance from focus to graph (in either 7 direction) is ⫽ 2.3. From the left focus ⫺9.4 9.4 3 1⫺3, 42 , we can now graph the additional 1⫺3, 4 ⫹ 2.32 ⫽ 1⫺3, 6.32 , and points 1⫺3, 4 ⫺ 2.32 ⫽ 1⫺3, 1.62 . From the right ⫺2.2 focus (5, 4), we obtain (5, 6.3) and (5, 1.6). Graphical verification is provided in Figure 8.35. Also see Exercise 80. As with the ellipse, if any two of the values for a, b, and c are known, the relationship between them can be used to construct the equation of the hyperbola. See Exercises 71 through 78.

to the graph (along the focal chord) is

C. You’ve just seen how we can locate the foci of a hyperbola and use the foci and other features to write its equation

D. Applications Involving Foci Applications involving the foci of a conic section can take many forms. As before, only partial information about the hyperbola may be available, and we’ll determine a solution by manipulating a given equation or constructing an equation from given facts. EXAMPLE 7

䊳

Applying the Properties of a Hyperbola—The Path of a Comet Comets with a high velocity cannot be captured by the Sun’s gravity, and are slung around the Sun in a hyperbolic path with the Sun at one focus. If the path illustrated by the graph shown is modeled by the equation 2116x2 ⫺ 400y2 ⫽ 846,400, how close did the comet get to the Sun? Assume units are in millions of miles and round to the nearest million.

Solution

䊳

We are essentially asked to find the distance between a vertex and focus. Begin by writing the equation in standard form: 2116x2 ⫺ 400y2 ⫽ 846,400 given y2 x2 ⫺ ⫽1 divide by 846,400 400 2116 write denominators in y2 x2 ⫺ ⫽1 squared form 2 2 20 46 This is a horizontal hyperbola with a ⫽ 20 1a2 ⫽ 4002 and b ⫽ 46 1b2 ⫽ 21162. Use the foci formula to find c2 and c.

y

(0, 0) x

c2 ⫽ a2 ⫹ b2 c2 ⫽ 400 ⫹ 2116 c2 ⫽ 2516 c ⬇ ⫺50 and c ⬇ 50 Since a ⫽ 20 and 冟c冟 ⬇ 50, the comet came within about 50 ⫺ 20 ⫽ 30 million miles of the Sun. Now try Exercises 81 through 84

䊳

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EXAMPLE 8

䊳

Applying the Properties of a Hyperbola — The Location of a Storm Two amateur meteorologists, living 4 km apart (4000 m), see a storm approaching. The one farthest from the storm hears a loud clap of thunder 9 sec after the one nearest. Assuming the speed of sound is 340 m/sec, determine an equation that models possible locations for the storm at this time.

Solution

D. You’ve just seen how we can solve applications involving foci

䊳

Let M1 represent the meteorologist nearest the storm and M2 the farthest. Since M2 heard the thunder 9 sec after M1, M2 must be 9 # 340 3060 m farther away from the storm S. In other words, from our definition of a hyperbola, we have 冟d1 d2冟 3060. The set of all points that satisfy this equation will be on the graph of a hyperbola, and we’ll use this fact to develop an equation model for possible locations of the storm. Let’s place the information on a coordinate grid. For convenience, we’ll use the straight y S line distance between M1 and M2 as 2 the x-axis, with the origin an equal distance from each. With the 1 constant difference equal to 3060, M2 M1 we have 2a 3060, a 1530 from 3 2 1 1 2 3 x in 1000s the definition of a hyperbola, giving 2 2 1 y x 2 1. With c 2000 m 2 1530 b 2 (the distance from the origin to M1 or M2), we find the value of b using the equation c2 a2 b2: 20002 15302 b2 or b2 120002 2 115302 2 1,659,100 ⬇ 12882. The equation that models possible y2 x2 ⬇ 1. locations of the storm is 15302 12882 Now try Exercises 85 and 86

8.3 EXERCISES 䊳



1. The line that passes through the vertices of a hyperbola is called the axis.

2. The center of a hyperbola is located between the vertices.

3. The conjugate axis is axis and contains the of the hyperbola.

4. The center of the hyperbola defined by

to the

1x 22 2 42

1y 32 2 52

1 is at

.

䊳

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5. Compare/Contrast the two methods used to find the asymptotes of a hyperbola. Include an example illustrating both methods.

䊳

741

6. Explore/Explain why A1x h2 2 B1y k2 2 F, 1A, B 7 02 results in a hyperbola regardless of whether A B or A B. Illustrate with an example.


Graph each hyperbola. Label the center, vertices, and any additional points used.

7.

y2 x2 1 9 4

8.

y2 x2 9. 1 4 9 11. 13. 15. 17.

12.

y2 x2 1 36 16

14.

y2 x2 1 9 1

16.

y2 x2 1 12 4

18.

2

19. 21.

2

y x 1 9 9 2

27.

1y 12 2

1x 22 2 y2 x2 1 28. 1 4 25 4 9 1x 32 2 1y 22 2 1 29. 36 49 30.

1x 22 2

1y 12 2

31.

1y 12 2

1x 52 2

y2 x2 1 1 4

32.

1y 32 2

1x 22 2

y2 x2 1 9 18

33. 1x 22 2 41y 12 2 16

y2 x2 1 16 9

y2 x2 10. 1 25 16

y2 x2 1 49 16

20.

2

y x 1 36 25

22.

Sketch a complete graph of each equation, including the asymptotes. Be sure to identify the center and vertices.

y2 x2 1 25 9 y2 x2 1 81 16

9 7

16

4 9 5

1 1 1

2

2

34. 91x 12 2 1y 32 2 81

2

2

36. 91y 42 2 51x 32 2 45

y x 1 4 4

35. 21y 32 2 51x 12 2 50

y x 1 16 4

37. 121x 42 2 51y 32 2 60 38. 81x 42 2 31y 32 2 24

For the graphs given, state the location of the vertices and the equation of the transverse axis. Then identify the location of the center and the equation of the conjugate axis. Note the scale used on each axis.

23.

24.

y 10

10

10

25.

10 x

26.

y

41. 9y2 4x2 36

42. 25y2 4x2 100

43. 12x2 9y2 72

44. 36x2 20y2 180

Graph each hyperbola by writing the equation in standard form. Label the center and vertices, and sketch the asymptotes. Then use a graphing calculator to graph each relation and locate four additional points whose coordinates are rational.

45. 4x2 y2 40x 4y 60 0

10

10

40. 16x2 25y2 400

y 10

10

10 x

39. 16x2 9y2 144

46. x2 4y2 12x 16y 16 0

y 10

47. 4y2 x2 24y 4x 28 0 48. 9x2 4y2 18x 24y 9 0

10

10 x

10

10

10 x

10

Classify each equation as that of a circle, ellipse, or hyperbola. Justify your response (assume all are nondegenerate).

49. 4x2 4y2 24 50. 9y2 4x2 36

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51. x2 y2 2x 4y 4

67. 4y2 16x2 24y 28 0

52. x2 y2 6y 7

68. 4y2 81x2 162x 405 0

53. 2x2 4y2 8

69. 9x2 3y2 54x 12y 33 0

54. 36x2 25y2 900

70. 10x2 60x 5y2 20y 20 0

55. x2 5 2y2

Find the equation of the hyperbola (in standard form) that satisfies the following conditions:

56. x y2 3x2 9 57. 2x2 2y2 x 20

71. vertices at (6, 0) and (6, 0); foci at (8, 0) and (8, 0)

58. 2y 3 6x 8 2

2

72. vertices at (4, 0) and (4, 0); foci at (6, 0) and (6, 0)

59. 16x2 5y2 3x 4y 538 60. 9x2 9y2 9x 12y 4 0 Use the definition of a hyperbola to find the distance between the vertices and the dimensions of the rectangle centered at (h, k). Figures are not drawn to scale. Note that Exercises 63 and 64 are vertical hyperbolas.

61.

62.

y

(5, 2.25) (a, 0) (5, 0)

(5, 0)

(15, 0)

64.

y

(0, b)

75.

(0, 13) (6, 7.5)

(9, 6.25)

x

y 5 4 3 2 1 54321 1 2 3 4 5

y

(0, 10) (0, b)

(a, 0) (15, 0) x

(a, 0)

(a, 0)

63.

74. foci at (5, 2) and (7, 2); length of conjugate axis: 8 units Use the characteristics of a hyperbola and the graph given to write the related equation and state the location of the foci (75 and 76) or the dimensions of the central rectangle (77 and 78).

y

(15, 6.75)

x

73. foci at 12, 3122 and 12, 3122; length of conjugate axis: 6 units

76.

1 2 3 4 5 x

y 10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

(0, b) (0, b)

(0, 10) (0, 13)

Write each equation in standard form to find and list the coordinates of the (a) center, (b) vertices, (c) foci, and (d) dimensions of the central rectangle. Then (e) sketch the graph, including the asymptotes.

x

77.

y 5 4 3 2 1 54321 1 2 3 4 5

1 2 3 4 5 x

78.

y 10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

65. 4x2 9y2 24x 72y 144 0 66. 4x2 36y2 40x 144y 188 0 䊳

WORKING WITH FORMULAS 36 ⴚ 4x2 B ⴚ9 The “upper half” of a certain hyperbola is given by the equation shown. (a) Simplify the radicand, (b) state the domain of the expression, and (c) enter the expression as Y1 on a graphing calculator and graph. What is the equation for the “lower half” of this hyperbola?

79. Equation of a semi-hyperbola: y ⴝ

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743


2m2 n The focal chords of a hyperbola are line segments parallel to the conjugate axis with 10 8 endpoints on the hyperbola, and containing points f1 and f2 (see grid). The length of the 6 chord is given by the formula shown, where n is the distance from center to vertex and m is 4 f 2 the distance from center to one side of the central rectangle. Use the formula to find the length of the focal chord for the hyperbola indicated, then compare the calculated value with 1086422 4 6 the length estimated from the given graph:

80. Focal chord of a hyperbola: L ⴝ

1

1x 22 2 4

䊳

1y 12 2 5

1.

y

f2 2 4 6 8 10 x

8 10

APPLICATIONS

81. Stunt pilots: At an air show, a stunt plane dives along a hyperbolic path whose vertex is directly over the grandstands. If the plane’s flight path can be modeled by the hyperbola 25y2 1600x2 40,000, what is the minimum altitude of the plane as it passes over the stands? Assume x and y are in yards. 82. Flying clubs: To test their skill as pilots, the members of a flight club attempt to drop sandbags on a target placed in an open field, by diving along a hyperbolic path whose vertex is directly over the target area. If the flight path of the plane flown by the club’s president is modeled by 9y2 16x2 14,400, what is the minimum altitude of her plane as it passes over the target? Assume x and y are in feet. 83. Charged particles: It has been shown that when like particles with a common charge are hurled at each other, they deflect and travel along paths that are hyperbolic. Suppose the paths of two such particles is modeled by the hyperbola x2 9y2 36. What is the minimum distance between the particles as they approach each other? Assume x and y are in microns. 84. Nuclear cooling towers: The natural draft cooling towers for nuclear power stations are called hyperboloids of one sheet. The perpendicular cross sections of these hyperboloids form two branches of a hyperbola. Suppose the central cross section of one such tower is modeled by the hyperbola 1600x2 4001y 502 2 640,000. What is the minimum distance between the sides of the tower? Assume x and y are in feet.

85. Locating a ship using radar: Under certain conditions, the properties of a hyperbola can be used to help locate the position of a ship. Suppose two radio stations are located 100 km apart along a straight shoreline. A ship is sailing parallel to the shore and is 60 km out to sea. The ship sends out a distress call that is picked up by the closer station in 0.4 milliseconds (msec — one-thousandth of a second), while it takes 0.5 msec to reach the station that is farther away. Radio waves travel at a speed of approximately 300 km/msec. Use this information to find the equation of a hyperbola that will help you find the location of the ship, then find the coordinates of the ship. (Hint: Draw the hyperbola on a coordinate system with the radio stations on the x-axis at the foci, then use the definition of a hyperbola.)

86. Locating a plane using radar: Two radio stations are located 80 km apart along a straight shoreline, when a “mayday” call (a plea for immediate help) is received from a plane that is about to ditch in the ocean (attempt a water landing). The plane was flying at low altitude, parallel to the shoreline, and 20 km out when it ran into trouble. The plane’s distress call is picked up by the closer station in 0.1 msec, while it takes 0.3 msec to reach the other. Use this information to construct the equation of a hyperbola that will help you find the location of the ditched plane, then find the coordinates of the plane. Also see Exercise 85.

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8–38



87. For a greater understanding as to why the branches y2 x2 of a hyperbola are asymptotic, solve 2 2 1 a b for y, then consider what happens as x S q (note that x2 k ⬇ x2 for large x).

88. Which has a greater area: (a) The central rectangle of the hyperbola given by 1x 52 2 1y 42 2 57, (b) the circle given by 1x 52 2 1y 42 2 57, or (c) the ellipse given by 91x 52 2 101y 42 2 570?

89. It is possible for the plane to intersect only the vertex of the cone or to be tangent to the sides. These are called degenerate cases of a conic section. Many times we’re unable to tell if the equation represents a degenerate case until it’s written in standard form. Write the following equations in standard form and comment. a. 4x2 32x y2 4y 60 0 b. x2 4x 5y2 40y 84 0

䊳


90. (2.5) Graph the piecewise-defined function: f 1x2 e

4x 5

2

2 x 6 3 x3

91. (4.1) Use synthetic division and the remainder theorem to determine if x 2 is a zero of g1x2 x5 5x4 4x3 16x2 32x 16. If yes, find its multiplicity. 92. (4.2) The number z 1 i 12 is a solution to two out of the three equations given. Which two? a. x4 4 0 b. x3 6x2 11x 12 0 c. x2 2x 3 0

8.4

93. (6.4) A government-approved company is licensed to haul toxic waste. Each container of solid waste weighs 800 lb and has a volume of 100 ft3. Each container of liquid waste weighs 1000 lb and is 60 ft3 in volume. The revenue from hauling solid waste is $300 per container, while the revenue from liquid waste is $350 per container. The truck used by this company has a weight capacity of 39.8 tons and a volume capacity of 6960 ft3. What combination of solid and liquid waste containers will produce the maximum revenue?

The Analytic Parabola; More on Nonlinear Systems


A. Graph parabolas with a horizontal axis of symmetry B. Identify and use the focus-directrix form of the equation of a parabola C. Solve nonlinear systems involving the conic sections D. Solve applications of the analytic parabola

In previous coursework, you likely learned that the Figure 8.36 graph of a quadratic function was a parabola. Parabolas Parabola are actually the fourth and final member of the family of conic sections, and as we saw in Section 8.1, the Axis graph can be obtained by observing the intersection of Element a plane and a cone. If the plane is parallel to the generator of the cone (shown as a dark line in Figure 8.36), the intersection of the plane with one nappe forms a parabola. In this section we develop the general equation of a parabola from its analytic definition, opening a new realm of applications that extends far beyond those involving only zeroes and extreme values.

A. Parabolas with a Horizontal Axis An introductory study of parabolas generally involves those with a vertical axis, defined by the equation y ax2 bx c. Unlike the previous conic sections, this equation has only one second-degree (squared) term in x and defines a function. As a

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Section 8.4 The Analytic Parabola; More on Nonlinear Systems

Figure 8.37 1. Opens upward

y

4. Axis of symmetry

745

review, the primary characteristics are listed here and illustrated in Figure 8.37. See Exercises 7 through 12. Vertical Parabolas For a second-degree equation of the form y ax2 bx c, the graph is a vertical parabola with these characteristics: 1. opens upward if a 7 0, downward if a 6 0. 2. y-intercept: (0, c) (substitute 0 for x) 3. x-intercept(s): substitute 0 for y and solve. b 4. axis of symmetry: x 2a b 4ac b2 5. vertex: 1h, k2 a , b 2a 4a

2. y-intercept

3. x-intercepts

x

5. Vertex

Horizontal Parabolas Similar to our study of horizontal and vertical hyperbolas, the graph of a parabola can open to the right or left, as well as up or down. After interchanging the variables x and y in the standard equation, we obtain the parabola x ay2 by c, noting the resulting graph will be a reflection about the line y x. Here, the axis of symmetry is a horizontal line and factoring or the quadratic formula is used to find the y-intercepts (if they exist). Note that although the graph is still a parabola—it is not the graph of a function. Horizontal Parabolas For a second-degree equation of the form x ay2 by c, the graph is a horizontal parabola with these characteristics: 1. opens right if a 7 0, left if a 6 0. 2. x-intercept: (c, 0) (substitute 0 for y) 3. y-intercepts(s): substitute 0 for x and solve. b 4. axis of symmetry: y 2a 2 4ac b b 5. vertex: a , b 4a 2a

EXAMPLE 1

䊳

Graphing a Horizontal Parabola Graph the relation whose equation is x y2 3y 4, then state the domain and range of the relation. y

Solution

䊳

10 Since the equation has a single squared term in y, the graph will be a horizontal parabola. With a 7 0 1a 12, the parabola opens to the right. (0, 1) The x-intercept is 14, 02. Factoring shows the (4, 0) y-intercepts are y 4 and y 1. The axis of 10 10 symmetry is y 3 (6.25, 1.5) y 1.5 2 1.5, and substituting this value into the original equation gives (0, 4) x 6.25. The coordinates of the vertex are 16.25, 1.52. Using horizontal and vertical 10 boundary lines we find the domain for this relation is x 僆 3 6.25, q 2 and the range is y 僆 1q, q 2. The graph is shown.


x

䊳

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As with the vertical parabola, the equation of a horizontal parabola can be written as a transformation: x a1y k2 2 h by completing the square. Note that in this case, the vertical shift is k units opposite the sign, with a horizontal shift of h units in the same direction as the sign. EXAMPLE 2

䊳

Graphing a Horizontal Parabola by Completing the Square Graph by completing the square: x 2y2 8y 9, then state the domain and range.

Solution

䊳

Using the original equation, we note the graph will be a horizontal parabola opening to the left 1a 22 and have an x-intercept of 19, 02. Completing the square gives x 21y2 4y 42 9 8, so the equation of this parabola in shifted form is

y 10

(9, 0) 10

y 2

x 21y 22 1 2

The vertex is at 11, 22 and y 2 is the axis of symmetry. This means there are no y-intercepts, a fact that comes to light when we attempt to solve the equation after substituting 0 for x: 21y 22 2 1 0

(0, 1) 10

x

(1, 2)

(9, 4)

substitute 0 for x

1 1y 22 2 2

no real roots

Using symmetry, the point 19, 42 is also on the graph. After plotting these points we obtain the graph shown. From the graph, the domain is x 僆 1q, 1 4 and the range is y 僆⺢. Now try Exercises 19 through 36

䊳

As with the other relations graphed in this chapter, a horizontal parabola also fails the vertical line test and we must graph the relation in two pieces. Using the equation from Example 2 we have x 21y 22 2 1

Figure 8.38

x 1 21y 22

4.2

x1 1y 22 2 2 13.4

5.4

8.2

A. You’ve just seen how we can graph parabolas with a horizontal axis of symmetry

x1 y2 B 2 2

x1 y B 2

Y1 2

2

shifted form isolate y-term divide by 2

take square roots

solve for y

X1 X1 , Y2 2 B 2 B 2

The graph is given in Figure 8.38, and shows that (3, 1) is also a point on the graph.

B. The Focus-Directrix Form of the Equation of a Parabola As with the ellipse and hyperbola, many significant applications of the parabola rely on its analytical definition rather than its algebraic form. From the construction of radio telescopes to the manufacture of flashlights, the location of the focus of a parabola is

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critical. To understand these and other applications, we use the analytic definition of a parabola first introduced in Section 8.1. Definition of a Parabola

Given a fixed point f and fixed line D in the plane, a parabola is the set of all points (x, y) such that the distance from f to (x, y) is equal to the distance from line D to (x, y). The fixed point f is the focus of the parabola, and the fixed line is the directrix.

d1

(x, y)

f d2 Vertex D

d1 d2

WORTHY OF NOTE For the analytic parabola, we use p to designate the focus since c is so commonly used as the constant term in y ax2 bx c.

The general equation of a parabola can be obtained by combining this definition with the distance formula. With no loss of generality, we can assume the parabola shown in the definition box is oriented in the plane with the vertex at (0, 0) and the focus at (0, p). As the diagram in Figure 8.39 indicates, this gives the directrix an equation of y p with all points on D having coordinates of 1x, p2. Using d1 d2 the distance formula yields

Figure 8.39 y

1x 02 1y p2 1x x2 1y p2 2

2

2

2

2

d2

y p (0, 0) D

2

x y 2py p 0 y 2py p 2

P(x, y)

F

21x 02 2 1y p2 2 21x x2 2 1y p2 2 2

d1

(0, p)

2

x 2py 2py 2

(0, p)

x (x, p)

from the definition square both sides simplify; expand binomials subtract p 2 and y 2

x2 4py

isolate x 2

The resulting equation is called the focus-directrix form of a vertical parabola with center at (0, 0). If we had begun by orienting the parabola so it opened to the right, we would have obtained the equation of a horizontal parabola with center (0, 0): y2 4px. The Equation of a Parabola in Focus-Directrix Form Vertical Parabola

Horizontal Parabola

x 4py

y2 4px

2

focus (0, p), directrix: y p If p 7 0, opens upward. If p 6 0, opens downward.

focus at ( p, 0), directrix: x p If p 7 0, opens to the right. If p 6 0, opens to the left.

For a parabola, note there is only one second-degree term.

EXAMPLE 3

䊳

Locating the Focus and Directrix of a Parabola Find the vertex, focus, and directrix for the parabola defined by x2 12y. Then sketch the graph, including the focus and directrix.

Solution

䊳

Since the x-term is squared and no shifts have been applied, the graph will be a vertical parabola with a vertex of (0, 0). Use a direct comparison between the given

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equation and the focus-directrix form to determine the value of p: x2 12y T x2 4py

y 10

x0

4p 12 p 3

y3 (0, 0)

10

f

focus-directrix form

This shows:

D

(6, 3)

given equation

10 x (6, 3)

(0, 3)

10

Since p 3 1p 6 02, the parabola opens downward, with the focus at 10, 32 and directrix y 3. To complete the graph we need a few additional points. Since 36 is divisible by 12, we can use inputs of x 6 and x 6 3 162 2 62 36 4, giving the points 16, 32 and 16, 32. Note the axis of symmetry is x 0. The graph is shown. Now try Exercises 37 through 48

Figure 8.40 y 2p f d1 Vertex y p

P

(x, y) d2 p x p D

EXAMPLE 4

䊳

䊳

As an alternative to calculating additional points to sketch the graph, we can use what is called the focal chord of the parabola. Similar to the ellipse and hyperbola, the focal chord is the line segment that contains the focus, is parallel to the directrix, and has its endpoints on the graph. Using the definition of a parabola and the diagram in Figure 8.40, we note the vertical distance from (x, y) to the directrix y p is 2p. Since d1 d2 a line segment parallel to the directrix from the focus to the graph will also have a length of 冟2p冟, and the focal chord of any parabola has a total length of 冟4p冟. Note that in Example 3, the points we happened to choose were actually the endpoints of the focal chord. Finally, if the vertex of a vertical parabola is shifted to (h, k), the equation will have the form 1x h2 2 4p1y k2. As with the other conic sections, both the horizontal and vertical shifts are “opposite the sign.” Locating the Focus and Directrix of a Parabola Find the vertex, focus, and directrix for the parabola whose equation is given, then sketch the graph, including the focus, focal chord, and directrix: x2 6x 12y 15 0.

Solution

䊳

Since only the x-term is squared, the graph will be a vertical parabola. To find the end-behavior, vertex, focus, and directrix, we complete the square in x and use a direct comparison between the shifted form and the focus-directrix form: x2 6x 12y 15 0 x2 6x ___ 12y 15 x2 6x 9 12y 24 1x 32 2 121y 22

y 10

x3 y5 (3, 2)

(3, 1)

(9, 1)

10

10

(3, 1)

10

x

given equation complete the square in x add 9 factor

Notice the parabola has been shifted 3 units right and 2 up, so all features of the parabola will likewise be shifted. Since we have 4p 12 (the coefficient of the linear term), we know p 3 1p 6 02 and the parabola opens downward. If the parabola were in standard position, the vertex would be at (0, 0), the focus at (0, 3) and the directrix a horizontal line at y 3. But since the parabola is shifted 3 right and 2 up, we add 3 to all x-values and 2 to all y-values to locate the features of the shifted parabola. The vertex is at 10 3, 0 22 13, 22. The focus is 10 3, 3 22 13, 12 and the directrix is y 3 2 5. Finally, the horizontal distance from the focus to the graph is 冟2p冟 6 units (since 冟4p冟 12), giving us the additional points 13, 12 and 19, 12 as endpoints of the focal chord. The graph is shown. Now try Exercises 49 through 60

䊳

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In many cases, we need to construct the equation of the parabola when only partial information in known, as illustrated in Example 5. EXAMPLE 5

䊳

Constructing the Equation of a Parabola Find the equation of the parabola with vertex (4, 4) and focus (1, 4). Then graph the parabola using the equation and focal chord.

Solution

䊳

B. You’ve just seen how we can identify and use the focus-directrix form of the equation of a parabola

As the vertex and focus are on a horizontal line, we have a horizontal parabola with general equation 1y k2 2 4p1x h2 . The distance p from vertex to focus is 3 units, and with the focus to the left of the vertex, the parabola opens left so p 3. Using the focal chord, the vertical distance from (1, 4) to the graph is 冟2p冟冟2132 冟 6, giving points (1, 10) and (1, 2). The vertex is shifted 4 units right and 4 units up from (0, 0), showing h 4 and k 4, and the equation of the parabola must be 1y 422 121x 42 , with directrix x 7. The graph is shown.

y 14

10

10

x

6


䊳

C. Nonlinear Systems and the Conic Sections Similar to our work with nonlinear systems in Section 6.3, the graphing, substitution, or elimination method can still be used when the system involves a conic section. When both equations in the system have at least one second degree term, it is generally easier to use the elimination method. EXAMPLE 6

䊳

Solving a System of Nonlinear Equations Solve the system using elimination: e

Solution

y 5

(1, 3)

䊳

The first equation represents a vertical and central hyperbola, while the second represents a horizontal and central ellipse. After writing the system with the x- and 5x2 2y2 13 y-terms in the same order, we obtain e . Using 2R1 R2 will 3x2 4y2 39 eliminate the y-term. 10x2 4y2 26 3x2 4y2 39 13x2 0 13 x2 1 x 1 or x 1 e

5x2 2y2 13 (1, 3)

5

5

x

3x 4y2 39 (1, 3) 2

(1, 3) 5

2y2 5x2 13 3x2 4y2 39

2R1 R2 sum divide by 13 square root property

Substituting x 1 and x 1 into the second equation we obtain 3112 2 4y2 39 3 4y2 39 4y2 36 y2 9 y 3 or y 3

3112 2 4y2 39 3 4y2 39 4y2 36 y2 9 y 3 or y 3

Since 1 and 1 each generated two outputs, there are a total of four ordered pair solutions: (1, 3), (1, 3), (1, 3), and (1, 3). The graph is shown and supports our results. Now try Exercises 77 through 82

䊳

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Nonlinear systems like the one in Example 6 can also be solved by graphing the system on a graphing calculator and looking for points of intersection. Solving each equation for y yields the following results. 2y2 5x2 13 2y2 13 5x2 y2

13 5x2 2

y Y1

13 5x2 B 2

3x2 4y2 39

original equation

4y2 39 3x2

isolate y-term

y2

divide by coefficient

y

take square roots

13 5X2 13 5X2 , Y2 B 2 B 2

Y3

39 3x2 4 39 3x2 B 4

39 3X2 39 3X2 , Y4 B 4 B 4

The equations and graphs are shown in Figures 8.41 and 8.42, and verify that the point (1, 3) is a solution to the system. Using symmetry, the other solutions are (1, 3), (1, 3), and (1, 3). See Exercises 83 through 86. Figure 8.42

Figure 8.41

6.2

9.4

C. You’ve just seen how we can solve nonlinear systems involving the conic sections

9.4

6.2

D. Application of the Analytic Parabola Here is just one of the many ways the analytic definition of a parabola can be applied. There are several others in the Exercise Set. Many applications use the parabolic property that light or sound coming in parallel to the axis of a parabola will be reflected to the focus. EXAMPLE 7

䊳

Locating the Focus of a Parabolic Receiver The diagram shows the cross section of a radio antenna dish. Engineers have located a point on the cross section that is 0.75 m above and 6 m to the right of the vertex. At what coordinates should the engineers build the focus of the antenna?

Solution

䊳

Focus

(6, 0.75) (0, 0)

By inspection we see this is a vertical parabola with center at (0, 0). This means its equation must be of the form x2 4py. Because we know (6, 0.75) is a point on this graph, we can substitute (6, 0.75) in this equation and solve for p: x2 4py 162 2 4p10.752 36 3p p 12

equation for vertical parabola, vertex at (0, 0) substitute 6 for x and 0.75 for y simplify result

With p 12, we see that the focus must be located at (0, 12), or 12 m directly above the vertex. Now try Exercises 89 through 96 D. You’ve just seen how we can solve applications of the analytic parabola

䊳

Note that in many cases, the focus of a parabolic dish may be above the rim of the dish.

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8.4 EXERCISES 䊳



1. The equation x ay2 by c is that of a(n) parabola, opening to the if a 7 0 and to the left if .

2. If point P is on the graph of a parabola with directrix D, the distance from P to line D is equal to the distance between P and the .

3. Given y2 4px, the focus is at equation of the directrix is

4. Given x2 16y, the value of p is the coordinates of the focus are

and the .

5. Discuss/Explain how to find the vertex, directrix, and focus from the equation 1x h2 2 4p1y k2.

䊳

and .

6. If a horizontal parabola has a vertex of 12, 3) with a 7 0, what can you say about the y-intercepts? Will the graph always have an x-intercept? Explain.


Find the x- and y-intercepts (if they exist) and the vertex of the parabola. Then sketch the graph by using symmetry and a few additional points or completing the square and shifting a parent function. Scale the axes as needed to comfortably fit the graph and state the domain and range.

7. y x2 2x 3

8. y x2 6x 5

9. y 2x2 8x 10 10. y 3x2 12x 15 11. y 2x2 5x 7

12. y 2x2 7x 3

Find the x- and y-intercepts (if they exist) and the vertex of the graph. Then sketch the graph using symmetry and a few additional points (scale the axes as needed). Finally, state the domain and range of the relation.

31. y 1x 22 2 3

32. y 1x 22 2 4

35. x 21y 32 2 1

36. x 21y 32 2 5

33. x 1y 32 2 2

34. x 1y 12 2 4

Find the vertex, focus, and directrix for the parabolas defined by the equations given, then use this information to sketch a complete graph (illustrate and name these features). For Exercises 49 to 60, also include the focal chord.

37. x2 8y

38. x2 16y

39. x2 24y

40. x2 20y

41. x2 6y

42. x2 18y

13. x y2 2y 3

14. x y2 4y 12

43. y2 4x

44. y2 12x

15. x y2 6y 7

16. x y2 8y 12

45. y2 18x

46. y2 20x

17. x y2 8y 16

18. x y2 6y 9

47. y2 10x

48. y2 14x

Sketch by completing the square and using symmetry and shifts of a basic function. Be sure to find the x- and y-intercepts (if they exist) and the vertex of the graph, then state the domain and range of the relation.

49. x2 8x 8y 16 0 50. x2 10x 12y 25 0 51. x2 14x 24y 1 0

19. x y2 6y

20. x y2 8y

52. x2 10x 12y 1 0

21. x y2 4

22. x y2 9

53. 3x2 24x 12y 12 0

23. x y2 2y 1

24. x y2 4y 4

54. 2x2 8x 16y 24 0

25. x y2 y 6

26. x y2 4y 5

55. y2 12y 20x 36 0

27. x y2 10y 4

28. x y2 12y 5

56. y2 6y 16x 9 0

29. x 3 8y 2y2

30. x 2 12y 3y2

57. y2 6y 4x 1 0

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58. y2 2y 8x 9 0

75.

76.

y

y 2

6

59. 2y2 20y 8x 2 0

(4, 0) 4

4

60. 3y 18y 12x 3 0 2

(4, 2)

2

For Exercises 61–72, find the equation of the parabola in standard form that satisfies the conditions given.

4

2 2

4

6

8

4 2

x y 6

2

61. focus: (0, 2) directrix: y 2


63. focus: (4, 0) directrix: x 4




77. e

x2 y2 25 2x2 3y2 5

78. e

y2 x2 12 x2 y2 20

67. vertex: (2, 2) focus: (1, 2)

68. vertex: (4, 1) focus: (1, 1)

79. e

80. e

69. vertex: (4, 7) focus: (4, 4)

70. vertex: (3, 4) focus: (3, 1)

x2 y 4 y2 x2 16

2x2 3y2 38 x2 5y 35

5x2 2y2 75 2x2 3y2 125

82. e



81. e

3x2 7y2 20 4x2 9y2 45

Solve using substitution or elimination, then verify your solutions by graphing the system on a graphing calculator.

Solve the following systems using a graphing calculator. Round approximate solutions to three decimal places.

For the graphs in Exercises 73–76, only two of the following four features are displayed: vertex, focus, directrix, and endpoints of the focal chord. Find the remaining two features and the equation of the parabola.

73.

74.

y 4

(1, 4) 6

y5

4 2

2

4

6

x

2

2 4

x 3

䊳

(1, 4)

4

1x 22 2 y2 20 83. • x2 y8 4 84. e

y

2

(2, 2)

2

4

6

10 x

2

(2, 2) 6

2

8

10 x

2

85. e 86. e

4x2 1y 122 2 441 x2 1y 122 2 1764

1x 102 2 1y 102 2 144 1x 42 2 y2 144

31x 242 2 y2 196 4x 4y 31


87. The area of a right parabolic segment: A ⴝ 23 ab

y

A right parabolic segment is that part of a parabola formed by a line perpendicular to its axis, which cuts the parabola. The area of this segment is given by the formula shown, where b is the length of the chord cutting the parabola and a is the perpendicular distance from the vertex to this chord. What is the area of the parabolic segment shown in the figure? 88. The arc length of a right parabolic segment: b2 4a ⴙ 2b2 ⴙ 16a2 1 2b2 ⴙ 16a2 ⴙ lna b 2 8a b Although a fairly simple concept, finding the length of the parabolic arc traversed by a projectile requires a good deal of computation. To find the length of the arc ABC shown, we use the formula given where a is the maximum height attained by the projectile, b is the horizontal distance it traveled, and “ln” represents the natural log function. Suppose a baseball thrown from centerfield reaches a maximum height of 20 ft and traverses an arc length of 340 ft. Will the ball reach the catcher 310 ft away without bouncing?

10 8 (3, 4) 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

B a A

C b

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APPLICATIONS

89. Parabolic car headlights: The cross section of a typical car headlight can be modeled by an equation similar to 25x 16y2, where x and y are in inches and x 僆 30, 4 4 . Use this information to graph the relation for the indicated domain. 90. Parabolic flashlights: The cross section of a typical flashlight reflector can be modeled by an equation similar to 4x y2, where x and y are in centimeters and x 僆 3 0, 2.254 . Use this information to graph the relation for the indicated domain. 91. Parabolic sound receivers: Sound technicians at professional sports events often use parabolic receivers as they move along the sidelines. If a two-dimensional cross section of the receiver is modeled by the equation y2 54x, and is 36 in. in diameter, how deep is the parabolic receiver? What is the location of the focus? [Hint: Graph the parabola on the coordinate grid (scale the axes).]

Exercise 91 y

x

92. Parabolic sound receivers: Private investigators will often use a smaller and less expensive parabolic receiver (see Exercise 91) to gather information for their clients. If a two-dimensional cross section of the receiver is modeled by the equation y2 24x, and the receiver is 12 in. in diameter, how deep is the parabolic dish? What is the location of the focus? 93. Parabolic radio wave receivers: The program known as S.E.T.I. (Search for Extra-Terrestrial Intelligence) involves a group of scientists using radio telescopes to look for radio signals from possible intelligent species in outer space. The radio telescopes are actually parabolic dishes that vary in size from a few feet to hundreds of feet in diameter. If a particular radio telescope is 100 ft in diameter 䊳

753


and has a cross section modeled by the equation x2 167y, how deep is the parabolic dish? What is the location of the focus? [Hint: Graph the parabola on the coordinate grid (scale the axes).]

y

94. Solar furnace: x Another form of technology that uses a parabolic dish is called a solar furnace. In general, the rays of the Sun are reflected by the dish and concentrated at the focus, producing extremely high temperatures. Suppose the dish of one of these parabolic reflectors has a 30-ft diameter and a cross section modeled by the equation x2 50y. How deep is the parabolic dish? What is the location of the focus? 95. Commercial flashlights: The reflector of a large, commercial flashlight has the shape of a parabolic dish, with a diameter of 10 cm and a depth of 5 cm. What equation will the engineers and technicians use for the manufacture of the dish? How far from the vertex (the lowest point of the dish) will the bulb be placed? (Hint: Analyze the information using a coordinate system.) 96. Industrial spotlights: The reflector of an industrial spotlight has the shape of a parabolic dish with a diameter of 120 cm. What is the depth of the dish if the correct placement of the bulb is 11.25 cm above the vertex (the lowest point of the dish)? What equation will the engineers and technicians use for the manufacture of the dish? (Hint: Analyze the information using a coordinate system.)


97. In a study of quadratic graphs from the equation y ax2 bx c, no mention is made of a parabola’s focus and directrix. Generally, when a 1, the focus of a parabola is very near its vertex. Complete the square of the function y 2x2 8x and write the result in the form 1x h2 2 4p1y k2 . What is the value of p? What are the coordinates of the vertex?

98. Like the ellipse and hyperbola, the focal chord of a parabola (also called the latus rectum) can be used to help sketch its graph. From our earlier work, we know the endpoints of the focal chord are 2p units from the focus. Write the equation 12y 15 x2 6x in the form 4p1y k2 1x h2 2, and use the endpoints of the focal chord to help graph the parabola.

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8–48



99. (6.2) Construct a system of three equations in three variables using the equation y ax2 bx c and the points 13, 32, (0, 6), and 11, 12. Then use a matrix equation to find the equation of the parabola containing these points. 100. (3.1/4.2) Find all roots (real and complex) of the equation x6 64 0. (Hint: Begin by factoring as the difference of two perfect squares.) 101. (2.1) What are the characteristics of an even function? What are the characteristics of an odd function?

102. (4.2/4.3) Use the function f 1x2 x5 2x4 17x3 34x2 18x 36 to comment and give illustrations of the tools available for working with polynomials: (a) synthetic division, (b) rational roots theorem, (c) the remainder and factor theorems, (d) the tests for x 1 and x 1, (e) the upper/lower bounds property, (f) Descartes’ rule of signs, and (g) roots of multiplicity (bounces, crosses, alternating intervals).

MAKING CONNECTIONS Making Connections: Graphically, Symbolically, Numerically, and Verbally Eight graphs (a) through (h) are given. Match the characteristics shown in 1 through 16 to one of the eight graphs. y

(a)

y

(b)

5

5

5

5 x

5

5 x

y

5

5 x

5

5 x

5

1. ____ 1x 12 2 1y 12 2 16 2. ____ y

5

5 x

1 1x 12 2 4

3. ____ foci at 11, 1 252

y

(h)

5

5

5 x

5

y

(g)

5

5

5

5

y

(f)

5

y

(d)

5

5

5

(e)

y

(c)

5

5 x

5

5

5

5 x

5

9. ____ vertices at (3, 1) and (5, 1) 10. ____ 41y 12 2 1x 12 2 16 11. ____ center at (0, 2)

4. ____ transverse axis y 1

12. ____ focus at (1, 1)

1 5. ____ x 1y 12 2 3 2

13. ____ 41x 22 2 1y 12 2 16

7. ____ 41x 12 2 1y 12 2 16

15. ____ axis of symmetry: y 1

6. ____ domain: x 僆 33, 5 4 , range: y 僆 3 3, 54 8. ____ x 1y 22 9 2

2

14. ____ 1x 12 2 41y 12 2 16 16. ____ domain: x 僆 34, 04 , range: y 僆 35, 34

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755


A Brief Introduction to Analytical Geometry

KEY CONCEPTS • The midpoint and distance formulas play important roles in the study of analytical geometry: x2 x1 y2 y1 midpoint: 1x, y2 a , b distance: d 21x2 x1 2 2 1y2 y1 2 2 2 2 • The perpendicular distance from a point to a line is the length of the line segment perpendicular to the given line with the given point and the point of intersection as endpoints. • Using these tools, we can verify or construct relationships between points, lines, and curves in the plane; verify properties of geometric figures; prove theorems from Euclidean geometry; and construct relationships that define the conic sections. EXERCISES 1. Verify the closed figure with vertices (3, 4), (5, 4), (3, 6), and (5, 2) is a square. 2. Find the equation of the circle that circumscribes the square in Exercise 1. 3. A theorem from Euclidean geometry states: If any two points are equidistant from the endpoints of a line segment, they are on the perpendicular bisector of the segment. Determine if the line through (3, 6) and (6, 9) is the perpendicular bisector of the segment through (5, 2) and (5, 4). 4. Four points are given here. Verify that the distance from each point to the line y 1 is the same as the distance from the given point to the fixed point (0, 1): (6, 9), (2, 1), (4, 4), and (8, 16).

SECTION 8.2

The Circle and the Ellipse

KEY CONCEPTS • The equation of a circle centered at (h, k) with radius r is 1x h2 2 1y k2 2 r2. 1x h2 2 1y k2 2 1, showing the horizontal and • Dividing both sides by r2, we obtain the standard form r2 r2 vertical distance from center to graph is r. 1x h2 2 1y k2 2 The equation of an ellipse in standard form is 1. The center of the ellipse is (h, k), with • a2 b2 horizontal distance a and vertical distance b from center to graph. y • Given two fixed points f1 and f2 in a plane (called the foci), an ellipse is the set of all (x, y) points (x, y) such that the distance from the first focus to (x, y), plus the distance from d d2 1 the second focus to (x, y), remains constant. (a, 0) (a, 0) x • For an ellipse, the distance from center to vertex is greater than the distance c from (c, 0) (c, 0) center to one focus. d1 d2 k • To find the foci of a horizontal ellipse, use: a2 b2 c2 (since a 7 c), or c2 |a2 b2 |. EXERCISES Sketch the graph of each equation in Exercises 5 through 9. 5. x2 y2 16 6. x2 4y2 36 7. 9x2 y2 18x 27 0 2 2 1x 32 1y 22 8. x2 y2 6x 4y 12 0 9. 1 16 9 10. Find the equation of the ellipse with minor axis of length 6 and foci at (4, 0) and (4, 0). Then graph the equation on a graphing calculator using a “friendly” window and use the TRACE feature to locate four additional points on the graph with coordinates that are rational.

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11. Find the equation of the ellipse with vertices at (a) (13, 0) and (13, 0), foci at (12, 0) and (12, 0); (b) foci at (0, 16) and (0, 16), major axis: 40 units. 12. Write the equation in standard form and sketch the graph, noting all of the characteristic features of the ellipse. 4x2 25y2 16x 50y 59 0

SECTION 8.3

The Hyperbola

KEY CONCEPTS

1x h2 2

1y k2 2

1. The center of the hyperbola a2 b2 is (h, k) with horizontal distance a from center to vertices, and vertical distance b from center to the midpoint of y the sides of the central rectangle. (x, y) • Given two fixed points f1 and f2 in a plane (called the foci), a hyperbola is the set of all d1 points (x, y) such that the distance from one focus to point (x, y), less the distance from d 2 (a, 0) (a, 0) the other focus to (x, y), remains a positive constant: |d1 d2 | k. (c, 0) (c, 0) x • For a hyperbola, the distance from center to one vertex is less than the distance from center to the focus c. • To find the foci of a hyperbola: c2 a2 b2 (since c 7 a). |d1 d2| k

• The equation of a horizontal hyperbola in standard form is

EXERCISES Sketch the graph of each equation in Exercises 13 through 17, indicating the center, vertices, and asymptotes. 1y 32 2 1x 22 2 1y 12 2 1x 22 2 13. 4y2 25x2 100 14. 15. 1 1 16 9 9 4 16. 9y2 x2 18y 72 0 17. x2 4y2 12x 8y 16 0 4 18. Find the equation of the hyperbola with vertices at (3, 0) and (3, 0), and asymptotes of y x. Then graph the 3 equation on a graphing calculator using a “friendly” window and use the TRACE feature to locate two additional points with rational coordinates. 19. Find the equation of the hyperbola with (a) vertices at (15, 0), foci at (17, 0), and (b) foci at (0, 5) with vertical dimension of central rectangle 8 units. 20. Write the equation in standard form and sketch the graph, noting all of the characteristic features of the hyperbola. 4x2 9y2 40x 36y 28 0

SECTION 8.4

The Analytic Parabola; More on Nonlinear Systems

KEY CONCEPTS • Horizontal parabolas have equations of the form x ay2 by c; a 0.

• A horizontal parabola will open to the right if a 7 0, and to the left if a 6 0. The axis of symmetry is y b or by completing the square and writing 2a 2 the equation in shifted form: 1x h2 a1y k2 . • Given a fixed point f (called the focus) and fixed line D (called the directrix) in the plane, a parabola is the set of all points (x, y) such that the distance from f to (x, y) is equal to the distance from (x, y) to line D. The equation x2 4py describes a vertical parabola, opening upward if p 7 0, and opening • downward if p 6 0. • The equation y2 4px describes a horizontal parabola, opening to the right if p 7 0, and opening to the left if p 6 0. with the vertex (h, k) found by evaluating at y

b , 2a

y

d 1 d2

f

d1

(x, y) d2

Vertex

x D

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Practice Test

757

• p is the distance from the vertex to the focus (or from the vertex to the directrix). • The focal chord of a parabola is a line segment that contains the focus and is parallel the directrix, with its

endpoints on the graph. It has a total length of 冟4p冟, meaning the distance from the focus to a point on the graph (as described) is 冟2p冟. It is commonly used to assist in drawing a graph of the parabola.

EXERCISES For Exercises 21 and 22, find the vertex and x- and y-intercepts if they exist. Then sketch the graph using symmetry and a few points or by completing the square and shifting a parent function. 21. x y2 4 22. x y2 y 6 For Exercises 23 and 24, find the vertex, focus, and directrix for each parabola. Then sketch the graph using this information and the focal chord. Also graph the directrix. 23. x2 20y 24. x2 8x 8y 16 0 25. Identify the conic sections in the system, then solve. Check solutions using the intersection-of-graphs method and a graphing calculator. e

1x 72 2 42 20 y2 7 x

PRACTICE TEST By inspection only (no graphing, completing the square, etc.), match each equation to its correct description. 1. x y 6x 4y 9 0 2

2

2. 4y2 x2 4x 8y 20 0 3. y x2 4x 20 0 4. x2 4y2 4x 12y 20 0 a. Parabola b. Hyperbola c. Circle

d. Ellipse

Graph each conic section, and label the center, vertices, foci, focal chords, asymptotes, and other important features where applicable. 5. 1x 42 2 1y 32 2 9 1x 22 2 1y 32 2 1 6. 16 1 1x 32 2 1y 42 2 1 7. 9 4

8. x2 y2 10x 4y 4 0 9. 9x2 4y2 18x 24y 9 0 10. 9x2 4y2 18x 24y 63 0 11. x 1y 32 2 2

12. y2 6y 12x 15 0

Solve each nonlinear system using any method. 2y2 x2 4 4x2 y2 16 b. e 2 yx2 x y2 8 14. A support bracket on the frame of a large ship is a steel right triangle with a hypotenuse of 25 ft and a perimeter of 60 ft. Find the lengths of the other sides using a system of nonlinear equations.

13. a. e

15. Find an equation for the circle whose center is at (2, 5) and whose graph goes through the point (0, 3). 16. Find the equation of the ellipse (in standard form) with vertices at (4, 0) and (4, 0) with foci located at (2, 0) and (2, 0). Then use a graphing calculator to determine where this ellipse and the circle x2 y2 13 intersect. 17. The orbit of Mars around the Sun is elliptical, with the Sun at one focus. When the orbit is expressed as a central ellipse on the coordinate grid, its equation y2 x2 1, with a and b in is 1141.652 2 1141.032 2 millions of miles. Use this information to find the aphelion of Mars (distance from the Sun at its farthest point), and the perihelion of Mars (distance from the Sun at its closest point).

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Determine the equation of each relation and state its domain and range. For the parabola and the ellipse, also give the location of the foci. 18.

19.

y 5 4 3 2 1 54321 1 2 3 4 5

(4, 1) 1 2 3 4 5 x

(1, 4)

20.

y 10 8 6 4 2

108642 2 4 6 8 10

y

(1, 6)

(3, 6) (6, 1)

(6, 0)

2 4 6 8 10 x

10 8 6 4 2

(0, 0)

108642 2 4 6 (3, 6)8 10

(1, 4)

2 4 6 8 10 x

CALCULATOR EXPLORATION AND DISCOVERY Elongation and Eccentricity Technically speaking, a circle is an ellipse with both foci at the center. As the distance between foci increases, the ellipse becomes more elongated. We saw other instances of elongation in stretches and compressions of parabolic graphs, and in hyperbolic graphs where the asymptotic slopes varied depending on the values a and b. The measure c used to quantify this elongation is called the eccentricity e, and is determined by the ratio e . For this Exploration and a Discovery, we’ll use the repeat graph feature of a graphing calculator to explore the eccentricity of the graph of a conic. The “repeat graph” feature enables you to graph a family of curves by enclosing changes in a parameter in braces “{ }.” For instance, entering 52, 1, 0, 1, 26 X 3 as Y1 on the Y= screen will automatically graph these five lines: y 2x 3, y x 3, y 3, y x 3, and y 2x 3. We’ll use this feature to graph a family of ellipses, observing the result and calculating the eccentricity for each y2 x2 curve in the family. The standard form is 2 2 1, which we’ll solve for y and enter as Y1 and Y2. After a b x2 simplification the result is y b 1 2 , but for this investigation we’ll use the constant b 2 and vary the B a x2 parameter a using the values a 2, 4, 6, and 8. The result is y 2 1 . Note from Figure 8.43 that B 54, 16, 36, 646 we’ve set Y2 Y1 to graph the lower half of the ellipse. Using the “friendly window” shown (Figure 8.44) gives the result shown in Figure 8.45, where we see the ellipse is increasingly elongated in the horizontal direction (note when a 2 the result is a circle since a b). Figure 8.45 Figure 8.43

Figure 8.44

6.2

9.4

9.4

6.2

Using a 2, 4, 6, and 8 with b 2 in the foci formula c 2a2 b2 gives c 0, 2 13, 4 12, and 2 115, respectively, with these eccentricities:

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Strengthening Core Skills

759

2115 0 2 13 412 e , , . While difficult to see in radical form, we find that the eccentricity of an ellipse always , and 2 4 6 8 satisfies the inequality 0 6 e 6 1 (excluding the circle ellipse case). To two decimal places, the values are e 0, 0.87, 0.94, and 0.97, respectively. c As a final note, it’s interesting how the e definition of eccentricity relates to our everyday use of the word a “eccentric.” A normal or “noneccentric” person is thought to be well-rounded, and sure enough e 0 produces a well-rounded figure—a circle. A person who is highly eccentric is thought to be far from the norm, deviating greatly from the center, and greater values of e produce very elongated ellipses. Exercise 1: Perform a similar exploration using a family of hyperbolas. What do you notice about the eccentricity? Exercise 2: Perform a similar exploration using a family of parabolas. What do you notice about the eccentricity?

STRENGTHENING CORE SKILLS Ellipses and Hyperbolas with Rational/Irrational Values of a and b Using the process known as completing the square, we were able to convert from the polynomial form of a conic section to the standard form. However, for some equations, values of a and b are somewhat difficult to identify, since the coefficients are not factors. Consider the equation 20x2 120x 27y2 54y 192 0, the equation of an ellipse. 20x2 120x 27y2 54y 192 0 201x2 6x ___ 2 271y2 2y ___ 2 192 201x2 6x 92 271y2 2y 12 192 27 180 201x 32 2 271y 12 2 15 41x 32 2 91y 12 2 1 3 5

original equation subtract 192 complete the square in x and y factor and simplify standard form

Unfortunately, we cannot easily identify the values of a and b, since the coefficients of each binomial square are not “1.” In these cases, we can write the equation in standard form by using a simple property of fractions—the numerator and denominator of any fraction can be divided by the same quantity to obtain an equivalent fraction. 1y 12 2 1x 32 2 1. Although the result may look odd, it can nevertheless be applied here, giving a result of 3 5 4 9 We can now identify a and b by writing these denominators in squared form, which gives the following expression: 1x 32 2 1y 12 2 15 13 1. The values of a and b are now easily seen as a ⬇ 0.866 and b ⬇ 0.745. 2 2 2 3 13 15 a b a b 2 3 Use this idea to complete the following exercises. Exercise 1: Write the equation in standard form, then identify the values of a and b and use them to graph the ellipse. 41x 32 2 49

251y 12 2 36

1

Exercise 2: Write the equation in standard form, then identify the values of a and b and use them to graph the hyperbola. 91x 32 2 80

41y 12 2 81

1

Exercise 3: Identify the values of a and b by writing the equation 100x2 400x 18y2 108y 230 0 in standard form. Exercise 4: Identify the values of a and b by writing the equation 28x2 56x 48y2 192y 195 0 in standard form.

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CUMULATIVE REVIEW CHAPTERS R–8 Solve each equation. 1. x ⫺ 2x ⫹ 4x ⫺ 8 ⫽ 0 2. 2冟n ⫹ 4冟 ⫹ 3 ⫽ 13 3

2

3. 2x ⫺ 3 ⫹ 5 ⫽ x 3

4. x2 ⫹ 8 ⫽ 0 5. x2 ⫺ 6x ⫹ 13 ⫽ 0 1 6. 4 # 2x⫹1 ⫽ 8 7. 3x⫺2 ⫽ 7 8. ln x ⫽ 2

Solve each system of equations with a graphing calculator. Use a matrix equation for Exercise 22, and the intersection-of-graphs method for Exercise 23.

9. log x ⫹ log 1x ⫺ 32 ⫽ 1 Graph each relation. Include vertices, x- and y-intercepts, asymptotes, and other features. 2 10. y ⫽ x ⫹ 2 3

11. y ⫽ 冟 x ⫺ 2 冟 ⫹ 3

1 ⫹2 12. y ⫽ x⫺1

13. y ⫽ 1x ⫺ 3 ⫹ 1

14. a. g1x2 ⫽ 1x ⫺ 321x ⫹ 121x ⫹ 42 b. f 1x2 ⫽ x4 ⫹ x3 ⫺ 13x2 ⫺ x ⫹ 12 x⫺2 x2 ⫺ 9 17. f 1x2 ⫽ log2 1x ⫹ 12 15. h1x2 ⫽

y 21. Determine the following for 10 the indicated graph (write all 8 6 answers in interval notation): (⫺1, 4) 4 2 (a) the domain, (b) the range, (⫺4, 0) (2, 0) x ⫺10⫺8⫺6⫺4⫺2 ⫺2 2 4 6 8 10 (c) interval(s) where f(x) is ⫺4 increasing or decreasing, ⫺6 ⫺8 (d) interval(s) where f(x) is ⫺10 constant, (e) location of any maximum or minimum value(s), (f) interval(s) where f(x) is positive, and (g) interval(s) where f(x) is negative.

16. q1x2 ⫽ 2x ⫹ 3 18. x ⫽ y2 ⫹ 4y ⫹ 7

19. x2 ⫹ y2 ⫹ 10x ⫺ 4y ⫹ 20 ⫽ 0 20. 41x ⫺ 12 2 ⫺ 361y ⫹ 22 2 ⫽ 144

4x ⫹ 3y ⫽ 13 22. • ⫺9y ⫹ 5z ⫽ 19 x ⫺ 4z ⫽ ⫺4

23. e

x2 ⫹ y2 ⫽ 25 64x2 ⫹ 12y2 ⫽ 768

24. If a person invests $5000 at 9% compounded quarterly, how long would it take for the money to grow to $12,000? 25. A radiator contains 10 L of liquid that is 40% antifreeze. How much should be drained off and replaced with pure antifreeze for a 60% mixture? Solve each equation using a graphing calculator. 26.

1 3 1 x ⫺ 4x ⫹ 3 ⫽ x2 ⫺ 5 8 4

27. 0x ⫹ 4 0 ⫽ 8 ⫺ 0x 0 28. e2x ⫺ 3ex ⫽ 4 29. 3x⫺1 ⫽ 22⫺x 30.

x⫹3 ⱖ3 x⫺4

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CHAPTER CONNECTIONS

Additional Topics in Algebra CHAPTER OUTLINE 9.1 Sequences and Series 762 9.2 Arithmetic Sequences 773 9.3 Geometric Sequences 782 9.4 Mathematical Induction 796 9.5 Counting Techniques 804

For a corporation of any size, decisions made by upper management often depend on a large number of factors, with the desired outcome attainable in many different ways. For instance, consider a legal firm that specializes in family law, with a support staff of 15 employees—6 paralegals and 9 legal assistants. Due to recent changes in the law, the firm wants to send some combination of five support staff to a conference dedicated to the new changes. In Chapter 9, we’ll see how counting techniques and probability can be used to determine the various ways such a group can be randomly formed, even if certain constraints are imposed. This application appears as Exercise 34 in Section 9.6.

9.6 Introduction to Probability 816


9.7 The Binomial Theorem 829

䊳

䊳

䊳

䊳

Determining the Effects of Inflation (Section 9.1, Exercise 86) Calculating Possible Movements of a Computer Animation (Section 9.2, Exercise 77) Counting the Number of Possible Area Codes and Phone Numbers (Section 9.5, Exercises 84 and 85) Tracking and Improving Customer Service Using Probability (Section 9.6, Exercise 53) 761

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9.1

Sequences and Series

LEARNING OBJECTIVES

A sequence can be thought of as a pattern of numbers listed in a prescribed order. A series is the sum of the numbers in a sequence. Sequences and series come in countless varieties, and we’ll introduce some general forms here. In following sections we’ll focus on two special types: arithmetic and geometric sequences. These are used in a number of different fields, with a wide variety of significant applications.


A. Write out the terms of a

B.

C. D.

E.

sequence given the general or n th term Work with recursive sequences and sequences involving a factorial Find the partial sum of a series Use summation notation to write and evaluate series Use sequences to solve applications

WORTHY OF NOTE Sequences can actually start with any natural number. For instance, 2 the sequence an ⫽ must start n⫺1 at n ⫽ 2 to avoid division by zero. In addition, we will sometimes use a0 to indicate a preliminary or inaugural element, as in a0 ⫽ $10,000 for the amount of money initially held, prior to investing it.

EXAMPLE 1A

䊳

A. Finding the Terms of a Sequence Given the General Term Suppose a person had $10,000 to invest, and decided to place the money in government bonds that guarantee an annual return of 7%. From our work in Chapter 5, we know the amount of money in the account after x years can be modeled by the function f 1x2 ⫽ 10,00011.072 x. If you reinvest your earnings each year, the amount in the account would be (rounded to the nearest dollar): Year: Value:

f (2)

f (3)

f (4)

T

T

T

T

f (5) p T

$10,700

$11,449

$12,250

$13,108

$14,026 p

Note the relationship (year, value) is a function that pairs 1 with $10,700, 2 with $11,449, 3 with $12,250, and so on. This is an example of a sequence. To distinguish sequences from other algebraic functions, we commonly name the functions a instead of f, use the variable n instead of x, and employ a subscript notation. The function f 1x2 ⫽ 10,00011.072 x would then be written an ⫽ 10,00011.072 n. Using this notation a1 ⫽ 10,700, a2 ⫽ 11,449, and so on. The values a1, a2, a3, a4, p are called the terms of the sequence. If the account were closed after a certain number of years (for example, after the fifth year) we have a finite sequence. If we let the investment grow indefinitely, the result is called an infinite sequence. The expression an that defines the sequence is called the general or nth term and the terms immediately preceding it are called the 1n ⫺ 12st term, the 1n ⫺ 22nd term, and so on. Sequences A finite sequence is a function an whose domain is the set of natural numbers from 1 to n. The terms of the sequence are labeled a1, a2, a3, p , ak, ak⫹1, p , an⫺1, an where ak represents an arbitrary “interior” term and an also represents the last term of the sequence. An infinite sequence is a function an whose domain is the set of all natural numbers. Computing Specified Terms of a Sequence For an ⫽

762

f (1)

n⫹1 , find a1, a3, a6, and a7. n2

1⫹1 ⫽2 12 6⫹1 7 a6 ⫽ ⫽ 2 36 6

3⫹1 4 ⫽ 2 9 3 7⫹1 8 a7 ⫽ ⫽ 2 49 7

Solution

䊳

a1 ⫽

a3 ⫽

EXAMPLE 1B

䊳

Computing the First k Terms of a Sequence

Find the first four terms of the sequence an ⫽ 1⫺12 n2n. Write the terms of the sequence as a list. 9–2

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Solution

䊳

WORTHY OF NOTE When the terms of a sequence alternate in sign as in Example 1B, we call it an alternating sequence.

a1 ⫽ 1⫺12 121 ⫽ ⫺2 a3 ⫽ 1⫺12 323 ⫽ ⫺8

a2 ⫽ 1⫺12 222 ⫽ 4 a4 ⫽ 1⫺12 424 ⫽ 16

The sequence can be written ⫺2, 4, ⫺8, 16, p , or more generally as ⫺2, 4, ⫺8, 16, p , 1⫺12 n2n, p to show how each term was generated. Now try Exercises 7 through 22

䊳

Much of the beauty and power of studying sequences comes from patterns detected within the sequence, or the ability to find a particular term quickly. Here, the calculator becomes an invaluable tool, aiding computations to be sure p but also enabling explorations not generally possible with paper and pencil alone. Most graphing calculators offer a “seq(” feature (often in a STAT or LIST menu), Figure 9.1 where the left parenthesis indicates we need to enter the following four items of information: the nth term formula defining the sequence, the variable in use, the starting value, and the ending value. The sequence generated can be seen on the home screen, or stored in a list for future use. In Figure 9.1, this feature was used to generate the first four terms of the sequence defined in Example 1(B) using the keystrokes 2nd STAT (LIST) (OPS) 5:seq(. In addition to the seq( feature, most calculators Figure 9.2A have a sequence MODE , which is especially useful when working with several sequences simultaneously. In sequence MODE (Figure 9.2A), the Y= screen presents a very different look (Figure 9.2B), first of all showing that the minimum value of n is 1, then naming the functions u(n), v(n), and w(n) instead of Y1, Y2, and Y3. The entries following each, for example “u(nMin) ⫽,” are used in a study of the recursive sequences, which are covered later. Note that function names for the sequences (u, v, and w) are the 2nd function for the numbers 7, 8, and 9, respectively.

Figure 9.2B

EXAMPLE 2

763

䊳

Finding the Terms of a Sequence Using Technology Use a calculator in sequence MODE to a. Find the sixth term (as a fraction) for the sequence defined in Example 1A. b. Generate the first five terms for the sequence defined in Example 1B and store the results in a list.

Solution Figure 9.3

䊳

a. On the

screen, define the sequence u(n) n⫹1 as u1n2 ⫽ (Figure 9.3), leaving the n2 second line blank. Then go to the home screen and access the “seq(” feature as before, and supply the information required. Since we only want the sixth term (a6), we enter seq(u(n), n, 6, 6) 䉴Frac, and after pressing , we obtain 7 the expected result (Figure 9.4). 36 Y=

ENTER

Figure 9.4

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CHAPTER 9 Additional Topics in Algebra

Figure 9.6

b. For the first five terms, we change u(n) to u1n2 ⫽ 1⫺12 n # 2n then go to the home screen and use the “5:seq(” command once again, this time for terms one through five. Knowing that we want the results stored in a list, we follow the command with STO 2nd 1 (L1) (Figure 9.5). The list is shown in Figure 9.6.

A. You’ve just seen how we can write out the terms of a sequence given the general or nth term

Figure 9.5


䊳

B. Recursive Sequences and Factorial Notation Sometimes the formula defining a sequence uses the preceding term or terms to generate those that follow. These are called recursive sequences and are particularly useful in writing computer programs. Because of how they are defined, recursive sequences must give an inaugural term or seed element(s), to begin the recursion process. Perhaps the most famous recursive sequence is associated with the work of Leonardo of Pisa (A.D. 1180–1250), better known to history as Fibonacci. In the Fibonacci sequence, each successive term is the sum of the previous two, beginning with 1, 1, p .

EXAMPLE 3

䊳

Computing the Terms of a Recursive Sequence Write out the first eight terms of the Fibonacci sequence, which is defined by c1 ⫽ 1, c2 ⫽ 1, and cn ⫽ cn⫺1 ⫹ cn⫺2.

Solution WORTHY OF NOTE One application of the Fibonacci sequence involves the Fibonacci spiral, found in the growth of many ferns and the spiral shell of many mollusks.

䊳

The first two terms are given, so we begin with n ⫽ 3. c3 ⫽ c3⫺1 ⫹ c3⫺2 c4 ⫽ c4⫺1 ⫹ c4⫺2 c5 ⫽ c5⫺1 ⫹ c5⫺2 ⫽ c2 ⫹ c1 ⫽ c3 ⫹ c2 ⫽ c4 ⫹ c3 ⫽1⫹1 ⫽2⫹1 ⫽3⫹2 ⫽2 ⫽3 ⫽5 At this point we can simply use the fact that each successive term is simply the sum of the preceding two, and find that c6 ⫽ 3 ⫹ 5 ⫽ 8, c7 ⫽ 5 ⫹ 8 ⫽ 13, and c8 ⫽ 8 ⫹ 13 ⫽ 21. The first eight terms are 1, 1, 2, 3, 5, 8, 13, and 21. Now try Exercises 33 through 38

Figure 9.8

䊳

Since a recursive sequence is defined using a preceding term or terms, the first term(s) must be given in order to determine those that follow. To generate the sequence using technology, we enter these initial terms as “u1nMin2 ⫽” on the Y= screen, enclosed in braces. For the Fibonacci sequence Figure 9.7 from Example 3, we would enter u1nMin2 ⫽ 51, 16” as shown in Figure 9.7, with the result shown in Figure 9.8 (use the right arrow to view any remaining terms). Some sequences may involve the computation of a factorial, which is the product of a given natural number with all those that precede it. The expression 5! is read, “five factorial,” and is evaluated as: 5! ⫽ 5 # 4 # 3 # 2 # 1 ⫽ 120. Factorials For any natural number n,

n! ⫽ n # 1n ⫺ 12 # 1n ⫺ 22 # p # 3 # 2 # 1

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9–5


765

Rewriting a factorial in equivalent forms often makes it easier to simplify certain expressions. For example, we can rewrite 5! as 5 # 4! or 5! ⫽ 5 # 4 # 3!, and so on. Consider Example 4. EXAMPLE 4

䊳

Simplifying Expressions Using Factorial Notation Simplify by writing the numerator in an equivalent form. 9! 11! 6! a. b. c. 7! 8!2! 3!5!

Solution

䊳

a.

9! 9 # 8 # 7! ⫽ 7! 7! ⫽9#8 ⫽ 72

b.

11 # 10 # 9 # 8! 11! ⫽ 8!2! 8!2! 990 ⫽ 2 ⫽ 495

c.

6! 6 # 5! ⫽ 3!5! 3!5! 6 ⫽ 6 ⫽1


䊳

Most calculators have a factorial option or key. On many calculator models it is located on a submenu of the MATH key: MATH PRB 4: !. EXAMPLE 5

䊳

Computing a Specified Term from a Sequence Defined Using Factorials Find and simplify the third term of each sequence. a. an ⫽

Algebraic Solution

䊳

Technology Solution

䊳

n! 2n

3! 23 6 3 ⫽ ⫽ 8 4

a. a3 ⫽

For this exercise we enter 1⫺1 2n 12n ⫺ 1 2!

b. cn ⫽

1⫺12 n 12n ⫺ 12!

b. c3 ⫽

1⫺12 32132 ⫺ 1 4 !

n!

3

3! 1⫺12 3 5 # 4 # 3! 4 1⫺1215!2 ⫽ ⫽ 3! 3! ⫽ ⫺20

n! as u(n), and 2n

Figure 9.9, 9.10

as v(n). Once these functions n! have been defined on the Y= screen, we can evaluate them on the home screen as with other functions (Figure 9.9), or use the TABLE feature (Figure 9.10).

B. You’ve just seen how we can work with recursive sequences and sequences involving a factorial

Figure 9.11


䊳

C. Series and Partial Sums Sometimes the terms of a sequence are dictated by context rather than a formula. Consider the stacking of large pipes in a storage yard. If there are 10 pipes in the bottom row, then 9 pipes, then 8 (see Figure 9.11), how many pipes are in the stack if there is a single pipe at the top? The sequence generated is 10, 9, 8, p , 3, 2, 1 and to answer the question we would have to compute the sum of all terms in the sequence. When the terms of a finite sequence are added, the result is called a finite series.

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Finite Series Given the sequence a1, a2, a3, a4, p , an, the sum of the terms is called a finite series or partial sum and is denoted Sn: Sn ⫽ a1 ⫹ a2 ⫹ a3 ⫹ p ⫹ an⫺1 ⫹ an

EXAMPLE 6

䊳

Computing a Partial Sum Given an ⫽ 2n, find the value of

Solution

䊳

a. S4

b. S7.

Since we eventually need the sum of the first seven terms, begin by writing out these terms: 2, 4, 6, 8, 10, 12, and 14. a. S4 ⫽ a1 ⫹ a2 ⫹ a3 ⫹ a4 b. S7 ⫽ a1 ⫹ a2 ⫹ a3 ⫹ a4 ⫹ a5 ⫹ a6 ⫹ a7 ⫽2⫹4⫹6⫹8 ⫽ 2 ⫹ 4 ⫹ 6 ⫹ 8 ⫹ 10 ⫹ 12 ⫹ 14 ⫽ 20 ⫽ 56 Now try Exercises 51 through 56

Figure 9.12

䊳

Figure 9.13 There are several ways of computing a partial sum using technology, with the most common being (1) storing the sequence in a list then computing the sum of the list elements, or (2) computing the sum of a sequence directly on the home screen using the appropriate commands. On many calculators, the “sum(” option is in a MATH subSTAT menu, accessed using 2nd (LIST) (MATH) 5:sum(. For the sum in Example 6(b), we have u1n2 ⫽ 2n, with option (1) demonstrated in Figure 9.12 and option (2) in Figure 9.13.

C. You’ve just seen how we can find the partial sum of a series

D. Summation Notation When the general term of a sequence is known, the Greek letter sigma © can be used to write the related series as a formula. For instance, to indicate the sum of the first four term 4

of an ⫽ 3n ⫹ 2, we write

兺 13i ⫹ 22 with this notation indicating we are to compute

i⫽1

the sum of all terms generated as i cycles from 1 through 4. This result is called summation or sigma notation and the letter i is called the index of summation. The letters j, k, l, and m are also used as index numbers, and the summation need not start at 1. EXAMPLE 7

䊳

Computing a Partial Sum Compute each sum: 4

6 1 c. 1⫺12 kk2 i⫽1 j⫽1 j k⫽3 d. Check each sum using a graphing calculator.

a.

兺

5

13i ⫹ 22

b.

兺

兺

4

Solution

䊳

a.

兺 13i ⫹ 22 ⫽ 13 # 1 ⫹ 22 ⫹ 13 # 2 ⫹ 22 ⫹ 13 # 3 ⫹ 22 ⫹ 13 # 4 ⫹ 22

i⫽1 5

b.

⫽ 5 ⫹ 8 ⫹ 11 ⫹ 14 ⫽ 38

1

1

1

1

1

1

兺j ⫽1⫹2⫹3⫹4⫹5

j⫽1

⫽

60 30 20 15 12 137 ⫹ ⫹ ⫹ ⫹ ⫽ 60 60 60 60 60 60

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Section 9.1 Sequences and Series 6

c.

兺 1⫺12 k

k 2

767

⫽ 1⫺12 3 # 32 ⫹ 1⫺12 4 # 42 ⫹ 1⫺12 5 # 52 ⫹ 1⫺12 6 # 62

⫽ ⫺9 ⫹ 16 ⫹ 1⫺252 ⫹ 36 ⫽ 18 d. Begin by entering the functions in parts (a), (b), and (c) as u(n), v(n), and w(n) respectively on the Y= screen (Figure 9.14). Note that while different indices are used in this example, all are entered into the calculator using the variable n. The results are shown in Figures 9.15 and 9.16. k⫽3

Figure 9.14

Figure 9.16

Figure 9.15


䊳

If a definite pattern is noted in a given series expansion, this process can be reversed, with the expanded form being expressed in summation notation using the nth term. EXAMPLE 8

䊳

Writing a Sum in Sigma Notation Write each of the following sums in summation (sigma) notation. a. 1 ⫹ 3 ⫹ 5 ⫹ 7 ⫹ 9 b. 6 ⫹ 9 ⫹ 12 ⫹ 15 ⫹ p

Solution

䊳

a. The series has five terms and each term is an odd number, or 1 less than a 5

multiple of 2. The general term is an ⫽ 2n ⫺ 1, and the series is

兺 12n ⫺ 12.

n⫽1

WORTHY OF NOTE

b. The raised ellipsis “ p ” indicates the sum continues infinitely. Since the terms are multiples of 3, we identify the general term as an ⫽ 3n, while noting the series starts at n ⫽ 2 (instead of n ⫽ 1). Since the sum continues indefinitely, we use the infinity symbol q as the “ending” value in sigma notation. The q

By varying the function given and/or where the sum begins, more than one acceptable form is possible.

series is

兺 3n.

n⫽2

For Example 8(b) 13 ⫹ 3k2 k⫽1 also works. q

兺


䊳

Since the commutative and associative laws hold for the addition of real numbers, summations have the following properties: Properties of Summation Given any real number c and natural number n, n

(I)

兺 c ⫽ cn

i⫽1

If you add a constant c “n” times the result is cn. n

(II)

兺

n

cai ⫽ c

i⫽1

兺a

i

i⫽1

A constant can be factored out of a sum.

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768

9–8


n

(III)

n

n

兺 1a ⫾ b 2 ⫽ 兺 a ⫾ 兺 b i

i

i

i⫽1

i

i⫽1

i⫽1

A summation can be distributed to two (or more) sequences. m

(IV)

兺

n

i⫽1

n

兺

ai ⫹

ai ⫽

i⫽m⫹1

兺a; 1 ⱕ m 6 n i

i⫽1

A summation is cumulative and can be written as a sum of smaller parts. The verification of property II depends solely on the distributive property. n

Proof:

兺 ca ⫽ ca i

i⫽1

1

⫹ ca2 ⫹ ca3 ⫹ p ⫹ can

expand sum

⫽ c1a1 ⫹ a2 ⫹ a3 ⫹ p ⫹ an 2

factor out c

⫽c

write series in summation form

n

兺a

i

i⫽1

The verifications of properties III and IV simply use the commutative and associative properties. You are asked to prove property III in Exercise 94. EXAMPLE 9

䊳

Computing a Sum Using Summation Properties 4

Recompute the sum

兺 13i ⫹ 22 from Example 7(a) using summation properties.

i⫽1 4

Solution

䊳

兺

i⫽1

13i ⫹ 22 ⫽

4

兺

4

兺2

property III

兺i ⫹ 兺2

property II

3i ⫹

i⫽1 4

⫽3

i⫽1

i⫽1 4 i⫽1

⫽ 31102 ⫹ 2142 ⫽ 38

D. You’ve just seen how we can use summation notation to write and evaluate series

1 ⫹ 2 ⫹ 3 ⫹ 4 ⫽ 10; property I result


䊳

E. Applications of Sequences To solve applications of sequences, (1) identify where the sequence begins (the initial term), (2) write out the first few terms to help identify the nth term, and (3) decide on an appropriate approach or strategy. EXAMPLE 10

䊳

Solving an Application — Accumulation of Stock Hydra already owned 1420 shares of stock when her company began offering employees the opportunity to purchase 175 discounted shares per year. If she made no purchases other than these discounted shares each year, how many shares will she have 9 yr later? If this continued for the 25 yr she will work for the company, how many shares will she have at retirement?

Solution

䊳

To begin, it helps to simply write out the first few terms of the sequence. Since she already had 1420 shares before the company made this offer, we let a0 ⫽ 1420 be the inaugural element, showing a1 ⫽ 1595 (after 1 yr, she owns 1420 ⫹ 175 ⫽ 1595 shares). The first few terms are 1595, 1770, 1945, 2120, and so on. This supports a general term of an ⫽ 1595 ⫹ 1751n ⫺ 12.

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After 9 years

After 25 years

a9 ⫽ 1595 ⫹ 175182 ⫽ 2995

a25 ⫽ 1595 ⫹ 1751242 ⫽ 5795

769

After 9 yr she would have 2995 shares. Upon retirement she would own 5795 shares of company stock. Now try Exercises 85 through 90

䊳

Surprisingly, sequences and series have a number of other interesting properties, applications, and mathematical connections. For instance, some of the most celebrated numbers in mathematics can be approximated using a series, as demonstrated in Example 11. EXAMPLE 11

䊳

Use a calculator to find the partial sums S4, S8, and S12 for the sequences given. If any sum seems to be approaching a fixed number, name that number. a. an ⫽

Solution

䊳

1 n!

b. an ⫽

1 3n

1 as u(n) on the Y= screen. Using the “sum(” and “seq(” n! commands as before produces the results shown in Figures 9.17 through 9.19, where it appears the sum becomes very close to e ⫺ 1 for larger values of n.

a. Begin by entering

Figure 9.17

Figure 9.18

Figure 9.19

1 as v(n) on the Y= screen, we once again compute the sums 3n indicated as shown in Figures 9.20 through 9.22. It appears the sum becomes 1 very close to for larger values of n. 2 Figure 9.20 Figure 9.22 Figure 9.21 b. After entering

E. You’ve just seen how we can use sequences to solve applications


䊳

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9.1 EXERCISES 䊳



1. A sequence is a(n) specific .

䊳

of numbers listed in a

2. A series is the given sequence.

of the numbers from a

3. A sequence that uses the preceding term(s) to generate those that follow is called a sequence.

4. The notation n! represents the natural number n, with all those

5. Describe the characteristics of a recursive sequence and give one example.

6. Describe the characteristics of an alternating sequence and give one example.

of the n.


Find the first four terms, then find the 8th and 12th term for each nth term given.

1 n 29. an ⫽ a1 ⫹ b ; a10 n

1 n 30. an ⫽ an ⫹ b ; a9 n

7. an ⫽ 2n ⫺ 1

8. an ⫽ 2n ⫹ 3

9. an ⫽ 3n ⫺ 3

10. an ⫽ 2n3 ⫺ 12

31. an ⫽

1 ; a4 n12n ⫹ 12

12. an ⫽

32. an ⫽

1 ; a5 12n ⫺ 12 12n ⫹ 12

2

11. an ⫽ 1⫺12 nn 13. an ⫽

n n⫹1

1 n 15. an ⫽ a b 2 17. an ⫽ 19. an ⫽

1 n

1 n 14. an ⫽ a1 ⫹ b n 2 n 16. an ⫽ a b 3 18. an ⫽

1⫺12 n

n1n ⫹ 12

21. an ⫽ 1⫺12 n2n

1⫺12 n n

20. an ⫽

1 n2 1⫺12 n⫹1 2n ⫺ 1 2

22. an ⫽ 1⫺12 n2⫺n

Use a calculator to: (a) find the indicated term for each sequence, and (b) generate the first five terms of each sequence and store the results in a list. Use fractions if possible; round to tenths when necessary.

23. an ⫽ n2 ⫺ 2; a9

24. an ⫽ 1n ⫺ 22 2; a9

25. an ⫽

26. an ⫽

1⫺12 n⫹1 ; a5 n

1 n⫺1 27. an ⫽ 2a b ; a7 2

1⫺12 n⫹1 2n ⫺ 1

; a5

1 n⫺1 28. an ⫽ 3a b ; a7 3

Find the first five terms of each recursive sequence.

33. e

a1 ⫽ 2 an ⫽ 5an⫺1 ⫺ 3

34. e

a1 ⫽ 3 an ⫽ 2an⫺1 ⫺ 3

35. e

a1 ⫽ ⫺1 an ⫽ 1an⫺1 2 2 ⫹ 3

36. e

a1 ⫽ ⫺2 an ⫽ an⫺1 ⫺ 16

38. e

c1 ⫽ 1, c2 ⫽ 2 cn ⫽ cn⫺1 ⫹ 1cn⫺2 2 2

c1 ⫽ 64, c2 ⫽ 32 37. • cn⫺2 ⫺ cn⫺1 cn ⫽ 2

Simplify each factorial expression.

39.

8! 5!

40.

12! 10!

41.

9! 7!2!

42.

6! 3!3!

43.

8! 2!6!

44.

10! 3!7!

Write out the first four terms in each sequence.

45. an ⫽

n! 1n ⫹ 12!

46. an ⫽

n! 1n ⫹ 32!

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1n ⫹ 12!

47. an ⫽

48. an ⫽

13n2!

nn n!

49. an ⫽

50. an ⫽

1n ⫹ 32!

70. a. ⫺1 ⫹ 4 ⫺ 9 ⫹ 16 ⫺ 25 ⫹ 36 b. 1 ⫺ 8 ⫹ 27 ⫺ 64 ⫹ 125 ⫺ 216 71. a. 1 ⫹ 3 ⫹ 5 ⫹ 7 ⫹ 9 ⫹ 11 ⫹ p

12n2!

2n n!

1 1 1 1 1 ⫹ ⫹ ⫹ ⫹ ⫹p 2 4 8 16 32 72. a. 0.1 ⫹ 0.01 ⫹ 0.001 ⫹ 0.0001 ⫹ p 1 1 1 1 1 b. 1 ⫹ ⫹ ⫹ ⫹ ⫹ ⫹p 2 6 24 120 720 b. 1 ⫹

Find the indicated partial sum for each sequence.

51. an ⫽ n; S5

52. an ⫽ n2; S7

53. an ⫽ 2n ⫺ 1; S8

54. an ⫽ 3n ⫺ 1; S6

1 55. an ⫽ ; S5 n

56. an ⫽

n ; S4 n⫹1

For the given general term an, write the indicated sum using sigma notation.

Expand and evaluate each series. Verify results using a graphing calculator. 4

57. 60.

兺 13i ⫺ 52

5

58.

59.

兺 12k ⫺ 32 2

i⫽1

i⫽1

k⫽1

5

7

5

兺 1k

2

⫹ 12

61.

k⫽1 4

兺

2

7

j

i 63. i⫽1 2 66.

5

兺 12i ⫺ 32 k

j

62.

k⫽1

64.

兺i

i⫽2

67.

兺 1⫺12 2

k k

k⫽1

4

8

兺2

j⫽3

兺 1⫺12 k

7

2

65.

兺 2j

j⫽3

1⫺12 k

6

兺 k1k ⫺ 22

68.

k⫽3

1⫺12 k⫹1

兺k

k⫽2

2

⫺1

Write each sum using sigma notation. Answers are not necessarily unique.

69. a. 4 ⫹ 8 ⫹ 12 ⫹ 16 ⫹ 20 b. 5 ⫹ 10 ⫹ 15 ⫹ 20 ⫹ 25

73. an ⫽ n ⫹ 3; S5 74. an ⫽

n2 ⫹ 1 ; S4 n⫹1

75. an ⫽

n2 ; third partial sum 3

76. an ⫽ 2n ⫺ 1; sixth partial sum 77. an ⫽

Compute each sum by applying properties of summation. 5

79.

兺

i⫽1

80.

13k2 ⫹ k2

82.

4

81.

兺

6

14i ⫺ 52

兺 13 ⫹ 2i2

i⫽1 4

兺 12k

3

⫹ 52

k⫽1


83. Sum of an ⴝ 3n ⴚ 2: Sn ⴝ

n13n ⴚ 12 2

The sum of the first n terms of the sequence defined by an ⫽ 3n ⫺ 2 ⫽ 1, 4, 7, 10, p , 13n ⫺ 22, p is given by the formula shown. Find S5 using the formula, then verify by direct calculation. 䊳

n ; sum for n ⫽ 3 to 7 2n

78. an ⫽ n2; sum for n ⫽ 2 to 6

k⫽1

䊳

771

n13n ⴙ 12 2 The sum of the first n terms of the sequence defined by an ⫽ 3n ⫺ 1 ⫽ 2, 5, 8, 11, p , 13n ⫺ 12, p is given by the formula shown. Find S8 using the formula, then verify by direct calculation. Observing the formulas from Exercises 83 and 84, can you now state the sum formula for an ⫽ 3n ⫺ 0?

84. Sum of an ⴝ 3n ⴚ 1: Sn ⴝ

APPLICATIONS

Use the information given in each exercise to determine the nth term an for the sequence described. Then use the nth term to list the specified number of terms.

85. Wage increases: Latisha gets $7.25 an hour for filling candy machines for Archtown Vending. Each year she receives a $0.50 hourly raise. List Latisha’s hourly wage for the first 5 yr. How much will she make in the fifth year if she works 8 hr per day for 240 working days? 86. Average birth weight: The average birth weight of a certain animal species is 900 g, with the baby gaining 125 g each day for the first 10 days. List the infant’s weight for the first 10 days. How much does the infant weigh on the 10th day?

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87. Blue-book value: Steve’s car has a blue-book value of $6000. Each year it loses 20% of its value (its value each year is 80% of the year before). List the value of Steve’s car for the next 5 yr. (Hint: a0 ⫽ 6000.) 88. Effects of inflation: Suppose inflation (an increase in value) will average 4% for the next 5 yr. List the growing cost (year by year) of a DVD that costs $15 right now. (Hint: a0 ⫽ 15.) 89. Stocking a lake: A local fishery stocks a large lake with 1500 bass and then adds an additional 100 mature bass per month until the lake nears maximum capacity. If the bass population grows at a rate of 5% per month through natural reproduction, the number of bass in the pond after n months is given by the recursive sequence b0 ⫽ 1500, bn ⫽ 1.05bn⫺1 ⫹ 100. How many bass will be in the lake after 6 months? 䊳

9–12


90. Species preservation: The Interior Department introduces 50 wolves (male and female) into a large wildlife area in an effort to preserve the species. Each year about 12 additional adult wolves are added from capture and relocation programs. If the wolf population grows at a rate of 10% per year through natural reproduction, the number of wolves in the area after n years is given by the recursive sequence w0 ⫽ 50, wn ⫽ 1.10wn⫺1 ⫹ 12. How many wolves are in the wildlife area after 6 years? Use your calculator to find the partial sums for n ⴝ 4, n ⴝ 8, and n ⴝ 12 for the summations given, and attempt to name the number the summation approximates: n

91.

2k ⫹ 3k 6k k⫽1 n

1 k k⫽1 2

兺

92.

兺


93. Verify that a constant can be factored out of a sum. That is, verify that the following statement is true: n

兺

n

caj ⫽ c

j⫽1

兺a

j

j⫽1

94. Verify that a summation may be distributed to two (or more) sequences. That is, verify that the following statement is true: n

兺

i⫽1

1ai ⫾ bi 2 ⫽

n

兺

n

ai ⫾

i⫽1

兺b . i

i⫽1

Regarding Exercises 91 and 92, sometimes a series will approach a fixed number very slowly, and many more terms must be added before this value is recognized. Use your graphing calculator to compute the sums S10, S25, and S50 for the following sequences to see if you can recognize the number. Add more terms if necessary.

95. an ⫽

䊳

1 n1n ⫹ 12 1n ⫹ 22

96. an ⫽

1 12n ⫺ 12 12n ⫹ 12


1 97. (5.3) Write log381 ⫽ ⫺x in exponential form, then solve by equating bases.

98. (3.6) Set up the difference quotient for f 1x2 ⫽ 1x, then rationalize the numerator.

99. (8.4) Solve the nonlinear system. e

x2 ⫹ y2 ⫽ 9 9y2 ⫺ 4x2 ⫽ 16

100. (7.3) Solve the system using a matrix equation. 25x ⫹ y ⫺ 2z ⫽ ⫺14 • 2x ⫺ y ⫹ z ⫽ 40 ⫺7x ⫹ 3y ⫺ z ⫽ ⫺13

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9.2

Arithmetic Sequences

LEARNING OBJECTIVES

Similar to the way polynomials fall into certain groups or families (linear, quadratic, cubic, etc.), sequences and series with common characteristics are likewise grouped. In this section, we focus on sequences where each successive term is generated by adding a constant value, as in the sequence 1, 8, 15, 22, 29, p , where 7 is added to a given term in order to produce the next term.


A. Identify an arithmetic sequence and its common difference B. Find the n th term of an arithmetic sequence C. Find the n th partial sum of an arithmetic sequence D. Solve applications involving arithmetic sequences

A. Identifying an Arithmetic Sequence and Finding the Common Difference An arithmetic sequence is one where each successive term is found by adding a fixed constant to the preceding term. For instance 3, 7, 11, 15, p is an arithmetic sequence, since adding 4 to any given term produces the next term. This also means if you take the difference of any two consecutive terms, the result will be 4 and in fact, 4 is called the common difference d for this sequence. Using the notation developed earlier, we can write d ⫽ ak⫹1 ⫺ ak, where ak represents any term of the sequence and ak⫹1 represents the term that follows ak. Arithmetic Sequences Given a sequence a1, a2, a3, p , ak, ak⫹1, p , an, where k, n 僆⺞ and k 6 n, if there exists a common difference d such that ak⫹1 ⫺ ak ⫽ d for all k, then the sequence is an arithmetic sequence. The difference of successive terms can be rewritten as ak⫹1 ⫽ ak ⫹ d (for k ⱖ 12 to highlight that each following term is found by adding d to the previous term.

EXAMPLE 1

䊳

Identifying an Arithmetic Sequence Determine if the given sequence is arithmetic. If yes, name the common difference. If not, try to determine the pattern that forms the sequence. 77 29 a. 2, 5, 8, 11, p b. 12, 56, 13 12 , 60 , 20 , p

Solution

䊳

a. Begin by looking for a common difference d ⫽ ak⫹1 ⫺ ak. Checking each pair of consecutive terms we have 5⫺2⫽3

8⫺5⫽3

11 ⫺ 8 ⫽ 3 and so on.

This is an arithmetic sequence with common difference d ⫽ 3. b. Checking each pair of consecutive terms yields 5 1 5 3 ⫺ ⫽ ⫺ 6 2 6 6 2 1 ⫽ ⫽ 6 3

13 5 13 10 ⫺ ⫽ ⫺ 12 6 12 12 3 1 ⫽ ⫽ 12 4

13 77 65 77 ⫺ ⫽ ⫺ 60 12 60 60 12 1 ⫽ ⫽ 60 5

Since the difference is not constant, this is not an arithmetic sequence. It appears the sequence is formed by adding 1k to each previous term, for natural numbers k. Now try Exercises 7 through 18

9–13

䊳

773

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774

9–14


EXAMPLE 2

䊳

Writing the First k Terms of an Arithmetic Sequence Write the first five terms of the arithmetic sequence, given the first term a1 and the common difference d. a. a1 ⫽ 12 and d ⫽ ⫺4 b. a1 ⫽ 12 and d ⫽ 13

Solution

䊳

A. You’ve just seen how we can identify an arithmetic sequence and its common difference

a. a1 ⫽ 12 and d ⫽ ⫺4. Starting at a1 ⫽ 12, add ⫺4 to each new term to generate the sequence: 12, 8, 4, 0, ⫺4. b. a1 ⫽ 12 and d ⫽ 13. Starting at a1 ⫽ 12 and adding 13 to each new term will generate the sequence: 12, 56, 76, 32, 11 6 . Note that since the common denominator is 6, terms of the sequence can quickly be found by adding 13 ⫽ 26 to the previous term and reducing if possible. Now try Exercises 19 through 30

䊳

B. Finding the n th Term of an Arithmetic Sequence If the values a1 and d from an arithmetic sequence are known, we could generate the terms of the sequence by adding multiples of d to the first term, instead of adding d to each new term. For example, we can generate the sequence 3, 8, 13, 18, 23 by adding multiples of 5 to the first term a1 ⫽ 3: 3 ⫽ 3 ⫹ 1025

a1 ⫽ a1 ⫹ 0d

13 ⫽ 3 ⫹ 1225

a3 ⫽ a1 ⫹ 2d

8 ⫽ 3 ⫹ 1125

a2 ⫽ a1 ⫹ 1d

18 ⫽ 3 ⫹ 1325

a4 ⫽ a1 ⫹ 3d

23 ⫽ 3 ⫹ 1425

current term

initial term

S

S

a5 ⫽ a1 ⫹ 4d coefficient of common difference

It’s helpful to note the coefficient of d is 1 less than the subscript of the current term (as shown): 5 ⫺ 1 ⫽ 4. This observation leads us to a formula for the nth term. The n th Term of an Arithmetic Sequence The nth term of an arithmetic sequence is given by

an ⫽ a1 ⫹ 1n ⫺ 12d

where d is the common difference.

EXAMPLE 3

䊳

Finding a Specified Term in an Arithmetic Sequence Find the 24th term of the sequence 0.1, 0.4, 0.7, 1, p .

Solution

䊳

Instead of creating all terms up to the 24th, we determine the constant d and use the nth term formula. By inspection we note a1 ⫽ 0.1 and d ⫽ 0.3. an ⫽ a1 ⫹ 1n ⫺ 12d ⫽ 0.1 ⫹ 1n ⫺ 120.3 ⫽ 0.1 ⫹ 0.3n ⫺ 0.3 ⫽ 0.3n ⫺ 0.2

n th term formula substitute 0.1 for a1 and 0.3 for d eliminate parentheses simplify

To find the 24th term we substitute 24 for n: a24 ⫽ 0.31242 ⫺ 0.2 ⫽ 7.0

substitute 24 for n result


䊳

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Section 9.2 Arithmetic Sequences

EXAMPLE 4

䊳

Determining the Number of Terms in an Arithmetic Sequence Find the number of terms in the arithmetic sequence 2, ⫺5, ⫺12, ⫺19, p , ⫺411.

Solution

䊳

By inspection we see that a1 ⫽ 2 and d ⫽ ⫺7. As before, an ⫽ a1 ⫹ 1n ⫺ 12d ⫽ 2 ⫹ 1n ⫺ 12 1⫺72 ⫽ 2 ⫺ 7n ⫹ 7 ⫽ ⫺7n ⫹ 9

n th term formula substitute 2 for a1 and ⫺7 for d distribute ⫺7 simplify

Although we don’t know the number of terms in the sequence, we do know the last or nth term is ⫺411. Substituting ⫺411 for an gives ⫺411 ⫽ ⫺7n ⫹ 9 60 ⫽ n

substitute ⫺411 for an solve for n

There are 60 terms in this sequence. Now try Exercises 43 through 50

䊳

Note that in both Examples 3 and 4, the nth term had the form of a linear equation 1y ⫽ mx ⫹ b2 after simplifying: an ⫽ 0.3n ⫺ 0.2 and an ⫽ ⫺7n ⫹ 9. This is a characteristic of arithmetic sequences, with the common difference d corresponding to the slope m. This means the graph of an arithmetic sequence will always be a set of discrete points that lie on a straight line. After entering u1n2 ⫽ 0.3n ⫺ 0.2 (from Example 3), Figure 9.23 shows the table of values for an ⫽ 0.3n ⫺ 0.2, with the graph in Figure 9.24A. Figure 9.24A

Figure 9.23

5

10

0

⫺1.5

Figure 9.24B In sequence MODE , we still set the size of the viewing window as before, but we can also stipulate the range of values of n to be used (nMin and nMax), which term we want to plot as a beginning (PlotStart), and whether we want all following terms graphed (PlotStep ⫽ 1), every second term graphed (PlotStep ⫽ 2), and so on. The graph in Figure 9.24A was generated using the values shown in Figure 9.24B (Ymin ⫽ ⫺1.5, Ymax ⫽ 5, and Yscl ⫽ 1 cannot be seen). To see the graph of a sequence more distinctly, we can enter the natural numbers 1 through 10 in L1, then define L2 as u(L1) (Figures 9.25 and 9.26) and plot these points (Figure 9.27).

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Figure 9.27 Figure 9.25

Figure 9.26

5

10

0

⫺1.5

One additional advantage of this approach is that we can easily verify the common difference is d using the “ ¢ List(” feature, which automatically computes the difference between each successive term in a specified list. This option is located in the same submenu as the “seq(” option, accessed using 2nd STAT (LIST) (OPS) 7: ¢ List(L2) . See Figures 9.28 and 9.29. ENTER

Figure 9.28

EXAMPLE 5

䊳

Figure 9.29

Graphing Arithmetic Sequences Enter the natural numbers 1 through 6 in L1, and the terms of the sequence ⫺0.45, ⫺0.1, 0.25, 0.6, 0.95, 1.3 in L2. Then verify the sequence in L2 is arithmetic by a. Graphing the related points to see if they appear linear. b. Using the ¢ List( feature. c. Finding the nth term that defines the sequence and graph the sequence.

Solution

䊳

a. The plotted points are shown in Figure 9.30 and appear to be linear. b. The ¢ List( feature shows there is a common difference of 0.35 (Figure 9.31). c. With a1 ⫽ ⫺0.45 and d ⫽ 0.35, the nth term must be an ⫽ a1 ⫹ 1n ⫺ 12d ⫽ ⫺0.45 ⫹ 1n ⫺ 12 10.352 ⫽ 0.35n ⫺ 0.8

n th term formula substitute for a1 and d simplify

The nth term for this sequence is an ⫽ 0.35n ⫺ 0.8. The graph is shown in Figure 9.32. Figure 9.32 Figure 9.30 Figure 9.31 2 2

7

0

⫺1

7

0

⫺1


䊳

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If the term a1 is unknown but a term ak is given, the nth term can be written an ⫽ ak ⫹ 1n ⫺ k2d

(the subscript of the term ak and coefficient of d sum to n). EXAMPLE 6

䊳

Finding the First Term of an Arithmetic Sequence Given an arithmetic sequence where a6 ⫽ 0.55 and a13 ⫽ 0.9, find the common difference d and the value of a1.

Solution

䊳

At first it seems that not enough information is given, but recall we can express a13 as the sum of any earlier term and the appropriate multiple of d. Since a6 is known, we write a13 ⫽ a6 ⫹ 7d (note 13 ⫽ 6 ⫹ 7 as required). a1 is unknown a13 ⫽ a6 ⫹ 7d 0.9 ⫽ 0.55 ⫹ 7d substitute 0.9 for a13 and 0.55 for a6 0.35 ⫽ 7d subtract 0.55 d ⫽ 0.05 solve for d Having found d, we can now solve for a1. a13 ⫽ a1 ⫹ 12d 0.9 ⫽ a1 ⫹ 1210.052 0.9 ⫽ a1 ⫹ 0.6 a1 ⫽ 0.3

B. You’ve just seen how we can find the nth term of an arithmetic sequence

n th term formula for n ⫽ 13 substitute 0.9 for a13 and 0.05 for d simplify solve for a1

The first term is a1 ⫽ 0.3 and the common difference is d ⫽ 0.05. Now try Exercises 55 through 60

䊳

C. Finding the n th Partial Sum of an Arithmetic Sequence Using sequences and series to solve applications often requires computing the sum of a given number of terms. To develop and understand the approach used, consider the sum of the first 10 natural numbers. Using S10 to represent this sum, we have S10 ⫽ 1 ⫹ 2 ⫹ 3 ⫹ 4 ⫹ 5 ⫹ 6 ⫹ 7 ⫹ 8 ⫹ 9 ⫹ 10. We could just use brute force, but if we rewrite the sum a second time but in reverse order, then add it to the first, we find that each column adds to 11. S10 ⫽ 1 ⫹ 2 ⫹ 3 ⫹ 4 ⫹ 5 ⫹ 6 ⫹ 7 ⫹ 8 ⫹ 9 ⫹ 10 S10 ⫽ 10 ⫹ 9 ⫹ 8 ⫹ 7 ⫹ 6 ⫹ 5 ⫹ 4 ⫹ 3 ⫹ 2 ⫹ 1 2S10 ⫽ 11 ⫹ 11 ⫹ 11 ⫹ 11 ⫹ 11 ⫹ 11 ⫹ 11 ⫹ 11 ⫹ 11 ⫹ 11 Since there are 10 columns, the total is 11 ⫻ 10 ⫽ 110 but this is twice the actual sum and we find that 2S10 ⫽ 110, so S10 ⫽ 55. Now consider the sequence a1, a2, a3, a4, p , an with common difference d. Use Sn to represent the sum of the first n terms and write the original series, then the series in reverse order underneath. Since one row increases at the same rate the other decreases, the sum of each column remains constant, and for simplicity’s sake we choose a1 ⫹ an to represent this sum. Sn ⫽ a1 ⫹ a2 ⫹ a3 ⫹ p ⫹ an⫺2 ⫹ an⫺1 ⫹ an add p columns Sn ⫽ an ⫹ an⫺1 ⫹ an⫺2 ⫹ ⫹ a3 ⫹ a2 ⫹ a1 p 2Sn ⫽ 1a1 ⫹ an 2 ⫹ 1a1 ⫹ an 2 ⫹ 1a1 ⫹ an 2 ⫹ ⫹ 1a1 ⫹ an 2 ⫹ 1a1 ⫹ an 2 ⫹ 1a1 ⫹ an 2 ↓ vertically To understand why each column adds to a1 ⫹ an, consider the sum in the second column: a2 ⫹ an⫺1. From a2 ⫽ a1 ⫹ d and an⫺1 ⫽ an ⫺ d, we obtain a2 ⫹ an⫺1 ⫽ 1a1 ⫹ d2 ⫹ 1an ⫺ d2 by adding the equations, which gives a result of a1 ⫹ an. Since there are n columns, we end up with 2Sn ⫽ n1a1 ⫹ an 2, and solving for Sn gives the formula for the first n terms of an arithmetic sequence.

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The nth Partial Sum of an Arithmetic Sequence Given an arithmetic sequence with first term a1, the nth partial sum is given by Sn ⫽

n 1a1 ⫹ an 2. 2

In words: The sum of an arithmetic sequence is one-half the number of terms times the sum of the first and last term.

EXAMPLE 7

䊳

Computing the Sum of an Arithmetic Sequence Use the summation formula to find the sum of the first 75 positive odd integers: 75

兺 12n ⫺ 12 . Verify the result using a graphing calculator.

n⫽1

Solution

䊳

The initial terms of the sequence are 1, 3, 5, p and we note a1 ⫽ 1, d ⫽ 2, and n ⫽ 75. To use the sum formula, we need the value of a75: 21752 ⫺ 1 ⫽ 149. formula shows a75 ⫽ a1 ⫹ 74d ⫽ 1 ⫹ 74122, so a75 ⫽ 149. n Sn ⫽ 1a1 ⫹ an 2 2 75 S75 ⫽ 1a1 ⫹ a75 2 2 75 ⫽ 11 ⫹ 1492 2 ⫽ 5625

sum formula

substitute 75 for n

substitute 1 for a1, 149 for a75 result

The sum of the first 75 positive odd integers is 5625. To verify, we enter u1n2 ⫽ 2n ⫺ 1 on the Y= screen, and find the sum of the first 75 terms of the sequence on the home screen as before. See figure.


䊳

By substituting the nth term formula directly into the formula for partial sums, we’re able to find a partial sum without actually having to find the nth term:

C. You’ve just seen how we can find the nth partial sum of an arithmetic sequence

n Sn ⫽ 1a1 ⫹ an 2 2 n ⫽ 1a1 ⫹ 3a1 ⫹ 1n ⫺ 12d 4 2 2 n ⫽ 3 2a1 ⫹ 1n ⫺ 12d 4 2

sum formula substitute a1 ⫹ 1n ⫺ 12d for an

alternative formula for the nth partial sum

See Exercises 67 through 72 for more on this alternative formula.

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D. Applications In the evolution of certain plants and shelled animals, sequences and series seem to have been one of nature’s favorite tools. The sprials found on many ferns and other plants are excellent examples of sequences in nature, as are the size of the chambers in nautilus shells (see Figures 9.33 and 9.34). Sequences and series also provide a good mathematical model for a variety of other situations as well.

Figure 9.33

spiral fern

EXAMPLE 8

Figure 9.34

nautilus

䊳

Solving an Application of Arithmetic Sequences: Seating Capacity Cox Auditorium is an amphitheater that has 40 seats in the first row, 42 seats in the second row, 44 in the third, and so on. If there are 75 rows in the auditorium, what is the auditorium’s seating capacity?

Solution

䊳

D. You’ve just seen how we can solve applications involving arithmetic sequences

The number of seats in each row gives the terms of an arithmetic sequence with a1 ⫽ 40, d ⫽ 2, and n ⫽ 75. To find the seating capacity, we need to find the total number of seats, which is the sum of this arithmetic sequence. Since the value of n a75 is unknown, we opt for the alternative formula Sn ⫽ 3 2a1 ⫹ 1n ⫺ 12d4 . 2 n sum formula Sn ⫽ 3 2a1 ⫹ 1n ⫺ 12d 4 2 75 3 21402 ⫹ 175 ⫺ 12122 4 substitute 40 for a1, 2 for d, and 75 for n S75 ⫽ 2 75 simplify ⫽ 12282 2 ⫽ 8550 result The seating capacity of Cox Auditorium is 8550. Now try Exercises 75 through 80

䊳

9.2 EXERCISES 䊳



1. Consecutive terms in an arithmetic sequence differ by a constant called the .

2. The sum of the first n terms of an arithmetic sequence is called the nth .

3. The formula for the nth partial sum of an

4. The nth term formula for an arithmetic sequence is an ⫽ term , where a1 is the and d is the .

arithmetic sequence is Sn ⫽ is the term.

, where an

5. Discuss how the terms of an arithmetic sequence can be written in various ways using the relationship an ⫽ ak ⫹ 1n ⫺ k2d.

6. Describe how the formula for the nth partial sum was derived, and illustrate its application using a sequence from the exercise set.

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9–20



Determine if the sequence given is arithmetic. If yes, name the common difference. If not, try to determine the pattern that forms the sequence.

7. ⫺5, ⫺2, 1, 4, 7, 10, p

41. a1 ⫽ ⫺0.025, d ⫽ 0.05; find a50 42. a1 ⫽ 3.125, d ⫽ ⫺0.25; find a20 Find the number of terms in each sequence.

8. 1, ⫺2, ⫺5, ⫺8, ⫺11, ⫺14, p

43. a1 ⫽ 2, an ⫽ ⫺22, d ⫽ ⫺3

9. 0.5, 3, 5.5, 8, 10.5, p

44. a1 ⫽ 4, an ⫽ 42, d ⫽ 2

10. 1.2, 3.5, 5.8, 8.1, 10.4, p

45. a1 ⫽ 0.4, an ⫽ 10.9, d ⫽ 0.25

11. 2, 3, 5, 7, 11, 13, 17, p

46. a1 ⫽ ⫺0.3, an ⫽ ⫺36, d ⫽ ⫺2.1

12. 1, 4, 8, 13, 19, 26, 34, p

47. ⫺3, ⫺0.5, 2, 4.5, 7, p , 47

13.

1 1 1 1 5 24 , 12 , 8 , 6 , 24 ,

14.

1 1 1 1 1 12 , 15 , 20 , 30 , 60 ,

48. ⫺3.4, ⫺1.1, 1.2, 3.5, p , 38

p

49.

p

15. 1, 4, 9, 16, 25, 36, p 16. ⫺125, ⫺64, ⫺27, ⫺8, ⫺1, p 17. ␲,

5␲ 2␲ ␲ ␲ ␲ , , , , ,p 6 3 2 3 6

18. ␲,

7␲ 3␲ 5␲ ␲ , , , ,p 8 4 8 2

1 1 1 5 1 12 , 8 , 6 , 24 , 4 ,

p , 98

50.

1 1 1 1 12 , 15 , 20 , 30 ,

p , ⫺14

For Exercises 51 through 54, enter the natural numbers 1 through 6 in L1 on a graphing calculator, and the terms of the given sequence in L2. Then determine if the sequence is arithmetic by (a) graphing the related points to see if they appear linear, and (b) using the ¢ List( feature. If an arithmetic sequence, (c) find the nth term and graph the sequence.

51. 1.5, 2.25, 3, 3.75, 4.5, 5.25, p Write the first four terms of the arithmetic sequence with the given first term and common difference.

19. a1 ⫽ 2, d ⫽ 3

20. a1 ⫽ 8, d ⫽ 3

21. a1 ⫽ 7, d ⫽ ⫺2

22. a1 ⫽ 60, d ⫽ ⫺12

23. a1 ⫽ 0.3, d ⫽ 0.03

24. a1 ⫽ 0.5, d ⫽ 0.25

25. a1 ⫽

1 26. a1 ⫽ 15, d ⫽ 10

3 2,

d⫽

1 2

27. a1 ⫽ 34, d ⫽ ⫺18

28. a1 ⫽ 16, d ⫽ ⫺13

29. a1 ⫽ ⫺2, d ⫽ ⫺3

30. a1 ⫽ ⫺4, d ⫽ ⫺4

Identify the first term and the common difference, then write the expression for the general term an and use it to find the 6th, 10th, and 12th terms of the sequence.

31. 2, 7, 12, 17, p

32. 7, 4, 1, ⫺2, ⫺5, p

33. 5.10, 5.25, 5.40, p

34. 9.75, 9.40, 9.05, p

35. 32, 94, 3, 15 4,p

3 36. 57, 14 , ⫺27, ⫺11 14 , p

52.

53. 9, 8, 6, 3, ⫺1, ⫺6, p

55. a3 ⫽ 7, a7 ⫽ 19

40. a1 ⫽

d⫽

1 ⫺10 ;

find a9

56. a5 ⫽ ⫺17, a11 ⫽ ⫺2

58. a6 ⫽ ⫺12.9, a30 ⫽ 1.5 59. a10 ⫽

13 18 ,

a24 ⫽ 27 2

60. a4 ⫽ 54, a8 ⫽ 94

Evaluate each sum. For Exercises 65 and 66, use the summation properties from Section 9.1. Verify all results on a graphing calculator. 30

兺

n⫽1 37

63.

38. a1 ⫽ 9, d ⫽⫺2; find a17 12 25 ,

1 1 1 1 1 1 , , , , , ,p 1 2 3 4 5 6

57. a2 ⫽ 1.025, a26 ⫽ 10.025

Find the indicated term using the information given.

1 39. a1 ⫽ 32, d ⫽ ⫺12 ; find a7

54.

Find the common difference d and the value of a1 using the information given.

61. 37. a1 ⫽ 5, d ⫽ 4; find a15

47 19 29 10 11 1 , , , , , ,p 18 9 18 9 18 9

13n ⫺ 42

兺 a 4 n ⫹ 2b 3

29

62.

20

64.

n⫽1 15

65.

兺

n⫽4

13 ⫺ 5n2

兺 14n ⫺ 12

n⫽1

兺 a 2 n ⫺ 3b 5

n⫽1 20

66.

兺 17 ⫺ 2n2

n⫽7

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Use the alternative formula for the nth partial sum to compute the sums indicated.

67. The sum S15 for the sequence ⫺12 ⫹ 1⫺9.52 ⫹ 1⫺72 ⫹ 1⫺4.52 ⫹ p

68. The sum S20 for the sequence 92 ⫹ 72 ⫹ 52 ⫹ 32 ⫹ p 69. The sum S30 for the sequence 0.003 ⫹ 0.173 ⫹ 0.343 ⫹ 0.513 ⫹ p

䊳

71. The sum S20 for the sequence 12 ⫹ 2 12 ⫹ 3 12 ⫹ 4 12 ⫹ p 72. The sum S10 for the sequence 12 13 ⫹ 10 13 ⫹ 8 13 ⫹ 613 ⫹ p


73. Sum of the first n natural numbers: Sn ⴝ

n1n ⴙ 12 2

The sum of the first n natural numbers can be found using the formula shown, where n represents the number of terms in the sum. Verify the formula by adding the first six natural numbers by hand, and then evaluating S6. Then find the sum of the first 75 natural numbers.

䊳

70. The sum S50 for the sequence 1⫺22 ⫹ 1⫺72 ⫹ 1⫺122 ⫹ 1⫺172 ⫹ p

74. Sum of the squares of the first n natural n1n ⴙ 1212n ⴙ 12 numbers: Sn ⴝ 6 If the first n natural numbers are squared, the sum of these squares can be found using the formula shown, where n represents the number of terms in the sum. Verify the formula by computing the sum of the squares of the first six natural numbers by hand, and then evaluating S6. Then find the sum of the squares of the first 20 natural numbers: 112 ⫹ 22 ⫹ 32 ⫹ p ⫹ 202 2.

APPLICATIONS

75. Temperature fluctuation: At 5 P.M. in Coldwater, the temperature was a chilly 36°F. If the temperature decreased by 3°F every half-hour for the next 7 hr, at what time did the temperature hit 0°F? 76. Arc of a baby swing: When Mackenzie’s baby swing is started, the first swing (one way) is a 30-in. arc. As the swing slows down, each successive arc is 32 in. less than the previous one. Find (a) the length of the tenth swing and (b) how far Mackenzie has traveled during the 10 swings. 77. Computer animations: The animation on a new computer game initially allows the hero of the game to jump a (screen) distance of 10 in. over booby traps and obstacles. Each successive jump is limited to 34 in. less than the previous one. Find (a) the length of the seventh jump and (b) the total distance covered after seven jumps. 78. Seating capacity: The Fox Theater creates a “theater in the round” when it shows any of Shakespeare’s plays. The first row has 80 seats, the second row has 88,

the third row has 96, and so on. How many seats are in the 10th row? If there is room for 25 rows, how many chairs will be needed to set up the theater? 79. Sales goals: At the time that I was newly hired, 100 sales per month was what I required. Each following month — the last plus 20 more, as I work for the goal of top sales award. When 2500 sales are thusly made, it’s Tahiti, Hawaii, and piña coladas in the shade. How many sales were made by this person in the seventh month? What were the total sales after the 12th month? Was the goal of 2500 total sales met? 80. Bequests to charity: At the time our mother left this Earth, she gave $9000 to her children of birth. This we kept and each year added $3000 more, as a lasting memorial from the children she bore. When $42,000 is thusly attained, all goes to charity that her memory be maintained. What was the balance in the sixth year? In what year was the goal of $42,000 met?

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9–22



81. From a study of numerical analysis, a function is known to be linear if its “first differences” (differences between successive outputs) are constant. Likewise, a function is known to be quadratic if its “first differences” form an arithmetic sequence. Use this information to determine if the following sets of output come from a linear or quadratic function: a. 19, 11.8, 4.6, ⫺2.6, ⫺9.8, ⫺17, ⫺24.2, p b. ⫺10.31, ⫺10.94, ⫺11.99, ⫺13.46, ⫺15.35, p 䊳

82. From elementary geometry it is known that the interior angles of a triangle sum to 180°, the interior angles of a quadrilateral sum to 360°, the interior angles of a pentagon sum to 540°, and so on. Use the pattern created by the relationship between the number of sides and the number of angles to develop a formula for the sum of the interior angles of an n-sided polygon. The interior angles of a decagon (10 sides) sum to how many degrees?


83. (5.5) Solve for t: 2530 ⫽ 500e0.45t

84. (3.2) Graph by completing the square. Label all important features: y ⫽ x2 ⫺ 2x ⫺ 3.

85. (1.3) In 2000, the deer population was 972. By 2005 it had grown to 1217. Assuming the growth is linear, find the function that models this data and use it to estimate the deer population in 2008.

86. (2.6) Given y varies inversely with x and directly with w. If y ⫽ 14 when x ⫽ 15 and w ⫽ 52.5, find the value of y when x ⫽ 32 and w ⫽ 208.

9.3

Geometric Sequences


A. Identify a geometric

B. C. D. E.

sequence and its common ratio Find the n th term of a geometric sequence Find the n th partial sum of a geometric sequence Find the sum of an infinite geometric series Solve application problems involving geometric sequences and series

Recall that arithmetic sequences are those where each term is found by adding a constant value to the preceding term. In this section, we consider geometric sequences, where each term is found by multiplying the preceding term by a constant value. Geometric sequences have many interesting applications, as do geometric series.

A. Geometric Sequences A geometric sequence is one where each successive term is found by multiplying the preceding term by a fixed constant. Consider growth of a bacteria population, where a single cell splits in two every hour over a 24-hr period. Beginning with a single bacterium 1a0 ⫽ 12, after 1 hr there are 2, after 2 hr there are 4, and so on. Writing the number of bacteria as a sequence we have: hours: bacteria:

a1 T 2

a2 T 4

a3 T 8

a4 T 16

a5 T 32

p p

The sequence 2, 4, 8, 16, 32, p is a geometric sequence since each term is found by multiplying the previous term by the constant factor 2. This also means that the ratio of any two consecutive terms must be 2 and in fact, 2 is called the common ratio r ak⫹1 , where for this sequence. Using the notation from Section 9.1 we can write r ⫽ ak ak represents any term of the sequence and ak⫹1 represents the term that follows ak.

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Section 9.3 Geometric Sequences

783

Geometric Sequences Given a sequence a1, a2, a3, p , ak, ak⫹1, p , an, where k, n 僆⺞ and k 6 n, ak⫹1 if there exists a common ratio r such that ⫽ r for all k, ak then the sequence is a geometric sequence. The ratio of successive terms can be rewritten as ak⫹1 ⫽ akr (for k ⱖ 12 to highlight that each term is found by multiplying the preceding term by r. EXAMPLE 1

Solution

䊳

䊳

Testing a Sequence for a Common Ratio Determine if the given sequence is geometric. If not geometric, try to determine the pattern that forms the sequence. 120 a. 1, 0.5, 0.25, 0.125, p b. 17, 27, 67, 24 7, 7 ,p ak⫹1 Apply the definition to check for a common ratio r ⫽ . ak a. For 1, 0.5, 0.25, 0.125, p , the ratio of consecutive terms gives 0.5 0.25 0.125 ⫽ 0.5, ⫽ 0.5, ⫽ 0.5, and so on. 1 0.5 0.25 This is a geometric sequence with common ratio r ⫽ 0.5. 120 b. For 17, 27, 67, 24 7 , 7 , p , we have: 2 1 2 7 6 2 6 7 6 24 # 7 24 and so on. ⫼ ⫽ # ⫼ ⫽ # ⫼ ⫽ 7 7 7 1 7 7 7 2 7 7 7 6 ⫽2 ⫽3 ⫽4 Since the ratio is not constant, this is not a geometric sequence. The sequence n! appears to be formed by dividing n! by 7: an ⫽ . 7 Now try Exercises 7 through 24

EXAMPLE 2

䊳

䊳

Writing the Terms of a Geometric Sequence Write the first five terms of the geometric sequence, given the first term a1 ⫽ ⫺16 and the common ratio r ⫽ 0.25.

Solution

䊳

Given a1 ⫽ ⫺16 and r ⫽ 0.25. Starting at a1 ⫽ ⫺16, multiply each term by 0.25 to generate the sequence. a2 ⫽ ⫺16 # 0.25 ⫽ ⫺4 a4 ⫽ ⫺1 # 0.25 ⫽ ⫺0.25

A. You’ve just seen how we can identify a geometric sequence and its common ratio

a3 ⫽ ⫺4 # 0.25 ⫽ ⫺1 a5 ⫽ ⫺0.25 # 0.25 ⫽ ⫺0.0625

The first five terms of this sequence are ⫺16, ⫺4, ⫺1, ⫺0.25, and ⫺0.0625. Now try Exercises 25 through 32

䊳

B. Find the n th Term of a Geometric Sequence If the values a1 and r from a geometric sequence are known, we could generate the terms of the sequence by applying additional factors of r to the first term, instead of multiplying each new term by r. If a1 ⫽ 3 and r ⫽ 2, we simply begin at a1, and

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continue applying additional factors of r for each successive term. a1 ⫽ a1r0

6 ⫽ 3 # 21

a2 ⫽ a1r1

12 ⫽ 3 # 22

a3 ⫽ a1r2

24 ⫽ 3 # 23

a4 ⫽ a1r3

48 ⫽ 3 # 24

a5 ⫽ a1r4

current term

initial term

S

3 ⫽ 3 # 20

S

784

exponent on common ratio

From this pattern, we note the exponent on r is always 1 less than the subscript of the current term: 5 ⫺ 1 ⫽ 4, which leads us to the formula for the nth term of a geometric sequence. The n th Term of a Geometric Sequence The nth term of a geometric sequence is given by an ⫽ a1rn⫺1 where r is the common ratio.

EXAMPLE 3

䊳

Finding a Specific Term in a Sequence Identify the common ratio r, and use it to write the expression for the nth term. Then find the 10th term of the sequence: 3, ⫺6, 12, ⫺24, p .

Solution

䊳

By inspection we note that a1 ⫽ 3 and r ⫽ ⫺2. This gives an ⫽ a1rn⫺1 ⫽ 31⫺22 n⫺1

n th term formula substitute 3 for a1 and ⫺2 for r

To find the 10th term we substitute n ⫽ 10: a10 ⫽ 31⫺22 10⫺1 ⫽ 31⫺22 9 ⫽ ⫺1536

substitute 10 for n simplify


EXAMPLE 4

䊳

䊳

Determining the Number of Terms in a Geometric Sequence 1 . Find the number of terms in the geometric sequence 4, 2, 1, p , 64

Solution

䊳

Observing that a1 ⫽ 4 and r ⫽ 12, we have an ⫽ a1 rn⫺1 1 n⫺1 ⫽ 4a b 2

n th term formula substitute 4 for a1 and

1 for r 2

Although we don’t know the number of terms in the sequence, we do know the last 1 1 . Substituting an ⫽ 64 or nth term is 64 gives 1 1 n⫺1 ⫽ 4a b 64 2 1 1 n⫺1 ⫽a b 256 2

substitute

1 for an 64

1 divide by 4 amultiply by b 4

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785

From our work in Chapter 5, we attempt to write both sides as exponentials with a like base, or apply logarithms. Since 256 ⫽ 28, we equate bases. 1 8 1 n⫺1 a b ⫽a b 2 2 S8 ⫽ n ⫺ 1 9⫽n

write

1 1 8 as a b 256 2

like bases imply exponents must be equal solve for n

This shows there are nine terms in the sequence. Now try Exercises 47 through 58

䊳

Note that in both Examples 3 and 4, the nth term had the form of an exponential 1 n⫺1 equation 1y ⫽ a # bn 2 after simplifying: an ⫽ 31⫺22 n⫺1 and an ⫽ 4a b . This is in 2 fact a characteristic of geometric sequences, with the common ratio r corresponding to the base b. This means the graph of a geometric sequence will always be a set of discrete points that lie on an exponential graph (see Worthy of Note). After entering 1 n⫺1 (from Example 4), Figure 9.35 shows the table of values for this u1n2 ⫽ 4a b 2 sequence, with the graph in Figure 9.36. As before, we can see the graph of the sequence more distinctly by entering the natural numbers 1 through 8 in L1, then defining L2 as u(L1). The resulting graph is shown in Figure 9.37.

WORTHY OF NOTE The sequence from Example 3 is an alternating sequence, and the exponential characteristics of its graph can be seen by taking a look at the graph of an ⫽ |31⫺22 n⫺1 | .

Figure 9.35

Figure 9.36

Figure 9.37

5

5

10

0

⫺1.5

10

0

⫺1.5

One additional advantage of using a list is that we can easily verify whether or not a common ratio r exists. We do this by duplicating L2 in L3 (define L3 ⫽ L2), then deleting the first entry of L3 and the last entry of L2 (to keep the same number of terms ak⫹1 in each list). See Figure 9.38. This enables us to find the ratio for the entire list ak by defining L4 as the ratio L3/L2 (which automatically computes the ratio of successive terms). See Figures 9.39 and 9.40. Figure 9.38

Figure 9.39

Figure 9.40

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EXAMPLE 5

䊳

Graphing Geometric Sequences Enter the natural numbers 1 through 6 in L1, and the terms of the sequence 0.25, 1.25, 6.25, 31.25, 156.25, 781.25 in L2. Then determine if the sequence is geometric by a. Graphing the related points to see if they appear to form along an exponential graph. b. Finding the successive ratios between terms. c. If the sequence is geometric, find the nth term and graph the sequence.

Solution

䊳

a. The plotted points are shown in Figure 9.41 and appear to lie along an exponential graph. Figure 9.41 Figure 9.42

1000

8

0

⫺250

b. Using the approach described prior to Example 5, we find there is a common ratio of r ⫽ 5 (Figure 9.42). c. With a1 ⫽ 0.25 and r ⫽ 5, the nth term must be an ⫽ a1rn⫺1 ⫽ 0.25152 n⫺1

Figure 9.43 1000

n th term formula 8

0

substitute for a1 and r

The nth term for this sequence is an ⫽ 0.25152 n⫺1. The graph is shown in Figure 9.43.

⫺250

Now try Exercises 59 through 62 If the term a1 is unknown but a term ak is given, the nth term can be written an ⫽ akrn⫺k, (the subscript on the term ak and the exponent on r sum to n). EXAMPLE 6

䊳

Finding the First Term of a Geometric Sequence Given a geometric sequence where a4 ⫽ 0.075 and a7 ⫽ 0.009375, find the common ratio r and the value of a1.

Solution

䊳

Since a1 is not known, we express a7 as the product of a known term and the appropriate number of common ratios: a7 ⫽ a4r3 17 ⫽ 4 ⫹ 3, as required). a7 ⫽ a4 # r3 0.009375 ⫽ 0.075r3 0.125 ⫽ r3 r ⫽ 0.5

a1 is unknown substitute 0.009375 for a7 and 0.075 for a4 divide by 0.075 solve for r

䊳

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787


Having found r, we can now solve for a1 a7 ⫽ a1r6 0.009375 ⫽ a1 10.52 6 0.009375 ⫽ a1 10.0156252 a1 ⫽ 0.6

n th term formula substitute 0.009375 for a7 and 0.5 for r simplify solve for a1

The first term is a1 ⫽ 0.6 and the common ratio is r ⫽ 0.5.

B. You’ve just seen how we can find the nth term of a geometric sequence


䊳

C. Find the n th Partial Sum of a Geometric Sequence As with arithmetic series, applications of geometric series often involve computing sums of consecutive terms. We can adapt the method for finding the sum of an arithmetic sequence to develop a formula for adding the first n terms of a geometric sequence. For the nth term an ⫽ a1rn⫺1, we have Sn ⫽ a1 ⫹ a1r ⫹ a1r2 ⫹ a1r3 ⫹ p ⫹ n⫺1 a1r . If we multiply Sn by ⫺r then add the original series, the “interior terms” sum to zero. ⫺ rSn ⫽ ⫺a1r ⫹ 1⫺a1r2 2 ⫹ 1⫺a1r 3 2 ⫹ p ⫹ 1⫺a1rn⫺1 2 ⫹ 1⫺a1rn 2 ------S --S --S S a rn⫺1 ⫹ Sn ⫽ a1 ⫹ a1r ⫹S a1r 2 ⫹ p ⫹ a1rn⫺2 ⫹ 1 Sn ⫺ rSn ⫽

a1 ⫹

0

⫹

0

⫹ 0 ⫹

0

⫹

0

⫹ 1⫺a1rn 2

We then have Sn ⫺ rSn ⫽ a1 ⫺ a1rn, and can now solve for Sn: Sn 11 ⫺ r2 ⫽ a1 ⫺ a1rn

factor out Sn

a1 ⫺ a1r solve for Sn (divide by 1 ⫺ r ) 1⫺r The result is a formula for the nth partial sum of a geometric sequence. n

Sn ⫽

The n th Partial Sum of a Geometric Sequence Given a geometric sequence with first term a1 and common ratio r, the nth partial sum (the sum of the first n terms) is a1 11 ⫺ rn 2 a1 ⫺ a1rn Sn ⫽ ⫽ ,r⫽1 1⫺r 1⫺r In words: The sum of the first n terms of a geometric sequence is the difference of the first and 1n ⫹ 12st term, divided by 1 minus the common ratio.

EXAMPLE 7

Solution

䊳

Computing a Partial Sum

䊳

3i (the first nine powers Use the preceding summation formula to find the sum: i⫽1 of 3). Verify the result on a graphing calculator. The initial terms of this series are 3 ⫹ 9 ⫹ 27 ⫹ p , and we note a1 ⫽ 3, r ⫽ 3, and n ⫽ 9. We could find the first nine terms and add, but using the partial sum formula is much faster and gives Sn ⫽ S9 ⫽

a1 11 ⫺ rn 2 1⫺r 311 ⫺ 39 2 1⫺3

9

兺

sum formula

substitute 3 for a1, 9 for n, and 3 for r

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⫽

31⫺19,6822

⫺2 ⫽ 29,523

simplify result

To verify, we enter u1n2 ⫽ 3 on the Y= screen, and find the sum of the first nine terms of the sequence on the home screen as before. See figure. n

C. You’ve just seen how we can find the nth partial sum of a geometric sequence


䊳

D. The Sum of an Infinite Geometric Series To this point we’ve considered only partial sums of a geometric series. While it is impossible to add an infinite number of these terms, some of these “infinite sums” appear to have what is called a limiting value. The sum appears to get ever closer to this value but never exceeds it—much like the asymptotic behavior of some graphs. We will define the sum of this infinite geometric series to be this limiting value, if it exists. Consider the illustration in Figure 9.44, where a standard sheet of typing paper is cut in half. One of the halves is again cut in half and the process is continued indefinitely, as shown. 1 1 p , 32, with a1 ⫽ 12 and r ⫽ 12. Notice the “halves” create an infinite sequence 21, 14, 18, 16 1 1 1 1 1 The corresponding infinite series is 2 ⫹ 4 ⫹ 8 ⫹ 16 ⫹ 32 ⫹ p ⫹ 21n ⫹ p . Figure 9.44 1 2

1 8 1 2

1 4

1 4 1 8

1 16

1 16

1 32 1 32

1 64

1 64

and so on

Figure 9.45 1 64

1 16

1 32

1 4 1 8

1 2

WORTHY OF NOTE The formula for the sum of an infinite geometric series can also be derived by noting that Sq ⫽ a1 ⫹ a1r ⫹ a1r2 ⫹ a1r3 ⫹ p can be rewritten as Sq ⫽ a1 ⫹ r1a1 ⫹ a1r ⫹ a1r2 ⫹ a1r3 ⫹ p 2 ⫽ a1 ⫹ rSq. Sq ⫺ rSq ⫽ a1 Sq 11 ⫺ r2 ⫽ a1 a1 . Sq ⫽ 1⫺r

If we arrange one of the halves from each stage as shown in Figure 9.45, we would be rebuilding the original sheet of paper. As we add more and more of these halves together, we get closer and closer to the size of the original sheet. We gain an intuitive sense that this series must add to 1, because the pieces of the original sheet of paper must add to 1 whole sheet. To explore this idea further, consider what happens to 1 12 2 n as n becomes large. 1 4 n ⫽ 4: a b ⫽ 0.0625 2

1 8 n ⫽ 8: a b ⬇ 0.004 2

1 12 n ⫽ 12: a b ⬇ 0.0002 2

Further exploration with a calculator seems to support the idea that as n S q, 1 12 2 n S 0, although a definitive proof is left for a future course. In fact, it can be shown that for any 冟r冟 6 1, rn becomes very close to zero as n becomes large. a1 ⫺ a1rn a1 a1rn n In symbols: as n S q, r S 0. For Sn ⫽ ⫽ ⫺ , note that if 1⫺r 1⫺r 1⫺r 冟r冟 6 1 and “we sum an infinite number of terms,” the second term becomes zero, leaving a1 . only the first term. In other words, the limiting value (represented by Sq) is Sq ⫽ 1⫺r Infinite Geometric Series

Given a geometric sequence with first term a1 and 0 r 0 6 1, the sum of the related infinite series is given by a1 Sq ⫽ ;r⫽1 1⫺r If 冟r冟 7 1, no finite sum exists.

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EXAMPLE 8

䊳

789

Computing an Infinite Sum Find the limiting value of each infinite geometric series (if it exists). a. 1 ⫹ 2 ⫹ 4 ⫹ 8 ⫹ p b. 3 ⫹ 2 ⫹ 43 ⫹ 89 ⫹ p c. 0.185 ⫹ 0.000185 ⫹ 0.000000185 ⫹ p

Solution

䊳

Begin by determining if the infinite series is geometric with 冟r冟 6 1. If so, a1 use Sq ⫽ . 1⫺r a. Since r ⫽ 2 (by inspection), a finite sum does not exist. b. Using the ratio of consecutive terms we find r ⫽ 23 and the infinite sum exists. With a1 ⫽ 3, we have 3 3 Sq ⫽ ⫽ 1 ⫽9 2 1⫺3 3 c. This series is equivalent to the repeating decimal 0.185185185 p ⫽ 0.185. The common ratio is r ⫽ 0.000185 0.185 ⫽ 0.001 and the infinite sum exists: Sq ⫽

0.185 5 ⫽ 1 ⫺ 0.001 27 Now try Exercises 93 through 108

䊳

While it is impossible for us to sum an infinite number of terms, we can often see a pattern that strongly suggests a limiting value exists. As in Section 9.2, we do this by increasing the number terms we sum, until such a limiting value seems apparent. But be very careful to note that such a calculator approach is far from an actual proof, and a more definitive means must be used to actually prove a limiting value exists. For the 2 n⫺1 sequence in Example 8(b), we enter 3a b as u(n), and compute the sums S10, S20, 3 S30, and S40 (Figures 9.46 and 9.47), where it indeed appears the limiting value is 9. Figure 9.46

Figure 9.47

D. You’ve just seen how we can find the sum of an infinite geometric series

E. Applications Involving Geometric Sequences and Series Here are a few of the ways these ideas can be put to use. EXAMPLE 9

䊳

Solving an Application of Geometric Sequences: Pendulums A pendulum is any object attached to a fixed point and allowed to swing freely under the influence of gravity. Suppose each swing is 0.9 the length of the previous one. Gradually the swings become shorter and shorter and at some point the pendulum will appear to have stopped (although theoretically it never does). a. How far does the pendulum travel on its eighth swing, if the first was 2 m? b. What is the total distance traveled by the pendulum for these eight swings? c. How many swings until the length of each swing falls below 0.5 m? d. What total distance does the pendulum travel before coming to rest? Verify your response to parts (a) through (c) using a graphing calculator.

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Solution

䊳

a. The lengths of each swing form the terms of a geometric sequence with a1 ⫽ 2 and r ⫽ 0.9. The first few terms are 2, 1.8, 1.62, 1.458, and so on. For the 8th term we have: an ⫽ a1rn⫺1 a8 ⫽ 210.92 8⫺1 ⬇ 0.957

n th term formula substitute 8 for n, 2 for a1, and 0.9 for r

The pendulum travels about 0.957 m on its 8th swing. See Figure 9.48. b. For the total distance traveled after eight swings, we compute the value of S8. Sn ⫽ S8 ⫽

a1 11 ⫺ rn 2

1⫺r 211 ⫺ 0.98 2

1 ⫺ 0.9 ⬇ 11.4

Figure 9.48

Figure 9.49

n th partial sum formula substitute 2 for a1, 0.9 for r, and 8 for n

The pendulum has traveled about 11.4 m by the end of the 8th swing. See Figure 9.49. c. To find the number of swings until the length of each swing is less than 0.5 m, we solve for n in the equation 0.5 ⫽ 210.92 n⫺1. This yields 0.25 ⫽ 10.92 n⫺1 ln 0.25 ⫽ 1n ⫺ 12ln 0.9 ln 0.25 ⫹1⫽n ln 0.9 14.16 ⬇ n

Figure 9.50

divide by 2 take the natural log, apply power property solve for n (exact form) solve for n (approximate form)

After the 14th swing, each successive swing will be less than 0.5 m. See Figure 9.50. d. For the total distance traveled before coming Figure 9.51 to rest, we consider the related infinite geometric series, with a1 ⫽ 2 and r ⫽ 0.9. a1 infinite sum formula Sq ⫽ 1⫺r 2 substitute 2 for a1 and 0.9 for r Sq ⫽ 1 ⫺ 0.9 result ⫽ 20 The pendulum would travel 20 m before coming to rest. Note that summing a larger number of terms on a calculator takes an increasing amount of time. The values for S15 and S150 are shown in Figure 9.51. Now try Exercises 111 and 112

䊳

As mentioned in Section 9.1, sometimes the sequence or series for a particular application will use the preliminary or inaugural term a0, as when an initial amount of money is deposited before any interest is earned, or the efficiency of a new machine after purchase—prior to any wear and tear.

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EXAMPLE 10

䊳

791

Equipment Efficiency—Furniture Manufacturing The manufacturing of mass-produced furniture requires robotic machines to drill numerous holes for the bolts used in the assembly process. When new, the drill bits are capable of drilling through hardwood at a rate of 6 cm/sec. As the bit becomes worn, it loses 4% of its drilling speed per day. a. How many cm/sec can the bit drill through after a 5-day workweek? b. When the drilling speed falls below 3.6 cm/sec, the bit must be replaced. After how many days must the bit be replaced?

Solution

䊳

The efficiency of a new drill bit (prior to use) is given as a0 ⫽ 6 cm/sec. Since the bit loses 4% ⫽ 0.04 of its efficiency per day, it maintains 96% ⫽ 0.96 of its efficiency, showing that after 1 day of use a1 ⫽ 0.96162 ⫽ 5.76. This means the nth term formula will be an ⫽ 5.7610.962 n⫺1. a. At the end of day 5 we have a5 ⫽ 5.7610.962 5⫺1 ⫽ 5.7610.962 4 ⬇ 4.9 After 5 days, the bit can drill through the hardwood at about 4.9 cm/sec. b. To find the number of days until the efficiency falls below 3.6 cm/sec, we replace an with 3.6 and solve for n.

E. You’ve just seen how we can solve application problems involving geometric sequences and series

3.6 ⫽ 5.7610.962 n⫺1 0.625 ⫽ 0.96n⫺1 ln 0.625 ⫽ ln 0.96n⫺1 ln 0.625 ⫽ 1n ⫺ 12 ln 0.96 ln 0.625 ⫽n⫺1 ln 0.96 ln 0.625 ⫹1⫽n ln 0.96 12.5 ⬇ n

substitute 3.6 for an divide take the natural log of both sides power property divide

solve for n (exact form) solution (approximate form)

The drill bit must be replaced after 12 full days of use. Now try Exercises 113 through 126

䊳

9.3 EXERCISES 䊳



1. In a geometric sequence, each successive term is found by the preceding term by a fixed value r. 2. In a geometric sequence, the common ratio r can be found by computing the of any two consecutive terms. 3. The nth term of a geometric sequence is given by an ⫽ , for any n ⱖ 1.

4. For the general sequence a1, a2, a3, p , ak, p , the fifth partial sum is given by S5 ⫽ . 5. Describe/Discuss how the formula for the nth partial sum is related to the formula for the sum of an infinite geometric series. 6. Describe the difference(s) between an arithmetic and a geometric sequence. How can a student prevent confusion between the formulas?

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9–32



Determine if the sequence given is geometric. If yes, name the common ratio. If not, try to determine the pattern that forms the sequence.

7. 4, 8, 16, 32, p

14. 12, 0.12, 0.0012, 0.000012, p 15. ⫺1, 3, ⫺12, 60, ⫺360, p 16. ⫺ 23, 2, ⫺8, 40, ⫺240, p

47. a1 ⫽ 9, an ⫽ 729, r ⫽ 3 48. a1 ⫽ 1, an ⫽ ⫺128, r ⫽ ⫺2 1 , r ⫽ 12 49. a1 ⫽ 16, an ⫽ 64 1 , r ⫽ 12 50. a1 ⫽ 4, an ⫽ 512

p

51. a1 ⫽ ⫺1, an ⫽ ⫺1296, r ⫽ 16

18. ⫺36, 24, ⫺16, 32 3,p

52. a1 ⫽ 2, an ⫽ 1458, r ⫽ ⫺ 13

1 19. 12, 14, 18, 16 ,p

53. 2, ⫺6, 18, ⫺54, p , ⫺4374

8 16 20. 23, 49, 27 , 81, p

54. 3, ⫺6, 12, ⫺24, p , ⫺6144

12 48 192 21. 3, , 2 , 3 , p x x x

55. 64, 3212, 32, 1612, p , 1 56. 243, 81 13, 81, 27 13, p , 1

10 20 40 22. 5, , 2 , 3 , p a a a

57. 38, ⫺34, 32, ⫺3, p , 96 5 5 , 9, ⫺53, ⫺5, p , ⫺135 58. ⫺27

23. 240, 120, 40, 10, 2, p 24. ⫺120, ⫺60, ⫺20, ⫺5, ⫺1, p Write the first four terms of the sequence, given a1 and r.

25. a1 ⫽ 5, r ⫽ 2 27. a1 ⫽ ⫺6, r ⫽

26. a1 ⫽ 2, r ⫽ ⫺4 ⫺12

28. a1 ⫽ 23, r ⫽ 15

29. a1 ⫽ 4, r ⫽ 13

30. a1 ⫽ 15, r ⫽ 15

31. a1 ⫽ 0.1, r ⫽ 0.1

32. a1 ⫽ 0.024, r ⫽ 0.01

Write the expression for the nth term, then find the indicated term for each sequence.

For Exercises 59 through 62, enter the natural numbers 1 through 6 in L1 on a graphing calculator, and the terms of the given sequence in L2. Then determine if the sequence is geometric by (a) graphing the related points to see if they appear to lie on an exponential curve, and (b) computing the successive ratios of all terms. If a geometric sequence, find the nth term and graph the sequence.

59. 131.25, 26.25, 5.25, 1.05, 0.21, 0.042, p 60. 2, 2 25, 10, 10 25, 50, 50 15, p 61. 20, 16, 12, 8, 4, 0, p

33. a1 ⫽ ⫺24, r ⫽ 12; find a7

36. a1 ⫽

1, 12, 2, p

Find the number of terms in each sequence.

13. 3, 0.3, 0.03, 0.003, p

3 20 ,

42. 625, 125, 25, 5, 1, p

46. 0.5, ⫺0.35, 0.245, ⫺0.1715, p

12. ⫺13, ⫺9, ⫺5, ⫺1, 3, p

35. a1 ⫽

12 2 ,

40. ⫺78, 74, ⫺72, 7, ⫺14, p

45. 0.2, 0.08, 0.032, 0.0128, p

11. 2, 5, 10, 17, 26, p

1 ⫺20 ,

⫺19, 13, ⫺1, 3, p

44. 3613, 36, 1213, 12, 4 13, p

10. 128, ⫺32, 8, ⫺2, p

34. a1 ⫽ 48, r ⫽

1 27 ,

43. 12,

9. 3, ⫺6, 12, ⫺24, 48, p

17. 25, 10, 4,

39.

41. 729, 243, 81, 27, 9, p

8. 2, 6, 18, 54, 162, p

8 5,

Identify a1 and r, then write the expression for the nth term an ⴝ a1rnⴚ1 and use it to find a6, a10, and a12.

⫺13;

find a6

r ⫽ ⫺5; find a4

r ⫽ 4; find a5

62.

1 1 1 2 5 , , , , , 1, p 6 3 2 3 6

Find the common ratio r and the value of a1 using the information given (assume r 7 0).

37. a1 ⫽ 2, r ⫽ 12; find a7

63. a3 ⫽ 324, a7 ⫽ 64

64. a5 ⫽ 6, a9 ⫽ 486

38. a1 ⫽ 13, r ⫽ 13; find a8

65. a4 ⫽

66. a2 ⫽ 16 81 , a5 ⫽

4 9,

a8 ⫽

9 4

67. a4 ⫽ 32 3 , a8 ⫽ 54

2 3

68. a3 ⫽ 16 25 , a7 ⫽ 25

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Find the partial sum indicated.

9 89. a3 ⫽ 49, a7 ⫽ 64 ; find S6

69. a1 ⫽ 8, r ⫽ ⫺2; find S12

2 90. a2 ⫽ 16 81 , a5 ⫽ 3 ; find S8

70. a1 ⫽ 2, r ⫽ ⫺3; find S8

91. a3 ⫽ 2 12, a6 ⫽ 8; find S7

71. a1 ⫽ 96, r ⫽ 13; find S5

92. a2 ⫽ 3, a5 ⫽ 9 13; find S7

72. a1 ⫽ 12, r ⫽ 73. a1 ⫽ 8, r ⫽

1 2;

3 2;

find S8

Determine whether the infinite geometric series has a finite sum. If so, find the limiting value.

find S7

74. a1 ⫽ ⫺1, r ⫽ ⫺32; find S10 75. 2 ⫹ 6 ⫹ 18 ⫹ p ; find S6

93. 9 ⫹ 3 ⫹ 1 ⫹ p

76. 2 ⫹ 8 ⫹ 32 ⫹ p ; find S7 77. 16 ⫺ 8 ⫹ 4 ⫺ p ; find S8

95. 3 ⫹ 6 ⫹ 12 ⫹ 24 ⫹ p 96. 4 ⫹ 8 ⫹ 16 ⫹ 32 ⫹ p

78. 4 ⫺ 12 ⫹ 36 ⫺ p ; find S8 1 79. 43 ⫹ 29 ⫹ 27 ⫹ p ; find S9

97. 25 ⫹ 10 ⫹ 4 ⫹ 85 ⫹ p 2 98. 10 ⫹ 2 ⫹ 25 ⫹ 25 ⫹p

80.

1 18

94. 36 ⫹ 24 ⫹ 16 ⫹ p

⫺ 16 ⫹ 12 ⫺ p ; find S7

99. 6 ⫹ 3 ⫹ 32 ⫹ 34 ⫹ p

Find the partial sum indicated, and verify the result using a graphing calculator. For Exercises 85 and 86, use the summation properties from Section 9.1. 5

81.

兺4

10

j

82.

兺2

k

j⫽1

k⫽1

8

7

2 k⫺1 83. 5a b 3 k⫽1

1 j⫺1 84. 3a b 5 j⫽1

兺

兺

10

1 i⫺1 85. 9 a⫺ b 2 i⫽4

1 i⫺1 86. 5 a⫺ b 4 i⫽3

100. ⫺49 ⫹ 1⫺72 ⫹ 1⫺17 2 ⫹ p 101. 6 ⫺ 3 ⫹ 32 ⫺ 34 ⫹ p 102. 10 ⫺ 5 ⫹ 52 ⫺ 54 ⫹ p 103. 0.3 ⫹ 0.03 ⫹ 0.003 ⫹ p 104. 0.63 ⫹ 0.0063 ⫹ 0.000063 ⫹ p 105.

3 2 k a b k⫽1 4 3

兺

106.

1 i 5a b 2 i⫽1

107.

5 j 9 a⫺ b 4 j⫽1

108.

4 k 12 a b 3 k⫽1

q

8

兺

793

兺

q

q

兺

兺 q

兺

Find the indicated partial sum using the information given. Write all results in simplest form.

87. a2 ⫽ ⫺5, a5 ⫽

1 25 ;

find S5

88. a3 ⫽ 1, a6 ⫽ ⫺27; find S6 䊳


109. Sum of the cubes of the first n natural numbers: n2 1n ⴙ 12 2 Sn ⴝ 4 3 Compute 1 ⫹ 23 ⫹ 33 ⫹ p ⫹ 83 using the formula given. Then confirm the result by direct calculation.

110. Student loan payment: An ⴝ P11 ⴙ r2 n If P dollars is borrowed at an annual interest rate r with interest compounded annually, the amount of money to be paid back after n years is given by the indicated formula. Find the total amount of money that the student must repay to clear the loan, if $8000 is borrowed at 4.5% interest and the loan is paid back in 10 yr.

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794 䊳


9–34

APPLICATIONS

Write the nth term formula for each application, then solve.

111. Pendulum movement: On each swing, a pendulum travels only 80% as far as it did on the previous swing. If the first swing is 24 ft, how far does the pendulum travel on the 7th swing? What total distance is traveled before the pendulum comes to rest? 112. Tire swings: Ernesto is swinging to and fro on his backyard tire swing. Using his legs and body, he pumps each swing until reaching a maximum height, then suddenly relaxes until the swing comes to a stop. With each swing, Ernesto travels 75% as far as he did on the previous swing. If the first arc (or swing) is 30 ft, find the distance Ernesto travels on the 5th arc. What total distance will he travel before coming to rest?

Identify the inaugural term and write the nth term formula for each application, then solve.

113. Depreciation— automobiles: A certain new SUV depreciates in value about 20% per year (meaning it holds 80% of its value each year). If the SUV is purchased for $46,000, how much is it worth 4 yr later? How many years until its value is less than $5000? 114. Depreciation—business equipment: A new photocopier under heavy use will depreciate about 25% per year (meaning it holds 75% of its value each year). If the copier is purchased for $7000, how much is it worth 4 yr later? How many years until its value is less than $1246? 115. Equipment aging— industrial oil pumps: Tests have shown that the pumping power of a heavyduty oil pump decreases by 3% per month. If the pump can move 160 gallons per minute (gpm) new, how many gpm can the pump move 8 months later? If the pumping rate falls below 118 gpm, the pump must be replaced. How many months until this pump is replaced?

116. Equipment aging—lumber production: At the local mill, a certain type of saw blade can saw approximately 2 log-feet/sec when it is new. As time goes on, the blade becomes worn, and loses 6% of its cutting speed each week. How many logfeet/sec can the saw blade cut after 6 weeks? If the cutting speed falls below 1.2 log-feet/sec, the blade must be replaced. During what week of operation will this blade be replaced? 117. Population growth—United States: At the beginning of the year 2000, the population of the United States was approximately 277 million. If the population is growing at a rate of 2.3% per year, what was the population in 2010, 10 yr later? 118. Population growth—space colony: The population of the Zeta Colony on Mars is 1000 people. Determine the population of the Colony 20 yr from now, if the population is growing at a constant rate of 5% per year. 119. Creating a vacuum: To create a vacuum, a hand pump is used to remove the air from an air-tight cube with a volume of 462 in3. With each stroke of the pump, two-fifths of the air that remains in the cube is removed. How much air remains inside after the 5th stroke? How many strokes are required to remove all but 12.9 in3 of the air? 120. Atmospheric pressure: In 1654, scientist Otto Von Guericke performed his famous demonstration of atmospheric pressure and the strength of a vacuum in front of Emperor Ferdinand III of Hungary. After joining two hemispheres with mating rims, he used a vacuum pump to remove all of the air from the sphere formed. He then attached a team of 15 horses to each hemisphere and despite their efforts, they could not pull the hemispheres apart. If the sphere held a volume of 4200 in3 of air and one-tenth of the remaining air was removed with each stroke of the pump, how much air was still in the sphere after the 11th stroke? How many strokes were required to remove 85% of the air? 121. Treating swimming pools: In preparation for the summer swim season, chlorine is added to swimming pools to control algae and bacteria. However, careful measurements must be taken as levels above 5 ppm (parts per million) can be highly irritating to the eyes and throat, while levels below 1 ppm will be ineffective (3.0 to 3.5 ppm is ideal). In addition, the water must be treated daily since within a 24-hr period, about 25% of the chlorine will dissipate into the air. If the chlorine

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level in a swimming pool is 8 ppm after its initial treatment, how many days should the County Pool Supervisor wait before opening it up to the public? If left untreated, how many days until the chlorine level drops below 1 ppm? 122. Venting landfill gases: The gases created from the decomposition of waste in landfills must be carefully managed, as their release can cause terrible odors, harm the landfill structure, damage vegetation, or even cause an explosion. Suppose the accumulated volume of gas is 50,000 ft3, and civil engineers are able to vent 2.5% of this gas into the atmosphere daily. What volume of gas remains after 21 days? How many days until the volume of gas drops below 10,000 ft3? 123. Population growth—bacteria: A biologist finds that the population of a certain type of bacteria doubles each half-hour. If an initial culture has 50 bacteria, what is the population after 5 hr? How long will it take for the number of bacteria to reach 204,800? 䊳

124. Population growth—boom towns: Suppose the population of a “boom town” in the old west doubled every 2 months after gold was discovered. If the initial population was 219, what was the population 8 months later? How many months until the population exceeded 28,000? 125. Elastic rebound—super balls: Megan discovers that a rubber ball dropped from a height of 2 m rebounds four-fifths of the distance it has previously fallen. How high does it rebound on the 7th bounce? How far does the ball travel before coming to rest? 126. Elastic rebound—computer animation: The screen saver on my computer is programmed to send a colored ball vertically down the middle of the screen so that it rebounds 95% of the distance it last traversed. If the ball always begins at the top and the screen is 36 cm tall, how high does the ball bounce on its 8th rebound? How far does the ball travel before coming to rest (and a new screen saver starts)?


127. A standard piece of typing paper is approximately 0.001 in. thick. Suppose you were able to fold this piece of paper in half 26 times. How thick would the result be? As tall as a hare, as tall as a hen, as tall as a horse, as tall as a house, or over 1 mi high? Find the actual height by computing the 27th term of a geometric sequence. Discuss what you find. 128. As part of a science experiment, identical rubber balls are dropped from a given height onto these surfaces: slate, cement, and asphalt. When dropped onto slate, the ball rebounds 80% of the height from which it last fell. Onto cement the figure is 75% and onto asphalt the figure is 70%. The ball is dropped from 130 m onto the slate, 175 m onto the cement, and 200 m onto the asphalt. Which ball has traveled the shortest total distance at the time of the fourth bounce? Which ball will travel farthest before coming to rest?

䊳

795


129. Consider the following situation. A person is hired at a salary of $40,000 per year, with a guaranteed raise of $1750 per year. At the same time, inflation is running about 4% per year. How many years until this person’s salary is overtaken and eaten up by the actual cost of living? 130. Find an alternative formula for the sum n

Sn ⫽

兺 log k, that does not use the sigma notation.

k⫽1

131. Verify the following statements: a. If a1, a2, a3, p , an is a geometric sequence with r and a1 greater than zero, then log a1, log a2, log a3, p , log an is an arithmetic sequence. b. If a1, a2, a3, p , an is an arithmetic sequence, then 10a1, 10a2, p , 10an, is a geometric sequence.


132. (3.2) Find the zeroes of f using the quadratic formula: f 1x2 ⫽ x2 ⫹ 5x ⫹ 9. 133. (R.5) Solve for x:

4 3 1 ⫺ ⫽ x ⫺ 5 x ⫹ 2 x ⫺ 3x ⫺ 10 2

134. (4.5) Graph the rational function: h1x2 ⫽

x2 x⫺1

135. (5.6) Given the logistic function shown, find p(50), p(75), p(100), and p(150): 4200 p1t2 ⫽ 1 ⫹ 10e⫺0.055t

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9.4

Mathematical Induction


A. Use subscript notation to evaluate and compose functions B. Apply the principle of mathematical induction to sum formulas involving natural numbers C. Apply the principle of mathematical induction to general statements involving natural numbers

EXAMPLE 1

䊳

Solution

䊳

A. You’ve just seen how we can use subscript notation to evaluate and compose functions

Since middle school (or even before) we have accepted that, “The product of two negative numbers is a positive number.” But have you ever been asked to prove it? It’s not as easy as it seems. We may think of several patterns that yield the result, analogies that indicate its truth, or even number line illustrations that lead us to believe the statement. But most of us have never seen a proof (see www.mhhe.com/coburn). In this section, we introduce one of mathematics’ most powerful tools for proving a statement, called proof by induction.

A. Subscript Notation and Function Notation One of the challenges in understanding a proof by induction is working with the notation. Earlier in the chapter, we introduced subscript notation as an alternative to function notation, since it is more commonly used when the functions are defined by a sequence. But regardless of the notation used, the functions can still be simplified, evaluated, composed, and even graphed. Consider the function f 1x2 ⫽ 3x2 ⫺ 1 and the sequence defined by an ⫽ 3n2 ⫺ 1. Both can be evaluated and graphed, with the only difference being that f(x) is continuous with domain x 僆⺢, while an is discrete (made up of distinct points) with domain n 僆⺞. Using Subscript Notation for a Composition

For f 1x2 ⫽ 3x2 ⫺ 1 and an ⫽ 3n2 ⫺ 1, find f 1k ⫹ 12 and ak⫹1.

f 1k ⫹ 12 ⫽ 31k ⫹ 12 2 ⫺ 1 ⫽ 31k2 ⫹ 2k ⫹ 12 ⫺ 1 ⫽ 3k2 ⫹ 6k ⫹ 2

ak⫹1 ⫽ 31k ⫹ 12 2 ⫺ 1 ⫽ 31k2 ⫹ 2k ⫹ 12 ⫺ 1 ⫽ 3k2 ⫹ 6k ⫹ 2 Now try Exercises 7 through 18

䊳

No matter which notation is used, every occurrence of the input variable is replaced by the new value or expression indicated by the composition.

B. Mathematical Induction Applied to Sums Consider the sum of odd numbers 1 ⫹ 3 ⫹ 5 ⫹ 7 ⫹ 9 ⫹ 11 ⫹ 13 ⫹ p . The sum of the first four terms is 1 ⫹ 3 ⫹ 5 ⫹ 7 ⫽ 16, or S4 ⫽ 16. If we now add a5 (the next term in line), would we get the same answer as if we had simply computed S5? Common sense would say, “Yes!” since S5 ⫽ 1 ⫹ 3 ⫹ 5 ⫹ 7 ⫹ 9 ⫽ 25 and S4 ⫹ a5 ⫽ 16 ⫹ 9 ⫽ 25✓. In diagram form, we have add next term a5 ⫽ 9 to S4

>

1 ⫹ 3 ⫹ 5 ⫹ 7 ⫹ 9 ⫹ 11 ⫹ 13 ⫹ 15 ⫹ p c c S4 S5

sum of 4 terms sum of 5 terms

Our goal is to develop this same degree of clarity in the notational scheme of things. For a given series, if we find the kth partial sum Sk (shown next) and then add the next term ak⫹1, would we get the same answer if we had simply computed Sk⫹1? In other words, is Sk ⫹ ak⫹1 ⫽ Sk⫹1 true?

796

9–36

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Section 9.4 Mathematical Induction

797

add next term ak1

>

a1 a2 a3 p ak1 ak ak1 p an–1 an c c sum of k terms Sk sum of k 1 terms

Sk1

Now, let’s return to the sum 1 3 5 7 p 12n 12. This is an arithmetic series with a1 1, d 2, and nth term an 2n 1. Using the sum formula for an arithmetic sequence, an alternative formula for this sum can be established. Sn

n1a1 an 2

2 n11 2n ⴚ 12 2

No matter how distant the city or how many relay stations are involved, if the generating plant is working and the kth station relays to the (k 1)st station, the city will get its power.

substitute 1 for a1 and 2nⴚ1 for an

n12n2

n2

WORTHY OF NOTE

summation formula for an arithmetic sequence

2

simplify result

This shows that the sum of the first n positive odd integers is given by Sn n2. As a check we compute S5 1 3 5 7 9 25 and compare to S5 52 25✓. We also note S6 62 36, and S5 a6 25 11 36, showing S6 S5 a6. For more on this relationship, see Exercises 19 through 24. While it may seem simplistic now, showing S5 a6 S6 and Sk ak1 Sk1 (in general) is a critical component of a proof by induction. Unfortunately, general summation formulas for many sequences cannot be established from known formulas. In addition, just because a formula works for the first few values of n, we cannot assume that it will hold true for all values of n (there are infinitely many). As an illustration, the formula an n2 n 41 yields a prime number for every natural number n from 1 to 40, but fails to yield a prime for n 41. This helps demonstrate the need for a more conclusive proof, particularly when a relationship appears to be true, and can be “verified” in a finite number of cases, but whether it is true in all cases remains in doubt. Proof by induction is based on a relatively simple idea. To help understand how it works, consider n relay stations that are used to transport electricity from a generating plant to a distant city. If we know the generating plant is operating, and if we assume that the kth relay station (any station (k 1)st kth Generating plant in the series) is making the transrelay relay fer to the 1k 12st station (the next station in the series), then we’re sure the city will have electricity. This idea can be applied mathematically as follows. Consider the statement, “The sum of the first n positive even integers is n2 n.” In other words, 2 4 6 8 p 2n n2 n. We can certainly verify the statement for the first few even numbers: 112 2 1 2 The first even number is 2 and p 122 2 2 6 The sum of the first two even numbers is 2 4 6 and p The sum of the first three even numbers is 132 2 3 12 2 4 6 12 and p The sum of the first four even numbers is 2 4 6 8 20 and p 142 2 4 20

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While we could continue this process for a very long time (or even use a computer), no finite number of checks can prove a statement is universally true. To prove the statement true for all positive integers, we use a reasoning similar to that applied in the relay stations example. If we are sure the formula works for n 1 (the generating station is operating), and if the truth of n k implies that n k 1 is true [the kth relay station is transferring electricity to the 1k 12st station], then the statement is true for all n (the city will get its electricity). The case where n 1 is called the base case of an inductive proof, and the assumption that the formula is true for n k is called the induction hypothesis. When the induction hypothesis is applied to a sum formula, we attempt to show that Sk ak1 Sk1. Since k and k 1 are arbitrary, the statement must be true for all n. Mathematical Induction Applied to Sums Let Sn be a sum formula involving positive integers. If 1. S1 is true, and 2. the truth of Sk implies that Sk1 is true, then Sn must be true for all positive integers n.

WORTHY OF NOTE To satisfy our finite minds, it might help to show that Sn is true for the first few cases, prior to extending the ideas to the infinite case.

EXAMPLE 2

䊳

Both parts 1 and 2 must be verified for the proof to be complete. Since the process requires the terms Sk, ak1, and Sk1, we will usually compute these first.

Proving a Statement Using Mathematical Induction Use induction to prove that the sum of the first n perfect squares is given by n1n 12 12n 12 . 1 4 9 16 25 p n2 6

Solution

䊳

Given an n2 and Sn

n1n 12 12n 12 6

, the needed components are p

ak1 1k 12 2 For an n2: ak k2 and n1n 12 12n 12 k1k 12 12k 12 1k 12 1k 2212k 32 : Sk For Sn and Sk1 6 6 6 1. Show Sn is true for n 1. Sn S1

n1n 12 12n 12 1122 132

sum formula

6

base case: n 1

6 1✓

result checks, the first term is 1

2. Assume Sk is true, 1 4 9 16 p k2

k1k 1212k 12 6

and use it to show the truth of Sk1 follows. That is,

1k 121k 2212k 32

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

⎫ ⎬ ⎭

1 4 9 16 p k2 1k 12 2

induction hypothesis: Sk is true

6

Sk

ak1

Sk1

Working with the left-hand side, we have 1 4 9 16 p k2 1k 12 2 ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

k1k 12 12k 12

1k 12 6 k1k 12 12k 12 61k 12 2 6

2

induction hypothesis: substitute

k1k 1212k 12

for 1 4 9 16 25 p k2 common denominator

6

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1k 12 3 k12k 12 61k 12 4

factor out k 1

6 1k 12 3 2k2 7k 64

multiply and combine terms

6 1k 121k 2212k 32

799

factor the trinomial, result is Sk1

6

Since the truth of Sk1 follows from Sk, the formula is true for all n.

B. You’ve just seen how we can apply the principle of mathematical induction to sum formulas involving natural numbers


䊳

C. The General Principle of Mathematical Induction Proof by induction can be used to verify many other kinds of relationships involving a natural number n. In this regard, the basic principles remain the same but are stated more broadly. Rather than using Sn to represent a sum, we will use Pn to represent any proposed statement or relationship we might wish to verify. This broadens the scope of the proof and makes it more widely applicable, while maintaining its connection to the sum formulas verified earlier. The General Principle of Mathematical Induction Let Pn be a statement involving natural numbers. If 1. P1 is true, and 2. the truth of Pk implies that Pk1 is also true then Pn must be true for all natural numbers n.

EXAMPLE 3

䊳

Proving a Statement Using the General Principle of Mathematical Induction Use the general principle of mathematical induction to show the statement Pn is true for all natural numbers n. Pn: 2n n 1

Solution

䊳

The statement Pn is defined as 2n n 1. This means that Pk is represented by 2k k 1 and Pk1 by 2k1 k 2. 1. Show Pn is true for n 1: Pn: P1:

2n n 1 21 1 1 2 2✓

given statement base case: n 1 true

Although not a part of the formal proof, a table of values can help to illustrate the relationship we’re trying to establish. It appears that the statement is true. n

1

2

3

4

5

2n

2

4

8

16

32

n1

2

3

4

5

6

2. Assume that Pk is true. Pk:

2k k 1

induction hypothesis

and use it to show the truth of Pk1. That is, Pk1:

2k1 1k 12 1 k2

Begin by working with the left-hand side of the inequality, 2k1.

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2k1 212k 2

21k ⴙ 12 2k 2

properties of exponents

induction hypothesis: substitute k 1 for 2k (symbol changes since k 1 is less than or equal to 2k) distribute

Since k is a positive integer, 2k1 2k 2 k 2, showing 2k1 k 2.

WORTHY OF NOTE Note there is no reference to an, ak, or ak+1 in the statement of the general principle of mathematical induction.

EXAMPLE 4

Since the truth of Pk1 follows from Pk, the formula is true for all n. Now try Exercises 39 through 42

䊳

䊳

Proving Divisibility Using Mathematical Induction Let Pn be the statement, “4n 1 is divisible by 3 for all positive integers n.” Use mathematical induction to prove that Pn is true.

Solution

䊳

If a number is evenly divisible by three, it can be written as the product of 3 and some positive integer we will call p. 1. Show Pn is true for n 1: Pn: 4n 1 3p P1: 4112 1 3p 3 3p ✓

given statement, p 僆⺪ substitute 1 for n statement is true for n 1

2. Assume that Pk is true. Pk:

4k 1 3p 4k 3p 1

induction hypothesis isolate 4k

and use it to show the truth of Pk1. That is, Pk1:

4k1 1 3q for q 僆⺪ is also true.

Beginning with the left-hand side we have: 4k1 1 4 # 4k 1 4 # 13p ⴙ 12 1 12p 3 314p 12 3q

properties of exponents induction hypothesis: substitute 3p 1 for 4k distribute and simplify factor

The last step shows 4 1 is divisible by 3. Since the original statement is true for n 1, and the truth of Pk implies the truth of Pk1, the statement, “4n 1 is divisible by 3” is true for all positive integers n. k1


C. You’ve just seen how we can apply the principle of mathematical induction to general statements involving natural numbers

䊳

We close this section with some final notes. Although the base step of a proof by induction seems trivial, both the base step and the induction hypothesis are necessary 1 1 parts of the proof. For example, the statement n 6 is false for n 1, but true for 3 3n all other positive integers. Finally, for a fixed natural number p, some statements are false for all n 6 p, but true for all n p. By modifying the base case to begin at p, we can use the induction hypothesis to prove the statement is true for all n greater than p. For example, n 6 13n2 is false for n 6 4, but true for all n 4.

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801


9.4 EXERCISES 䊳



1. No statement

number of verifications can prove a true.

3. Assuming that a statement/formula is true for n ⫽ k is called the . 5. Explain the equation Sk ⫹ ak⫹1 ⫽ Sk⫹1. Begin by saying, “Since the kth term is arbitrary p ” (continue from here). 䊳

7. an ⫽ 10n ⫺ 6 9. an ⫽ n 11. an ⫽ 2n⫺1

8. an ⫽ 6n ⫺ 4 10. an ⫽ 7n

12. an ⫽ 213n⫺1 2

For the given sum formula Sn, find S4, S5, Sk, and Skⴙ1.

13. Sn ⫽ n15n ⫺ 12 15. Sn ⫽

n1n ⫹ 12

2 17. Sn ⫽ 2n ⫺ 1

14. Sn ⫽ n13n ⫺ 12 16. Sn ⫽

7n1n ⫹ 12

2 18. Sn ⫽ 3n ⫺ 1

Verify that S4 ⴙ a5 ⴝ S5 for each exercise. Note that each Sn is identical to those in Exercises 13 through 18.

19. an ⫽ 10n ⫺ 6; Sn ⫽ n15n ⫺ 12 20. an ⫽ 6n ⫺ 4; Sn ⫽ n13n ⫺ 12 21. an ⫽ n; Sn ⫽

n1n ⫹ 12

22. an ⫽ 7n; Sn ⫽ 23. an ⫽ 2

n⫺1

2 7n1n ⫹ 12 2

; Sn ⫽ 2 ⫺ 1 n

24. an ⫽ 213n⫺1 2; Sn ⫽ 3n ⫺ 1


25. Sum of the first n cubes (alternative form): (1 ⴙ 2 ⴙ 3 ⴙ 4 ⴙ p ⴙ n)2 Earlier we noted the formula for the sum of the n2 1n ⫹ 12 2 . An alternative is first n cubes was 4 given by the formula shown. a. Verify the formula for n ⫽ 1, 5, and 9. 䊳

4. The graph of a sequence is , meaning it is made up of distinct points. 6. Discuss the similarities and differences between mathematical induction applied to sums and the general principle of mathematical induction.


For the given nth term an, find a4, a5, ak, and akⴙ1.

䊳

2. Showing a statement is true for n ⫽ 1 is called the of an inductive proof.

b. Verify the formula using 1⫹2⫹3⫹p⫹n⫽

n1n ⫹ 12 2

.

26. Powers of the imaginary unit: in ⴙ 4 ⴝ in, where i ⴝ 1 ⴚ1 Use a proof by induction to prove that powers of the imaginary unit are cyclic. That is, that they cycle through the numbers i, ⫺1, ⫺i, and 1 for consecutive powers.

APPLICATIONS

Use mathematical induction to prove the indicated sum formula is true for all natural numbers n.

27. 2 ⫹ 4 ⫹ 6 ⫹ 8 ⫹ 10 ⫹ p ⫹ 2n; an ⫽ 2n, Sn ⫽ n1n ⫹ 12 28. 3 ⫹ 7 ⫹ 11 ⫹ 15 ⫹ 19 ⫹ p ⫹ 14n ⫺ 12; an ⫽ 4n ⫺ 1, Sn ⫽ n12n ⫹ 12

29. 5 ⫹ 10 ⫹ 15 ⫹ 20 ⫹ 25 ⫹ p ⫹ 5n; 5n1n ⫹ 12 an ⫽ 5n, Sn ⫽ 2 30. 1 ⫹ 4 ⫹ 7 ⫹ 10 ⫹ 13 ⫹ p ⫹ 13n ⫺ 22; n13n ⫺ 12 an ⫽ 3n ⫺ 2, Sn ⫽ 2

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31. 5 9 13 17 p 14n 12; an 4n 1, Sn n12n 32 32. 4 12 20 28 36 p 18n 42; an 8n 4, Sn 4n2

37.

1 1 1 1 p ; 1132 3152 5172 12n 1212n 12 1 n an , Sn 2n 1 12n 12 12n 12

1 1 1 1 p ; 1122 2132 3142 n1n 12 1 n , Sn an n 1 n1n 12 Use the principle of mathematical induction to prove that each statement is true for all natural numbers n.

33. 3 9 27 81 243 p 3n; 313n 12 an 3n, Sn 2 34. 5 25 125 625 p 5n; 515n 12 an 5n, Sn 4 35. 2 4 8 16 32 64 p 2n; an 2n, Sn 2n1 2

38.

39. 3n 2n 1

40. 2n n 1

41. 3 # 4n1 4n 1

42. 4 # 5n1 5n 1

43. n2 7n is divisible by 2

36. 1 8 27 64 125 216 p n3; n2 1n 12 2 an n3, Sn 4

44. n3 n 3 is divisible by 3 45. n3 3n2 2n is divisible by 3 46. 5n 1 is divisible by 4 47. 6n 1 is divisible by 5

䊳


48. You may have noticed that the sum formula for the first n integers was quadratic, and the formula for the first n integer squares was cubic. Is the formula for the first n integer cubes, if it exists, a quartic (degree four) function? Use your calculator to run a quartic regression on the first five perfect cubes (enter 1 through 5 in L1 and the cumulative sums in L2). What did you find? How is this exercise related to Exercise 36? xn 1 11 x x2 x3 p xn1 2. x1 50. Use mathematical induction to prove that for 14 24 34 p n4, where an n4,

49. Use mathematical induction to prove that

Sn 䊳

n1n 12 12n 12 13n2 3n 12 30

.


51. (7.2) Given the matrices A c

1 3

53. (1.1) State the equation of the circle whose graph is shown here.

2 d and 1

2 1 d , find A B, A B, 2A 3B, 4 3 AB, BA, and B1. B c

52. (2.5) State the domain and range of the piecewise function shown here.

y 10 (1, 7) 8 6 (4, 3) 4 2 108642 2 4 6 8 10

y 5 4 3 (1, 1) 2 1 54321 1 2 3 4 5

1 2 3

(3, 2)

5 x

2 4 6 8 10 x

54. (5.5) Solve: 3e2x1 5 17. Answer in exact form.

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MID-CHAPTER CHECK

1. an 7n 4

Find the number of terms in each series and then find the sum. Verify results on a graphing calculator. 12. 2 5 8 11 p 74

2. an n2 3

13.

In Exercises 1 through 3, the nth term is given. Write the first three terms of each sequence and find a9.

3. an 112 n 12n 12

兺3

32 52 72 p 31 2

14. For an arithmetic series, a3 8 and a7 4. Find S10.

4

4. Evaluate the sum

1 2

n1

15. For a geometric series, a3 81 and a6 3. Find S10.

n1

5. Rewrite using sigma notation. 1 4 7 10 13 16

16. Identify a1 and the common ratio r. Then find an expression for the general term an. a. 2, 6, 18, 54, p

Match each formula to its correct description. n1a1 an 2 6. Sn 7. an a1rn1 2 a1 8. Sq 9. an a1 1n 12d 1r a1 11 rn 2 10. Sn 1r a. sum of an infinite geometric series

1 b. 12, 14, 18, 16 ,p

17. Find the number of terms in the series then compute the sum. 541 181 16 p 812 18. Find the infinite sum (if it exists). 49 172 112 117 2 p 19. Barrels of toxic waste are stacked at a storage facility in pyramid form, with 60 barrels in the first row, 59 in the second row, and so on, until there are 10 barrels in the top row. How many barrels are in the storage facility? Verify results using a graphing calculator.

b. nth term formula for an arithmetic series c. sum of a finite geometric series d. summation formula for an arithmetic series

20. As part of a conditioning regimen, a drill sergeant orders her platoon to do 25 continuous standing broad jumps. The best of these recruits was able to jump 96% of the distance from the previous jump, with a first jump distance of 8 ft. Use a sequence/ series to determine the distance the recruit jumped on the 15th try, and the total distance traveled by the recruit after all 25 jumps. Verify results using a graphing calculator.

e. nth term formula for a geometric series 11. Identify a1 and the common difference d. Then find an expression for the general term an. a. 2, 5, 8, 11, p b. 32, 94, 3, 15 4,p

REINFORCING BASIC CONCEPTS Applications of Summation The properties of summation play a large role in the development of key ideas in a first semester calculus course, and the following summation formulas are an integral part of these ideas. The first three formulas were verified in Section 9.4, while proof of the fourth was part of Exercise 48 on page 802. n

(1)

兺

n

c cn

i1

(2)

兺

i

i1

n1n 12 2

n

(3)

兺

i2

n1n 1212n 12

i1

6

n

(4)

兺

i3

i1

n2 1n 12 2 4

To see the various ways they can be applied consider the following. Illustration 1 䊳 Over several years, the owner of Morgan’s LawnCare has noticed that the company’s monthly profits (in thousands) can be approximated by the sequence an 0.0625n3 1.25n2 6n, with the points plotted in Figure 9.52 (the continuous graph is shown for effect only). Find the company’s approximate annual profit.

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Solution 䊳 The most obvious approach would be to simply compute terms a1 through a12 (January through December) and find their sum: sum(seq(Y1, X, 1, 12), which gives a result of 35.75 or $35,750. As an alternative, we could add the amount of profit earned by the company in the first 8 months, then add the amount the company lost (or broke even) during the last 4 months. In other words, we could apply summation property IV: 12

兺a

n

8

Figure 9.52 12

0

12

兺a

12

n

兺a

n

[(see Figure 9.53), which gives the same result:

42 16.252 35.75 or $35,750]. As a third option, we could use summation properties along with the appropriate summation formulas, and compute the result manually. Note the function is now written in terms of “i.” Distribute summation and factor out constants (properties II and III): i1

i1

12

兺

i1

i9

10.0625i3 1.25i2 6i2 0.0625

12

兺

12

i3 1.25

i1

兺

5

Figure 9.53

12

i2 6

i1

兺i

i1

Replace each summation with the appropriate summation formula, then substitute 12 for n: 0.0625 c 0.0625 c

n2 1n 12 2 4 2 1122 1132 2

d 1.25 c

d 1.25 c

n1n 12 12n 12 6 1122 1132 1252

4 6 0.0625160842 1.2516502 61782 35.75

d 6c

d 6c

n1n 12

1122 1132 2

2

d

d

As we expected, the result shows profit was $35,750. While some approaches seem “easier” than others, all have great value, are applied in different ways at different times, and are necessary to adequately develop key concepts in future classes. Exercise 1: Repeat Illustration 1 if the profit sequence is an 0.125x3 2.5x2 12x.

9.5

Counting Techniques


A. Count possibilities using B.

C.

D.

E.

lists and tree diagrams Count possibilities using the fundamental principle of counting Quick-count distinguishable permutations Quick-count nondistinguishable permutations Quick-count using combinations

How long would it take to estimate the number of fans sitting shoulder-to-shoulder at a sold-out basketball game? Well, it depends. You could actually begin counting 1, 2, 3, 4, 5, p , which would take a very long time, or you could try to simplify the process by counting the number of fans in the first row and multiplying by the number of rows. Techniques for “quick-counting” the objects in a set or various subsets of a large set play an important role in a study of probability.

A. Counting by Listing and Tree Diagrams Consider the simple spinner shown in Figure 9.54, which is divided into three equal parts. What are the different possible outcomes for two spins, spin 1 followed by spin 2? We might begin by organizing the possibilities using a tree diagram. As the name implies, each choice or possibility appears as the branch of a tree, with the total possibilities being equal to the number of (unique)

Figure 9.54 B A

C

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Section 9.5 Counting Techniques

Figure 9.55

paths from the beginning point to the end of a branch. Figure 9.55 shows how the spinner exercise would appear (possibilities for two spins). Moving from top to bottom we can trace nine possible paths: AA, AB, AC, BA, BB, BC, CA, CB, and CC.

Begin

A A

EXAMPLE 1

䊳

B

B C

A

B

C C

A

B

C

Listing Possibilities Using a Tree Diagram A basketball player is fouled and awarded three free throws. Let H represent the possibility of a hit (basket is made), and M the possibility of a miss. Determine the possible outcomes for the three shots using a tree diagram.

Solution

䊳

Each shot has two possibilities, hit (H) or miss (M), so the tree will branch in two directions at each level. As illustrated in the figure, there are a total of eight possibilities: HHH, HHM, HMH, HMM, MHH, MHM, MMH, and MMM. Begin

H

M

H H

M M

H

H M

H

M M

H

M


WORTHY OF NOTE Sample spaces may vary depending on how we define the experiment, and for simplicity’s sake we consider only those experiments having outcomes that are equally likely.

䊳

To assist our discussion, an experiment is any task that can be done repeatedly and has a well-defined set of possible outcomes. Each repetition of the experiment is called a trial. A sample outcome is any potential outcome of a trial, and a sample space is a set of all possible outcomes. In our first illustration, the experiment was spinning a spinner, there were three sample outcomes (A, B, or C), the experiment had two trials (spin 1 and spin 2), and there were nine elements in the sample space. Note that after the first trial, each of the three sample outcomes will again have three possibilities (A, B, and C). For two trials we have 32 9 possibilities, while three trials would yield a sample space with 33 27 possibilities. In general, for N equally likely outcomes we have A “Quick-Counting” Formula for a Sample Space If an experiment has N sample outcomes that are equally likely and the experiment is repeated t times, the number of elements in the sample space is N t.

EXAMPLE 2

䊳

Counting the Outcomes in a Sample Space Many combination locks have the digits 0 through 39 arranged along a circular dial. Opening the lock requires stopping at a sequence of three numbers within this range, going counterclockwise to the first number, clockwise to the second, and counterclockwise to the third. How many three-number combinations are possible?

5

35

10

30 25

15 20

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Solution

䊳

A. You’ve just seen how we can count possibilities using lists and tree diagrams

There are 40 sample outcomes 1N 402 in this experiment, and three trials 1t 32. The number of possible combinations is identical to the number of elements in the sample space. The quick-counting formula gives 403 64,000 possible combinations. Now try Exercises 11 and 12

䊳

B. Fundamental Principle of Counting The number of possible outcomes may differ depending on how the event is defined. For example, some security systems, license plates, and telephone numbers exclude certain numbers. For example, phone numbers cannot begin with 0 or 1 because these are reserved for operator assistance, long distance, and international calls. Constructing a three-digit area code is like filling in three blanks with three digit digit digit

digits. Since the area code must start with a number between 2 and 9, there are eight choices for the first blank. Since there are 10 choices for the second digit and 10 choices for the third, there are 8 # 10 # 10 800 possibilities in the sample space. EXAMPLE 3

䊳

Counting Possibilities for a Four-Digit Security Code A digital security system requires that you enter a four-digit PIN (personal identification number), using only the digits 1 through 9. How many codes are possible if a. Repetition of digits is allowed? b. Repetition is not allowed? c. The first digit must be even and repetitions are not allowed?

Solution

䊳

a. Consider filling in the four blanks

digit digit digit digit

with the number of

ways the digit can be chosen. If repetition is allowed, the experiment is similar to that of Example 2 and there are N t 94 6561 possible PINs. b. If repetition is not allowed, there are only eight possible choices for the second digit of the PIN, then seven for the third, and six for the fourth. The number of possible PIN numbers decreases to 9 # 8 # 7 # 6 3024. c. There are four choices for the first digit (2, 4, 6, 8). Once this choice has been made there are eight choices for the second digit, seven for the third, and six for the last: 4 # 8 # 7 # 6 1344 possible codes. Now try Exercises 13 through 16

䊳

Given any experiment involving a sequence of tasks, if the first task can be completed in p possible ways, the second task has q possibilities, and the third task has r possibilities, a tree diagram will show that the number of possibilities in the sample space for task1–task2–task3 is p # q # r. This situation is simply a variation of the previous quick-counting formula. Even though the examples we’ve considered to this point have varied a great deal, this idea was fundamental to counting all possibilities in a sample space and is, in fact, known as the fundamental principle of counting (FPC). Fundamental Principle of Counting (Applied to Three Tasks) Given any experiment with three defined tasks, if there are p possibilities for the first task, q possibilities for the second, and r possibilities for the third, the total number of ways the experiment can be completed is p # q # r. This fundamental principle can be extended to include any number of tasks, and can be applied in many different ways. See Exercises 17 through 20.

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EXAMPLE 4

807

Counting Possibilities for Seating Arrangements Adrienne, Bob, Carol, Dax, Earlene, and Fabian bought tickets to see The Marriage of Figaro. Assuming they sat together in a row of six seats, how many different seating arrangements are possible if a. Bob and Carol are sweethearts and must sit together? b. Bob and Carol are enemies and must not sit together?

䊳

Solution Figure 9.56 Bob 1

Carol 2

3

4

5

6

1

Bob 2

Carol 3

4

5

6

1

2

Bob 3

Carol 4

5

6

1

2

3

Bob 4

Carol 5

6

1

2

3

4

Bob 5

Carol 6

B. You’ve just seen how we can count possibilities using the fundamental principle of counting

a. Since a restriction has been placed on the seating arrangement, it will help to divide the experiment into a sequence of tasks: task 1: they sit together; task 2: either Bob is on the left or Bob is on the right; and task 3: the other four are seated. Bob and Carol can sit together in five different ways, as shown in Figure 9.56, so there are five possibilities for task 1. There are two ways they can be side-by-side: Bob on the left and Carol on the right, as shown, or Carol on the left and Bob on the right. The remaining four people can be seated randomly, so task 3 has 4! ⫽ 24 possibilities. Under these conditions they can be seated 5 # 2 # 4! ⫽ 240 ways. b. This is similar to part (a), but now we have to count the number of ways they can be separated by at least one seat: task 1: Bob and Carol are in nonadjacent seats; task 2: either Bob is on the left or Bob is on the right; and task 3: the other four are seated. For tasks 1 and 2, be careful to note there is no multiplication involved, just a simple counting. If Bob sits in seat 1 (to the left of Carol), there are four nonadjacent seats on the right. If Bob sits in seat 2, there are three nonadjacent seats on the right. With Bob in seat 3, there are two nonadjacent seats to his right. Similar reasoning for the remaining seats shows there are 10 # 2 ⫽ 20 possibilities for Bob and Carol not sitting together (by symmetry, Bob could also sit to the right of Carol). Multiplying by the number of ways the other four can be seated task 3 gives 20 # 4! ⫽ 480 possible seating arrangements. We could also reason that since there are 6! ⫽ 720 random seating arrangements and 240 of them consist of Bob and Carol sitting together, the remaining 720 ⫺ 240 ⫽ 480 must consist of Bob and Carol not sitting together. More will be said about this type of reasoning in Section 9.6. Now try Exercises 21 through 28

䊳

C. Distinguishable Permutations In the game of Scrabble® (Milton Bradley), players attempt to form words by rearranging letters. Suppose a player has the letters P, S, T, and O at the end of the game. These letters could be rearranged or permuted to form the words POTS, SPOT, TOPS, OPTS, POST, or STOP. These arrangements are called permutations of the four letters. A permutation is any new arrangement, listing, or sequence of objects obtained by changing an existing order. A distinguishable permutation is a permutation that produces a result different from the original. For example, a distinguishable permutation of the digits in the number 1989 is 8199. Example 4 considered six people, six seats, and the various ways they could be seated. But what if there were fewer seats than people? By the FPC, with six people and four seats there could be 6 # 5 # 4 # 3 ⫽ 360 different arrangements, with six people and three seats there are 6 # 5 # 4 ⫽ 120 different arrangements, and so on. These rearrangements are called distinguishable permutations. You may have noticed that for six people and six seats, we will use all six factors of 6!, while for six people and four seats we used the first four, six people and three seats required only the first three, and so on. Generally, for n people and r seats, the first r factors of n! will be used. The notation and formula for distinguishable permutations of n objects taken r at a time is n! . By defining 0! ⫽ 1, the formula includes the case where all n objects nPr ⫽ 1n ⫺ r2! n! n! n! ⫽ ⫽ ⫽ n!. are selected, which of course results in nPn ⫽ 0! 1 1n ⫺ n2!

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Distinguishable Permutations: Unique Elements

If r objects are selected from a set containing n unique elements 1r n2 and placed in an ordered arrangement, the number of distinguishable permutations is nPr

EXAMPLE 5

䊳

n! 1n r2!

or

nP r

n1n 121n 22 # # # 1n r 12

Computing a Permutation Compute each value of nPr using the methods described previously. a. 7P4 b. 10P3

Solution

䊳

n! , noting the Begin by evaluating each expression using the formula nPr 1n r2! third line (in bold) gives the first r factors of n!. 7! 10! a. 7P4 b. 10P3 17 42! 110 32! 7 # 6 # 5 # 4 # 3! 10 # 9 # 8 # 7! 3! 7! 7#6#5#4 10 # 9 # 8 840 720 Now try Exercises 29 through 36

䊳

Figure 9.57 When the number of objects is very large, the formula for permutations can become somewhat unwieldy and the computed result is often a very large number. When needed, most graphing calculators have the ability to compute permutations, with this option accessed using MATH (PRB) 2:nPr. Figure 9.57 verifies the computation for Example 5(b), and also shows that if there were 15 people and 7 chairs, the number of possible seating arrangements exceeds 32 million! Note that the value of n is entered first, followed by the nPr command and the value of r.

EXAMPLE 6

䊳

Counting the Possibilities for Finishing a Race As part of a sorority’s initiation process, the nine new inductees must participate in a 1-mi race. Assuming there are no ties, how many first- through fifth-place finishes are possible if it is well known that Mediocre Mary will finish fifth and Lightning Louise will finish first?

Solution

䊳

To help understand the situation, we can diagram the possibilities for finishing first through fifth. Since Louise will finish first, this slot can be filled in only one way, by Louise herself. The same goes for Mary and her fifth-place finish: Mary

Louise 1st

C. You’ve just seen how we can quick-count distinguishable permutations

2nd

3rd

4th

5th

The remaining three slots can be filled in 7P3 7 # 6 # 5 different ways, indicating that under these conditions, there are 1 # 7 # 6 # 5 # 1 210 different ways to finish. Now try Exercises 37 through 42

䊳

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809

D. Nondistinguishable Permutations As the name implies, certain permutations are nondistinguishable, meaning you cannot tell one apart from another. Such is the case when the original set contains elements or sample outcomes that are identical. Consider a family with four children, Lyddell, Morgan, Michael, and Mitchell, who are at the photo studio for a family picture. Michael and Mitchell are identical twins and cannot be told apart. In how many ways can they be lined up for the picture? Since this is an ordered arrangement of four children taken from a group of four, there are 4P4 24 ways to line them up. A few of them are Lyddell Morgan Michael Mitchell Lyddell Michael Morgan Mitchell Michael Lyddell Morgan Mitchell

Lyddell Morgan Mitchell Michael Lyddell Mitchell Morgan Michael Mitchell Lyddell Morgan Michael

But of these six arrangements, half will appear to be the same picture, since the difference between Michael and Mitchell cannot be distinguished. In fact, of the 24 total permutations, every picture where Michael and Mitchell have switched places will be nondistinguishable. To find the distinguishable permutations, we need to take the total permutations (4P4) and divide by 2!, the number of ways the twins can be 24 4P 4 12 distinguishable pictures. permuted: 2 122! These ideas can be generalized and stated in the following way. Nondistinguishable Permutations: Nonunique Elements In a set containing n elements where one element is repeated p times, another is repeated q times, and another is repeated r times 1p q r n2, the number of nondistinguishable permutations is n! nP n p!q!r! p!q!r! The idea can be extended to include any number of repeated elements. EXAMPLE 7

䊳

Counting Nondistinguishable Permutations A Scrabble player starts the game with the seven letters S, A, O, O, T, T, and T in her rack. How many distinguishable arrangements can be formed as she attempts to play a word?

Solution

䊳

D. You’ve just seen how we can quick-count nondistinguishable permutations

Essentially the exercise asks for the number of distinguishable permutations of the seven letters, given T is repeated three times and O is repeated twice (for S and 7P 7 420 distinguishable permutations. A, 1! 1). There are 3!2! Now try Exercises 43 through 54

䊳

E. Combinations WORTHY OF NOTE In Example 7, if a Scrabble player is able to play all seven letters in one turn, he or she “bingos” and is awarded 50 extra points. The player in Example 7 did just that. Can you determine what word was played?

Similar to nondistinguishable permutations, there are other times the total number of permutations must be reduced to quick-count the elements of a desired subset. Consider a vending machine that offers a variety of 40¢ candies. If you have a quarter (Q), dime (D), and nickel (N), the machine wouldn’t care about the order the coins are deposited. Even though QDN, QND, DQN, DNQ, NQD, and NDQ give the 3P3 6 possible permutations, the machine considers them as equal and will vend your snack. Using sets, this is similar to saying the set A 5X, Y, Z6 has only one subset with three elements, since {X, Z, Y}, {Y, X, Z}, {Y, Z, X}, and so on, all represent the same set. Similarly, there are six two-letter permutations of X, Y, and Z 1 3P2 62: XY, XZ, YX,

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YZ, ZX, and ZY, but only three two-letter subsets: {X, Y}, {X, Z} and {Y, Z}. When permutations having the same elements are considered identical, the result is the number of possible combinations and is denoted nCr. Since the r objects can be selected in r! nP r , ways, we divide nPr by r! to “quick-count” the number of possibilities: nCr r! which can be thought of as the first r factors of n!, divided by r!. By substituting n! for nPr in this formula, we find an alternative method for computing nCr is 1n r2! n! . Take special note that when r objects are selected from a set with n elements r!1n r2! and the order they’re listed is unimportant (because you end up with the same subset), the result is a combination, not a permutation. Combinations The number of combinations of n objects taken r at a time is given by nC r

EXAMPLE 8

䊳

nPr

r!

or

nCr

n! r!1n r2!

Computing Combinations Using a Formula Compute each value of nCr given. a. 7C4 b. 8C3 c. 5C2

Solution

䊳

7#6#5#4 4! 35

a. 7C4

8#7#6 3! 56

5#4 2! 10

b. 8C3

c. 5C2

Now try Exercises 55 through 64 As with permutations, when the number of objects is very large, the formula for combinations can also become somewhat cumbersome. Most graphing calculators have the ability to compute combinations, with this option accessed on the same submenu as nPr: MATH (PRB) 3:nCr. Figure 9.58 verifies the computation from Example 8(b), and also shows that in a Political Science class with 30 students, 5 can be picked at random to attend a seminar in the nation’s capitol 142,506 ways. EXAMPLE 9

䊳

䊳

Figure 9.58

Applications of Combinations-Lottery Results A small city is getting ready to draw five Ping-Pong balls of the nine they have numbered 1 through 9 to determine the winner(s) for its annual raffle. If a ticket holder has the same five numbers, they win. In how many ways can the winning numbers be drawn?

Solution

䊳

Since the winning numbers can be drawn in any order, we have a combination of 9 things taken 5 at a time. The five numbers can be drawn in 9C5

9#8#7#6#5 126 ways. 5! Now try Exercises 65 and 66

䊳

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811

Somewhat surprisingly, there are many situations where the order things are listed is not important. Such situations include • The formation of committees, since the order people volunteer is unimportant • Card games with a standard deck, since the order cards are dealt is unimportant • Playing BINGO, since the order the winning numbers are called is unimportant When the order in which people or objects are selected from a group is unimportant, the number of possibilities is a combination, not a permutation. Another way to tell the difference between permutations and combinations is the following memory device: Permutations have Priority or Precedence; in other words, the Position of each element matters. By contrast, a Combination is like a Committee of Colleagues or Collection of Commoners; all members have equal rank. For permutations, a-b-c is different from b-a-c. For combinations, a-b-c is the same as b-a-c. EXAMPLE 10

䊳

Applications of Quick-Counting — Committees and Governance The Sociology Department of Lakeside Community College has 12 dedicated faculty members. (a) In how many ways can a three-member textbook selection committee be formed? (b) If the department is in need of a Department Chair, Curriculum Chair, and Technology Chair, in how many ways can the positions be filled?

Solution

䊳

a. Since textbook selection depends on a Committee of Colleagues, the order members are chosen is not important. This is a Combination of 12 people taken 3 at a time, and there are 12C3 220 ways the committee can be formed. b. Since those selected will have Position or Priority, this is a Permutation of 12 people taken 3 at a time, giving 12P3 1320 ways the positions can be filled. Now try Exercises 67 through 78

E. You’ve just seen how we can quick-count using combinations

䊳

The Exercise Set contains a wide variety of additional applications. See Exercises 81 through 107.

9.5 EXERCISES 䊳



1. A(n) and has a(n)

is any task that can be repeated set of possible outcomes.

2. When unique elements of a set are rearranged, the result is called a(n) permutation.

3. If an experiment has N equally likely outcomes and is repeated t times, the number of elements in the sample space is given by .

4. If some elements of a group are identical, certain rearrangements are identical and the result is a(n) permutation.

5. A three-digit number is formed from digits 1 to 9. Explain how forming the number with repetition differs from forming it without repetition.

6. Discuss/Explain the difference between a permutation and a combination. Try to think of new ways to help remember the distinction.

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812 䊳

9–52


DEVELOPING YOUR SKILLS 7. For the spinner shown here, (a) draw a tree diagram illustrating all possible outcomes for two spins and (b) create an ordered list showing all possible outcomes for two spins.

Z

W

Y

X

Heads

8. For the fair coin shown here, (a) draw a tree diagram illustrating all possible outcomes for four flips and (b) create an ordered list showing the possible outcomes for four flips.

Tails

9. A fair coin is flipped five times. If you extend the tree diagram from Exercise 8, how many possibilities are there? 10. A spinner has the two equally likely outcomes A or B and is spun four times. How is this experiment related to the one in Exercise 8? How many possibilities are there? 11. An inexpensive lock uses the numbers 0 to 24 for a three-number combination. How many different combinations are possible? 12. Grades at a local college consist of A, B, C, D, F, and W. If four classes are taken, how many different report cards are possible? License plates. In a certain (English-speaking) country, license plates for automobiles consist of two letters followed by one of four symbols (■, ◆, ❍, or ●), followed by three digits. How many license plates are possible if

13. Repetition is allowed? 14. Repetition is not allowed? 15. A remote access door opener requires a five-digit (1–9) sequence. How many sequences are possible if (a) repetition is allowed? (b) repetition is not allowed? 16. An instructor is qualified to teach Math 020, 030, 140, and 160. How many different four-course schedules are possible if (a) repetition is allowed? (b) repetition is not allowed? Use the fundamental principle of counting and other quick-counting techniques to respond.

17. Menu items: At Joe’s Diner, the manager is offering a dinner special that consists of one choice of entree (chicken, beef, soy meat, or pork), two vegetable

servings (corn, carrots, green beans, peas, broccoli, or okra), and one choice of pasta, rice, or potatoes. How many different meals are possible? 18. Getting dressed: A frugal businessman has five shirts, seven ties, four pairs of dress pants, and three pairs of dress shoes. Assuming that all possible arrangements are appealing, how many different shirt-tie-pants-shoes outfits are possible? 19. Number combinations: How many four-digit numbers can be formed using the even digits 0, 2, 4, 6, 8, if (a) no repetitions are allowed; (b) repetitions are allowed; (c) repetitions are not allowed and the number must be less than 6000 and divisible by 10. 20. Number combinations: If I was born in March, April, or May, after the 19th but before the 30th, and after 1949 but before 1981, how many different MM–DD–YYYY dates are possible for my birthday? Seating arrangements: William, Xayden, York, and Zelda decide to sit together at the movies. How many ways can they be seated if

21. They sit in random order? 22. York must sit next to Zelda? 23. York and Zelda must be on the outside? 24. William must have the aisle seat? Course schedule: A college student is trying to set her schedule for the next semester and is planning to take five classes: English, art, math, fitness, and science. How many different schedules are possible if

25. The classes can be taken in any order. 26. She wants her science class to immediately follow her math class. 27. She wants her English class to be first and her fitness class to be last. 28. She can’t decide on the best order and simply takes the classes in alphabetical order. Find the value of nPr in two ways: (a) compute r factors n! of n! and (b) use the formula nPr . 1n r2!

29. 10P3

30.

12P2

32. 5P3

33. 8P7

31. 9P4 34. 8P1

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Determine the number of three-letter permutations of the letters given, then use an organized list to write them all out. How many of them are actually words or common names?

35. T, R, and A

36. P, M, and A

37. The regional manager for an office supply store needs to replace the manager and assistant manager at the downtown store. In how many ways can this be done if she selects the personnel from a group of 10 qualified applicants? 38. The local chapter of Mu Alpha Theta will soon be electing a president, vice-president, and treasurer. In how many ways can the positions be filled if the chapter has 15 members? 39. The local school board is going to select a principal, vice-principal, and assistant viceprincipal from a pool of eight qualified candidates. In how many ways can this be done? 40. From a pool of 32 applicants, a board of directors must select a president, vice-president, labor relations liaison, and a director of personnel for the company’s day-to-day operations. Assuming all applicants are qualified and willing to take on any of these positions, how many ways can this be done? 41. A hugely popular chess tournament now has six finalists. Assuming there are no ties, (a) in how many ways can the finalists place in the final round? (b) In how many ways can they finish first, second, and third? (c) In how many ways can they finish if it’s sure that Roberta Fischer is going to win the tournament and that Geraldine Kasparov will come in sixth? 42. A field of 10 horses has just left the paddock area and is heading for the gate. Assuming there are no ties in the big race, (a) in how many ways can the horses place in the race? (b) In how many ways can they finish in the win, place, or show positions? (c) In how many ways can they finish if it’s sure that John Henry III is going to win, Seattle Slew III will come in second (place), and either Dumb Luck II or Calamity Jane I will come in tenth? Assuming all multiple births are identical and the children cannot be told apart, how many distinguishable photographs can be taken of a family of six, if they stand in a single row and there is

43. one set of twins 44. one set of triplets 45. one set of twins and one set of triplets

813

46. one set of quadruplets 47. How many distinguishable numbers can be made by rearranging the digits of 105,001? 48. How many distinguishable numbers can be made by rearranging the digits in the palindrome 1,234,321? How many distinguishable permutations can be formed from the letters of the given word?

49. logic

50. leave

51. lotto

52. levee

A Scrabble player (see Example 7) has the six letters shown remaining in her rack. How many distinguishable, six-letter permutations can be formed? (If all six letters are played, what was the word?)

53. A, A, A, N, N, B 54. D, D, D, N, A, E Find the value of nCr: (a) using nCr

nPr

r!

(r factors of n!

n! . r!1n r2!

over r!) and (b) using nCr

55. 9C4

56.

10C3

58. 6C3

59. 6C6

57. 8C5 60. 6C0

Use a calculator to verify that each pair of combinations is equal.

61. 9C4, 9C5

62.

10C3, 10C7

63. 8C5, 8C3

64. 7C2, 7C5

65. A platoon leader needs to send four soldiers to do some reconnaissance work. There are 12 soldiers in the platoon and each soldier is assigned a number between 1 and 12. The numbers 1 through 12 are placed in a helmet and drawn randomly. If a soldier’s number is drawn, then that soldier goes on the mission. In how many ways can the reconnaissance team be chosen? 66. Seven colored balls (red, indigo, violet, yellow, green, blue, and orange) are placed in a bag and three are then withdrawn. In how many ways can the three colored balls be drawn? 67. When the company’s switchboard operators went on strike, the company president asked for three volunteers from among the managerial ranks to temporarily take their place. In how many ways can the three volunteers “step forward,” if there are 14 managers and assistant managers in all?

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68. Becky has identified 12 books she wants to read this year and decides to take four with her to read while on vacation. She chooses Pastwatch by Orson Scott Card for sure, then decides to randomly choose any three of the remaining books. In how many ways can she select the four books she’ll end up taking? 69. A new garage band has built up their repertoire to 10 excellent songs that really rock. Next month they’ll be playing in a Battle of the Bands contest, with the winner getting some guaranteed gigs at the city’s most popular hot spots. In how many ways can the band select 5 of their 10 songs to play at the contest? 70. Pierre de Guirré is an award-winning chef and has just developed 12 delectable, new main-course recipes for his restaurant. In how many ways can he select three of the recipes to be entered in an international culinary competition? For each exercise, determine whether a permutation, a combination, counting principles, or a determination of the number of subsets is the most appropriate tool for obtaining a solution, then solve. Some exercises can be completed using more than one method.

72. If you flip a fair coin five times, how many different outcomes are possible? 73. Eight sprinters are competing for the gold, silver, and bronze medals. In how many ways can the medals be awarded? 74. Motorcycle license plates are made using two letters followed by three numbers. How many plates can be made if repetition of letters (only) is allowed? 75. A committee of five students is chosen from a class of 20 to attend a seminar. How many different ways can this be done? 76. If onions, cheese, pickles, and tomatoes are available to dress a hamburger, how many different hamburgers can be made? 77. A caterer offers eight kinds of fruit to make various fruit trays. How many different trays can be made using four different fruits? 78. Eighteen females try out for the basketball team, but the coach can only place 15 on her roster. How many different teams can be formed?

71. In how many ways can eight second-grade children line up for lunch?

䊳


79. Stirling’s Formula: n! ⬇ 12␲ # 1nnⴙ0.5 2 # eⴚn Values of n! grow very quickly as n gets larger (13! is already in the billions). For some applications, scientists find it useful to use the approximation for n! shown, called Stirling’s Formula. a. Compute the value of 7! on your calculator, b. Compute the value of 10! on your calculator, then use Stirling’s Formula with n 7. By then use Stirling’s Formula with n 10. By what percent does the approximate value what percent does the approximate value differ from the true value? differ from the true value? 80. Factorial formulas: For n, k 僆⺧, where n 7 k, a. Verify the formula for n 7 and k 5. 䊳

n! ⴝ n1n ⴚ 12 1n ⴚ 22 p 1n ⴚ k ⴙ 12 1n ⴚ k2! b. Verify the formula for n 9 and k 6.

APPLICATIONS

81. Yahtzee: In the game of “Yahtzee”® (Milton Bradley) five dice are rolled simultaneously on the first turn in an attempt to obtain various arrangements (worth various point values). How many different arrangements are possible?

82. Twister: In the game of “Twister”® (Milton Bradley) a simple spinner is divided into four quadrants designated Left Foot (LF), Right Hand (RH), Right Foot (RF), and Left Hand (LH), with four different color possibilities in each quadrant (red, green, yellow, blue). Determine the number of possible outcomes for three spins.

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83. Clue: In the game of “Clue”® (Parker Brothers) a crime is committed in one of nine rooms, with one of six implements, by one of six people. In how many different ways can the crime be committed? Phone numbers in North America have 10 digits: a threedigit area code, a three-digit exchange number, and the four final digits that make each phone number unique. Neither area codes nor exchange numbers can start with 0 or 1. Prior to 1994 the second digit of the area code had to be a 0 or 1. Sixteen area codes are reserved for special services (such as 911 and 411). In 1994, the last area code was used up and the rules were changed to allow the digits 2 through 9 as the middle digit in area codes.

815

Seating arrangements: In how many different ways can eight people (six students and two teachers) sit in a row of eight seats if

96. the teachers must sit on the ends 97. the teachers must sit together Television station programming: A television station needs to fill eight half-hour slots for its Tuesday evening schedule with eight programs. In how many ways can this be done if

98. there are no constraints 99. Seinfeld must have the 8:00 P.M. slot

84. How many different area codes were possible prior to 1994?

100. Seinfeld must have the 8:00 P.M. slot and The Drew Carey Show must be shown at 6:00 P.M.

85. How many different exchange numbers were possible prior to 1994?

101. Friends can be aired at 7:00 or 9:00 P.M. and Everybody Loves Raymond can be aired at 6:00 or 8:00 P.M.

86. How many different phone numbers were possible prior to 1994? 87. How many different phone numbers were possible after 1994? Aircraft N-Numbers: In the United States, private aircraft are identified by an “N-Number,” which is generally the letter “N” followed by five characters and includes these restrictions: (1) the N-Number can consist of five digits, four digits followed by one letter, or three digits followed by two letters; (2) the first digit cannot be a zero; (3) to avoid confusion with the numbers zero and one, the letters O and I cannot be used; and (4) repetition of digits and letters is allowed. How many unique N-Numbers can be formed

88. that have four digits and one letter? 89. that have three digits and two letters?

Scholarship awards: Fifteen students at Roosevelt Community College have applied for six available scholarship awards. How many ways can the awards be given if

102. there are six different awards given to six different students 103. there are six identical awards given to six different students Committee composition: The local city council has 10 members and is trying to decide if they want to be governed by a committee of three people or by a president, vice-president, and secretary.

104. If they are to be governed by committee, how many unique committees can be formed? 105. How many different president, vice-president, and secretary possibilities are there?

90. that have five digits? 91. that have three digits, two letters with no repetitions of any kind allowed? Seating arrangements: Eight people would like to be seated. Assuming some will have to stand, in how many ways can the seats be filled if the number of seats available is

92. eight

93. five

94. three

95. one

106. Team rosters: A soccer team has three goalies, eight defensive players, and eight forwards on its roster. How many different starting line-ups can be formed (one goalie, three defensive players, and three forwards)? 107. e-mail addresses: A business wants to standardize the e-mail addresses of its employees. To make them easier to remember and use, they consist of two letters and two digits (followed by @esmtb.com), with zero being excluded from use as the first digit and no repetition of letters or digits allowed. Will this provide enough unique addresses for their 53,000 employees worldwide?

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9–56



Tic-Tac-Toe: In the game Tic-Tac-Toe, players alternately write an “X” or an “O” in one of nine squares on a 3 3 grid. If either player gets three in a row horizontally, vertically, or diagonally, that player wins. If all nine squares are played with neither person winning, the game is a draw. Assuming “X” always goes first, 108. How many different “ending boards” are possible if the game ends after five plays? 䊳

109. How many different “ending boards” are possible if the game ends after six plays?


110. (6.4) Solve the given system of linear inequalities by graphing. Shade the feasible region. 2x y 6 6 x 2y 6 6 μ x0 y0

112. (7.2/7.3) Given matrices A and B shown, use a calculator to find A B, AB, and A1. 1 A £ 2 2

3 1§ 4

0.5 B £ 9 1.2

0.2 0.1 0

7 8§ 6

113. (8.3) Graph the hyperbola that is defined by

111. (9.2) For the series 1 5 9 13 p 197, state the nth term formula then find the 35th term and the sum of the first 35 terms.

9.6

0 5 1

1x 22 2 4

1y 32 2 9

1.

Introduction to Probability

LEARNING OBJECTIVES

There are few areas of mathematics that give us a better view of the world than probability and statistics. Unlike statistics, which seeks to analyze and interpret data, probability (for our purposes) attempts to use observations and data to make statements concerning the likelihood of future events. Such predictions of what might happen have found widespread application in such diverse fields as politics, manufacturing, gambling, opinion polls, product life, and many others. In this section, we develop the basic elements of probability.


A. Define an event on a sample space

B. Compute elementary probabilities C. Use certain properties of probability D. Compute probabilities using quick-counting techniques E. Compute probabilities involving nonexclusive events

EXAMPLE 1

A. Defining an Event In Section 9.5 we defined the following terms: experiment and sample outcome. Flipping a coin twice in succession is an experiment, and two sample outcomes are HH and HT. An event E is any designated set of sample outcomes, and is a subset of the sample space. One event might be E1: (two heads occur), another possibility is E2: (at least one tail occurs).

䊳

Stating a Sample Space and Defining an Event Consider the experiment of rolling one standard, six-sided die (plural is dice). State the sample space S and define any two events relative to S.

Solution A. You’ve just seen how we can define an event on a sample space

䊳

S is the set of all possible outcomes, so S 51, 2, 3, 4, 5, 66. Two possible events are E1: (a 5 is rolled) and E2: (an even number is rolled). Now try Exercises 7 through 10

䊳

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Section 9.6 Introduction to Probability

817

B. Elementary Probability When rolling a die, we know the result can be any of the six equally likely outcomes in the sample space, so the chance of E1:(a five is rolled) is 16. Since three of the elements in S are even numbers, the chance of E2:(an even number is rolled) is 36 12. This suggests the following definition. The Probability of an Event E Given S is a sample space of equally likely events and E is an event relative to S, the probability of E, written P(E), is computed as n1E2 P1E2 n1S2 where n(E) represents the number of elements in E, and n(S) represents the number of elements in S.

WORTHY OF NOTE Our study of probability will involve only those sample spaces with events that are equally likely.

A standard deck of playing cards consists of 52 cards divided in four groups or suits. There are 13 hearts (♥), 13 diamonds 1䉬2, 13 spades (♠), and 13 clubs (♣). As you can see in Figure 9.59, each of the 13 cards in a suit is labeled A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, and K. Also notice that 26 of the cards are red (hearts and diamonds), 26 are black (spades and clubs), and 12 of the cards are “face cards” (J, Q, K of each suit).

EXAMPLE 2

䊳

Figure 9.59

Stating a Sample Space and the Probability of a Single Outcome A single card is drawn from a well-shuffled deck. Define S and state the probability of any single outcome. Then define E as a King is drawn and find P(E).

Solution

䊳

Sample space: S 5the 52 cards6 . There are 52 equally likely outcomes, 1 so the probability of any one outcome is 52 . Since S has four Kings, n1E2 4 or about 0.077. P1E2 52 n1S2


EXAMPLE 3

䊳

䊳

Stating a Sample Space and the Probability of a Single Outcome A family of five has two girls and three boys named Sophie, Maria, Albert, Isaac, and Pythagoras. Their ages are 21, 19, 15, 13, and 9, respectively. One is to be selected randomly. Find the probability a teenager is chosen.

Solution B. You’ve just seen how we can compute elementary probabilities

䊳

The sample space is S 59, 13, 15, 19, 216. Three of the five are teenagers, meaning the probability is 35, 0.6, or 60%.


䊳

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C. Properties of Probability A study of probability necessarily includes recognizing some basic and fundamental properties. For example, when a fair die is rolled, what is P(E) if E is defined as a 1, 2, 3, 4, 5, or 6 is rolled? The event E will occur 100% of the time, since 1, 2, 3, 4, 5, 6 are the only possibilities. In symbols we write P(outcome is in the sample space) or simply P1S2 ⫽ 1 (100%). What percent of the time will a result not in the sample space occur? Since the die has only the six sides numbered 1 through 6, the probability of rolling something else is zero. In symbols, P1outcome is not in sample space2 ⫽ 0 or simply P1~S2 ⫽ 0.

WORTHY OF NOTE In probability studies, the tilde “~” acts as a negation symbol. For any event E defined on the sample space, ~E means the event does not occur.

Properties of Probability Given sample space S and any event E defined relative to S. 1. P1S2 ⫽ 1

EXAMPLE 4

䊳

2. P1~S2 ⫽ 0

3. 0 ⱕ P1E2 ⱕ 1

Determining the Probability of an Event A game is played using a spinner like the one shown. Determine the probability of the following events: E1: A nine is spun.

Solution

䊳

2 8

P1E2 2 ⫽

5 7

The sample space consists of eight equally likely outcomes. 0 ⫽0 8

4

1

E2: An integer greater than 0 and less than 9 is spun. P1E1 2 ⫽

3

6

8 ⫽ 1. 8

Technically, E1: A nine is spun is not an “event,” since it is not in the sample space and cannot occur, while E2 contains the entire sample space and must occur. Now try Exercises 17 and 18

䊳

Because we know P1S2 ⫽ 1 and all sample outcomes are equally likely, the probabilities of all single events defined on the sample space must sum to 1. For the experiment of rolling a fair die, the sample space has six outcomes that are equally likely. Note that P112 ⫽ P122 ⫽ P132 ⫽ P142 ⫽ P152 ⫽ P162 ⫽ 16, and 16 ⫹ 16 ⫹ 16 ⫹ 16 ⫹ 16 ⫹ 16 ⫽ 1. Probability and Sample Outcomes Given a sample space S with n equally likely sample outcomes s1, s2, s3, p , sn. n

兺 P1s 2 ⫽ P1s 2 ⫹ P1s 2 ⫹ P1s 2 ⫹ p ⫹ P1s 2 ⫽ 1 i

1

2

3

n

i⫽1

The complement of an event E is the set of sample outcomes in S not contained in E. Symbolically, ~E is the complement of E. Probability and Complementary Events Given sample space S and any event E defined relative to S, the complement of E, written ~E, is the set of all outcomes not in E and: 1. P1E2 ⫽ 1 ⫺ P1~E2

2. P1E2 ⫹ P1~E2 ⫽ 1

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EXAMPLE 5

䊳

819

Stating a Probability Using Complements Use complementary events to answer the following questions: a. A single card is drawn from a well-shuffled deck. What is the probability that it is not a diamond? b. A single letter is picked at random from the letters in the word “divisibility.” What is the probability it is not an “i”?

Solution

䊳

WORTHY OF NOTE Probabilities can be written in fraction form, decimal form, or as a percent. For P(~D) from Example 5(a), the probability could be written 34 , 0.75, or 75%.

EXAMPLE 6

a. Since there are 13 diamonds in a standard 52-card deck: 39 P1~D2 ⫽ 1 ⫺ P1D2 ⫽ 1 ⫺ 13 52 ⫽ 52 ⫽ 0.75. b. Of the 12 letters in d-i-v-i-s-i-b-i-l-i-t-y, 5 are “i’s.” This means 5 7 P1~i2 ⫽ 1 ⫺ P1i2, or 1 ⫺ 12 ⫽ 12 . The probability of choosing a letter other than i is 0.583. Now try Exercises 19 through 22

䊳

䊳

Stating a Probability Using Complements Inter-Island Waterways has just opened hydrofoil service between several islands. The hydrofoil is powered by two engines, one forward and one aft, and will operate if either of its two engines is functioning. Due to testing and past experience, the company knows the probability of the aft engine failing is P1aft engine fails2 ⫽ 0.05, the probability of the forward engine failing is P1forward engine fails2 ⫽ 0.03, and the probability that both fail is P1both engines simultaneously fail2 ⫽ 0.012. What is the probability the hydrofoil completes its next trip?

Solution

䊳

Although the answer may seem complicated, note that P(trip is completed) and P(both engines simultaneously fail) are complements. P1trip is completed2 ⫽ 1 ⫺ P1both engines simultaneously fail2 ⫽ 1 ⫺ 0.012 ⫽ 0.988 There is close to a 99% probability the trip will be completed. Now try Exercises 23 and 24

䊳

The chart in Figure 9.60 shows all 36 possible outcomes (the sample space) from the experiment of rolling two fair dice. Figure 9.60

EXAMPLE 7

䊳

Stating a Probability Using Complements Two fair dice are rolled. What is the probability the sum of both dice is greater than or equal to 5, P1sum ⱖ 52?

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Solution

䊳

C. You’ve just seen how we can use certain properties of probability

See Figure 9.60. For P 1sum ⱖ 52 it may be easier to use complements as there are far fewer possibilities: P1sum ⱖ 52 ⫽ 1 ⫺ P1sum 6 52 , which gives 6 1 5 1⫺ ⫽ 1 ⫺ ⫽ ⫽ 0.83. 36 6 6 Now try Exercises 25 and 26

䊳

D. Probability and Quick-Counting Quick-counting techniques were introduced earlier to help count the number of elements in a large or more complex sample space, and the number of sample outcomes in an event. EXAMPLE 8A

䊳

Stating a Probability Using Combinations Five cards are drawn from a shuffled 52-card deck. Calculate the probability of E1:(all five cards are face cards) or E2:(all five cards are hearts).

Solution

䊳

The sample space for both events consists of all five-card groups that can be formed from the 52 cards or 52C5. For E1 we are to select five face cards from the 12 that are available (three from each suit), or 12C5. The probability of five face n1E2 792 12C5 , which gives ⫽ cards is ⬇ 0.0003. For E2 we are to select five 2,598,960 n1S2 52C5 n1E2 13C5 , ⫽ hearts from the 13 available, or 13C5. The probability of five hearts is n1S2 52C5 1287 ⬇ 0.0005. which is 2,598,960

䊳

Stating a Probability Using Combinations and the Fundamental Principle of Counting

WORTHY OF NOTE It seems reasonable that the probability of 5 hearts is slightly higher, as 13 of the 52 cards are hearts, while only 12 are face cards.

EXAMPLE 8B

Of the 42 seniors at Jacoby High School, 23 are female and 19 are male. A group of five students is to be selected at random to attend a conference in Reno, Nevada. What is the probability the group will have exactly three females?

Solution

䊳

The sample space consists of all five-person groups that can be formed from the 42 seniors or 42C5. The event consists of selecting 3 females from the 23 available (23C3) and 2 males from the 19 available (19C2). Using the fundamental principle of counting n1E2 ⫽ 23C3 # 19C2 and the probability the group has 3 females is # n1E2 302,841 23C3 19C2 , which gives ⬇ 0.356. There is approximately a ⫽ 850,668 n1S2 42C5 35.6% probability the group will have exactly 3 females. Now try Exercises 27 through 34 䊳 While the fundamentals of probability are usually introduced using dice, cards, and very basic applications, this should not take away from its true power and utility. We use probability to help us to analyze things quantitatively (with numbers) so that important decisions can be made. In manufacturing, the probability a product becomes defective during its warranty period is crucial to the company’s financial stability. Health organizations rely heavily on probability to plan against the anticipated spread

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821

of a symptom through a given population. In these and many other situations, the sample spaces are very large and the defined events likewise large, making the use of technology an integral part of probability studies. The computed result of Example 8A is shown in Figure 9.61 (in scientific notation—shift the decimal four places to the left). The computed result of Example 8B appears in Figure 9.62. Figure 9.61

Figure 9.62

D. You’ve just seen how we can compute probabilities using quick-counting techniques

E. Probability and Nonexclusive Events Figure 9.63 Sometimes the way events are defined S causes them to share sample outcomes. Using a standard deck of playE1 E2 J♠ A♣ ing cards once again, if we define the 2♣ Q♠ 3♣ J♣ events E1:(a club is drawn) and E2:(a J♦ K♠ 4♣ 5♣ face card is drawn), they share the outQ♦ Q♣ 6♣ comes J♣, Q♣, and K♣ as shown in K♦ K♣ J♥ 8♣ 7♣ Figure 9.63. This overlapping region Q♥ 9♣ K♥ is the intersection of the events, or 10♣ E1 傽 E2. If we compute n1E1 ´ E2 2 as n1E1 2 ⫹ n1E2 2 as before, this intersecting region gets counted twice! In cases where the events are nonexclusive (not mutually exclusive), we maintain the correct count by subtracting one of the two intersections, obtaining n1E1 ´ E2 2 ⫽ n1E1 2 ⫹ n1E2 2 ⫺ n1E1 傽 E2 2. This leads to the following calculation for the probability of nonexclusive events:

WORTHY OF NOTE This can be verified by simply counting the elements involved: n1E1 2 ⫽ 13 and n1E2 2 ⫽ 12 so n1E1 2 ⫹ n1E2 2 ⫽ 25. However, there are only 22 possibilities in E1 ´ E2—the J♣, Q♣, and K♣ got counted twice.

P1E1 ´ E2 2 ⫽ ⫽

n1E1 2 ⫹ n1E2 2 ⫺ n1E1 傽 E2 2 n1E1 2 n1S2

n1S2

⫹

n1E1 2 n1S2

⫺

n1E1 傽 E2 2 n1S2

⫽ P1E1 2 ⫹ P1E2 2 ⫺ P1E1 傽 E2 2

definition of probability

property of rational expressions definition of probability

Probability and Nonexclusive Events Given sample space S and nonexclusive events E1 and E2 defined relative to S, the probability of E1 or E2 is given by P1E1 ´ E2 2 ⫽ P1E1 2 ⫹ P1E2 2 ⫺ P1E1 傽 E2 2

EXAMPLE 9A

䊳

Stating the Probability of Nonexclusive Events What is the probability that a club or a face card is drawn from a standard deck of 52 well-shuffled cards?

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Solution

䊳

As before, define the events E1:(a club is drawn) and E2:(a face card is drawn). 12 Since there are 13 clubs and 12 face cards, P1E1 2 ⫽ 13 52 and P1E2 2 ⫽ 52 . But three 3 . This leads to of the face cards are clubs, so P1E1 傽 E2 2 ⫽ 52 P1E1 ´ E2 2 ⫽ P1E1 2 13 ⫹ ⫽ 52 22 ⫽ ⬇ 52

⫹ P1E2 2 ⫺ P1E1 傽 E2 2 12 3 ⫺ 52 52

0.423

nonexclusive events substitute

combine terms

There is about a 42% probability that a club or face card is drawn.

EXAMPLE 9B

䊳

Stating the Probability of Nonexclusive Events A survey of 100 voters was taken to gather information on critical issues and the demographic information collected is shown in the table. One out of the 100 voters is to be drawn at random to be interviewed on the 5 P.M. News. What is the probability the person is a woman (W) or a Republican (R)?

Solution

䊳

Women

Men

Totals

Republican

17

20

37

Democrat

22

17

39

Independent

8

7

15

Green Party

4

1

5

Tax Reform

2

2

4

53

47

100

Totals

Since there are 53 women and 37 Republicans, P1W2 ⫽ 0.53 and P1R2 ⫽ 0.37. The table shows 17 people are both female and Republican so P1W 傽 R2 ⫽ 0.17. P1W ´ R2 ⫽ P1W2 ⫹ P1R2 ⫺ P1W 傽 R2 ⫽ 0.53 ⫹ 0.37 ⫺ 0.17 ⫽ 0.73

nonexclusive events substitute combine

There is a 73% probability the person is a woman or a Republican. Now try Exercises 35 through 48 䊳 Two events that have no common outcomes are called mutually exclusive events (one excludes the other and vice versa). For example, in rolling one die, E1:(a 2 is rolled) and E2:(an odd number is rolled) are mutually exclusive, since 2 is not an odd number. For the probability of E3:(a 2 is rolled or an odd number is rolled), we note that n1E1 傽 E2 2 ⫽ 0 and the previous formula simply reduces to P1E1 2 ⫹ P1E2 2 . See Exercises 49 and 50. There is a large variety of additional applications in the Exercise Set. See Exercises 53 through 68. When probability calculations require a repeated use of permutations or combinations, tables can make the work more efficient and help to explore the patterns they generate. For instance, to choose r children from a group of six children 1n ⫽ 62, we can set TBLSET to AUTO, then press Y= and enter 6 nCr X as Y1 (Figure 9.64). Access the TABLE ( 2nd GRAPH ) and note that the calculator has automatically computed the value of 6C0, 6C1, 6C2, p , 6C6 (Figure 9.65) and the pattern of outputs is symmetric. For calculations like those required in Example 8B 1 23C3 # 19C2 2, we can enter Y1 ⫽ 23 nCr X, Y2 ⫽ 19 nCr 15 ⫺ X2, and Y3 ⫽ Y1 # Y2 to further explore the number of ways groups with different numbers of females can be chosen. Also see Exercise 70.

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Figure 9.64

823

Figure 9.65

E. You’ve just seen how we can compute probabilities involving nonexclusive events

9.6 EXERCISES 䊳



1. Given a sample space S and an event E defined relative to S, P1E2 ⫽

䊳

n1S2

.

4. The of an event E is the set of sample outcomes in S which are not contained in E.

2. In elementary probability, we consider all events in the sample space to be likely.

5. Discuss/Explain the difference between mutually exclusive events and nonexclusive events. Give an example of each.

3. Given a sample space S and an event E defined relative to S: , P1S2 ⫽ , ⱕ P1E2 ⱕ and P1~S2 ⫽ .

6. A single die is rolled. With no calculations, explain why the probability of rolling an even number is greater than rolling a number greater than four.


State the sample space S and the probability of a single outcome. Then define any two events E relative to S (many answers possible). Exercise 8

7. Two fair coins (heads and tails) are flipped. 8. The simple spinner shown is spun.

1 4

2 3

9. The head coaches for six little league teams (the Patriots, Cougars, Angels, Sharks, Eagles, and Stars) have gathered to discuss new changes in the rule book. One of them is randomly chosen to ask the first question. 10. Experts on the planets Mercury, Venus, Mars, Jupiter, Saturn, Uranus, Neptune, and the Kuiper object formerly known as the planet Pluto have gathered at a space exploration conference. One group of experts is selected at random to speak first.

Find P(E) for the events defined.

11. Nine index cards numbered 1 through 9 are shuffled and placed in an envelope, then one of the cards is randomly drawn. Define event E as the number drawn is even. 12. Eight flash cards used for studying basic geometric shapes are shuffled and one of the cards is drawn at random. The eight cards include information on circles, squares, rectangles, kites, trapezoids, parallelograms, pentagons, and triangles. Define event E as a quadrilateral is drawn. 13. One card is drawn at random from a standard deck of 52 cards. What is the probability of a. drawing a Jack b. drawing a spade c. drawing a black card d. drawing a red three

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14. Pinochle is a card game played with a deck of 48 cards consisting of 2 Aces, 2 Kings, 2 Queens, 2 Jacks, 2 Tens, and 2 Nines in each of the four standard suits [hearts (♥), diamonds (䉬), spades (♠), and clubs (♣)]. If one card is drawn at random from this deck, what is the probability of a. drawing an Ace b. drawing a club c. drawing a red card d. drawing a face card (Jack, Queen, King)

Use the complementary events to complete Exercises 19 through 22.

15. A group of finalists on a game show consists of three males and five females. Hank has a score of 520 points, with Harry and Hester having 490 and 475 points, respectively. Madeline has 532 points, with Mackenzie, Morgan, Maggie, and Melanie having 495, 480, 472, and 470 points, respectively. One of the contestants is randomly selected to start the final round. Define E1 as Hester is chosen, E2 as a female is chosen, and E3 as a contestant with fewer than 500 points is chosen. Find the probability of each event.

22. A corporation will be moving its offices to Los Angeles, Miami, Atlanta, Dallas, or Phoenix. If the site is randomly selected, what is the probability Dallas is not chosen?

16. Soccer coach Maddox needs to fill the last spot on his starting roster for the opening day of the season and has to choose between three forwards and five defenders. The forwards have jersey numbers 5, 12, and 17, while the defenders have jersey numbers 7, 10, 11, 14, and 18. Define E1 as a forward is chosen, E2 as a defender is chosen, and E3 as a player whose jersey number is greater than 10 is chosen. Find the probability of each event. 17. A game is played using a spinner like the one shown. For each spin, a. What is the probability the 1 2 arrow lands in a shaded region? 4 3 b. What is the probability your spin is less than 5? c. What is the probability you spin a 2? d. What is the probability the arrow points to prime number? 18. A game is played using a spinner like the one shown here. For each spin, a. What is the probability the 2 1 arrow lands in a lightly 3 6 shaded region? 5 4 b. What is the probability your spin is greater than 2? c. What is the probability the arrow lands in a shaded region? d. What is the probability you spin a 5?

19. One card is drawn from a standard deck of 52. What is the probability it is not a club? 20. Four standard dice are rolled. What is the probability the sum is less than 24? 21. A single digit is randomly selected from among the digits of 10!. What is the probability the digit is not a 2?

23. A large manufacturing plant can remain at full production as long as one of its two generators is functioning. Due to past experience and the age difference between the systems, the plant manager estimates the probability of the main generator failing is 0.05, the probability of the secondary generator failing is 0.01, and the probability of both failing is 0.009. What is the probability the plant remains in full production today? 24. A fire station gets an emergency call from a shopping mall in the mid-afternoon. From a study of traffic patterns, Chief Nozawa knows the probability the most direct route is clogged with traffic is 0.07, while the probability of the secondary route being clogged is 0.05. The probability both are clogged is 0.02. What is the probability they can respond to the call unimpeded using one of these routes? 25. Two fair dice are rolled (see Figure 9.60). What is the probability of a. a sum less than four b. a sum less than eleven c. the sum is not nine d. a roll is not a “double” (both dice the same) “Double-six” dominos is a game played with the 28 numbered tiles shown in the diagram.

26. The 28 dominos are placed in a bag, shuffled, and then one domino is randomly drawn. What is the probability the total number of dots on the domino a. is three or less b. is greater than three c. does not have a blank half d. is not a “double” (both sides the same)

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Find P(E) given the values for n(E) and n(S) shown.

# 4C2; n1S2 ⫽ 10C5 n1E2 ⫽ 12C9 # 8C7; n1S2 ⫽ 20C16 n1E2 ⫽ 9C6 # 5C3; n1S2 ⫽ 14C9 n1E2 ⫽ 7C6 # 3C2; n1S2 ⫽ 10C8

27. n1E2 ⫽ 6C3 28. 29. 30.

31. Five cards are drawn from a well-shuffled, standard deck of 52 cards. Which has the greater probability: (a) all five cards are red or (b) all five cards are numbered cards? How much greater? 32. Five cards are drawn from a well-shuffled pinochle deck of 48 cards (see Exercise 14). Which has the greater probability, (a) all five cards are face cards (King, Queen, or Jack) or (b) all five cards are black? How much greater?

825

41. Two fair dice are rolled. What is the probability the sum of the dice is a. a multiple of 3 and an odd number b. a sum greater than 5 and a 3 on one die c. an even number and a number greater than 9 d. an odd number and a number less than 10 42. Eight Ball is a game played on a pool table with 15 balls numbered 1 through 15 and a cue ball that is solid white. Of the 15 numbered balls, 8 are a solid (nonwhite) color and numbered 1 through 8, and seven are striped balls numbered 9 through 15. The fifteen numbered pool balls (no cueball) are placed in a large bowl and mixed, then one is drawn out. What is the probability of drawing

33. A dietetics class has 24 students. Of these, 9 are vegetarians and 15 are not. The instructor receives enough funding to send six students to a conference. If the students are selected randomly, what is the probability the group will have a. exactly two vegetarians b. exactly four nonvegetarians c. at least three vegetarians 34. A large law firm has a support staff of 15 employees: six paralegals and nine legal assistants. Due to recent changes in the law, the firm wants to send five of them to a forum on the new changes. If the selection is done randomly, what is the probability the group will have a. exactly three paralegals b. exactly two legal assistants c. at least two paralegals Find the probability indicated using the information given.

35. Given P1E1 2 ⫽ 0.7, P1E2 2 ⫽ 0.5, and P1E1 ¨ E2 2 ⫽ 0.3, compute P1E1 ´ E2 2. 36. Given P1E1 2 ⫽ 0.6, P1E2 2 ⫽ 0.3, and P1E1 ¨ E2 2 ⫽ 0.2, compute P1E1 ´ E2 2.

37. Given P1E1 2 ⫽ 38, P1E2 2 ⫽ 34, and P1E1 ´ E2 2 ⫽ 56; compute P1E1 ¨ E2 2.

38. Given P1E1 2 ⫽ 12, P1E2 2 ⫽ 35, and P1E1 ´ E2 2 ⫽ 17 20 ; compute P1E1 ¨ E2 2. 39. Given P1E1 ´ E2 2 ⫽ 0.72, P1E2 2 ⫽ 0.56, and P1E1 ¨ E2 2 ⫽ 0.43; compute P(E1).

40. Given P1E1 ´ E2 2 ⫽ 0.85, P1E1 2 ⫽ 0.4, and P1E1 ¨ E2 2 ⫽ 0.21; compute P(E2).

a. b. c. d. e. f. g. h.

the eight ball a number greater than fifteen an even number a multiple of three a solid color and an even number a striped ball and an odd number an even number and a number divisible by 3 an odd number and a number divisible by 4

43. A survey of 50 veterans was taken to gather information on their service career and what life is like out of the military. A breakdown of those surveyed is shown in the table. One out of the 50 will be selected at random for an interview and a biographical sketch. What is the probability the person chosen is Women

Men

Totals

6

9

15

Corporal

10

8

18

Sergeant

4

5

9

Private

Lieutenant

2

1

3

Captain

2

3

5

24

26

50

Totals

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a. b. c. d. e.

a woman and a sergeant a man and a private a private and a sergeant a woman and an officer a person in the military

44. Referring to Exercise 43, what is the probability the person chosen is a. a woman or a sergeant b. a man or a private c. a woman or a man d. a woman or an officer e. a captain or a lieutenant A computer is asked to randomly generate a three-digit number. What is the probability the

45. ten’s digit is odd or the one’s digit is even 46. first digit is prime and the number is a multiple of 10 A computer is asked to randomly generate a four-digit number. What is the probability the number is

49. Two fair dice are rolled. What is the probability of a. boxcars (sum of 12) or snake eyes (sum of 2) b. a sum of 7 or a sum of 11 c. an even-numbered sum or a prime sum d. an odd-numbered sum or a sum that is a multiple of 4 e. a sum of 15 or a multiple of 12 f. a sum that is a prime number 50. Suppose all 16 balls from a game of pool (see Exercise 42) are placed in a large leather bag and mixed, then one is drawn out. Consider the cue ball as “0.” What is the probability of drawing a. a striped ball b. a solid-colored ball c. a polka-dotted ball d. the cue ball e. the cue ball or the eight ball f. a striped ball or a number less than five g. a solid color or a number greater than 12 h. an odd number or a number divisible by 4

47. at least 4000 or a multiple of 5 48. less than 7000 and an odd number 䊳


51. Games involving a fair spinner (with numbers 1 through 4): P1n2 ⴝ 1 14 2 n Games that involve moving pieces around a board using a fair spinner are fairly common. If a fair spinner has the numbers 1 through 4, the probability that any one number is spun n times in succession is given by the formula shown, where n represents the number of spins. What is the probability (a) the first player spins a two? (b) all four players spin a two? (c) Discuss the graph of P(n) and explain the connection between the graph and the probability of consistently spinning a two. 䊳

52. Games involving a fair coin (heads and tails): P1n2 ⴝ 1 12 2 n When a fair coin is flipped, the probability that heads (or tails) is flipped n times in a row is given by the formula shown, where n represents the number of flips. What is the probability (a) the first flip is heads? (b) the first four flips are heads? (c) Discuss the graph of P(n) and explain the connection between the graph and the probability of consistently flipping heads.

APPLICATIONS

53. To improve customer service, a company tracks the number of minutes a caller is “on hold” and waiting for a customer service representative. The table shows the probability that a caller will wait m minutes. Based on the table, what is the probability a caller waits a. at least 2 minutes b. less than 2 minutes c. 4 minutes or less d. over 4 minutes

e. less than 2 or more than 4 minutes f. 3 or more minutes Wait Time (minutes m)

Probability

0

0.07

0 6 m 6 1

0.28

1ⱕm 6 2

0.32

2ⱕm 6 3

0.25

3ⱕm 6 4

0.08

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54. To study the impact of technology on American families, a researcher first determines the probability that a family has n computers at home. Based on the table, what is the probability a home a. has at least one computer b. has two or more computers c. has fewer than four computers d. has five computers e. has one, two, or three computers f. does not have two computers Number of Computers

Probability

0

9%

1

51%

2

28%

3

9%

4

3%

55. Jolene is an experienced markswoman and is able to hit a 10 in. by 20 in. target 100% of the time at a range of 100 yd. Assuming the 10 in. probability she hits a target is related to its area, what is the 20 in. probability she hits the shaded portions shown? a. isosceles triangle b. right triangle

c. isosceles right triangle

56. a. square

b. circle

c. isosceles trapezoid with b ⫽

B 2

827

57. A circular dartboard has a 2 total radius of 8 in., with 4 circular bands that are 2 in. 6 wide, as shown. You are 8 skilled enough to hit this board 100% of the time so you always score at least two points each time you throw a dart. Assuming the probabilities are related to area, on the next dart that you throw what is the probability you a. score at least a 4? b. score at least a 6? c. hit the bull’s-eye? d. score exactly 4 points? 58. Three red balls, six blue balls, and four white balls are placed in a bag. What is the probability the first ball you draw out is a. red b. blue c. not white d. purple e. red or white f. red and white 59. Three red balls, six blue balls, and four white balls are placed in a bag, then two are drawn out and placed in a rack. What is the probability the balls drawn are a. first red, second blue b. first blue, second red c. both white d. first blue, second not red e. first white, second not blue f. first not red, second not blue 60. Consider the 210 discrete points found in the first and second quadrants where ⫺10 ⱕ x ⱕ 10, 1 ⱕ y ⱕ 10, and x and y are integers. The coordinates of each point is written on a slip of paper and placed in a box. One of the slips is then randomly drawn. What is the probability the point (x, y) drawn a. is on the graph of y ⫽ 冟x冟 b. is on the graph of y ⫽ 2冟x冟 c. is on the graph of y ⫽ 0.5冟x冟 d. has coordinates 1x, y 7 ⫺22 e. has coordinates 1x ⱕ 5, y 7 ⫺22 f. is between the branches of y ⫽ x2

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61. Your instructor surprises you with a True/False quiz for which you are totally unprepared and must guess randomly. What is the probability you pass the quiz with an 80% or better if there are a. three questions b. four questions c. five questions 62. A robot is sent out to disarm a timed explosive device by randomly changing some switches from a neutral position to a positive flow or negative flow position. The problem is, the switches are independent and unmarked, and it is unknown which direction is positive and which direction is negative. The bomb is harmless if a majority of the switches yield a positive flow. All switches must be thrown. What is the probability the device is disarmed if there are a. three switches b. four switches c. five switches 63. A survey of 100 retirees was taken to gather information concerning how they viewed the Vietnam War back in the early 1970s. A breakdown of those surveyed is shown in the table. One out of the hundred will be selected at random for a personal, taped interview. What is the probability the person chosen had a a. career of any kind and opposed the war b. medical career and supported the war c. military career and opposed the war d. legal or business career and opposed the war e. academic or medical career and supported the war Career

Support

Military

9

3

12

Medical

8

16

24

Legal

15

12

27

Business

18

6

24

3

10

13

53

47

100

Academics Totals

Opposed

Total

64. Referring to Exercise 63, what is the probability the person chosen a. had a career of any kind or opposed the war b. had a medical career or supported the war

c. supported the war or had a military career d. had a medical or a legal career e. supported or opposed the war 65. The Board of Directors for a large hospital has 15 members. There are six doctors of nephrology (kidneys), five doctors of gastroenterology (stomach and intestines), and four doctors of endocrinology (hormones and glands). Eight of them will be selected to visit the nation’s premier hospitals on a 3-week, expenses-paid tour. What is the probability the group of eight selected consists of exactly a. four nephrologists and four gastroenterologists b. three endocrinologists and five nephrologists 66. A support group for hodophobics (an irrational fear of travel) has 32 members. There are 15 aviophobics (fear of air travel), eight siderodrophobics (fear of train travel), and nine thalassophobics (fear of ocean travel) in the group. Twelve of them will be randomly selected to participate in a new therapy. What is the probability the group of 12 selected consists of exactly a. two aviophobics, six siderodrophobics, and four thalassophobics b. five thalassophobics, four aviophobics, and three siderodrophobics 67. A trained chimpanzee is given a box containing eight wooden cubes with the letters p, a, r, a, l, l, e, l printed on them (one letter per block). Assuming the chimp can’t read or spell, what is the probability he draws the eight blocks in order and actually forms the word “parallel”? 68. A number is called a “perfect number” if the sum of its proper factors is equal to the number itself. Six is the first perfect number since the sum of its proper factors is six: 1 ⫹ 2 ⫹ 3 ⫽ 6. Twenty-eight is the second since: 1 ⫹ 2 ⫹ 4 ⫹ 7 ⫹ 14 ⫽ 28. A young child is given a box containing eight wooden blocks with the following numbers (one per block) printed on them: four 3’s, two 5’s, one 0, and one 6. What is the probability she draws the eight blocks in order and forms the fifth perfect number: 33,550,336?

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9–69 䊳

Section 9.7 The Binomial Theorem


69. The function f 1x2 ⫽ 1 12 2 x gives the probability that x number of flips will all result in heads (or tails). Compute the probability that 20 flips results in 20 heads in a row, then use the Internet or some other resource to find the probability of winning a state lottery. Which is more likely to happen (which has the greater probability)? Were you surprised? 70. Recall that a function is a relation in which each element of the domain is paired with only one element from the range. Is the relation defined by C1x2 ⫽ nCx (n is a constant) a function? To 䊳

829

investigate, plot the points generated by C1x2 ⫽ 8Cx for x ⫽ 0 to x ⫽ 8 using a “friendly” window 1x 僆 30, 9.4 4 , y 僆 3 0, 93 4 2 and answer the following questions: a. Is the resulting graph continuous or discrete (made up of distinct points)? b. Does the resulting graph pass the vertical line test? c. Discuss the features of the relation and its graph, including the domain, range, maximum or minimum values, and symmetries observed.


71. (7.1) Solve the system using matrices and x ⫺ 2y ⫹ 3z ⫽ 10 row reduction: • 2x ⫹ y ⫺ z ⫽ 18 3x ⫺ 2y ⫹ z ⫽ 26 73. (4.6) Solve the inequality by graphing the function and labeling the appropriate interval(s): x2 ⫺ 1 ⱖ 0. x

9.7


A. Use Pascal’s triangle to

C. D. E.

logb b ⫽

logb 1 ⫽

logb b ⫽

blogb n ⫽

n

74. (9.3) A rubber ball is dropped from a height of 25 ft onto a hard surface. With each bounce, it rebounds 60% of the height from which it last fell. Use sequences/series to find (a) the height of the sixth bounce, (b) the total distance traveled up to the sixth bounce, and (c) the distance the ball will travel before coming to rest.

The Binomial Theorem

LEARNING OBJECTIVES

B.

72. (5.4) Complete the following logarithmic properties:

find 1a ⫹ b2 n Find binomial coefficients n using a b notation k Use the binomial theorem to find 1a ⫹ b2 n Find a specific term of a binomial expansion Solve applications of binomial powers

Strictly speaking, a binomial is a polynomial with two terms. This limits us to terms with real number coefficients and whole number powers on variables. In this section, we will loosely regard a binomial as the sum or difference of any two terms. Hence 1 13 1 3x2 ⫺ y4, 1x ⫹ 4, x ⫹ , and ⫺ ⫹ i are all “binomials.” Our goal is to develop x 2 2 an ability to raise a binomial to any natural number power, with the results having important applications in genetics, probability, polynomial theory, and other areas. The tool used for this purpose is called the binomial theorem.

A. Binomial Powers and Pascal’s Triangle Much of our mathematical understanding comes from a study of patterns. One area where the study of patterns has been particularly fruitful is Pascal’s triangle (Figure 9.66), named after the French scientist Blaise Pascal (although the triangle was well known before his time). It begins with a “1” at the vertex of the triangle, with 1’s extending diagonally downward to the left and right as shown. The entries on the interior of the triangle are found by adding the two entries directly above and to the left and right of each new position.

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Figure 9.66 1

First row

1 1 1 1 1

?

Third row

1

2 3

?

Second row

1

6

Fourth row

1

3

1

?

? ? and so on

1

?

There are a variety of patterns hidden within the triangle. In this section, we’ll use the horizontal rows of the triangle to help us raise a binomial to various powers. To begin, recall that 1a ⫹ b2 0 ⫽ 1 and 1a ⫹ b2 1 ⫽ 1a ⫹ 1b (unit coefficients are included for emphasis). In our earlier work, we saw that a binomial square (a binomial raised to the second power) always followed the pattern 1a ⫹ b2 2 ⫽ 1a2 ⫹ 2ab ⫹ 1b2. Observe the overall pattern that is developing as we include 1a ⫹ b2 3: 1a ⫹ b2 0 1a ⫹ b2 1 1a ⫹ b2 2 1a ⫹ b2 3

1 1a ⫹ 1b 1a2 ⫹ 2ab ⫹ 1b2 1a3 ⫹ 3a2b ⫹ 3ab2 ⫹ 1b3

row 1 row 2 row 3 row 4

Apparently the coefficients of 1a ⫹ b2 will occur in row n ⫹ 1 of Pascal’s triangle. Also observe that in each term of the expansion, the exponent of the first term a decreases by 1 as the exponent on the second term b increases by 1, keeping the degree of each term constant (recall the degree of a term with more than one variable is the sum of the exponents). n

1a3b0 ⫹ 3a2b1 ⫹ 3a1b2 ⫹ 1a0b3 3⫹0 degree 3

2⫹1 degree 3

1⫹2 degree 3

0⫹3 degree 3

These observations help us to quickly expand a binomial power. EXAMPLE 1

䊳

Expanding a Binomial Using Pascal’s Triangle

Solution

䊳

Working step-by-step we have

Use Pascal’s triangle and the patterns noted to expand 1x ⫹ 12 2 4. 1. The coefficients will be in the fifth row of Pascal’s triangle. 1 4 6 4 1 2. The exponents on x begin at 4 and decrease, while the exponents on 21 begin at 0 and increase. 1 0 1 1 1 2 1 3 1 4 1x4a b ⫹ 4x3a b ⫹ 6x2 a b ⫹ 4x1 a b ⫹ 1x0 a b 2 2 2 2 2 3. Simplify each term. 3 1 1 The result is x4 ⫹ 2x3 ⫹ x2 ⫹ x ⫹ . 2 2 16 Now try Exercises 7 through 10

䊳

If the exercise involves a difference rather than a sum, we simply rewrite the expression using algebraic addition and proceed as before.

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EXAMPLE 2

䊳

Raising a Complex Number to a Power Using Pascal’s Triangle

Solution

䊳

Begin by rewriting 13 ⫺ 2i2 5 as 33 ⫹ 1⫺2i2 4 5.

831

Use Pascal’s triangle and the patterns noted to compute 13 ⫺ 2i2 5.

1. The coefficients will be in the sixth row of Pascal’s triangle. 5 10 10 5 1 1 2. The exponents on 3 begin at 5 and decrease, while the exponents on 1⫺2i2 begin at 0 and increase. 1135 2 1⫺2i2 0 ⫹ 5134 21⫺2i2 1 ⫹ 10133 21⫺2i2 2 ⫹ 10132 21⫺2i2 3 ⫹ 5131 21⫺2i2 4 ⫹ 1130 21⫺2i2 5 3. Simplify each term. 243 ⫺ 810i ⫺ 1080 ⫹ 720i ⫹ 240 ⫺ 32i The result is ⫺597 ⫺ 122i. Now try Exercises 11 and 12

䊳

Expanding Binomial Powers 1a ⴙ b2 n

1. The coefficients will be in row n ⫹ 1 of Pascal’s triangle. 2. The exponents on the first term begin at n and decrease, while the exponents on the second term begin at 0 and increase. 3. For any binomial difference 1a ⫺ b2 n, rewrite the base as 3a ⫹ 1⫺b2 4 n using algebraic addition and proceed as before, then simplify each term.

A. You’ve just seen how we can use Pascal’s triangle to find 1a ⴙ b2 n

B. Binomial Coefficients and Factorials

Pascal’s triangle can easily be used to find the coefficients of 1a ⫹ b2 n, as long as the exponent is relatively small. If we needed to expand 1a ⫹ b2 25, writing out the first 26 rows of the triangle would be rather tedious. To overcome this limitation, we introduce a formula that enables us to find the coefficients of any binomial expansion. The Binomial Coefficients

n For natural numbers n and r where n ⱖ r, the expression a b, read “n choose r,” r is called the binomial coefficient and evaluated as: n n! a b⫽ r r!1n ⫺ r2!

Notice the formula for determining binomial coefficients is identical to that for and it turns out the coefficients are actually found using a combination, with the new notation used primarily as a convenience. In Example 1, we found the coefficients of 1a ⫹ b2 4 using the fifth or 1n ⫹ 12st row of Pascal’s triangle. In Example 3, these coefficients are found using the formula for binomial coefficients. nCr,

EXAMPLE 3

䊳

Computing Binomial Coefficients n n! Evaluate a b ⫽ as indicated: r r!1n ⫺ r2! 4 a. a b 1

4 b. a b 2

4 c. a b 3

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Solution

䊳

4 # 3! 4 4! ⫽ a. a b ⫽ ⫽4 1 1!3! 1!14 ⫺ 12! 4 # 3 # 2! 4 4! 4#3 ⫽ ⫽ ⫽6 b. a b ⫽ 2 2!2! 2 2!14 ⫺ 22! 4 4! 4 # 3! ⫽ ⫽4 c. a b ⫽ 3 3!1! 3!14 ⫺ 32! Now try Exercises 13 through 20

䊳

4 4 4 Note a b ⫽ 4, a b ⫽ 6, and a b ⫽ 4 give the interior entries in the fifth row of 1 2 3 Pascal’s triangle: 1 4 6 4 1. For consistency and symmetry, we define 0! ⫽ 1, which enables the formula to generate all entries of the triangle, including the “1’s.” 4! 4 4! 4 4! ⫽ apply formula a b ⫽ apply formula a b⫽ 4 0 4! # 0! 0!14 ⫺ 02! 4!14 ⫺ 42! 4! 4! ⫽ 0! ⫽ 1 ⫽ 0! ⫽ 1 ⫽1 ⫽1 # 1 4! 4! # 1 n The formula for a b with 0 ⱕ r ⱕ n now gives all coefficients in the 1n ⫹ 12st r row. For n ⫽ 5, we have 5 a b 0 1 EXAMPLE 4

䊳

5 a b 1 5

5 a b 2 10

5 a b 3 10

5 a b 4 5

5 a b 5 1

Computing Binomial Coefficients Compute the binomial coefficients: 9 a. a b 0

Solution

䊳

9 b. a b 1

9 9! a. a b ⫽ 0 0!19 ⫺ 02! 9! ⫽ ⫽1 9! 6 6! c. a b ⫽ 5 5!16 ⫺ 52! 6! ⫽ ⫽6 5!

6 c. a b 5

6 d. a b 6 9 9! b. a b ⫽ 1 1!19 ⫺ 12! 9! ⫽ ⫽9 8! 6 6! d. a b ⫽ 6 6!16 ⫺ 62! 6! ⫽ ⫽1 6! Now try Exercises 21 through 24

B. You’ve just seen how we can find binomial coefficients n using a b notation k

䊳

n As mentioned, the formulas for a b and nCr yield like results for given values of n r and r. For future use, it will help to commit the general results from Example 4 to n n n n memory: a b ⫽ 1, a b ⫽ n, a b ⫽ n, and a b ⫽ 1. 0 1 n⫺1 n

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833

C. The Binomial Theorem n Using a b notation and the observations made regarding binomial powers, we can now r state the binomial theorem. Binomial Theorem

For any binomial 1a ⫹ b2 and natural number n, n n n 1a ⫹ b2 n ⫽ a b anb0 ⫹ a b an⫺1b1 ⫹ a b an⫺2b2 ⫹ p 0 1 2 n n ⫹a b a1bn⫺1 ⫹ a b a0bn n⫺1 n The theorem can also be stated in summation form as 1a ⫹ b2 n ⫽

n

兺 arba n

n⫺r r

b

r⫽0

The expansion actually looks overly impressive in this form, and it helps to summarize the process in words, as we did earlier. The exponents on the first term a begin at n and decrease, while the exponents on the second term b begin at 0 and increase, n keeping the degree of each term constant. The a b notation simply gives the coefficients r n of each term. As a final note, observe that the r in a b gives the exponent on b. r EXAMPLE 5

䊳

Solution

䊳

Expanding a Binomial Using the Binomial Theorem Expand 1a ⫹ b2 6 using the binomial theorem.

6 6 6 6 6 6 6 1a ⫹ b2 6 ⫽ a b a6b0 ⫹ a b a5b1 ⫹ a b a4b2 ⫹ a b a3b3 ⫹ a b a2b4 ⫹ a b a1b5 ⫹ a b a0b6 0 1 2 3 4 5 6 6! 6 6! 5 1 6! 4 2 6! 3 3 6! 2 4 6! 1 5 6! 6 ⫽ a ⫹ ab ⫹ ab ⫹ ab ⫹ ab ⫹ ab ⫹ b 0!6! 1!5! 2!4! 3!3! 4!2! 5!1! 6!0! ⫽ 1a6 ⫹ 6a5b ⫹ 15a4b2 ⫹ 20a3b3 ⫹ 15a2b4 ⫹ 6ab5 ⫹ 1b6 Now try Exercises 25 through 32

EXAMPLE 6

䊳

Using the Binomial Theorem to Find the Initial Terms of an Expansion

Solution

䊳

Use the binomial theorem with a ⫽ 2x, b ⫽ y2, and n ⫽ 10.

䊳

Find the first three terms of 12x ⫹ y2 2 10.

12x ⫹ y2 2 10 ⫽ a

C. You’ve just seen how we can use the binomial theorem to find 1a ⴙ b2 n

10 10 10 b12x2 10 1y2 2 0 ⫹ a b12x2 9 1y2 2 1 ⫹ a b12x2 8 1y2 2 2 ⫹ p 0 1 2 10! ⫽ 1121024x10 ⫹ 1102512x9y2 ⫹ 256x8y4 ⫹ p 2!8! ⫽ 1024x10 ⫹ 5120x9y2 ⫹ 1452256x8y4 ⫹ p ⫽ 1024x10 ⫹ 5120x9y2 ⫹ 11,520x8y4 ⫹ p

first three terms a

10 10 b ⫽ 1, a b ⫽ 10 0 1 10! ⫽ 45 2!8! result


䊳

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D. Finding a Specific Term of the Binomial Expansion In some applications of the binomial theorem, our main interest is a specific term of the expansion, rather than the expansion as a whole. To find a specified term, it helps to consider that the expansion of 1a ⫹ b2 n has n ⫹ 1 terms: 1a ⫹ b2 0 has one term, n 1a ⫹ b2 1 has two terms, 1a ⫹ b2 2 has three terms, and so on. Because the notation a b r always begins at r ⫽ 0 for the first term, the value of r will be 1 less than the term we are seeking. In other words, for the seventh term of 1a ⫹ b2 9, we use r ⫽ 6. The k th Term of a Binomial Expansion

For the binomial expansion 1a ⫹ b2 n, the kth term is given by n a b an⫺rbr, where r ⫽ k ⫺ 1. r

EXAMPLE 7

䊳

Solution

䊳

Finding a Specific Term of a Binomial Expansion Find the eighth term in the expansion of 1x ⫹ 2y2 12.

By comparing 1x ⫹ 2y2 12 to 1a ⫹ b2 n we have a ⫽ x, b ⫽ 2y, and n ⫽ 12. Since we want the eighth term, k ⫽ 8 and r ⫽ 7. The eighth term of the expansion is a

D. You’ve just seen how we can find a specific term of a binomial expansion

12 5 12! 128x5y7 b x 12y2 7 ⫽ 7!5! 7 ⫽ 179221128x5y7 2 ⫽ 101,376x5y7

27 ⫽ 128 1 12 7 2 ⫽ 792 result


䊳

E. Applications One application of the binomial theorem involves a binomial experiment and binomial probability. For binomial probabilities, the following must be true: (1) The experiment must have only two possible outcomes, typically called success and failure, and (2) if the experiment has n trials, the probability of success must be constant for all n n trials. If the probability of success for each trial is p, the formula a b11 ⴚ p2 nⴚkpk k gives the probability that exactly k trials will be successful. Binomial Probability Given a binomial experiment with n trials, where the probability for success in each trial is p. The probability that exactly k trials are successful is given by n a b11 ⫺ p2 n⫺k pk. k

EXAMPLE 8

䊳

Applying the Binomial Theorem — Binomial Probability Paula Rodrigues has a free-throw shooting average of 85%. On the last play of the game, with her team behind by three points, she is fouled at the three-point line, and is awarded two additional free throws via technical fouls on the opposing coach (a total of five free-throws). What is the probability she makes at least three (meaning they at least tie the game)?

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Solution

䊳

835

Here we have p ⫽ 0.85, 1 ⫺ p ⫽ 0.15, and n ⫽ 5. The key idea is to recognize the phrase at least three means “3 or 4 or 5.” So P(at least 3) ⫽ P13 ´ 4 ´ 52. “or” implies a union P1at least 32 ⫽ P13 ´ 4 ´ 52 ⫽ P132 ⫹ P142 ⫹ P152 sum of probabilities (mutually exclusive events) 5 5 5 ⫽ a b 10.152 2 10.852 3 ⫹ a b 10.152 1 10.852 4 ⫹ a b 10.152 0 10.852 5 4 5 3 ⬇ 0.1382 ⫹ 0.3915 ⫹ 0.4437 ⫽ 0.9734

Paula’s team has an excellent chance 1⬇97.3% 2 of at least tying the game.


䊳

As you can see, calculations involving binomial probabilities can become quite extensive. Here again, a conceptual understanding of what the numbers mean can be combined with the use of technology to solve significant applications of the idea. Most graphing calculators provide a binomial probability distribution function, abbreviated “binompdf(” and accessed using 2nd VARS (DISTR) 0:binompdf(. The function requires three inputs: the number of trials n, the probability of success p for each trial, and the value of k. As with the evaluation of other functions, k can be a single value or a list of values enclosed in braces: “{ }.” The resulting calculation for Example 8 is shown in Figure 9.67, and verifies each of the individual probabilities, although we must use the right arrow to see them all. To find the sum of these probabilities, we simply precede the “binompdf(” command with the “sum(” feature used previously. The final result is shown in Figure 9.68, and verifies our earlier calculation. See Exercises 47 and 48. Figure 9.67

Figure 9.68

E. You’ve just seen how we can solve applications of binomial powers

9.7 EXERCISES 䊳



1. In any binomial expansion, there is always more term than the power applied. 2. In all terms in the expanded form of 1a ⫹ b2 n, the exponents on a and b must sum to .

3. To expand a binomial difference such as 1a ⫺ 2b2 5, we rewrite the binomial as and proceed as before.

4. In a binomial experiment with n trials, the probability there are exactly k successes is given by the formula .

5. Discuss why the expansion of 1a ⫹ b2 n has n ⫹ 1 terms. 6. For any defined binomial experiment, discuss the relationships between the phrases, “exactly k success,” and “at least k successes.”

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Use Pascal’s triangle and the patterns explored to write each expansion.

Use the binomial theorem to expand each expression. Write the general form first, then simplify.

10. 1x2 ⫹ 13 2 3

28. 1x ⫺ y2 7

7. 1x ⫹ y2 5

8. 1a ⫹ b2 6

11. 11 ⫺ 2i2 5

Evaluate each of the following

7 13. a b 4 9 16. a b 5 40 19. a b 3 5 22. a b 0

䊳

8 14. a b 2 20 17. a b 17 45 20. a b 3 15 23. a b 15

9. 12x ⫹ 32 4

12. 12 ⫺ 5i2 4 5 15. a b 3 30 18. a b 26 6 21. a b 0 10 24. a b 10

31. 11 ⫺ 2i2 3

26. 1v ⫹ w2 4

29. 12x ⫺ 32 4

32. 12 ⫹ i 132 5

27. 1a ⫺ b2 6

30. 1a ⫺ 2b2 5

Use the binomial theorem to write the first three terms.

33. 1x ⫹ 2y2 9

36. 1 12a ⫺ b2 2 10

34. 13p ⫹ q2 8

35. 1v2 ⫺ 12w2 12

Find the indicated term for each binomial expansion.

37. 1x ⫹ y2 7; 4th term

38. 1m ⫹ n2 6; 5th term

41. 12x ⫹ y2 12; 11th term

42. 13n ⫹ m2 9; 6th term

39. 1p ⫺ 22 8; 7th term

40. 1a ⫺ 32 14; 10th term


n 1 k 1 nⴚk 43. Binomial probability: P1k2 ⴝ a b a b a b k 2 2 The theoretical probability of getting exactly k heads in n flips of a fair coin is given by the formula above. What is the probability that you would get exactly 5 heads in 10 flips of the coin?

䊳

25. 1c ⫹ d2 5

n 1 k 4 nⴚk 44. Binomial probability: P1k2 ⴝ a b a b a b k 5 5 A multiple choice test has five options per question. The probability of guessing correctly k times out of n questions is found using the formula shown. What is the probability a person scores a 70% by guessing randomly (7 out of 10 questions correct)?

APPLICATIONS

45. Batting averages: Tony Gwynn (San Diego Padres) had a lifetime batting average of 0.347, ranking him as one of the greatest hitters of all time. Suppose he came to bat five times in any given game. a. What is the probability that he will get exactly three hits? b. What is the probability that he will get at least three hits?

47. Late rental returns: The manager of Victor’s DVD Rentals knows that 6% of all DVDs rented are returned late. Of the eight videos rented in the last hour, what is the probability that a. exactly five are returned on time b. exactly six are returned on time c. at least six are returned on time d. none of them will be returned late

46. Pollution testing: Erin suspects that a nearby iron smelter is contaminating the drinking water over a large area. A statistical study reveals that 83% of the wells in this area are likely contaminated. If the figure is accurate, find the probability that if another 10 wells are tested a. exactly 8 are contaminated b. at least 8 are contaminated

48. Opinion polls: From past experience, a research firm knows that 20% of telephone respondents will agree to answer an opinion poll. If 20 people are contacted by phone, what is the probability that a. exactly 18 refuse to be polled b. exactly 19 refuse to be polled c. at least 18 refuse to be polled d. none of them agree to be polled

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Making Connections


49. If you sum the entries in each row of Pascal’s triangle, a pattern emerges. Find a formula that generalizes the result for any row of the triangle, and use it to find the sum of the entries in the 12th row of the triangle. 䊳

837

50. The derived polynomial of f (x) is f 1x ⫹ h2 or the original polynomial evaluated at x ⫹ h. Use Pascal’s triangle or the binomial theorem to find the derived polynomial for f 1x2 ⫽ x3 ⫹ 3x2 ⫹ 5x ⫺ 11. Simplify the result completely.


51. (2.5) Graph the function shown and find f (3): f 1x2 ⫽ e

x⫹2 1x ⫺ 42 2

xⱕ2 x 7 2

52. (3.1) Show that x ⫽ ⫺1 ⫹ i is a solution to x4 ⫹ 2x3 ⫺ x2 ⫺ 6x ⫺ 6 ⫽ 0.

53. (4.3/4.6) Graph the function g1x2 ⫽ x3 ⫺ x2 ⫺ 6x. Clearly indicate all intercepts and intervals where g1x2 7 0. 54. (5.6) If $2500 is deposited at 6% compounded continuously, how much would be in the account 10 years later?

MAKING CONNECTIONS Making Connections: Graphically, Symbolically, Numerically, and Verbally Eight situations are described in (a) through (h) below. Match the characteristics, formulas, operations, or results indicated in 1 through 16 to one of the eight situations. (a)

(b) ⫺2 ⫹ 0.5 ⫹ 3 ⫹ 5.5 ⫹ 8 ⫹ 10.5 ⫹ p ⫹ 33

7

3i⫺1 i⫽1 18

兺

10

(c)

(d)

5.5

0

⫺6

(e)

16a4 ⫺ 32a3b ⫹ 24a2b2 ⫺ 8ab3 ⫹ b4

(f) ⫺29, ⫺23, ⫺17, ⫺11, p

(g)

1, 1, 2, 3, 5, 8, 13, …

(h) a4 ⫹ 8a3b ⫹ 24a2b2 ⫹ 32ab3 ⫹ 16b4

1. ____ alternating sequence

9. ____ a22 ⫽ ⫺11.8

2. ____ Fibonacci sequence

10. ____ geometric series

3. ____ 232.5

11. ____ r ⫽ 3

4. ____ d ⫽ ⫺0.7

12. ____ Sq ⫽

5. ____ an ⫽ 3.6 ⫺ 0.7n

13. ____ 12a ⫺ b2 4

6. ____ S39 ⫽ 3315

16 3

14. ____ an ⫽ 6n ⫺ 35

7. ____ arithmetic series

15. ____ recursively defined

8. ____ 1a ⫹ 2b2 4

16. ____

1093 18

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Sequences and Series

KEY CONCEPTS • A finite sequence is a function an whose domain is the set of natural numbers from 1 to n. • The terms of the sequence are labeled a1, a2, a3, p , ak⫺1, ak, ak⫹1, p , an⫺2, an⫺1, an. • The expression an, which defines the sequence (generates the terms in order), is called the nth term. • An infinite sequence is a function whose domain is the set of natural numbers. • When each term of a sequence is larger than the preceding term, it is called an increasing sequence. • When each term of a sequence is smaller than the preceding term, it is called a decreasing sequence. • When successive terms of a sequence alternate in sign, it is called an alternating sequence. • When the terms of a sequence are generated using previous term(s), it is called a recursive sequence. • Sequences are sometimes defined using factorials, which are the product of a given natural number with all natural numbers that precede it: n! ⫽ n # 1n ⫺ 12 # 1n ⫺ 22 # p # 3 # 2 # 1. • Given the sequence a1, a2, a3, a4, p , an the sum is called a finite series and is denoted Sn. • Sn ⫽ a1 ⫹ a2 ⫹ a3 ⫹ a4 ⫹ p ⫹ an. The sum of the first n terms is called a partial sum. k

• In sigma notation, the expression

兺a ⫽ a i

1

⫹ a2 ⫹ p ⫹ ak represents a finite series,

i⫽1

and the letter “i ” is called the index of summation.

EXERCISES Write the first four terms that are defined and the value of a10. 1. an ⫽ 5n ⫺ 4

2. an ⫽

n⫹1 n2 ⫹ 1

Find the general term an for each sequence, and the value of a6. 3. 1, 16, 81, 256, p 4. ⫺17, ⫺14, ⫺11, ⫺8, p Find the eighth partial sum (S8). 5. 12, 14, 18, p

6. ⫺21, ⫺19, ⫺17, p

Evaluate each sum. 7

7.

兺

5

n2

8.

n⫽1

兺 13n ⫺ 22

n⫽1

Write the first five terms that are defined. n! 9. an ⫽ 1n ⫺ 22!

10. e

a1 ⫽ 12 an⫹1 ⫽ 2an ⫺ 14

Write as a single summation and evaluate. 7

11.

兺i

i⫽1

7

2

⫹

兺 13i ⫺ 22

i⫽1

12. A large wildlife preserve brings in 40 rare hawks (male and female) in an effort to repopulate the species. Each year they are able to add an average of 10 additional hawks in cooperation with other wildlife areas. If the population of hawks grows at a rate of 12% through natural reproduction, the number of hawks in the preserve after x yr is given by the recursive sequence h0 ⫽ 40, hn ⫽ 1.12 hn⫺1 ⫹ 10. (a) How many hawks are in the wildlife preserve after 5 yr? (b) How many years before the number of hawks exceeds 200?

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SECTION 9.2

Arithmetic Sequences

KEY CONCEPTS • In an arithmetic sequence, successive terms are found by adding a fixed constant to the preceding term. • In a sequence, if there exists a number d, called the common difference, such that ak⫹1 ⫺ ak ⫽ d, then the sequence is arithmetic. Alternatively, ak⫹1 ⫽ ak ⫹ d for k ⱖ 1. • The nth term n of an arithmetic sequence is given by an ⫽ a1 ⫹ 1n ⫺ 12d, where a1 is the first term and d is the common difference. If • the initial term is unknown or is not a1 the nth term can be written an ⫽ ak ⫹ 1n ⫺ k2d, where the subscript of the term ak and the coefficient of d sum to n. • For an arithmetic sequence with first term a1, the nth partial sum (the sum of the first n terms) is given by n1a1 ⫹ an 2 Sn ⫽ . 2 EXERCISES Find the general term (an) for each arithmetic sequence. Then find the indicated term. 13. 2, 5, 8, 11, p ; find a40 14. 3, 1, ⫺1, ⫺3, p ; find a35 Find the sum of each series. 15. ⫺1 ⫹ 3 ⫹ 7 ⫹ 11 ⫹ p ⫹ 75 17. 3 ⫹ 6 ⫹ 9 ⫹ 12 ⫹ p ; S20

16. 1 ⫹ 4 ⫹ 7 ⫹ 10 ⫹ p ⫹ 88 18. 1 ⫹ 34 ⫹ 12 ⫹ 14 ⫹ p ; S15

25

19.

兺 13n ⫺ 42

n⫽1

20. From a point just behind the cockpit, the width of a modern fighter plane’s swept-back wings is 1.25 m. The width of the wings, measured in equal increments, increases according to the pattern 1.25, 2.15, 3.05, 3.95, p . Find the width of the wings on the eighth measurement.

SECTION 9.3

Geometric Sequences

KEY CONCEPTS • In a geometric sequence, successive terms are found by multiplying the preceding term by a nonzero constant. ak⫹1 • In other words, if there exists a number r, called the common ratio, such that a ⫽ r, then the sequence is k ⫽ a r for k ⱖ 1. geometric. Alternatively, we can write a k⫹1

k

• The nth term an of a geometric sequence is given by an ⫽ a1rn⫺1, where a1 is the first term and an represents the general term of a finite sequence. • If the initial term is unknown or is not a1, the nth term can be written an ⫽ akrn⫺k, where the subscript of the term ak and the exponent on r sum to n. a1 11 ⫺ rn 2 . • The nth partial sum of a geometric sequence is Sn ⫽ 1⫺r a1 . • If 冟r 冟 6 1, the sum of an infinite geometric series is Sq ⫽ 1⫺r

EXERCISES Find the indicated term for each geometric sequence. 21. a1 ⫽ 5, r ⫽ 3; find a7 22. a1 ⫽ 4, r ⫽ 12; find a7 Find the indicated sum, if it exists. 24. 16 ⫺ 8 ⫹ 4 ⫺ p ; find S7 27. 4 ⫹ 8 ⫹ 16 ⫹ 32 ⫹ p

23. a1 ⫽ 17, r ⫽ 17; find a8

25. 2 ⫹ 6 ⫹ 18 ⫹ p ; find S8

26.

28. 5 ⫹ 0.5 ⫹ 0.05 ⫹ 0.005 ⫹ p

29. 6 ⫺ 3 ⫹

4 5

⫹ 25 ⫹ 15 ⫹

1 p 10 ⫹ 3 3 p 2 ⫺ 4 ⫹

; find S12

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30.

9–80


2 k 5a b 3 k⫽1

兺

4 k 12a b 3 k⫽1 q

31.

兺

1 k 5a b 2 k⫽1 q

32.

兺

33. Sumpter reservoir contains 121,500 ft3 of water and is being drained in the following way. Each day one-third of the water is drained (and not replaced). Use a sequence/series to compute how much water remains in the pond after 7 days. 34. Credit-hours taught at Cody Community College have been increasing at 7% per year since it opened in 2001 and taught 1225 credit-hours. For the new faculty, the college needs to predict the number of credit-hours that will be taught in 2015. Use a sequence/series to compute the credit-hours for 2015 and to find the total number of credit hours taught through the 2015 school year.

SECTION 9.4

Mathematical Induction

KEY CONCEPTS • Functions written in subscript notation can be evaluated, graphed, and composed with other functions. • A sum formula involving only natural numbers n as inputs can be proven valid using a proof by induction. Given that Sn represents a sum formula involving natural numbers, if (1) S1 is true and (2) Sk ⫹ ak⫹1 ⫽ Sk⫹1, then Sn must be true for all natural numbers. • Proof by induction can also be used to validate other relationships, using a more general statement of the principle. Let Pn be a statement involving the natural numbers n. If (1) P1 is true (Pn for n ⫽ 12 and (2) the truth of Pk implies that Pk⫹1 is also true, then Pn must be true for all natural numbers n. EXERCISES Use the principle of mathematical induction to prove the indicated sum formula is true for all natural numbers n. 35. 1 ⫹ 2 ⫹ 3 ⫹ 4 ⫹ 5 ⫹ p ⫹ n; 36. 1 ⫹ 4 ⫹ 9 ⫹ 16 ⫹ 25 ⫹ 36 ⫹ p ⫹ n2; n1n ⫹ 12 n1n ⫹ 1212n ⫹ 12 an ⫽ n and Sn ⫽ . an ⫽ n2 and Sn ⫽ . 2 6 Use the principle of mathematical induction to prove that each statement is true for all natural numbers n. 37. 4n ⱖ 3n ⫹ 1 38. 6 # 7n⫺1 ⱕ 7n ⫺ 1 39. 3n ⫺ 1 is divisible by 2

SECTION 9.5

Counting Techniques

KEY CONCEPTS • An experiment is any task that can be repeated and has a well-defined set of possible outcomes. • Each repetition of an experiment is called a trial. • Any potential outcome of an experiment is called a sample outcome. • The set of all sample outcomes is called the sample space. • An experiment with N (equally likely) sample outcomes that is repeated t times, has a sample space with N t elements. • If a sample outcome can be used more than once, the counting is said to be with repetition. If a sample outcome can be used only once, the counting is said to be without repetition. • The fundamental principle of counting states: If there are p possibilities for a first task, q possibilities for the second, and r possibilities for the third, the total number of ways the experiment can be completed is pqr. This fundamental principle can be extended to include any number of tasks. • If the elements of a sample space have precedence or priority (order or rank is important), the number of elements is counted using a permutation, denoted nPr and read, “the distinguishable permutations of n objects taken r at a time.” n! . • To expand nPr, we can write out the first r factors of n! or use the formula nPr ⫽ 1n ⫺ r2!

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• If any of the sample outcomes are identical, certain permutations will be nondistinguishable. In a set containing n

elements where one element is repeated p times, another is repeated q times, and another r times 1p ⫹ q ⫹ r ⫽ n2, n! nP n the number of distinguishable permutations is given by ⫽ . p!q!r! p!q!r! • If the elements of a set have no rank, order, or precedence (as in a committee of colleagues) permutations with the n! same elements are considered identical. The result is the number of combinations, nCr ⫽ . r!1n ⫺ r2!

EXERCISES 40. Three slips of paper with the letters A, B, and C are placed in a box and randomly drawn one at a time. Show all possible ways they can be drawn using a tree diagram. 41. The combination for a certain bicycle lock consists of three digits. How many combinations are possible if (a) repetition of digits is not allowed and (b) repetition of digits is allowed. 42. Jethro has three work shirts, four pairs of work pants, and two pairs of work shoes. How many different ways can he dress himself (shirt, pants, shoes) for a day’s work? 43. From a field of 12 contestants in a pet show, three cats are chosen at random to be photographed for a publicity poster. In how many different ways can the cats be chosen? 44. Compute the following values by hand, showing all work: c. 7C4 a. 7! b. 7P4 45. Six horses are competing in a race at the McClintock Ranch. Assuming there are no ties, (a) how many different ways can the horses finish the race? (b) How many different ways can the horses finish first, second, and third place? (c) How many finishes are possible if it is well known that Nellie-the-Nag will finish last and Sea Biscuit will finish first? 46. How many distinguishable permutations can be formed from the letters in the word “tomorrow”? 47. Quality Construction Company has 12 equally talented employees. (a) How many ways can a three-member crew be formed to complete a small job? (b) If the company is in need of a Foreman, Assistant Foreman, and Crew Chief, in how many ways can the positions be filled?

SECTION 9.6

Introduction to Probability

KEY CONCEPTS • An event E is any designated set of sample outcomes. • Given S is a sample space of equally likely sample outcomes and E is an event relative to S, the probability of E, n1E2 written P(E), is computed as P1E2 ⫽ , where n(E) represents the number of elements in E, and n(S) n1S2 represents the number of elements in S. • The complement of an event E is the set of sample outcomes in S, but not in E and is denoted ⬃E. • Given sample space S and any event E defined relative to S: 112 P1⬃S2 ⫽ 0, 122 0 ⱕ P1E2 ⱕ 1, 132 P1S2 ⫽ 1, 142 P1E2 ⫽ 1 ⫺ P1⬃E2, and 152 P1E2 ⫹ P1⬃E2 ⫽ 1. • Two events that have no outcomes in common are said to be mutually exclusive. • If two events are not mutually exclusive, P1E1 or E2 2 S P1E1 ´ E2 2 ⫽ P1E1 2 ⫹ P1E2 2 ⫺ P1E1 ¨ E2 2. • If two events are mutually exclusive, P1E1 or E2 2 S P1E1 ´ E2 2 ⫽ P1E1 2 ⫹ P1E2 2 . EXERCISES 48. One card is drawn from a standard deck. What is the probability the card is a ten or a face card? 49. One card is drawn from a standard deck. What is the probability the card is a Queen or a face card? 50. One die is rolled. What is the probability the result is not a three? 51. Given P1E1 2 ⫽ 38, P1E2 2 ⫽ 34, and P1E1 ´ E2 2 ⫽ 56, compute P1E1 ¨ E2 2.

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52. Find P(E) given that n1E2 ⫽ 7C4

# 5C3 and n1S2 ⫽ 12C7.

53. To determine if more physicians should be hired, a medical clinic tracks the number of days between a patient’s request for an appointment and the actual appointment date. The table given shows the probability that a patient must wait d days. Based on the table, what is the probability a patient must wait a. at least 20 days c. 40 days or less e. less than 40 and more than 10 days

b. less than 20 days d. over 40 days f. 30 or more days

Wait (days d)

Probability

0

0.002

0 6 d 6 10

0.07

10 ⱕ d 6 20

0.32

20 ⱕ d 6 30

0.43

30 ⱕ d 6 40

0.178

The Binomial Theorem

SECTION 9.7

KEY CONCEPTS • To expand 1a ⫹ b2 n for n of “moderate size,” we can use Pascal’s triangle and observed patterns. n • For any natural numbers n and r, where n ⱖ r, the expression a b (read “n choose r”) is called the binomial r n n! . coefficient and evaluated as a b ⫽ r r!1n ⫺ r2! If n is large, it is more efficient to expand using the binomial coefficients and binomial theorem. • • The following binomial coefficients are useful/common and should be committed to memory: n n n n a b⫽n a b⫽n a b⫽1 a b⫽1 1 n⫺1 n 0 1 1 n n! ⫽ ⫽ ⫽ 1. • We define 0! ⫽ 1; for example a b ⫽ 0! 1 n n!1n ⫺ n2! n n n n n b a1bn⫺1 ⫹ a b a0bn. • The binomial theorem: 1a ⫹ b2 n ⫽ a b anb0 ⫹ a b an⫺1b1 ⫹ a b an⫺2b2 ⫹ p ⫹ a 0 1 2 n⫺1 n n • The kth term of 1a ⫹ b2 n can be found using the formula a b an⫺rbr, where r ⫽ k ⫺ 1. r EXERCISES 54. Evaluate each of the following: 7 8 a. a b b. a b 5 3

55. Use Pascal’s triangle to expand the expressions: a. 1x ⫺ y2 4

Use the binomial theorem to: 56. Write the first four terms of

b. 11 ⫹ 2i2 5

57. Find the indicated term of each expansion.

a. 1a ⫹ 132 b. 15a ⫹ 2b2 a. 1x ⫹ 2y2 7; fourth b. 12a ⫺ b2 14; 10th 58. Mark Leland is a professional bowler who is able to roll a strike (knocking down all 10 pins on the first ball) 91% of the time. (a) What is the probability he rolls at least four strikes in the first five frames? (b) What is the probability he rolls five strikes (and scares the competition)? 8

7

PRACTICE TEST 1. The general term of a sequence is given. Find the first four terms and the 8th term. 1n ⫹ 22! 2n a. an ⫽ b. an ⫽ n⫹3 n! c. an ⫽ e

a1 ⫽ 3 an⫹1 ⫽ 21an 2 2 ⫺ 1

2. Expand each series and evaluate. 6

a.

兺

k⫽2 5

c.

12k2 ⫺ 32

兺 1⫺22a 4 b

j⫽1

3

6

b.

兺 1⫺12 a j ⫹ 1 b j

j⫽2 q

j

d.

k

兺 7a 2 b

k⫽1

1

j

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Practice Test

3. Identify the first term and the common difference or common ratio. Then find the general term an. a. 7, 4, 1, ⫺2, p b. ⫺8, ⫺6, ⫺4, ⫺2, p c. 4, ⫺8, 16, ⫺32, p

d. 10, 4, 85, 16 25 , p

4. Find the indicated value for each sequence. a. a1 ⫽ 4, d ⫽ 5; find a40 b. a1 ⫽ 2, an ⫽ ⫺22, d ⫽ ⫺3; find n c. a1 ⫽ 24, r ⫽ 12; find a6 d. a1 ⫽ ⫺2, an ⫽ 486, r ⫽ ⫺3; find n 5. Find the sum of each series. a. 7 ⫹ 10 ⫹ 13 ⫹ p ⫹ 100 37

b.

兺

k⫽1

13k ⫹ 22

c. For 4 ⫺ 12 ⫹ 36 ⫺ 108 ⫹ p , find S7 d. 6 ⫹ 3 ⫹ 32 ⫹ 34 ⫹ p 6. Each swing of a pendulum (in one direction) is 95% of the previous one. If the first swing is 12 ft, (a) find the length of the seventh swing and (b) determine the distance traveled by the pendulum for the first seven swings. 7. A rare coin that cost $3000 appreciates in value 7% per year. Find the value after 12 yr. 8. A car that costs $50,000 decreases in value by 15% per year. Find the value of the car after 5 yr. 9. Use mathematical induction to verify that for 5n2 ⫺ n an ⫽ 5n ⫺ 3, the sum formula Sn ⫽ is true 2 for all natural numbers n.

Juliet (Shakespeare), four identical copies of Faustus (Marlowe), and four identical copies of The Faerie Queen (Spenser). If these books are to be arranged on a shelf, how many distinguishable permutations are possible? 16. A company specializes in marketing various cornucopia (traditionally a curved horn overflowing with fruit, vegetables, gourds, and ears of grain) for Thanksgiving table settings. The company has seven fruit, six vegetable, five gourd, and four grain varieties available. If two from each group (without repetition) are used to fill the horn, how many different cornucopia are possible? 17. Use Pascal’s triangle to expand/simplify: a. 1x ⫺ 2y2 4 b. 11 ⫹ i2 4 18. Use the binomial theorem to write the first three terms of (a) 1x ⫹ 122 10 and (b) 1a ⫺ 2b3 2 8. 19. Michael and Mitchell are attempting to make a nonstop, 100-mi trip on a tandem bicycle. The probability that Michael cannot continue pedaling for the entire trip is 0.02. The probability that Mitchell cannot continue pedaling for the entire trip is 0.018. The probability that neither one can pedal the entire trip is 0.011. What is the probability that they complete the trip? 20. The spinner shown is spun once. What is the probability of spinning a. a striped wedge b. a shaded wedge

11 10 9 8

c. a clear wedge

12 1 2 3 4 7 6

10. Use the principle of mathematical induction to verify that Pn: 2 # 3n⫺1 ⱕ 3n ⫺ 1 is true for all natural numbers n.

d. an even number

11. Three colored balls (aqua, brown, and creme) are to be drawn without replacement from a bag. List all possible ways they can be drawn using (a) a tree diagram and (b) an organized list.

g. a shaded wedge or a number greater than 12

12. Suppose that license plates for motorcycles must consist of three numbers followed by two letters. How many license plates are possible if zero and “Z” cannot be used and no repetition is allowed? 13. If one icon is randomly chosen from the following set, find the probability a mailbox is not chosen: {,,,,,}. 14. Compute the following values by hand, showing all work: (a) 6! (b) 6P3 (c) 6C3 15. An English major has built a collection of rare books that includes two identical copies of The Canterbury Tales (Chaucer), three identical copies of Romeo and

5

e. a two or an odd number f. a number greater than nine h. a shaded wedge and a number greater than 12 21. To improve customer service, a cable company tracks the number of days a customer must wait until their cable service is Wait (days d ) Probability installed. The table 0 0.02 shows the probability that a customer must 0 6 d 6 1 0.30 wait d days. Based on 1ⱕd 6 2 0.60 the table, what is the 2ⱕd 6 3 0.05 probability a customer 3ⱕd 6 4 0.03 waits a. at least 2 days b. less than 2 days c. 4 days or less

d. over 4 days

e. less than 2 or at least 3 days f. three or more days

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22. An experienced archer can hit the rectangular target shown 100% of the time at a range of 75 m. Assuming the probability the target is hit is 64 cm related to its area, what is the probability the archer hits within the a. triangle b. circle

in the right column. One out of the hundred will be selected at random for a personal interview. What is the probability the person chosen is a a. woman or a craftsman

48 cm

b. man or a contractor c. man and a technician d. journeyman or an apprentice 24. Cheddar is a 12-year-old male box turtle. Provolone is an 8-year-old female box turtle. The probability that Cheddar will live another 8 yr is 0.85. The probability that Provolone will live another 8 yr is 0.95. Find the probability that a. both turtles live for another 8 yr

c. circle but outside the triangle d. lower half-circle e. rectangle but outside the circle f. lower half-rectangle, outside the circle 23. A survey of 100 union workers was taken to register concerns to be raised at the next bargaining session. A breakdown of those surveyed is shown in the table Expertise Level

Women

Men

Total

Apprentice

16

18

34

Technician

15

13

28

Craftsman

9

9

18

Journeyman

7

6

13

Contractor

3

4

7

50

50

100

Totals

b. neither turtle lives for another 8 yr c. at least one of them will live another 8 yr 25. The quality control department at a lightbulb factory has determined that the company is losing money because their manufacturing process produces a defective bulb 12% of the time. If a random sample of 10 bulbs is tested, (a) what is the probability that none are defective? (b) What is the probability that no more than 3 bulbs are defective?

CALCULATOR EXPLORATION AND DISCOVERY Infinite Series, Finite Results Although there were many earlier flirtations with infinite processes, it may have been the paradoxes of Zeno of Elea (⬃450 B.C.) that crystallized certain questions that simultaneously frustrated and fascinated early mathematicians. The first paradox, called the dichotomy paradox, can be summarized by the following question: How can one ever finish a race, seeing that one-half the distance must first be traversed, then one-half the remaining distance, then one-half the distance that then remains, and so on an infinite number of times? Although we easily accept that races can be finished, the subtleties involved in this question stymied mathematicians for centuries and were not satisfactorily resolved until the 1 ⫹ p 6 1. This is a geometric series eighteenth century. In modern notation, Zeno’s first paradox says 12 ⫹ 14 ⫹ 18 ⫹ 16 1 1 with a1 ⫽ 2 and r ⫽ 2. 1 1 1 and r ⫽ , the nth term is an ⫽ n . Use the “sum(” and 2 2 2 “seq(” features of your calculator to compute S5, S10, and S15 (see Section 9.1). Does Figure 9.69 the sum appear to be approaching some “limiting value?” If so, what is this value? Now compute S20, S25, and S30. Does there still appear to be a limit to the sum? What happens when you have the calculator compute S35? Illustration 1 䊳 For the geometric sequence with a1 ⫽

Solution 䊳 the calculator and enter sum(seq (0.5^X, X, 1, 5)) on the home screen. Pressing gives S5 ⫽ 0.96875 (Figure 9.69). Press 2nd to recall the expression and overwrite the 5, changing it to a 10. Pressing shows S10 ⫽ 0.9990234375. Repeating these steps gives S15 ⫽ 0.9999694824, and it seems that “1” may be a limiting value. Our conjecture receives further support as S20, S25, and S30 are closer and closer to 1, but do not exceed it. CLEAR

ENTER

ENTER

ENTER

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Note that the sum of additional terms will create a longer string of 9’s. That the sum of an infinite number of these terms is 1 can be understood by converting the repeating decimal 0.9 to its fractional form (as shown). For x ⫽ 0.9, 10x ⫽ 9.9 and it follows that 10x ⫽ 9.9 ⫺x ⫽ ⫺0.9 9x ⫽ 9 x⫽ 1 a1 . However, there are 1⫺r many nongeometric, infinite series that also have a limiting value. In some cases these require many, many more terms before the limiting value can be observed. For a geometric sequence, the result of an infinite sum can be verified using Sq ⫽

Use a calculator to write the first five terms and to find S5, S10, and S15. Decide if the sum appears to be approaching some limiting value, then compute S20 and S25. Do these sums support your conjecture? Exercise 1: a1 ⫽ 13 and r ⫽ 13

Exercise 2: a1 ⫽ 0.2 and r ⫽ 0.2

Exercise 3: an ⫽

1 1n ⫺ 12!

Additional Insight: Zeno’s first paradox can also be “resolved” by observing that the “half-steps” needed to complete the race require increasingly shorter (infinitesimally short) amounts of time. Eventually the race is complete.

STRENGTHENING CORE SKILLS Probability, Quick-Counting, and Card Games The card game known as Five Card Stud is often played for fun and relaxation, using toothpicks, beans, or pocket change as players attempt to develop a winning “hand” from the five cards dealt. The various “hands” are given here with the higher value hands listed first (e.g., a full house is a better/higher hand than a flush). Five Card Hand

Description

Probability of Being Dealt

royal flush

five cards of the same suit in sequence from 10 to Ace

0.000 001 540

straight flush

any five cards of the same suit in sequence (exclude royal)

0.000 013 900

four of a kind

four cards of the same rank, any fifth card

full house

three cards of the same rank, with one pair

flush

five cards of the same suit, no sequence required

straight

five cards in sequence, regardless of suit

three of a kind

three cards of the same rank, any two other cards

two pairs

two cards of the one rank, two of another rank, one other card

0.047 500

one pair

two cards of the same rank, any three others

0.422 600

0.001 970

For this study, we will consider the hands that are based on suit (the flushes) and the sample space to be five cards dealt from a deck of 52, or 52C5.

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A flush consists of five cards in the same suit, a straight consists of five cards in sequence. Let’s consider that an Ace can be used as either a high card (as in 10, J, Q, K, A) or a low card (as in A, 2, 3, 4, 5). Since the dominant characteristic of a flush is its suit, we first consider choosing one suit of the four, then the number of ways that the straight can be formed (if needed). Illustration 1 䊳 What is the probability of being dealt a royal flush? Solution 䊳 For a royal flush, all cards must be of one suit. Since there are four suits, it can be chosen in 4C1 ways. A royal flush must have the cards A, K, Q, J, and 10 and once the suit has been decided, it can happen in only this (one) # 4C 1 1C 1 ⬇ 0.000 001 540. way or 1C1. This means P 1royal flush2 ⫽ 52C5 Illustration 2 䊳 What is the probability of being dealt a straight flush? Solution 䊳 Once again all cards must be of one suit, which can be chosen in 4C1 ways. A straight flush is any five cards in sequence and once the suit has been decided, this can happen in 10 ways (Ace on down, King on down, Queen on down, and so on). By the FCP, there are 4C1 # 10C1 ⫽ 40 ways this can happen, but four of these will be royal flushes that are of a higher value and must be subtracted from this total. So in the intended context we have # 4C1 10C1 ⫺ 4 ⬇ 0.000 013 900 P 1straight flush2 ⫽ 52C5 Using these examples, determine the probability of being dealt Exercise 1: a simple flush (no royal or straight flushes) Exercise 2: three cards of the same suit and any two other (nonsuit) cards Exercise 3: four cards of the same suit and any one other (nonsuit) card Exercise 4: a flush having no face cards

CUMULATIVE REVIEW CHAPTERS R–9 1. Robot Moe is assembling memory cards for computers. At 9:00 A.M., 52 cards had been assembled. At 11:00 A.M., a total of 98 had been made. Assuming the production rate is linear a. Find the slope of this line and explain what it means in this context. b. Find a linear equation model for this data. c. Determine how many cards Moe can assemble in an eight-hour day. d. Determine the approximate time that Moe began work this morning. 2. Verify by direct substitution that x ⫽ 2 ⫹ i is a solution to x2 ⫺ 4x ⫹ 5 ⫽ 0. 3. Solve using the quadratic formula: 3x2 ⫹ 5x ⫺ 7 ⫽ 0. State your answer in exact and approximate form. 4. Sketch the graph of y ⫽ 1x ⫹ 4 ⫺ 3 using transformations of a tool box function. Plot the x- and y-intercepts. 5. Write a variation equation and find the value of k: Y varies inversely with X and jointly with V and W. Y is equal to 10 when X ⫽ 9, V ⫽ 5, and W ⫽ 12.

6. Graph the piecewise-defined function and state the domain and range. ⫺2 y ⫽ •x x2

⫺3 ⱕ x ⱕ ⫺1 ⫺1 6 x 6 2 2ⱕxⱕ3

3 7. Verify that f 1x2 ⫽ x3 ⫺ 5 and g1x2 ⫽ 1x ⫹ 5 are inverse functions. How are the graphs of f and g related?

8. For the graph of g(x) shown, state where a. g1x2 ⫽ 0 b. g1x2 6 0 5 c. g1x2 7 0 d. g1x2c e. g1x2T f. local max g. local min h. g1x2 ⫽ 2 ⫺5 i. g(4) j. g1⫺12 k. as x S ⫺1 ⫹ , g1x2 S ____ ⫺5 l. as x S q, g1x2 S ____ m. the domain of g(x)

y

g(x)

9. Compute the difference quotient for each function given. 1 a. f 1x2 ⫽ 2x2 ⫺ 3x b. h1x2 ⫽ x⫺2

5 x

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10. Graph the polynomial function given. Clearly indicate all intercepts. f 1x2 ⫽ x3 ⫹ x2 ⫺ 4x ⫺ 4

2x2 ⫺ 8 . Clearly x2 ⫺ 1 indicate all asymptotes and intercepts.

11. Graph the rational function h1x2 ⫽

12. Write each expression in logarithmic form: 1 ⫽ 3⫺4 81 13. Write each expression in exponential form: a. 3 ⫽ logx 11252 b. ln12x ⫺ 12 ⫽ 5 a. x ⫽ 10 y

b.

14. What interest rate is required to ensure that $2000 will double in 10 yr if interest is compounded continuously? 15. Solve for x. a. e2x⫺1 ⫽ 217

b. log13x ⫺ 22 ⫹ 1 ⫽ 4

16. Solve using matrices and row reduction: 2a ⫹ 3b ⫺ 6c ⫽ 15 • 4a ⫺ 6b ⫹ 5c ⫽ 35 3a ⫹ 2b ⫺ 5c ⫽ 24 17. Solve using a calculator and inverse matrices. 0.7x ⫹ 1.2y ⫺ 3.2z ⫽ ⫺32.5 • 1.5x ⫺ 2.7y ⫹ 0.8z ⫽ ⫺7.5 2.8x ⫹ 1.9y ⫺ 2.1z ⫽ 1.5 18. Find the equation of the hyperbola with foci at 1⫺6, 02 and (6, 0) and vertices at 1⫺4, 02 and (4, 0). 19. Identify the center, vertices, and foci of the conic section defined by x2 ⫹ 4y2 ⫺ 24y ⫹ 6x ⫹ 29 ⫽ 0. 20. Use properties of sequences to determine a20 and S20. a. 262,144, 65,536, 16,384, 4096, p 7 27 19 11 b. , , , , p 8 40 40 40 21. Empty 55-gal drums are stacked at a storage facility in the form of a pyramid with 52 barrels in the first row, 51 barrels in the second row, and so on, until there are 10 barrels in the top row. Use properties of sequences to determine how many barrels are in this stack. 22. Three $20 bills, six $10 bills, and four $5 bills are placed in a box, then two bills are drawn out and placed in a savings account. What is the probability the bills drawn are a. first $20, second $10 b. first $10, second $20 c. both $5 d. first $5, second not $20 e. first $5, second not $10 f. first not $20, second $20

847

23. The manager of Tom’s Tool and Equipment Rentals knows that 4% of all tools rented are returned late. Of the 12 tools rented in the last hour, what is the probability that a. exactly ten will be returned on time b. at least eleven will be returned on time c. at least ten will be returned on time d. none of them will be returned on time 24. Use a proof by induction to verify 3 ⫹ 7 ⫹ 11 ⫹ 15 ⫹ p ⫹ 14n ⫺ 12 ⫽ n12n ⫹ 12 for all natural numbers n. 25. A park ranger tracks the number of campers at a popular park from March 1m ⫽ 32 to September 1m ⫽ 92 and collects the following data (month, number of campers): (3, 56), (5, 126), and (9, 98). Assuming the data is quadratic, draw a scatterplot and construct a 3 ⫻ 3 system of equations and solve to obtain a parabolic equation model. (a) What month had the maximum number of campers? (b) What was this maximum number? (c) How many campers might be expected in April? (d) Based on your model, what month(s) is the park apparently closed to campers (number of campers is zero or negative)? Exercises 26 through 30 require the use of a graphing calculator. 26. Solve the system. For this system x, y, and z are the variables, with ␲ and e the well-known constants. Round your answer to two decimal places. ␲x ⫹ 12y ⫹ ez ⫽ 5 • 12x ⫹ ey ⫹ ␲z ⫽ 5 ex ⫹ ␲y ⫹ 12z ⫽ 5 27. Find the solution region for the system of linear inequalities. Your answer should include a screen shot or facsimile, and the location of any points of intersection, rounded to four decimal places. 112x ⫹ 39y 7 438 57x ⫺ 64y 6 101 μ xⱖ0 yⱖ0 28. A triangle has vertices at (112.3, 98.5), (67.7, ⫺39), and (⫺27␲, 21.5). Use the determinant formula to determine its area, rounded to the nearest tenth. 29. A recursive sequence is defined by a1 ⫽ 0.3 and ak⫹1 ⫽ a2k ⫺ 0.4ak. Find S40, rounded to four decimal places. 30. Solve the equation. Round your answer to two decimal places. log x ⫺ 2 ln x ⫹ 3 log3 x ⫽ 6.


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Appendix I The Language, Notation, and Numbers of Mathematics LEARNING OBJECTIVES

The most fundamental requirement for learning algebra is mastering the words, symbols, and numbers used to express mathematical ideas. “Words are the symbols of knowledge, the keys to accurate learning” (Norman Lewis in Word Power Made Easy, Penguin Books).

In Section A.I you will review:

A. Sets of numbers, graphing real numbers, and set notation B. Inequality symbols and order relations C. The absolute value of a real number D. The order of operations

A. Sets of Numbers, Graphing Real Numbers, and Set Notation To effectively use mathematics as a problem-solving tool, we must first be familiar with the sets of numbers used to quantify (give a numeric value to) the things we investigate. Only then can we make comparisons and develop equations that lead to informed decisions.

Natural Numbers The most basic numbers are those used to count physical objects: 1, 2, 3, 4, and so on. These are called natural numbers and are represented by the capital letter , often written in the special font shown. We use set notation to list or describe a set of numbers. Braces { } are used to group members or elements of the set, commas separate each member, and three dots (called an ellipsis) are used to indicate a pattern that continues indefinitely. The notation 51, 2, 3, 4, 5, p6 is read, “ is the set of numbers 1, 2, 3, 4, 5, and so on.” To show membership in a set, the symbol 僆 is used. It is read “is an element of” or “belongs to.” The statements 6 僆 (6 is an element of ) and 0 僆 (0 is not an element of ) are true statements. A set having no elements is called the empty or null set, and is designated by empty braces { } or the symbol .

EXAMPLE 1

䊳

Writing Sets of Numbers Using Set Notation List the set of natural numbers that are a. greater than 100 b. negative c. greater than or equal to 5 and less than 12

Solution

䊳

a. 5101, 102, 103, 104, p6 b. { }; all natural numbers are positive. c. {5, 6, 7, 8, 9, 10, 11}


䊳

Whole Numbers Combining zero with the natural numbers produces a new set called the whole numbers 50, 1, 2, 3, 4, p6. We say that the natural numbers are a proper subset of the whole numbers, denoted ( , since every natural number is also a whole number. The symbol ( means “is a proper subset of.” A-1

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APPENDIX I The Language, Notation, and Numbers of Mathematics

EXAMPLE 2

䊳

Solution

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Determining Membership in a Set

Given A 51, 2, 3, 4, 5, 66, B 52, 46, and C 50, 1, 2, 3, 5, 86, determine whether the following statements are true or false. Justify your response. a. B ( A b. B ( C c. C ( d. C ( e. 104 僆 f. 0 僆 g. 2 僆 a. c. e. g.

b. False: 4 僆 C, so B 僆 C. d. False: 0 僆 , so C 僆 . f. False: 0 僆 .

True: Every element of B is in A. True: All elements are whole numbers. True: 104 is a whole number. False: 2 is a whole number.


䊳

Integers Numbers greater than zero are positive numbers. Every positive number has an opposite that is a negative number (a number less than zero). Combining zero and the natural numbers with their opposites produces the set of integers 5p , 3, 2, 1, 0, 1, 2, 3, p6 . We can illustrate the location and magnitude of a number (in relation to other numbers) using a number line (see Figure AI.1). Negative numbers . . . 5 4 3 2 1

Figure AI.1

Positive numbers 0 1 2 3 4 5

Negative 3 is the opposite of positive 3

. . .

Positive 3 is the opposite of negative 3

The number that corresponds to a given point on the number line is called the coordinate of that point. When we want to note a specific location on the line, a bold dot “•” is used and we have then graphed the number. Since we need only one coordinate to denote a location on the number line, it is referred to as a one-dimensional graph.

Rational Numbers Fractions and mixed numbers are part of a set called the rational numbers . A rational number is one that can be written as a fraction with an integer numerator and an integer denominator other than zero. In set notation we write 5 pq p, q 僆 ; q 06. The vertical bar “” is read “such that” and indicates that a description follows. In words, we say, “ is the set of numbers of the form p over q, such that p and q are integers and q is not equal to zero.”

WORTHY OF NOTE The integers are a subset of the rational numbers: ( , since any integer can be written as a fraction using a denominator of 0 one: 2 2 1 and 0 1 , etc.

EXAMPLE 3

䊳

Graphing Rational Numbers Graph the fractions by converting to decimal form and estimating their location between two integers: a. 213 b. 72

Solution

䊳

a. 213 2.3333333 p or 2.3

b.

7 2

3.5

2.3 4 3 2 1

3.5 0

1

2

3

4


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Since the division 27 terminated, the result is called a terminating decimal. The decimal form of 213 is called repeating and nonterminating (note that 2.3 2.3). Recall that a repeating decimal is written with a horizontal bar over the first block of digit(s) that repeat. For instance 119 55 2.1454545 p 2.145. When using a calculator for computations involving repeating decimals, you must either use the rational form or “fill the display” with the digits that repeat. As an exploration, suppose that you are to inherit 13 0.3 of a $90,000 estate. How many “repeating threes” (times 90,000) are needed until the calculator returns an answer of $30.000? See Exercises 19 and 20.

Irrational Numbers Although any fraction can be written in decimal form, not all decimal numbers can be written as a fraction. One example is the number represented by the Greek letter ␲ (pi), frequently seen in a study of circles. Although we often approximate ␲ using 3.14, its true value has a nonrepeating and nonterminating decimal form. Other numbers of this type include 2.101001000100001 p (there is no block of digits that repeat—the number of zeroes between each “1” is increasing), and 15 2.2360679 p (the decimal form never terminates). Numbers with a nonrepeating and nonterminating decimal form belong to the set of irrational numbers . EXAMPLE 4

䊳

Approximating Irrational Numbers Use a calculator as needed to approximate the value of each number given (round to 100ths), then graph them on the number line: 12 a. 23 b. ␲ c. 219 d. 2

Solution

䊳

a. 23 1.73

b. ␲ 3.14 2 2

WORTHY OF NOTE Checking the approximation for 15 shown, we obtain 2.23606792 4.999999653. While we can find better approximations by using more and more decimal places, we never obtain five exactly (although some calculators will say the result is 5 due to limitations in programming).

. . .

3 2 1

3 0

d. 12 2 0.71

c. 219 4.36

1

2

p 19 3

4

5

6

7

8

. . .


䊳

Real Numbers The set of rational numbers combined with the set of irrational numbers produces the set of real numbers . Figure AI.2 illustrates the relationship between the sets of numbers we’ve discussed so far. Notice how each subset appears “nested” in a larger set. R (real): All rational and irrational numbers Q (rational): {qp, where p, q z and q 0} Z (integer): {. . . , 2, 1, 0, 1, 2, . . .} W (whole): {0, 1, 2, 3, . . .} N (natural): {1, 2, 3, . . .}

Figure AI.2

H (irrational): Numbers that cannot be written as the ratio of two integers; a real number that is not rational. 2, 7, 10, 0.070070007... and so on.

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EXAMPLE 5

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Solution

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Identifying Members of a Number Set

List the numbers in set A 52, 0, 5, 17, 12, 23, 4.5, 121, ␲, 0.756 that belong to a. b. c. d. a. 2, 0, 5, 12, 23, 4.5, 0.75 僆 c. 0, 5, 12 僆

b. 17, 121, ␲ 僆 d. 2, 0, 5, 12 僆 Now try Exercises 25 through 28

EXAMPLE 6

䊳

䊳

Evaluating Statements about Sets of Numbers Determine whether the statements are true or false. Justify your response. a. ( b. ( c. ( d. (

Solution

䊳

A. You’ve just reviewed sets of numbers, graphing real numbers, and set notation

a. b. c. d.

True: All natural numbers can be written as fractions over 1. False: No irrational number can be written in fraction form. True: All whole numbers are integers. True: Every integer is a real number. Now try Exercises 29 through 40

䊳

B. Inequality Symbols and Order Relations We compare numbers of different size using inequality notation, known as the greater than () and less than () symbols. Note that 4 6 3 is the same as saying 4 is to the left of 3 on the number line. In fact, on a number line, any given number is smaller than any number to the right of it (see Figure AI.3). 4 3 2 1

Figure AI.3

a

0 1 4 3

2

3

4

b

ab

Order Property of Real Numbers Given any two real numbers a and b. 1. a 6 b if a is to the left of b on the number line. 2. a 7 b if a is to the right of b on the number line. Inequality notation is used with numbers and variables to write mathematical statements. A variable is a symbol, commonly a letter of the alphabet, used to represent an unknown quantity. Over the years x, y, and n have become most common, although any letter (or symbol) can be used. Often we’ll use variables that remind us of the quantities they represent, like L for length, and D for distance. EXAMPLE 7

䊳

Writing Mathematical Models Using Inequalities Use a variable and an inequality symbol to represent the statement: “To hit a home run out of Jacobi Park, the ball must travel over three hundred twenty-five feet.”

Solution

䊳

Let D represent distance: D 7 325 ft. Now try Exercises 41 through 44

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In Example 7, note the number 325 itself is not a possible value for D. If the ball traveled exactly 325 ft, it would hit the fence and stay in play. Numbers that mark the limit or boundary of an inequality are called endpoints. If the endpoint(s) are not included, the less than 162 or greater than 172 symbols are used. When the endpoints are included, the less than or equal to symbol 12 or the greater than or equal to symbol 12 is used. The decision to include or exclude an endpoint is often an important one, and many mathematical decisions (and real-life decisions) depend on a clear understanding of the distinction. See Exercises 45 through 50.

B. You’ve just reviewed inequality symbols and order relations

C. The Absolute Value of a Real Number Any nonzero real number “n” is either a positive number or a negative number. But in some applications, our primary interest is simply the size of n, rather than its sign. This is called the absolute value of n, denoted 冟n冟, and can be thought of as its distance from zero on the number line, regardless of the direction (see Figure AI.4). Since distance is always positive or zero, 冟n冟 0. 3 3

Figure AI.4

EXAMPLE 8

䊳

0

1

2

3

4

Absolute Value Reading and Reasoning In the table shown, the absolute value of a number is given in column 1. Complete the remaining columns.

Solution

䊳

Column 1 (In Symbols)

Column 2 (Spoken)

Column 3 (Result)

Column 4 (Reason)

冟7.5冟

“the absolute value of seven and five-tenths”

7.5

the distance between 7.5 and 0 is 7.5 units

冟2冟

“the absolute value of negative two”

2

the distance between 2 and 0 is 2 units

冟6冟

“the opposite of the absolute value of negative six”

6

the distance between 6 and 0 is 6 units, the opposite of 6 is 6


䊳

Example 8 illustrates that the absolute value of a positive number is the number itself, while the absolute value of a negative number is the opposite of that number (recall that n is positive if n itself is negative). For this reason the formal definition of absolute value is stated as follows. Absolute Value For any real number n, 0n 0 e

n n

if if

n0 n 6 0

The concept of absolute value can actually be used to find the distance between any two numbers on a number line. For instance, we know the distance between 2 and 8 is 6 (by counting). Using absolute values, we can write 08 2 0 0 6 0 6, or 0 2 8 0 0 6 0 6. Generally, if a and b are two numbers on the real number line, the distance between them is 0a b 0 , which is identical to 0 b a 0 .

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EXAMPLE 9

䊳

Using Absolute Value to Find the Distance between Points Find the distance between ⫺5 and 3 on the number line.

Solution

䊳

Substituting ⫺5 for a and 3 for b in the formula shown gives 0 ⫺5 ⫺ 3 0 ⫽ 0 ⫺8 0 ⫽ 8 or

C. You’ve just reviewed the absolute value of a real number

0 3 ⫺ 1⫺52 0 ⫽ 08 0 ⫽ 8.


䊳

TECHNOLOGY SUPPORT On many calculators, the absolute value function is accessed by pressing the MATH key, then navigating to “NUM” (number options) and selecting option 1: abs(. The calculator provides the left parenthesis, you must supply the right.

D. The Order of Operations The operations of addition, subtraction, multiplication, and division are defined for the set of real numbers, and the concept of absolute value plays an important role. Prior to our study of the order of operations, we will review fundamental concepts related to division and zero, exponential notation, and square roots/cube roots.

Division and Zero

# The quotient 36 9 ⫽ 4 can be checked using the related multiplication: 4 9 ⫽ 36✓. A similar check can be used to understand quotients involving zero. EXAMPLE 10

䊳

Understanding Division with Zero by Writing the Related Product Rewrite each quotient using the related product. 0 a. 0 ⫼ 8 ⫽ p b. 16 c. 12 ⫽n 0 ⫽ q

Solution

䊳

0 # a. 0 ⫼ 8 ⫽ p, if p # 8 ⫽ 0. b. 16 c. 12 ⫽ n, if n # 12 ⫽ 0. 0 ⫽ q, if q 0 ⫽ 16. This shows p ⫽ 0. There is no such number q. This shows n ⫽ 0.

Now try Exercises 67 through 70 WORTHY OF NOTE When a pizza is delivered to your home, it often has “8 parts to the whole,” and in fraction form we have 88 . When all 8 pieces are eaten, 0 pieces remain and the fraction form becomes 08 ⫽ 0. However, the expression 80 is meaningless (undefined), since it would indicate a pizza that has “0 parts to the whole (??).”

䊳

In Example 10(a), a dividend of 0 and a divisor of 8 means we are going to divide zero into eight groups. The related multiplication shows there will be zero in each group. As in Example 10(b), an expression with a divisor of 0 cannot be computed or checked. Although it seems trivial, division by zero has many implications in a study of mathematics, so make an effort to know the facts: The quotient of zero and any nonzero number is zero 1 0n ⫽ 02 but division by zero is undefined 1 n0 is undefined2 . The special case of 00 is said to be indeterminate, as 00 ⫽ n appears to be true for all real numbers n (since the check gives n # 0 ⫽ 0✓). The expression 00 is studied in greater detail in more advanced classes. Division and Zero

The quotient of zero and any real number n is zero 1n ⫽ 02: 0⫼n⫽0 The expressions n ⫼ 0

and

0 ⫽ 0. n n are undefined. 0

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TECHNOLOGY SUPPORT On a graphing calculator, we can evaluate the same expression simultaneously for multiple values using braces “{ }”. Note that when the divisor is any nonzero number x, the calculator shows that 0x 0 (see Figures AI.5 and AI.6), while x 0 returns an error message (Figure AI.7). Figure AI.5

Figure AI.6

Figure AI.7

Squares, Cubes, and Exponential Form When a number is repeatedly multiplied by itself as in (10)(10)(10)(10), we write it using exponential notation as 104. The number used for repeated multiplication (in this case 10) is called the base, and the superscript number is called an exponent. The exponent tells how many times the base occurs as a factor, and we say 104 is written in exponential form. Numbers that result from squaring an integer are called perfect squares, while numbers that result from cubing an integer are called perfect cubes. These are often collected into a table, such as Table AI.1, and students are strongly encouraged to memorize these values to help complete many common calculations mentally. Only the square and cube of selected positive integers are shown. Table AI.1 Perfect Cubes

Perfect Squares N

N2

N

N2

N

N3

1

1

7

49

1

1 8

2

4

8

64

2

3

9

9

81

3

27 64

4

16

10

100

4

5

25

11

121

5

125

144

6

216

6

EXAMPLE 11

䊳

36

12

Evaluating Numbers in Exponential Form Write each exponential in expanded form, then determine its value. a. 43 b. 162 2 c. 62 d. 1 23 2 3

Solution

䊳

a. 43 4 # 4 # 4 64

c. 62 16 # 62 36

b. 162 2 162 # 162 36 d. 1 23 2 3 23

# 23 # 23 278


䊳

Examples 11(b) and 11(c) illustrate an important distinction. The expression 162 2 gives a single operation, “the square of negative six” and the negative sign is included in both factors. The expression 62 gives two operations, “six is squared, and the result is made negative.” The square of six is calculated first, with the negative sign applied afterward.

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Square Roots and Cube Roots Index

2 For the square root operation, either the 1 or 1 notation can be used. The 1 symbol is called a radical, the number under the radical is called the radicand, and the small case number used is called the index (see figure). The index tells how many factors are needed to obtain the radicand. For example, 125 5, since 5 # 5 52 25 (when the 1 symbol is used, the index is understood to be 2). In general, 1a b only if b2 a. All numbers greater than zero have one positive and one negative square root. The positive or principal square root of 49 is 7 1 149 72 since 72 49. The negative square root of 49 is 7 1149 7). 3 The cube root of a number has the form 1 a b, where b3 a. This means 3 3 3 1 27 3 since 3 27, and 1 8 2 since 122 3 8. The cube root of a real number has one unique real value. In general, we have the following:

Radical

A 3

Radicand

Square Roots

Cube Roots

1a b if b a 1a 02 This indicates that

3 1 a b if b3 a 1a 僆 2 This indicates that

1a # 1a a or 1 1a2 2 a

3 3 3 1 a# 1 a# 1 aa 3 3 or 1 1 a2 a

2

WORTHY OF NOTE It is helpful to note that both 0 and 1 are their own square root, cube root, and nth root. That is, 10 0, n 3 1 0 0, p , 1 0 0; and 11 1, n 3 1 1 1, p , 1 1 1.

EXAMPLE 12

䊳

Evaluating Square Roots and Cube Roots Determine the value of each expression. 3 9 a. 149 b. 1 c. 216 125

Solution

䊳

d. 116

e. 125

3 a. 149 7 since 7 # 7 49 b. 1 125 5 since 5 # 5 # 5 125 9 3 # 3 9 3 c. 216 4 since 4 4 16 d. 116 4 since 116 4 e. 125 is not a real number [note that 5 # 5 152 152 25]


䊳

TECHNOLOGY SUPPORT On many graphing calculators, the cube root function is accessed by pressing the MATH key and using 3 option 4: 1 (. The properties stated in the tan box for cube roots are illustrated here using the cube root of 7. WORTHY OF NOTE Sometimes the acronym PEMDAS is used as a more concise way to recall the order of operations: Parentheses, Exponents, Multiplication, Division, Addition, and Subtraction. The idea has merit, so long as you remember that multiplication and division have an equal rank, as do addition and subtraction, and these must be computed in the order they occur (from left to right).

For square roots, if the radicand is a perfect square or has perfect squares in both the numerator and denominator, the result is a rational number as in Examples 12(a) and 12(c). If the radicand is not a perfect square, the result is an irrational number. Similar statements can be made regarding cube roots [Example 12(b)].

The Order of Operations When basic operations are combined into a larger mathematical expression, we use a specified priority or order of operations to evaluate them.

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The Order of Operations 1. Simplify within grouping symbols (parentheses, brackets, braces, etc.). If there are “nested” symbols of grouping, begin with the innermost group. If a fraction bar is used, simplify the numerator and denominator separately. 2. Evaluate all exponents and roots. 3. Compute all multiplications or divisions in the order they occur from left to right. 4. Compute all additions or subtractions in the order they occur from left to right. EXAMPLE 13

䊳

Evaluating Expressions Using the Order of Operations Simplify using the order of operations: a. 5 2 # 3 4.5182 3 c. 3 1 125 23

Solution WORTHY OF NOTE Many common tendencies are hard to overcome. For instance, let’s evaluate the expressions 3 4 # 5 and 24 6 # 2. For the first, the correct result is 23 (multiplication before addition), though some will get 35 by adding first. For the second, the correct result is 8 (multiplication or division in order), though some will get 2 by multiplying first.

䊳

a. 5 2 # 3 5 6 11

multiplication before addition result

b. 8 36 4112 3 2 8 36 4112 92 8 36 4132 8 9132 8 27 35 2

c.

4.5182 3

simplify within parentheses 12 9 3 division before multiplication multiply result original expression

3 2 125 23 36 3 58 39 13 3

d. 7500 a1

D. You’ve just reviewed the order of operations

b. 8 36 4112 32 2 # 0.075 12 15 d. 7500 a1 b 12

simplify terms in the numerator and denominator

combine terms result 12 # 15

0.075 b 12 # 750011.006252 12 15 750011.006252 180 750013.0694517272 23,020.88795

original expression simplify within the parenthesis (division before addition) simplify the exponent so it can be applied exponents before multiplication result


TECHNOLOGY SUPPORT While graphing and calculating technology give us enormous power and convenience, great care must still be taken in its use and application. In the case of Example 13(d), note that the product in the exponent must be parenthesized so that the calculator executes the operations as intended.

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A.I EXERCISES 䊳



1. The symbol ( means: is a and the symbol 僆 means: is an

of of.

2. A number corresponding to a point on the number line is called the of that point. 3. Every positive number has two square roots, one and one . The two square roots 149 represents the of 49 are ; and square root of 49.

䊳

4. The decimal form of 17 contains an infinite number of non and non digits. This means that 17 is a(n) number. 5. Discuss/Explain why the value of 12 and not 12.

# 13 23 is 423

6. Discuss/Explain (a) why 152 2 25, while 52 25; and (b) why 53 152 3 125.

DEVELOPING YOUR SKILLS 7. List the natural numbers that are a. less than 6. b. less than 1. 8. List the natural numbers that are a. between 0 and 1. b. greater than 50.

Identify each of the following statements as either true or false. If false, give an example that shows why.

9. (

11. 533, 35, 37, 396 (

10. 傺

12. 52.2, 2.3, 2.4, 2.56 (

Convert to decimal form and graph by estimating the number’s location between two integers.

16. 78

17. 259

22.

75 4

23. 3

24.

25␲ 2

For the sets in Exercises 25 through 28:

14. 1297 僆 50, 1, 2, 3, p6

4 3

Use a calculator to approximate the value of each number (round to hundredths as needed). Then graph each number by estimating its location between two integers.

21. 7

13. 6 僆 50, 1, 2, 3, p6

15.

20. An architect is reviewing the floor plan for a new office building that offers 36,000 ft2 of office 13 0.72 is space on the first floor. On this floor, 18 considered premium frontage space. (a) Using the 13 fraction 18 , how many square feet is considered “premium?” (b) Using a calculator, determine the number of “repeating 2’s” that are required (0.72, 0.722, 0.7222, etc) before the correct answer is returned.

18. 156

19. A Texas rancher has 120,000 acres of range land, and wants to use two-thirds of it for cattle and the remaining 31 0.3 for sheep. (a) Using the fraction 1 3 , how many acres will be set aside for sheep? (b) Using a calculator, determine the number of “repeating 3’s” that are required (0.3, 0.33, 0.333, etc.) before the correct answer is returned.

a. List all numbers that are elements of (i) , (ii) , (iii) , (iv) , (v) , and (vi) . b. Reorder the elements of each set from smallest to largest. c. Graph the elements of each set on a number line.

25. 51, 8, 0.75, 92, 5.6, 7, 35, 66

26. 57, 2.1, 5.73, 356, 0, 1.12, 78 6

27. 55, 149, 2,3, 6,1,13, 0, 4, ␲6

28. 58, 5, 235, 1.75, 22, 0.6, ␲, 72,2646

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State true or false. If false, state why.

29. (

30. (

31. (

32. (

33. 225 僆

34. 219 僆

Match each set with its correct symbol and description/illustration.

35. 36.

Irrational numbers Integers

a. b.

I. {1, 2, 3, 4, p} II. 5 ab, |a, b 僆 ; b 06

c. III. {0, 1, 2, 3, 4, p}

37.

Real numbers

38.

Rational numbers

39.

Whole numbers

e.

40.

Natural numbers

f.

d. IV. 5␲, 17, 113, etc.} V. 5. . . 3, 2, 1, 0, 1, 2, 3, p6 VI. , , , ,

Use a descriptive variable or the variable given with an inequality symbol (⬍, ⬎, ⱕ, ⱖ) to write a model for each statement.

50. To warn against trespassers, a new motion detector is installed. The detector’s range is from 2 m, to no more than 20 m. Evaluate/simplify each expression.

51. 2.75

52. 7.24

53. 4

54. 6

1 55. ` ` 2

2 56. ` ` 5

3 57. ` ` 4

3 58. ` ` 7

Use the concept of absolute value to complete Exercises 59 to 68.

59. Write the statement two ways, then simplify. “The distance between 7.5 and 2.5 is p” 60. Write the statement two ways, then simplify. “The distance between 1325 and 235 is p” 61. What two numbers on the number line are five units from negative three? 62. What two numbers on the number line are three units from two?

41. To spend the night at a friend’s house, Kylie must be at least 6 years old.

63. If n is positive, then n is

.

64. If n is negative, then n is 65. If n 6 0, then 0 n 0

.

42. Monty can spend at most $2500 on the purchase of a used automobile. 43. If Jerod gets no more than two words incorrect on his spelling test he can play in the soccer game this weekend. 44. Andy must weigh less than 112 lb to be allowed to wrestle in his weight class at the meet. 45. In order for the expression 12x 3 to represent a 3 real number, x must be greater than or equal to 2 46. In order for the expression 15 4x to represent a 5 real number, x must be less than or equal to . 4 1 47. In order for the expression to represent a 12 x real number, x must be less than 2. 1 to represent a 1x 7 real number, x must be greater than 7.

48. In order for the expression

49. In order for a weight sensor to function properly, an item must weigh at least 5 grams, but less than 32 grams.

A-11

.

66. If n 7 0, then 0 n 0

.

Determine which expressions are equal to zero and which are undefined. Justify your responses by writing the related multiplication.

67. 12 0 69.

7 0

68. 0 12 70.

0 7

Without computing the actual answer, state whether the result will be positive or negative. Be careful to note what power is used and whether the negative sign is included in parentheses.

71. a. 172 2 c. 172 5 72. a. 172 3 c. 172 4

b. 72 d. 75 b. 73 d. 74

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Evaluate without the aid of a calculator.

121 B 36

Evaluate without a calculator, using the order of operations.

25 B 49

73.

74.

3 75. 1 8

3 76. 1 64

95. 12 10 2 5 132 2 96. 15 22 2 16 4 # 2 1

77. What perfect square is closest to 78?

97.

78. What perfect cube is closest to 71? Perform the operation indicated without the aid of a calculator.

79. 24 1312

80. 45 1542

83. 456 112 2

84. 118 134 2

81. 7.045 9.23 85.

123 21358 2

87. 1122132102 89. 60 12 182

91.

4 5

93.

23

䊳

16 21

82. 0.0762 0.9034 86.

1821214 2

88. 112102152 90. 75 1152

99.

94.

4172 62

100.

6 149

5162 32 9 164

#

0.06 4 10 b 101. 2475 a1 4 #

0.078 52 20 102. 5100 a1 b 52 103. 104.

7 8

25 3 2 9 98. a b a b 2 4 B 64

Evaluate using a calculator (round to hundredths).

92. 15 12 34

9 3 5 2 #a b B 16 5 3

4 2142 2 41321392 2132 12 21122 2 41221322 2122


105. Pitch diameter: D ⴝ

d#n nⴙ2

106. Pediatric dosages and Clark’s rule: DC ⴝ

Mesh gears are used to transfer rotary motion and power from one shaft to another. The pitch diameter D of a drive gear is given by the formula shown, where d is the outer diameter of the gear and n is the number of teeth on the gear. Find the pitch diameter of a gear with 12 teeth and an outer diameter of 5 cm.

d

DA # W 150

The amount of medication prescribed for young children depends on their weight, height, age, body surface area and other factors. Clark’s rule is a formula that helps estimate the correct child’s dose DC based on the adult dose DA and the weight W of the child (an average adult weight of 150 lb is assumed). Compute a child’s dose if the adult dose is 50 mg and the child weighs 30 lb.

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APPENDIX I The Language, Notation, and Numbers of Mathematics 䊳

APPLICATIONS

Use positive and negative numbers to model the situation, then compute.

107. Temperature changes: At 6:00 P.M., the temperature was 50°F. A cold front moves through that causes the temperature to drop 3°F each hour until midnight. What is the temperature at midnight?

108. Air conditioning: Most air conditioning systems are designed to create a 2° drop in the air temperature each hour. How long would it take to reduce the air temperature from 86° to 71°?

109. Record temperatures: The state of California holds the record for the greatest temperature swing between a record high and a record low. The record high was 134°F and the record low was ⫺45°F. How many degrees difference are there between the record high and the record low?

110. Cold fronts: In Juneau, Alaska, the temperature was 17°F early one morning. A cold front later moved in and the temperature dropped 32°F by lunchtime. What was the temperature at lunchtime?

Use a calculator and the rational/irrational numbers given to compute.

111. An insect crawls 1512 cm along the diagonal of a square, 31 2 cm along the length of a line segment, and 10␲ cm around the circumference of a circle. What is the total distance traveled (to the nearest 100th of a cm)? 31 cm 2

10 cm

112. Find the distance between points A and B (rounded to the nearest 10th cm), given that the square and equilateral triangle have sides of length 10 cm, the circle has a circumference of 22 cm, and the line segment is 35 2 cm long.

10 cm

10 cm

15 cm A

䊳

10 √2 cm

5 √3 cm

22 cm ␲

35 cm 2


113. Here are some historical approximations for ␲. Which one is closest to the true value? Archimedes: 317 Aryabhata:

62,832 20,000

355 Tsu Ch’ung-chih: 113

Brahmagupta: 110

114. If A 7 0 and B 6 0, is the product A # 1⫺B2 positive or negative? 115. If A 6 0 and B 6 0, is the quotient ⫺1A ⫼ B2 positive or negative?

B

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Appendix II Geometry Review with Unit Conversions LEARNING OBJECTIVES In Section A.II you will review how to:

A. Find the perimeter and area of common geometric figures B. Compute the volume of common geometric solids C. Use unit conversion factors effectively

Developing the ability to use mathematics as a descriptive tool is a major goal of this text. Without a solid understanding of basic geometry, this goal would be difficult to achieve—as many of the tasks we perform daily are based on decisions regarding size, measurement, configuration, and the like.

A. Perimeter and Area Formulas Basic geometry plays an important role in the application of mathematics. For your convenience, the most common formulas are collected in Table A.II.1, and a focused effort should be made to commit them to memory. Note that some of the formulas use subscripted variables, or a variable with a small case number to the lower right (sl, s2, and so on). To help understand the table, we quickly review some fundamental terms Table A.II.1 Perimeter Formula (linear units or units)

Definition and Diagram

Area Formula (square units or units2)

a three sided polygon s1

triangle

s2

P ⫽ s1 ⫹ s2 ⫹ s3

h

s3

A⫽

1 bh 2

b

a quadrilateral with four right angles and opposite sides parallel rectangle

P ⫽ 2L ⫹ 2W

A ⫽ LW

P ⫽ 4s

A ⫽ s2

sum of all sides P ⫽ s 1 ⫹ s2 ⫹ s3 ⫹ s4

h A ⫽ 1b1 ⫹ b2 2 2

C ⫽ 2␲r or C ⫽ ␲d

A ⫽ ␲r2

W

L a rectangle with four equal sides square

s

a quadrilateral with one pair of parallel sides (called bases b1 and b2) s2

trapezoid s1

circle

A-14

s4

b1 s3

the set of all points lying in a plane that are an equal distance (called the radius r) from a given point (called the center C ).

h

b2

r C

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APPENDIX II Geometry Review with Unit Conversions

and their meaning. A plane is the infinite extension of length and width along a flat surface. Perimeter is the distance around a two-dimensional figure, or a closed figure that lies in a plane. Many times these figures are polygons, or closed figures composed of line segments. The general name for a four-sided polygon is a quadrilateral. A right angle is an angle measuring 90⬚. A quadrilateral with four right angles is called a rectangle. Area is a measure of the amount of surface covered by a plane figure, with the measurement given in square units. The formulas C ⫽ ␲d and C ⫽ 2␲r both use the symbol “␲,” which represents the ratio of a circle’s circumference to its diameter. We will use a two decimal approximation in calculations done by hand: ␲ ⬇ 3.14. On many calculators, ␲ is the 2nd function to the key and produces a much better approximation (see Figure A.II.1). When using a calculator, we most often use all displayed digits and round only the answer to the desired level of accuracy. If a problem or application uses a formula, begin by stating the formula rather than by immediately making any substitutions. This will help to prevent many careless errors. For Example 1, recall that a trapezoid is a quadrilateral with two parallel sides.

EXAMPLE 1

䊳

Finding the Area of a Window A basement window is shaped like an isosceles trapezoid (base angles equal, opposite sides equal in length), with a height of 10 in. and bases of 1.5 ft and 2 ft. What is the area of the glass in the window?

Solution

䊳

1.5 ft

10 in. 2 ft

Before applying the area formula, all measures must use the same unit. In inches we have 1.5 ft ⫽ 18 in. and 2 ft ⫽ 24 in. h A ⫽ 1b1 ⫹ b2 2 2 10 in. 118 in. ⫹ 24 in.2 ⫽ 2 ⫽ 15 in.2 142 in.2 ⫽ 210 in

2

given formula

substitute 10 in. for h, 18 in. for b1 and 24 in. for b2 simplify result

The area of the glass in the window is 210 in2. Now try Exercises 7 through 18

䊳

WORTHY OF NOTE In actual practice, most calculations are done without using the units of measure, with the correct units supplied in the final answer. When like units occur in an exercise, they are treated just as the numeric factors. If they are part of a product, we write the units with an appropriate exponent as in Example 1. If the like units occur in the numerator and denominator, they “cancel” as we will see in Example 5.

Composite Figures The largest part of geometric applications, whether in art, construction, or architecture, involve composite figures, or figures that combine basic shapes. In many cases we are able to partition or break the figure into more common shapes using an auxiliary line, or a dashed line drawn to highlight certain features of the diagram. When computing a perimeter, we use only the exposed, outer edges, much as a soldier would guard the base camp by marching along the outer edge—the perimeter. For composite figures, it’s helpful to verbally describe the situation given, creating an English language model that can easily be translated into an equation model.

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EXAMPLE 2

䊳

Determining the Perimeter and Area of a Composite Figure Figure A.II.2

Find the perimeter and area of the composite Figure A.II.2. Use ␲ ⬇ 3.14. 䊳

To compute the perimeter, use only the exposed (outer) edges as shown in Figure A.II.3.

3.4 ft

Solution

7.5 ft

Figure A.II.3 3.4 ft

7.5 ft

3.4 ft

7.5 ft

Perimeter does not include these lengths

• Perimeter ⫽ three sides of rectangle ⫹ one-half circle ␲d P ⫽ 2L ⫹ W ⫹ 2 ⬇ 217.52 ⫹ 3.4 ⫹

verbal model formula model

13.142 1 3.42

substitute 7.5 for L, 3.4 for W, and 3.14 for ␲

2

⬇ 15 ⫹ 3.4 ⫹ 5.338 ⬇ 23.738

simplify result

The perimeter of the figure is about 23.7 ft.

• Total Area ⫽ area of rectangle ⫹ one-half area of circle ␲r 2

verbal model

2

A ⫽ LW ⫹

⬇ 17.5213.42 ⫹

formula model

13.14211.72

2

substitute 1.7 for r (d/2)

2

⬇ 25.5 ⫹ 4.5373 ⬇ 30.0373 A. You’ve just reviewed how to find the perimeter and area of common geometric figures.

simplify result 2

The area of the figure is about 30.0 ft . Now try Exercises 19 through 30

䊳

B. Volume Volume is a measure of the amount of space occupied by a three-dimensional object and is measured in cubic units. Some of the more common formulas are given in Table AII.2.

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Table A.II.2 Volume Formula (cubic units or units3)

Illustration

rectangular solid

a six-sided solid figure with opposite faces congruent and adjacent faces meeting at right angles

a rectangular solid with six congruent, square faces

cube

sphere

the set of all points an equal distance (called the radius) from a given point (called the center)

right circular cylinder

union of all line segments connecting two congruent circles in parallel planes, meeting each at a right angle

right circular cone

union of all line segments connecting a given point (vertex) to a given circle (base) and whose altitude meets the center of the base at a right angle

right square pyramid

EXAMPLE 3

WORTHY OF NOTE It is again worth noting that units of measure are treated as though they were numeric factors. For the cylinder in Example 3: ␲ # 15 ft2 2 # 13 ft2 ⫽ ␲ # 25 ft2 # 3 ft ⫽ 75␲ ft3. This concept is an important part of the unit conversions often used in the application of mathematics.

䊳

union of all line segments connecting a given point (vertex) to a given polygon (base) and whose altitude meets the center of the base at a right angle

V ⫽ LWH

H L

W

V ⫽ S3

S

r

4 V ⫽ ␲r3 3

h

V ⫽ ␲r2h

r

1 V ⫽ ␲r2h 3

h r

1 V ⫽ s2h 3

h s

Determining the Volume of a Composite Figure Sand at a cement factory is being dumped from a conveyor belt into a pile shaped like a right circular cone atop a right circular cylinder (see figure). How many cubic feet of sand are there at the moment the cone is 6 ft high with a diameter of 10 ft?

6 ft

3 ft 10 ft

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Solution

䊳

Total Volume ⫽ volume of cylinder ⫹ volume of cone 1 V ⫽ ␲r2h1 ⫹ ␲r2h2 3 1 ⫽ ␲152 2 132 ⫹ ␲152 2 162 3 ⫽ 75␲ ⫹ 50␲ ⫽ 125␲

verbal model formula model substitute 5 for r, 3 for h1, and 6 for h2 simplify result (exact form)

There are about 392.7 ft3 of sand in the pile.

B. You’ve just reviewed how to compute the volume of common geometric solids.


䊳

C. Unit Conversion Factors Unit conversion factors are used to convert from one unit of measure to a related unit, like ounces to pounds. The procedure used is based on the fact that multiplying any quantity by the equivalent of 1 does not change its value. Several examples of unit conversion factors are shown here. a.

12 inches 1 foot

b.

2000 pounds 1 ton

c.

4 quarts 1 gallon

OR

OR

OR

1 foot 12 inches

1 ton 2000 pounds

1 gallon 4 quarts

A unit factor is set up so that the units you are converting from cancel out, leaving only the units you are converting to. When doing these conversions, it is helpful to set up the units first and then write the numeric values with their related unit. EXAMPLE 4

䊳

Unit Conversions — Feet to Miles Mt. Everest is 29,035 ft high and is the highest mountain in the world when measured from see level. Convert 29,035 ft to miles.

Solution

䊳

a. Set up the units of the conversion factor so that “feet” will cancel. 29,035 ft mi a b 1 ft b. Recall that 5280 ft ⫽ 1 mi, and write the numeric values with their related unit: “5280” with feet and “1” with mile. Then simplify the result and state your answer. 29,035 29,035 ft 1 mi a b⫽ mi 1 5280 ft 5280 ⬇ 5.49905303 Mount Everest is about 5.5 mi high. Now try Exercises 37 and 38

䊳

Volume and Capacity Strictly speaking, there is a distinction between the volume of a three dimensional object and its capacity. Volume is the amount of space occupied by the object, capacity is considered to be a measure of the object’s potential for holding or storing something.

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Some of the common relationships between units of volume and capacity are given in the following table. Volume and Capacity 1 gal ⫽ 231 in3

EXAMPLE 5

䊳

1 L ⫽ 1000 cm3 ⫽ 1 dm3

1 mL ⫽ 1 cm3

Unit Conversions — Cubic Inches to gallons

10.5 in.

While taking inventory of a warehouse, workers find a large cylindrical paint can with no label. The cylinder measures 13 in. high and has a diameter of 10.5 in. (see figure). How many gallons of paint does the container hold?

Solution

䊳

13 in.

a. Compute the volume in cubic inches. V ⫽ ␲r2h ⬇ 13.142 15.252 2 1132 ⬇ 1125 in3

volume of a cylinder 10.5 ⫽ 5.25 for r, 13 for h, and 3.14 for ␲ substitute 2 simplify

b. Convert cubic inches to gallons using a unit conversion factor. 1125 in3 ⫽

1125 in3 1 gal a b 1 231 in3

⬇ 4.87 gallons

set up units result

There are just under 4.9 gal of paint in the container. Now try Exercises 39 and 40

䊳

U.S. Units and Metric Units Since U.S. Customary units and metric units were developed independently of each other, there is no “direct link” between the two, and we usually look up equivalent values in a reference book or appendix. Since two equal quantities have a ratio of 1, any conversion factor needed can be formed as before. The length/distance conversions shown in Table AII.3 are rounded to three decimal places. Table A.II.3 U.S. to Metric 1 inch ⫽ 2.54 centimeters

Metric to U.S. 1 centimeter ⬇ 0.3937 inch

1 foot ⫽ 0.3048 meter

1 meter ⬇ 3.2808 feet

1 yard ⫽ 0.9144 meter

1 meter ⬇ 1.0936 yards

1 mile ⬇ 1.6093 kilometers

1 kilometer ⬇ 0.6214 mile

Using the left half of the table when doing U.S. to metric conversions and the right-half when doing metric to U.S. conversions helps to simplify the calculation. EXAMPLE 6

䊳

Unit Conversions — Kilometers to Miles The flight distance between New York City and London is 5536 kilometers. Approximately how many miles is the related flight?

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Solution

䊳

Set up the conversion factor, noting from the table that 1 km ⬇ 0.6214 mi 5536 km 0.6214 miles a b 5536 km ⫽ 1 1 km ⬇ 3440 mi

conversion factors set up units result

The flight distance is approximately 3440 mi. 䊳


Conversion factors for weight, volume, and capacity can be found online or in reference books and applied in the same manner.

Similar Triangles Another important geometric relationship is that of similar triangles. Two triangles are similar if corresponding angles are equal. The angles are usually named with capital letters and the side opposite each angle is named used the related lower case letter (see Figure A.II.4). Similar triangles have the following useful properties: Figure A.II.4 Similar Triangles B c

A

a

b

C

E

Given 䉭ABC and 䉭DEF as shown. If ⬔A ⬔D, ⬔B ⬔E and ⬔C ⬔F, then 䉭ABC and 䉭DEF are similar triangles and corresponding sides are in proportion. a b d e

b c e f

c a f d

f

D

d

F

e

The phrase corresponding sides means sides that are in the same relative position in each triangle. This property enables us to find the length of a missing side by setting up and solving a proportion. One important application of similar triangles involves ramps or other inclined planes. EXAMPLE 7

䊳

Determining the Height of a Ramp The ramp leading up to a loading dock has a base of 8 m with vertical braces placed every 2 m. If the first vertical support is 1.5 m high, how high is the dock?

Solution

䊳

Since the two triangles in question are similar, we set up and solve a proportion. Let h represent the height of the dock (see figure). dock height brace height ⫽ horizontal length dock length h 1.5 ⫽ 2 8 12 ⫽ 2h 6⫽h

h 1.5 m 2m

2m

2m

2m

corresponding sides are in proportion

substitute known values clear denominators solve for h

The dock is 6 ft high. Now try Exercises 47 and 48

䊳

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Additional Applications Cubes and rectangular boxes are part of a larger family of solids called prisms. A prism is any solid figure with bases of the same size and shape (the top and bottom are both called bases). A right prism has sides which are perpendicular to the bases, with all cross sections congruent. The volume of a right prism is the area of its base times its height. This is easily seen in the case of a rectangular box: V ⫽ LWH ⫽ 1LW2H ⫽ 1area of base2 # H. Volume of a Right Prism triangular prism

EXAMPLE 8

䊳

V ⫽ Bh, where B represents the area of the base and h represents the height.

trapezoidal prism

Determining the Volume of a Right Prism The feeding trough at a large cattle ranch is a right prism with trapezoidal bases (see figure). The trough is 1.4 ft deep and 38 ft long. If the bases of the trapezoid are 2 ft and 3 ft, how many loads of feed are needed to fill the trough if each load is 54 ft3? 1.4 ft

3 ft 2 ft 38 ft

Solution

䊳

Notice the trough is a right trapezoidal prism. a. Find the area of the trapezoidal bases: h 1b1 ⫹ b2 2 2 1.4 ⫽ 13 ⫹ 22 2 ⫽ 3.5 ft2

A⫽

area of a trapezoid

substitute 1.4 for h, 3 for b1, and 2 for b2 result

The area of each base is 3.5 ft2. b. Compute the volume of the trough (the right prism). V ⫽ Bh ⫽ 13.52 1382 ⫽ 133 ft3

volume ⫽ area # height substitute 3.5 for B, 38 for h result

The volume of the trough is 133 ft3. c. Find how many 54-ft3 loads are needed for 133 ft3 (divide). 133 ft3 54 ft3 ⬇ 2.463

loads ⬇

C. You’ve just reviewed how to use unit conversion factors effectively.

Approximately 2.5 loads of feed are needed to fill the trough. Now try Exercises 49 through 66

䊳

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A.II EXERCISES 䊳



䊳

1. A geometric figure composed of more than one basic shape is a figure.

2. A trapezoid is any sides.

that has exactly two

3. Volume is a measure of the amount of occupied by an object.

4. Capacity is a measure of an object’s potential to or something.

5. Discuss/Explain the methods available to find the perimeter, area, and volume of composite figures.

6. Discuss/Explain the formula for computing the volume of a prism. Does the concept apply to the formula for the volume of a cylinder? Explain.


Compute the area of each trapezoid using the dimensions given.

7.

88 mm

8.

124 in. 25 in.

3 cm 64 mm

7 ft

Use a ruler to estimate the area of each trapezoid using the actual measurements.

9. In square millimeters,

10. In square inches,

Graph paper can be an excellent aid in understanding why perimeter must be measured in linear units, while area is measured in square units. Suppose the graph paper shown has squares which are 1 cm by 1 cm. Use the grid to answer each question by counting and by computation.

11. What are the perimeter and area of the rectangle shown? Exercises 11 and 13

12. What are the perimeter and area of the trapezoid shown? Exercises 12 and 14

13. What are the perimeter and area of the triangle shown?

14. What are the perimeter and area of the triangle shown?

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15. What is the outer circumference of a circular gear with a radius (center to teeth) of 2.5 cm? Round to tenths.

16. If the inner radius of the gear from Exercise 15 (center to base of teeth) is 2.2 cm, what is the working depth of the gear (height of gear teeth)?

17. Find the missing length, then compute the area and perimeter.

18. Find the missing length, then compute the area and perimeter. b

c

27 cm 39 in.

89 in. 36 cm

In Exercises 19–26, a composite figure is shown. Construct an appropriate formula and use it to find the perimeter, area, or volume as indicated.

19. Perimeter

20. Perimeter

26. Find the perimeter and area of the skirt pattern shown.

36 in. 9 in.

7.5 in. 12 cm 7.5 in.

7.5 in. 12 cm

12 cm

7.5 in.

21. Area

27. How many square inches of paper are needed to cover the kite?

22. Area

12 in.

30 in.

12 in.

18.4 in. 11 ft 7 ft 18.4 in.

28. Find the area of the mainsail and the area of the jib sail of the sailboat. 9 ft

6 ft

23. Area

24. Area 9m

18 in.

6m

Main

Jib

8 ft

4 ft

29. Find the length of the missing side. 24 cm

25. Determine the outer perimeter and total area of the track and field shown.

Perimeter ⫽ 72 cm

120 ft

30. Find the height of the trapezoid. 24 in.

80 ft h

Area 720 in2 36 in.

W

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For each figure, (a) draw auxiliary lines that partition the figure into basic shapes and label the sides as needed. (b) Give a verbal model and a formula model that could be used to complete the exercise.

31. Area

32. Volume

33. Volume

34. Perimeter

36. Find the composite volume.

35. Find the composite volume. 5 in.

10 m

7 in.

6m

2 in. 10 in.

10 in.

37. Convert 625 quarts to gallons.

41. Convert 6500 m to yards.

38. Convert 52,600 pounds to tons.

42. A mountain has a height of 18,300 ft. What is the height in kilometers?

39. Convert 2887.5 in3 to gallons. 40. How many gallons in a container with a volume of 623.7 in3? 䊳

43. Convert 300 pounds to kilograms. 44. Convert 95 gallons to liters.


45. Matching: Place the correct letter in the corresponding blank without using any reference material. A. perimeter of a square

B. perimeter of a rectangle

C. perimeter of a triangle

D. area of a triangle

E. circumference of a circle

F. volume of a cube

G. area of a square

H. perimeter of a trapezoid

I. volume of a rectangular solid

J. area of a circle

K. area of a rectangle

L. area of a trapezoid

____ LW

____ S1 ⫹ S2 ⫹ S3

____ ␲r2

____ 4S

____ LWH

1 ____ BH 2

____ S1 ⫹ S2 ⫹ S3 ⫹ S4

____ 2␲r

____ S2

____ 2L ⫹ 2W

____ S3

____

h1b1 ⫹ b2 2 2

46. Surface Area of a Rectangular Box: SA ⴝ 2(LW ⴙ LH ⴙ WH) The surface area of a rectangular box is found by summing the areas of all six sides. Find the surface area of a box 15 in. tall, 8 in. wide, and 3 in. deep.

Exercise 46 3 in.

15 in.

8 in.

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APPENDIX II Geometry Review with Unit Conversions 䊳

APPLICATIONS (ROUND TO TENTHS AS NEEDED)

47. Similar triangles: To estimate the height of a flagpole, Mitchell reasons his shadow must be proportional to the shadow of the flagpole due to similar triangles. Mitchell is 4 feet tall and his shadow measures 7.5 ft. The shadow of the flagpole measures 60 ft. How high is the pole? Exercise 47

51. Length of a cable: A radio tower is secured by cables which are clamped 21.5 m up the tower and anchored in the ground 9 m from its base. If 30 cm lengths are needed to secure the cable at each end, how long are the cables? Round to hundredths of a meter.

Exercise 51 21.5 m

52. Height of a kite: Benjamin Franklin is flying his kite in a storm once again p and has let out 200 m of string. John Adams has walked to a position directly under the kite and is 150 m from Ben. How high is the kite to the nearest meter? 48. Similar triangles: A triangular image measuring 8 in. ⫻ 9 in. ⫻ 10 in. is projected on a screen using an overhead projector. If the smallest side of the projected image is 2 feet, what are the other dimensions of the projected image? 49. Volume of a prism: Using a sheet of canvas that is 10 ft by 16 ft, a simple tent is made using a long pole through the middle and pegging the sides into the ground as shown. If the tent is 4 feet high at the apex, what is the volume of space contained in the tent? Exercise 49 4 ft 16 ft

50. Volume of a prism: The feeding trough at a pig farm is a right prism with bases that are isosceles triangles (two equal sides). The trough is 3 ft wide, 2 ft deep, and 12 ft long. What is its volume? If each load is 7 ft3, how many loads of slop are needed to fill the trough? Exercise 50 3 ft 2 ft 12 ft

c

9m

Exercise 52

200 m

?

150 m

53. Unit conversions: A U.S. citizen living in Brazil wants a Brazilian carpenter to build a wooden chest 3 ft high, 2.5 ft wide, and 4 ft long. If the carpenter knows only the metric system, what dimensions should the carpenter be given? 54. Unit conversions: A baseball pitcher at the majorleague level can throw the ball around 145 ft/sec. What is the speed of the ball in kilometers per hour? Exercise 54

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55. Most economical purchase: Missy’s Famous Pizza Emporium is running a special — one large for $8.99 or two mediums for $13.49. If a large pizza has a 14-in. diameter and a medium pizza has a 12-in. diameter, which is the better buy (least expensive per square inch)?

cement cost $3.50 ft3 and a 5.65% tax was paid, what was the total cost of the materials used to complete the driveway? Exercise 60 12 ft

56. Most economical purchase: A large can of peaches costs 79¢ and is 15 cm high with a radius of 6 cm. A small can of peaches costs 39¢ and is 15 cm high with a radius of 4 cm. Which is the better buy (least expensive per cubic cm)?

32 ft 50 ft

57. Volume of a jacuzzi: A certain Jacuzzi tub is 84 in. long 60 in. wide and 30 in. deep. How many gallons of water will it take to fill this tub? 58. Volume of a birdbath: I am trying to fill an outdoor birdbath, and all I can find is a plastic dish pan 7 in. high, with a 12 in. by 15 in. base. If the birdbath holds 15 gal, how many trips will have to be made? Exercise 58

33 ft

61. Cost of drywall: After the studs are up, the wall shown in the figure must be covered in drywall. (a) How many square feet of drywall are needed? (b) If drywall is sold only in 4 ft by 8 ft sheets, about how many sheets are required? (c) If drywall costs $8.25 per sheet and a 5.75% tax must be paid, what is the total cost of the material? Exercise 61

3 ft 10 ft

15 ft

7 ft 19 ft

59. Paving a walkway: Current Exercise 59 plans call for building a circular fountain 6 m in diameter, surrounded by a 6m circular walkway 1.5 m wide. (a) What is the approximate area of the walkway? (b) If 1.5 m the concrete for the walkway is 6 cm deep, what volume of cement must be used? (c) If the cement costs $125/m3 with a 7% sales tax, what is the cost of the materials for this walkway? 60. Paving a driveway: A driveway and turn-about has the dimensions shown in the figure. (a) What is the area of the driveway? (b) If the concrete was poured to a depth of 4 in., what volume of cement was used to construct the driveway? (c) If the

62. Cost of baseboards: The dimensions for the living room/dining room of a home are shown, (a) How many feet and inches of molding are needed for the baseboards around the perimeter of the room? (b) If the molding is only sold in 8-ft lengths, how many are needed? (c) If the molding costs $1.74 per foot and sales tax is 5.75%, what is the total cost of the baseboards? Exercise 62 14' 4"

10' 9" 10' 9" 4' 3"

2' 3" 6' 8"

63. Dimensions of an index card: A popular-sized index card has a length that is 1 in. less than twice its width. If the card has an area of 15 in2, find the length and width.

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A-27


64. Dimensions of a ruler: The plastic ruler that Albert uses for graphing lines has a length that is 1 cm more than seven times its width. If the ruler has an area of 30 cm2, find its length and width. 65. Tracking an oil leak: At an oil storage facility, one of the tanks has a slow leak. If the tank shown was full to begin with, how many gallons have been lost at the moment the height of the oil in the tank is 24 ft? 66. Tracking a water leak: A cylindrical water tank has developed a slow leak. If the tank is standing vertically and is 4 ft tall with a radius of 9 in., how many gallons have leaked out at the moment the height of the water in the tank is 3 ft?

䊳

Exercise 65 50 ft 25 ft


67. The area (in cm2) of the region shown in the figure is a. 35 ⫺ xy b. 35 ⫹ xy d. 35 ⫹ xy ⫺ 7y e. 35 ⫺ xy ⫹ 5x

Exercise 67

c. 35 ⫹ 7y ⫺ 5x f. none of these

68. A 2-quart saucepan is 75% full of leftover soup. Which plastic container will best store the leftover soup (i.e., with minimum wasted volume)? a. a hemispheric bowl with radius 3.5 in. b. a cylinder with radius 3 in. and height 5 in. c. a rectangular container 4 in. ⫻ 6 in. ⫻ 3 in. d. None of these will hold the soup.

x cm y cm 5 cm

7 cm

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A-28

APPENDIX III More on Synthetic Division

Appendix III More on Synthetic Division As the name implies, synthetic division simulates the long division process, but in a condensed and more efficient form. It’s based on a few simple observations of long division, as noted in the division 1x3 ⫺ 2x2 ⫺ 13x ⫺ 172 ⫼ 1x ⫺ 52 shown in Figure AIII.1. Figure AIII.1

Figure AIII.2

x ⫹ 3x ⫹ 2 x ⫺ 5冄 x3 ⫺ 2x2 ⫺ 13x ⫺ 17 2

⫺ 1x3 ⫺ 5x2 2

x ⫺ 5冄 1

3 ⫺13

2 ⫺17

5

↓

3x ⫺ 13x 2

1 ⫺2 3

⫺ 13x ⫺ 15x2 2

15

↓

2x ⫺ 17

2

⫺ 12x ⫺ 102 ⫺7

10 ⫺7

remainder

remainder

A careful observation reveals a great deal of repetition, as any term in red is a duplicate of the term above it. In addition, since the dividend and divisor must be written in decreasing order of degree, the variable part of each term is unnecessary as we can let the position of each coefficient indicate the degree of the term. In other words, we’ll agree that 1

⫺2

⫺13 ⫺17

represents the polynomial 1x3 ⫺ 2x2 ⫺ 13x ⫺ 17.

Finally, we know in advance that we’ll be subtracting each partial product, so we can “distribute the negative,” shown at each stage. Removing the repeated terms and variable factors, then distributing the negative to the remaining terms produces Figure AIII.2. The entire process can now be condensed by vertically compressing the rows of the division so that a minimum of space is used (Figure AIII.3). Figure AIII.3 1 3

x ⫺ 5冄 1

Figure AIII.4

⫺2

⫺13

2 ⫺17

dividend

5

15

10

products

3

2

sums

⫺7

1 ⫺2

3 ⫺13

2 ⫺17

dividend

↓

5

15

10

products

1

3

2

quotient

x ⫺ 5冄1

⫺7

remainder

quotient

Further, if we include the lead coefficient in the bottom row (Figure AIII.4), the coefficients in the top row (in blue) are duplicated and no longer necessary, since the quotient and remainder now appear in the last row. Finally, note all entries in the product row (in red) are five times the sum from the prior column. There is a direct connection between this multiplication by 5 and the divisor x ⫺ 5, and in fact, it is the zero of the divisor that is used in synthetic division (x ⫽ 5 from x ⫺ 5 ⫽ 0). A simple change in format makes this method of division easier to use, and highlights the location of the A-28

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APPENDIX III More on Synthetic Division

A-29

divisor and remainder (the blue brackets in Figure AIII.5). Note the process begins by “dropping the lead coefficient into place” (shown in bold). The full process of synthetic division is shown in Figure AIII.6 for the same exercise. Figure AIII. 5 ⫺2 ⫺13

divisor (use 5, not ⫺5) →

5 1 ↓ drop lead coefficient into place → 1

⫺17

← coefficients of dividend ← quotient and remainder appear in this row

We then multiply this coefficient by the “divisor,” place the result in the next column and add. In a sense, we “multiply in the diagonal direction,” and “add in the vertical direction.” Continue the process until the division is complete. 5 1 multiply by 5

↓ 1

The result is x2 ⫹ 3x ⫹ 2 ⫹

⫺2 5 3

↓

cob19545_app_028-029.qxd

Figure AIII. 6 ⫺13 ⫺17 15 10 2 ⫺7

← coefficients of dividend ← quotient and remainder appear in this row

⫺7 , read from the last row. x⫺5

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A-30

APPENDIX IV More on Synthetic Division

Appendix IV More on Matrices Reduced Row-Echelon Form A matrix is in reduced row-echelon form if it satisfies the following conditions: 1. All null rows (zeroes for all entries) occur at the bottom of the matrix. 2. The first non-zero entry of any row must be a 1. 3. For any two consecutive, nonzero rows, the leading 1 in the higher row is to the left of the 1 in the lower row. 4. Every column with a leading 1 has zeroes for all other entries in the column. Matrices A through D are in reduced row-echelon form. 0 A ⫽ £0 0

1 0 0

0 0 0

0 1 0

0 0 1

5 3§ 2

1 B ⫽ £0 0

0 1 0

0 0 0

5 3§ 0

1 C ⫽ £0 0

0 1 0

0 3 0

5 ⫺2 § 0

1 D ⫽ £0 0

0 1 0

5 2 0

0 0§ 1

Where Gaussian elimination places a matrix in row-echelon form (satisfying the first three conditions), Gauss-Jordan elimination places a matrix in reduced row-echelon form. To obtain this form, continue applying row operations to the matrix until the fourth condition above is also satisfied. For a 3 ⫻ 3 system having a unique solution, the diagonal entries of the coefficient matrix will be 1’s, with 0’s for all other entries. To illustrate, we’ll extend Example 4 from Section 7.1 until reduced row-echelon form is obtained. EXAMPLE 4

䊳

Solving Systems Using the Augmented Matrix and Gauss-Jordan Elimination 2x ⫹ y ⫺ 2z ⫽ ⫺7 Solve using Gauss-Jordan elimination: • x ⫹ y ⫹ z ⫽ ⫺1 ⫺2y ⫺ z ⫽ ⫺3 2x ⫹ y ⫺ 2z ⫽ ⫺7 • x ⫹ y ⫹ z ⫽ ⫺1 ⫺2y ⫺ z ⫽ ⫺3 1 1 1 ⫺1 £2 1 ⫺2 ⫺7 § 0 ⫺2 ⫺1 ⫺3 1 1 1 ⫺1 £0 1 4 5§ 0 ⫺2 ⫺1 ⫺3 1 £0 0

1 1 0

1 ⫺1 4 5§ 1 1

matrix form →

2 1 ⫺2 ⫺7 £1 1 1 ⫺1 § 0 ⫺2 ⫺1 ⫺3

R 1 4 R2

1 1 1 ⫺1 £2 1 ⫺2 ⫺7 § 0 ⫺2 ⫺1 ⫺3

⫺2R1 ⫹ R2 S R2

1 1 1 ⫺1 £ 0 ⫺1 ⫺4 ⫺5 § 0 ⫺2 ⫺1 ⫺3

⫺1R2

1 1 1 ⫺1 £0 1 4 5§ 0 ⫺2 ⫺1 ⫺3

1 ⫺1 4 5§ 7 7

2R2 ⫹ R3 S R3

1 £0 0

1 1 0

⫺R2 ⫹ R1 S R1

1 £0 0

0 ⫺3 ⫺6 1 4 5§ 0 1 1

R3 S R3 7

3R3 ⫹ R1 S R1 ⫺4R3 ⫹ R2 S R2

1 £0 0

1 1 0

1 ⫺1 4 5§ 1 1

1 £0 0

0 1 0

0 ⫺3 0 1§ 1 1

The final matrix is in reduced row-echelon form with solution (⫺3, 1, 1) just as in Section 7.1. A-30

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A-31

APPENDIX IV More on Matrices

The Determinant of a General Matrix To compute the determinant of a general square matrix, we introduce the idea of a cofactor. For an n ⫻ n matrix A, Aij ⫽ 1⫺12 i⫹j冟Mij冟 is the cofactor of matrix element aij, where 冟Mij冟 represents the determinant of the corresponding minor matrix. Note that i ⫹ j is the sum of the row and column of the entry, and if this sum is even, 1⫺12 i⫹j ⫽ 1, while if the sum is odd, 1⫺12 i⫹j ⫽ ⫺1 (this is how the sign table for a 3 ⫻ 3 determinant was generated). To compute the determinant of an n ⫻ n matrix, multiply each element in any row or column by its cofactor and add. The result is a tier-like process in which the determinant of a larger matrix requires computing the determinant of smaller matrices. In the case of a 4 ⫻ 4 matrix, each of the minor matrices will be size 3 ⫻ 3, whose determinant then requires the computation of other 2 ⫻ 2 determinants. In the following illustration, two of the entries in the first row are zero for convenience. For ⫺2 1 A⫽ ≥ 3 0 we have: det1A2 ⫽ ⫺2 # 1⫺12

1⫹1

2 † ⫺1 ⫺3

0 2 ⫺1 ⫺3 0 4 2

3 0 4 2

0 ⫺2 ¥, 1 1

⫺2 1 1⫹3 1 † ⫹ 132 # 1⫺12 †3 1 0

2 ⫺1 ⫺3

⫺2 1† 1

Computing the first 3 ⫻ 3 determinant gives ⫺16, the second 3 ⫻ 3 determinant is 14. This gives: det1A2 ⫽ ⫺21⫺162 ⫹ 31142 ⫽ 74

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A-32

APPENDIX V Deriving the Equation of a Conic

Appendix V Deriving the Equation of a Conic The Equation of an Ellipse In Section 8.2, the equation 21x ⫹ c2 2 ⫹ y2 ⫹ 21x ⫺ c2 2 ⫹ y2 ⫽ 2a was developed using the distance formula and the definition of an ellipse. To find the standard form of the equation, we treat this result as a radical equation, isolating one of the radicals and squaring both sides. 21x ⫹ c2 2 ⫹ y2 ⫽ 2a ⫺ 21x ⫺ c2 2 ⫹ y2 1x ⫹ c2 2 ⫹ y2 ⫽ 4a2 ⫺ 4a 21x ⫺ c2 2 ⫹ y2 ⫹ 1x ⫺ c2 2 ⫹ y2

isolate one radical square both sides

We continue by simplifying the equation, isolating the remaining radical, and squaring again. expand

x2 ⫹ 2cx ⫹ c2 ⫹ y2 ⫽ 4a2 ⫺ 4a 21x ⫺ c2 2 ⫹ y2 ⫹ x2 ⫺ 2cx ⫹ c2 ⫹ y2 binomials 4cx ⫽ 4a2 ⫺ 4a 21x ⫺ c2 2 ⫹ y2 simplify 2 a 21x ⫺ c2 ⫹ y2 ⫽ a2 ⫺ cx isolate radical; divide by 4 2 2 2 4 2 2 2 a 3 1x ⫺ c2 ⫹ y 4 ⫽ a ⫺ 2a cx ⫹ c x square both sides a2x2 ⫺ 2a2cx ⫹ a2c2 ⫹ a2y2 ⫽ a4 ⫺ 2a2cx ⫹ c2x2 expand and distribute a2 on left add 2a2cx and rewrite equation a2x2 ⫺ c2x2 ⫹ a2y2 ⫽ a4 ⫺ a2c2 2 2 2 2 2 2 2 2 factor x 1a ⫺ c 2 ⫹ a y ⫽ a 1a ⫺ c 2 2 2 y x ⫹ 2 ⫽1 divide by a2 1a2 ⫺ c2 2 2 a a ⫺ c2 Since a 7 c, we know a2 7 c2 and a2 ⫺ c2 7 0. For convenience, let b2 ⫽ a2 ⫺ c2 (it also follows that a2 7 b2 and a 7 b, since c 7 0). Substituting b2 for a2 ⫺ c2 we obtain the standard form of the equation of an ellipse (major axis horizontal, since we y2 x2 stipulated a 7 b): 2 ⫹ 2 ⫽ 1. Note once again the x-intercepts are (⫾a, 0), while a b the y-intercepts are (0, ⫾b). For the foci, c2 ⫽ 冟a2 ⫺ b2冟 to allow for a major axis that may be vertical.

The Equation of a Hyperbola In Section 8.3, the equation 21x ⫹ c2 2 ⫹ y2 ⫺ 21x ⫺ c2 2 ⫹ y2 ⫽ 2a was developed using the distance formula and the definition of a hyperbola. To find the standard form of this equation, we apply the same procedures as before.

A-32

21x ⫹ c2 2 ⫹ y2 ⫽ 2a ⫹ 21x ⫺ c2 2 ⫹ y2 isolate one radical 2 2 2 2 2 1x ⫹ c2 ⫹ y ⫽ 4a ⫹ 4a 21x ⫺ c2 ⫹ y ⫹ 1x ⫺ c2 2 ⫹ y2 square both sides expand x2 ⫹ 2cx ⫹ c2 ⫹ y2 ⫽ 4a2 ⫹ 4a 21x ⫺ c2 2 ⫹ y2 ⫹ x2 ⫺ 2cx ⫹ c2 ⫹ y2 binomials 2 2 2 4cx ⫽ 4a ⫹ 4a 21x ⫺ c2 ⫹ y simplify isolate radical; divide by 4 cx ⫺ a2 ⫽ a 21x ⫺ c2 2 ⫹ y2 2 2 2 4 2 2 2 c x ⫺ 2a cx ⫹ a ⫽ a 3 1x ⫺ c2 ⫹ y 4 square both sides c2x2 ⫺ 2a2cx ⫹ a4 ⫽ a2x2 ⫺ 2a2cx ⫹ a2c2 ⫹ a2y2 expand and distribute a2 on the right add 2a2cx and rewrite equation c2x2 ⫺ a2x2 ⫺ a2y2 ⫽ a2c2 ⫺ a4 2 2 2 2 2 2 2 2 x 1c ⫺ a 2 ⫺ a y ⫽ a 1c ⫺ a 2 factor y2 x2 ⫺ 2 ⫽1 divide by a2 1c2 ⫺ a2 2 a2 c ⫺ a2

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APPENDIX V Deriving the Equation of a Conic

A-33

From the definition of a hyperbola we have 0 6 a 6 c, showing c2 7 a2 and c2 ⫺ a2 7 0. For convenience, let b2 ⫽ c2 ⫺ a2 and substitute to obtain the y2 x2 standard form of the equation of a hyperbola (transverse axis horizontal): 2 ⫺ 2 ⫽ 1. a b Note the x-intercepts are (0, ⫾ a) and there are no y-intercepts. For the foci, c2 ⫽ a2 ⫹ b2.

The Asymptotes of a Central Hyperbola From our work in Section 8.3, a central hyperbola with a horizontal axis will have b asymptotes at y ⫽ ⫾ x. To understand why, recall that for asymptotic behavior we a investigate what happens to the relation for large values of x, meaning as 冟x冟 S q . Starting y2 x2 with 2 ⫺ 2 ⫽ 1, we have a b b2x2 ⫺ a2y2 ⫽ a2b2

clear denominators

ay ⫽bx ⫺ab 2 2

2 2

2 2

a2y2 ⫽ b2x2 a1 ⫺

a b x2 b2 a2 y2 ⫽ 2 x2 a1 ⫺ 2 b a x b a2 y⫽⫾ x 1⫺ 2 a B x

As 冟x冟 S q,

isolate term with y

2

factor out b 2x 2 from right side

divide by a 2

take square roots of both sides

b a2 S 0, and we find that for large values of x, y ⬇ ⫾ x. 2 a x

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A-34

APPENDIX VI Proof Positive—A Selection of Proofs from College Algebra

Appendix VI Proof Positive — A Selection of Proofs from College Algebra Proofs from Chapter 4 The Remainder Theorem If a polynomial p(x) is divided by (x ⫺ c) using synthetic division, the remainder is equal to p(c). Proof of the Remainder Theorem From our previous work, any number c used in synthetic division will occur as the factor (x ⫺ c) when written as 1quotient21divisor2 ⫹ remainder: p1x2 ⫽ 1x ⫺ c2q1x2 ⫹ r. Here, q(x) represents the quotient polynomial and r is a constant. Evaluating p(c) gives p1x2 ⫽ 1x ⫺ c2q1x2 ⫹ r

p1c2 ⫽ 1c ⫺ c2q1c2 ⫹ r ⫽ 0 # q1c2 ⫹ r ⫽r✓

The Factor Theorem Given a polynomial p(x), (1) if p1c2 ⫽ 0, then x ⫺ c is a factor of p(x), and (2) if x ⫺ c is a factor of p(x), then p1c2 ⫽ 0. Proof of the Factor Theorem 1. Consider a polynomial p written in the form p1x2 ⫽ 1x ⫺ c2q1x2 ⫹ r. From the remainder theorem we know p1c2 ⫽ r, and substituting p(c) for r in the equation shown gives p1x2 ⫽ 1x ⫺ c2q1x2 ⫹ p1c2

and x ⫺ c is a factor of p(x), if p1c2 ⫽ 0,

p1x2 ⫽ 1x ⫺ c2q1x2 ✓

2. The steps from part 1 can be reversed, since any factor (x ⫺ c) of p(x), can be written in the form p1x2 ⫽ 1x ⫺ c2q1x2 . Evaluating at x ⫽ c produces a result of zero: p1c2 ⫽ 1c ⫺ c2q1x2 ⫽0✓

Complex Conjugates Theorem Given p(x) is a polynomial with real number coefficients, complex solutions must occur in conjugate pairs. If a ⫹ bi, b ⫽ 0 is a solution, then a ⫺ bi must also be a solution. To prove this for polynomials of degree n 7 2, we let z1 ⫽ a ⫹ bi and z2 ⫽ c ⫹ di be complex numbers, and let z1 ⫽ a ⫺ bi, and z2 ⫽ c ⫺ di represent their conjugates, and observe the following properties: A-34

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A-35

1. The conjugate of a sum is equal to the sum of the conjugates. sum of conjugates: z1 ⫹ z2 sum: z1 ⫹ z2 1a ⫹ bi2 ⫹ 1c ⫹ di2

1a ⫹ c2 ⫹ 1b ⫹ d2i

1a ⫺ bi2 ⫹ 1c ⫺ di2

1a ⫹ c2 ⫺ 1b ⫹ d2i ✓

→ conjugate of sum →

2. The conjugate of a product is equal to the product of the conjugates. product: z1 # z2

product of conjugates: z1 # z2

ac ⫹ adi ⫹ bci ⫹ bdi2

ac ⫺ adi ⫺ cbi ⫹ bdi2

1a ⫹ bi2 # 1c ⫹ di2

1ac ⫺ bd2 ⫹ 1ad ⫹ bc2i

1a ⫺ bi2 # 1c ⫺ di2

1ac ⫺ bd2 ⫺ 1ad ⫹ bc2i ✓

→ conjugate of product →

Since polynomials involve only sums and products, and the complex conjugate of any real number is the number itself, we have the following: Proof of the Complex Conjugates Theorem Given polynomial p1x2 ⫽ anxn ⫹ an⫺1xn⫺1 ⫹ p ⫹ a1x1 ⫹ a0, where an, an⫺1, p , a1, a0 are real numbers and z ⫽ a ⫹ bi is a zero of p, we must show that z ⫽ a ⫺ bi is also a zero. anzn ⫹ an⫺1zn⫺1 ⫹ p ⫹ a1z1 ⫹ a0 ⫽ p1z2 evaluate p (x ) at z anzn ⫹ an⫺1zn⫺1 ⫹ p ⫹ a1z1 ⫹ a0 ⫽ anzn ⫹ an⫺1zn⫺1 ⫹ p ⫹ a1z1 ⫹ a0 ⫽ anzn ⫹ an⫺1zn⫺1 ⫹ p ⫹ a1z1 ⫹ a0 ⫽ 1 n n⫺1 an 1z 2 ⫹ an⫺1 1z 2 ⫹ p ⫹ a1 1z 2 ⫹ a0 ⫽ n n⫺1 1 an 1z 2 ⫹ an⫺1 1z 2 ⫹ p ⫹ a1 1z 2 ⫹ a0 ⫽

0

p 1z2 ⫽ 0 given

0

conjugate both sides

0

property 1

0

property 2

0

conjugate of a real number is the number

p1z2 ⫽ 0

✓

result

An immediate and useful result of this theorem is that any polynomial of odd degree must have at least one real root.

Linear Factorization Theorem If p(x) is a complex polynomial of degree n ⱖ 1, then p has exactly n linear factors and can be written in the form p1x2 ⫽ an 1x ⫺ c1 2 1x ⫺ c2 2 # p # 1x ⫺ cn 2 , where an ⫽ 0 and c1, c2, p , cn are complex numbers. Some factors may have multiplicities greater than 1 (c1, c2, p , cn are not necessarily distinct). Proof of the Linear Factorization Theorem Given p1x2 ⫽ anxn ⫹ an⫺1xn⫺1 ⫹ p ⫹ a1x ⫹ a0 is a complex polynomial, the Fundamental Theorem of Algebra establishes that p(x) has a least one complex zero, call it c1. The factor theorem stipulates 1x ⫺ c1 2 must be a factor of P, giving p1x2 ⫽ 1x ⫺ c1 2q1 1x2

where q1(x) is a complex polynomial of degree n ⫺ 1. Since q1(x) is a complex polynomial in its own right, it too must also have a complex zero, call it c2. Then (x ⫺ c2) must be a factor of q1(x), giving p1x2 ⫽ 1x ⫺ c1 21x ⫺ c2 2q2 1x2

where q2(x) is a complex polynomial of degree n ⫺ 2. Repeating this rationale n times will cause p(x) to be rewritten in the form p1x2 ⫽ 1x ⫺ c1 21x ⫺ c2 2 # p # 1x ⫺ cn 2qn 1x2

where qn(x) has a degree of n ⫺ n ⫽ 0, a nonzero constant typically called an. The result is p1x2 ⫽ an 1x ⫺ c1 21x ⫺ c2 2 # p # 1x ⫺ cn 2 , and the proof is complete.

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A-36


Proofs from Chapter 5 The Product Property of Logarithms Given M, N, and b ⫽ 1 are positive real numbers, logb 1MN2 ⫽ logbM ⫹ logbN. Proof of the Product Property For P ⫽ logbM and Q ⫽ logbN, we have bP ⫽ M and bQ ⫽ N in exponential form. It follows that logb 1MN2 ⫽ logb 1bPbQ 2

⫽ logb 1bP⫹Q 2

substitute b P for M and b Q for N properties of exponents

⫽P⫹Q

log property 3

⫽ logb M ⫹ logb N

substitute logb M for P and logb N for Q

The Quotient Property of Logarithms Given M, N, and b ⫽ 1 are positive real numbers, M logba b ⫽ logbM ⫺ logbN. N Proof of the Quotient Property For P ⫽ logbM and Q ⫽ logbN, we have bP ⫽ M and bQ ⫽ N in exponential form. It follows that logb a

M bP b ⫽ logb a Q b N b

⫽ logb 1bP⫺Q 2

substitute b P for M and b Q for N properties of exponents

⫽P⫺Q

log property 3

⫽ logb M ⫺ logb N

substitute logb M for P and logb N for Q

The Power Property of Logarithms Given M, N, and b ⫽ 1 are positive real numbers and any real number x, logb Mx ⫽ x logbM. Proof of the Power Property For P ⫽ logb M, we have bP ⫽ M in exponential form. It follows that logb 1M2 x ⫽ logb 1bP 2 x ⫽ logb 1bPx 2 ⫽ Px

⫽ 1logb M2x

⫽ x logb M

substitute b P for M properties of exponents log property 3 substitute logb M for P rewrite factors

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Student Answer Appendix CHAPTER R Exercises R.1, pp. 8–10 1. constant 3. coefficient 5. Answers will vary 7. two; 3 and ⫺5 9. two; 2 and 14 11. three; ⫺2, 1, and ⫺5 13. one; ⫺1 15. n ⫺ 7 17. n ⫹ 4 19. 1n ⫺ 52 2 21. 2n ⫺ 13 23. n2 ⫹ 2n 25. 23 n ⫺ 5 27. 31n ⫹ 52 ⫺ 7 29. Let w represent the width in meters. Then 2w represents twice the width and 2w ⫺ 3 represents three meters less than twice the width. 31. Let b represent the speed of the bus. Then b ⫹ 15 represents 15 mph more than the speed of the bus. 33. h ⫽ b ⫹ 150 35. L ⫽ 2W ⫹ 20 37. M ⫽ 2.5N 39. T ⫽ 12.50g ⫹ 50 41. 14 43. 19 45. 0 47. 16 49. ⫺36 51. 51 53. 2 55. 144 57. ⫺41 59. 24 5 61. x 63. Output x Output 14 ⫺3 6 ⫺2 0 ⫺1 0 ⫺4 1 ⫺6 2 ⫺6 3 ⫺4 ⫺1 gives an output of 0.

⫺3 ⫺18 ⫺2 ⫺15 ⫺1 ⫺12 0 ⫺9 1 ⫺6 2 ⫺3 3 0 3 gives an output of 0.

Exercises R.3, pp. 35–39 1. identity; unknown; contradiction; unknown

13. b ⫽ ⫺15

Exercises R.2, pp. 21–24 1. power 3. 20x; 0 5. a. cannot be simplified, unlike terms b. can be simplified, like bases 7. 14n7 9. ⫺12p5q4 11. a14b7 17.

p2 4q2

21.

9 6 2 16 x y

25. a. V ⫽ 27x6

27 4

3. simple; compound 6 9. v ⫽ ⫺11 11. b ⫽ 5

17. x ⫽ 12 19. x ⫽ 12 20 27. n ⫽ 21

25. v ⫽ ⫺0.5

21. p ⫽ ⫺56

12 29. p ⫽ 5

11 10 35. identity; 5x 0 x 僆⺢6 37. 5x 0 x ⱖ ⫺26; 3⫺2, q 2 39. 5x 0 ⫺2 ⱕ x ⱕ 16; 3⫺2, 1 4 41. 5a |a ⱖ 26; 31. contradiction; { } 33. conditional; n ⫽ ⫺

[

⫺3 ⫺2 ⫺1

0

1

43. 5n |n ⱖ 16;

2

3

4

5

[ 0

45. 5x| x 6

1

2

⫺32 5 6;

3

)

⫺10 ⫺9 ⫺8 ⫺7 ⫺6 ⫺5 ⫺4⫺3 ⫺2 ⫺1 0

47. 53. 55. 57. 59.

; a 僆 32, q2 ; n 僆 31, q 2 ⫺32 ; x 僆 1⫺q, 5 2

{ } 49. 5x |x ⑀ ⺢6 51. 5p |p ⑀ ⺢6 526; 5⫺3, ⫺2, ⫺1, 0, 1, 2, 3, 4, 6, 86 5 6; 5⫺3, ⫺2, ⫺1, 0, 1, 2, 3, 4, 5, 6, 76 54, 66; 52, 4, 5, 6, 7, 86 x 僆 1⫺q,⫺22 ´ 11, q 2;

)

19. 49c14d 4

b. 1728 units3 27. 3w3 29. ⫺3ab 4p8 27 ⫺1 8x6 25m4n6 31. 33. 2h3 35. 37. ⫺8 39. 6 41. 43. 9 8 8 q 27y 4r8 2 2 2 12 25p q 3p 5 1 a ⫺12 45. 47. 49. 51. 3 53. 4 8 55. 4 ⫺4q2 3h7 a bc 5x4 7 ⫺2b 7 57. 59. 2 61. 10 63. 13 65. ⫺4 67. 6.77 ⫻ 109 9 27a9c3 69. 0.000 000 006 5 71. 26,571 hrs; 1107 days 73. polynomial, none of these, degree 3 75. nonpolynomial because exponents are not whole numbers, NA, NA 77. polynomial, binomial, degree 3 23. 94 x3y2

15. m ⫽ ⫺

23. a ⫽ ⫺3.6

x Output ⫺3 ⫺5 ⫺2 8 ⫺1 9 0 4 ⫺1 1 2 0 3 13 2 gives an output of 0. 67. a. 7 ⫹ 1⫺52 ⫽ 2 b. n ⫹ 1⫺22 c. a ⫹ 1⫺4.22 ⫹ 13.6 ⫽ a ⫹ 9.4 d. x ⫹ 7 ⫺ 7 ⫽ x 69. a. 3.2 b. 56 71. ⫺5x ⫹ 13 2 73. ⫺15 79. ⫺2a2 ⫹ 2a 81. 6x2 ⫺ 3x p ⫹ 6 75. ⫺2a 77. 17 12 x 38 83. 2a ⫹ 3b ⫹ 2c 85. 29 87. 7a2 ⫺ 13a ⫺ 5 89. 10 ohms 8n ⫹ 5 91. a. t ⫽ 12 j b. t ⫽ 275 mph 93. a. L ⫽ 2W ⫹ 3 b. 107 ft 95. t ⫽ c ⫹ 29; 44¢ 97. C ⫽ 25t ⫹ 43.50; $81 99. a. positive odd integer

15. 32.768h3k6

7. x ⫽ 3

5. Answers will vary.

65.

13. 216p3q6

79. ⫺w3 ⫺ 3w2 ⫹ 7w ⫹ 8.2; ⫺1 81. c3 ⫹ 2c2 ⫺ 3c ⫹ 6; 1 ⫺2 2 83. ⫺2 85. 3p3 ⫺ 3p2 ⫺ 12 87. 7.85b2 ⫺ 0.6b ⫺ 1.9 3 x ⫹ 12; 3 1 2 89. 4 x ⫺ 8x ⫹ 6 91. q6 ⫹ q5 ⫺ q4 ⫹ 2q3 ⫺ q2 ⫺ 2q 93. ⫺3x3 ⫹ 3x2 ⫹ 18x 95. 3r2 ⫺ 11r ⫹ 10 97. x3 ⫺ 27 99. b3 ⫺ b2 ⫺ 34b ⫺ 56 101. 21v2 ⫺ 47v ⫹ 20 103. 9 ⫺ m2 9 105. p2 ⫹ 1.1p ⫺ 9 107. x2 ⫹ 34 x ⫹ 18 109. m2 ⫺ 16 111. 6x2 ⫹ 11xy ⫺ 10y2 113. 12c2 ⫹ 23cd ⫹ 5d2 115. 2x4 ⫺ x2 ⫺ 15 117. 4m ⫹ 3; 16m2 ⫺ 9 119. 7x ⫹ 10; 49x2 ⫺ 100 121. 6 ⫺ 5k; 36 ⫺ 25k2 123. x ⫺ 16; x2 ⫺ 6 125. x2 ⫹ 8x ⫹ 16 127. 16g2 ⫹ 24g ⫹ 9 129. 16p2 ⫺ 24pq ⫹ 9q2 131. 16 ⫺ 8 1x ⫹ x 133. xy ⫹ 2x ⫺ 3y ⫺ 6 135. k3 ⫹ 3k2 ⫺ 28k ⫺ 60 137. a. 340 mg, 292.5 mg b. Less, amount is decreasing. c. after 5 hr 139. F ⫽ kPQd⫺2 141. 5x⫺3 ⫹ 3x⫺2 ⫹ 2x⫺1 ⫹ 4 143. $15 145. 6

)

⫺4 ⫺3 ⫺2 ⫺1

0

61. x 僆 3⫺2, 52 ;

1

2

[

⫺3 ⫺2 ⫺1

0

3

) 1

2

3

4

5

6

3

4

63. no solution 65. x 僆 1⫺q, q 2 ; ⫺4 ⫺3 ⫺2 ⫺1

67. x 僆 3⫺5, 0 4;

0

1

2

[

⫺6 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1

69. x 僆 ⫺1

1⫺13 ,

⫺0.5

⫺14 2;

[ 0

1

)(

0

0.5

1

SA-1

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Page SA-2

Student Answer Appendix

71. x 僆 1⫺q, q2; ⫺4 ⫺3 ⫺2 ⫺1

73. x 僆 3⫺4, 12;

0

1

2

[

0

75. x 僆 3⫺1.4, 0.8 4 ; ⫺1.4 ⫺2

⫺1

77. x 僆 3⫺16, 82;

1

3

v v a1 ⫹ b a1 ⫺ b, L ⫽ 12 211 ⫹ 0.75211 ⫺ 0.752 B c c ⫽ 3 27 in. ⬇ 7.94 in. 109. 11 in. by 13 in. 1 112b5 ⫺ 3b3 ⫹ 8b2 ⫺ 182 111. a. 18 14x4 ⫹ x3 ⫺ 6x2 ⫹ 322 b. 18 113. 2x116x ⫺ 272 16x ⫹ 52 115. 1x ⫹ 32 1x ⫺ 32 1x2 ⫹ 92 117. 1p ⫹ 121p2 ⫺ p ⫹ 121p ⫺ 12 1p2 ⫹ p ⫹ 12

107. L ⫽ L0

119. 1q ⫹ 521q ⫺ 52 1q ⫹ 2321q ⫺ 232

[ 1

2

Exercises R.5, pp. 61–64

[

79. 81. 85. 89. 93. 95.

2

0.8

0

⫺20 ⫺16⫺12 ⫺8 ⫺4

4

)

⫺6 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1

[

3

) 0

4

8

12

a. 8 ⫹ 6 ⫺ 12 ⫽ 2 ✓ b. 6 ⫹ 8 ⫺ 12 ⫽ 2 ✓ c. 20 vartices W 6 177.34 lb 83. 2f ⫹ 5 ⫽ 73; 34 fans 2c ⫹ 13 ⫽ 1467; 727 cal 87. 41 ⫽ 23 S ⫹ 1; 60 yr 91. 82 ⫹ 76 ⫹ 565 ⫹ 71 ⫹ x ⱖ 75, at least 81% 74 ⫽ 10 3 r ⫹ 34; 12 in. 1125 ⫹ 850 ⫹ 625 ⫹ 400 ⫹ x ⱖ 1000, at least $2000 5 0 6 20 w 6 150, 0 6 w 6 7.5 m

97. 45 6 95 C ⫹ 32 6 85, 7.2° 6 C 6 29.4° 99. S ⫽ 4.5 h ⫹ 20, K ⫽ 6 h ⫹ 11, more than 6 hr 101. a. 216.5 ft2 b. 7 sheets will be needed 103. about 337.4 in3 1 105. x ⫽ 107. Answers may vary. 109. 6 111. 6 113. 6 2 115. 7

Exercises R.4, pp. 49–53 1. product 3. binomial; conjugate 5. Answers will vary. 7. a. ⫺171x2 ⫺ 32 b. 7b13b2 ⫺ 2b ⫹ 82 c. ⫺3a2 1a2 ⫹ 2a ⫺ 32 9. a. 1a ⫹ 2212a ⫹ 32 b. 1b2 ⫹ 3213b ⫹ 22 c. 1n ⫹ 72 14m ⫺ 112 11. a. 13q ⫹ 2213q2 ⫹ 52 b. 1h ⫺ 1221h4 ⫺ 32 c. 1k2 ⫺ 72 1k3 ⫺ 52 13. a. ⫺11p ⫺ 721p ⫹ 22 b. prime c. 1n ⫺ 421n ⫺ 52 15. a. 13p ⫹ 221p ⫺ 52 b. 14q ⫺ 521q ⫹ 32 c. 15u ⫹ 32 12u ⫺ 52 17. a. 12s ⫹ 5212s ⫺ 52 b. 13x ⫹ 7213x ⫺ 72 c. 215x ⫹ 62 15x ⫺ 62 d. 111h ⫹ 122111h ⫺ 122 e. 1b ⫹ 2521b ⫺ 252 19. a. 1a ⫺ 32 2 b. 1b ⫹ 52 2 c. 12m ⫺ 52 2 d. 13n ⫺ 72 2 21. a. 12p ⫺ 3214p2 ⫹ 6p ⫹ 92 b. 1m ⫹ 12 21m2 ⫺ 12 m ⫹ 14 2 c. 1g ⫺ 0.321g2 ⫹ 0.3g ⫹ 0.092 d. ⫺2t1t ⫺ 321t2 ⫹ 3t ⫹ 92 23. a. 1x ⫹ 321x ⫺ 321x ⫹ 121x ⫺ 12 b. 1x2 ⫹ 921x2 ⫹ 42 c. 1x ⫺ 22 1x2 ⫹ 2x ⫹ 421x ⫹ 121x2 ⫺ x ⫹ 12 25. a. 1n ⫹ 121n ⫺ 12 b. 1n ⫺ 121n2 ⫹ n ⫹ 12 c. 1n ⫹ 121n2 ⫺ n ⫹ 12 d. 7x12x ⫹ 1212x ⫺ 12 27. 1a ⫹ 521a ⫹ 22 29. 21x ⫺ 221x ⫺ 102 31. ⫺113m ⫹ 8213m ⫺ 82 33. 1r ⫺ 32 1r ⫺ 62 35. 12h ⫹ 321h ⫹ 22 37. 13k ⫺ 42 2 39. ⫺3x12x ⫺ 721x ⫺ 32 41. 4m1m ⫹ 521m ⫺ 22 43. 1a ⫹ 52 1a ⫺ 122 45. 12x ⫺ 5214x2 ⫹ 10x ⫹ 252 47. prime 49. 1x ⫺ 52 1x ⫹ 321x ⫺ 32 51. a. H b. E c. C d. F e. B f. A g. I h. D i. G 53. a ⫽ ⫺1; b ⫽ 2; c ⫽ ⫺15 55. not quadratic 1a ⫽ 02 57. a ⫽ 14 ; b ⫽ ⫺6; c ⫽ 0 59. a ⫽ 2; b ⫽ 0; c ⫽ 7 61. not quadratic (degree 3) 63. a ⫽ 1; b ⫽ ⫺1; c ⫽ ⫺5 65. x ⫽ 5 or x ⫽ ⫺3 67. m ⫽ 4 69. p ⫽ 0 or p ⫽ 2 71. h ⫽ 0 or h ⫽ ⫺1 73. a ⫽ 3 or a ⫽ ⫺3 75. g ⫽ ⫺9 2 77. m ⫽ ⫺5 or m ⫽ ⫺3 or m ⫽ 3 79. c ⫽ ⫺3 or c ⫽ 15 81. r ⫽ 8 or r ⫽ ⫺3 83. t ⫽ ⫺13 or t ⫽ 2 85. x ⫽ 5 or x ⫽ ⫺3 87. w ⫽ ⫺12 or w ⫽ 3 89. x ⫽ ⫺2, x ⫽ 0, x ⫽ 11 91. x ⫽ ⫺3, x ⫽ 0, x ⫽ 23 93. p ⫽ ⫺7, p ⫽ ⫺3, p ⫽ 3 95. x ⫽ ⫺5, x ⫽ 2, x ⫽ 5 97. x ⫽ ⫺5, x ⫽ ⫺2, x ⫽ 2, x ⫽ 5 99. b ⫽ ⫺2, b ⫽ ⫺1, b ⫽ 4, b ⫽ 5 101. 2␲r 1r ⫹ h2,7000␲ cm2; 21,991 cm2 1 103. V ⫽ ␲h1R ⫹ r21R ⫺ r2; 6␲ cm3; 18.8 cm3 3 105. V ⫽ x1x ⫹ 521x ⫹ 32 a. 3 in. b. 5 in. c. V ⫽ 241292 1272 ⫽ 18,792 in3

1. 1; ⫺1 3. common denominator 5. F; numerator should be ⫺1 a⫺4 1 x⫹3 7. a. ⫺ b. 9. a. simplified b. 3 2x1x ⫺ 22 a⫺7 x⫹3 ⫺1 9 11. a. ⫺1 b. ⫺1 13. a. ⫺3ab b. c. ⫺11y ⫹ 32 d. 9 m 1a ⫺ 221a ⫹ 12 2n ⫹ 3 3x ⫹ 5 15. a. b. c. x ⫹ 2 d. n ⫺ 2 17. n 2x ⫹ 3 1a ⫹ 321a ⫹ 22 y 1p ⫺ 42 2 81a ⫺ 72 ⫺15 3 19. 1 21. 23. 25. 27. 29. 2 4 2 a⫺5 x p 1 n⫹ y⫹3 m x ⫹ 0.3 5 31. 33. 35. 37. m⫺4 3y1y ⫹ 42 x ⫺ 0.2 2 n⫹ 3 31a2 ⫹ 3a ⫹ 92 14y ⫺ x 2n ⫹ 1 3 ⫹ 20x 39. 41. 43. 45. 2 n 8x2 8x2y4 ⫺y ⫹ 11 2 ⫺3m ⫺ 16 ⫺5m ⫹ 37 47. 49. 51. 53. p⫹6 1m ⫹ 421m ⫺ 42 m⫺7 1y ⫹ 621y ⫺ 52 2a ⫺ 5 1 m2 ⫺ 6m ⫹ 21 55. 57. 59. 1a ⫹ 421a ⫺ 52 y⫹1 1m ⫹ 32 2 1m ⫺ 32 1 5 1 ⫺ 5p 63. a. 2 ⫺ ; 15y ⫹ 121y ⫹ 32 1y ⫺ 22 p p p2 2 x⫹2 1 4a x b. 2 ⫹ 3 ; 65. 67. p ⫺ 1 69. a ⫹ 20 313x ⫺ 42 x x x3 3 2 1⫹ 1⫹ 2 2 ⫺2 m m⫹3 x x ⫹2 ; ; 71. 73. a. b. 75. x ⫽ 1 y ⫹ 31 3 m⫺3 2 x2 ⫺ 2 1⫺ 1⫺ 2 m x 77. a ⫽ 32 79. y ⫽ 12 81. x ⫽ 3; x ⫽ 7 is extraneous 83. n ⫽ 7 85. a ⫽ ⫺1, a ⫽ ⫺8 87. a. $300 million; $2550 million 89. Price rises rapidly for first four b. It would require many resources. days, then begins a gradual c. No decrease. Yes, on the 35th day of trading. 450P P Day Price 100 ⫺ P 0 10 40 300 1 16.67 2 32.76 60 675 3 47.40 80 1800 4 53.51 5 52.86 90 4050 6 49.25 93 5979 7 44.91 95 8550 8 40.75 9 37.03 98 22050 10 33.81 100 ERROR 61.

y2 ⫹ 26y ⫺ 1

91. t ⫽ 8 weeks 93. P ⬇ 80% 12x ⫹ 3y2 95. b. 5

cob19545_saa_001-009.qxd

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Exercises R.6, pp. 76–80

1. even 3. 1164 2 3 5. Answers will vary. 7. a. 冟9冟 9 b. 冟10冟 10 9. a. 7冟p冟 b. 冟x 3冟 c. 9m2 d. 冟x 3冟 11. a. 4 b. 6x c. 6z4 v d. 13. a. 2 b. not a real number c. 3x2 d. 3x e. k 3 2 7冟v5冟 f. 冟h 2冟 15. a. 5 b. 3冟n3冟 c. not real number d. 6 9p4 64 125 1728 17. a. 4 b. c. d. 19. a. 125 8 4q2 1 256 32n10 1 b. not a real number c. d. 21. a. b. 1 9 81x4 p2 2y4 3 3 2 2 3 3 23. a. 3m12 b. 10pq 1q c. mn 2n d. 4pq 12p 2 x4 1y 9 e. 3 17 f. 12 25. a. 15a2 b. 4b 1b c. 2 3 3 3 15 18 1 3 2 3 2 d. 3u v1v 27. a. 2m b. 3n c. d. 4x z3 6 5 3 1 2 6 4 12 29. a. 2x2y3 b. x2 1 x c. 1 b d. 6 e. b4 6 16 31. a. 9 12 b. 14 13 c. 16 12m d. 5 17p 3 33. a. x 12x b. 2 13x 3 15 c. 6x22x 5 22 27x 3 23 35. a. 98 b. 115 121 c. n2 5 d. 39 12 13 37. a. 19 b. 110 165 2 17 1182 c. 1215 2 114 36 115 6 142 1

39. Verified 41. Verified 43. a.

23 2

b.

2 215x 2

c.

3 26b 10b

9x 3 6 1x 612 5 2a2 d. e. 45. a. 12 4 111; 1.27 b. 2p a x2 47. a. 130 2 15 3 13 3 12; 0.05 7 7 12 16 2 13 b. ; 7.60 49. a. x 14 3 3 b. x 8, x 1 is extraneous 51. a. m 3 b. x 5 c. m 64 d. x 16 53. a. x 25 b. x 7; x 2 is extraneous c. x 2, x 18 d. x 6; x 0 is extraneous 55. a. x 32 b. x 81 57. a. x 32, x 22 b. x 30, x 34 59. x 27, x 125 61. 8.33 ft 63. a. 8210 m b. about 25.3 m 65. a. 365.02 days b. 688.69 days c. 87.91 days 67. a. 36 mph b. 46.5 mph 69. 12␲ 234 ⬇ 219.82 m2 71. a. 36 million mi b. 67 million mi c. 93 million mi d. 142 million mi e. 484 million mi f. 887 million mi 73. a. 1x 1521x 152 b. 1n 11921n 1192 3 2 2p2

75. a. 13 13x 39 1x b. Answers will vary. 79. x 僆 3 1, 22 ´ 12, q2

322 77. 2

Practice Test, pp. 83–84 1. d. 3. 5. b.

a. False; parentheses first b. False; undefined c. True False; 2x 6 2. a. 11 b. 5 c. not a real number d. 20 a. 89 b. 7 c. 0.5 d. 4.6 4. a. 28 b. 0.9 c. 4 d. 7 6 3 ⬇ 4439.28 6. a. 0 b. undefined 7. a. 3; 2, 6, 5 2; 13 , 1 8. a. 13 b. ⬇ 7.29 9. a. x3 12x 92 n 2 b. 2n 3a b 10. a. Let r represent Earth’s radius. Then 11r 119 2 represents Jupiter’s radius. b. Let e represent this year’s earnings. Then 4e 1.2 million represents last year’s earnings. 11. a. 9v2 3v 7 b. 7b 8 c. x2 6x 12. a. 13x 4213x 42 b. v12v 32 2 m6 25 2 2 c. 1x 521x 321x 32 13. a. 5b3 b. 4a12b12 c. d. p q 4 8n3 14. a. 4ab

b. 6.4 102 0.064

c.

a12 b4c8

d. 6

15. a. 9x4 25y2 b. 4a2 12ab 9b2 16. a. 7a4 5a3 8a2 3a 18 b. 7x4 4x2 5x 17. a. 1 31m 72 2n x5 x5 b. c. x 3 d. e. f. 2n 3x 2 3x 1 51m 42 1m 32

2 64 1 12 c. d. e. 11 110 3v 125 2 2 110x f. x2 5 g. h. 21 16 122 19. 0.5x2 10x 1200; 5x a. 10 decreases of 0.50 or $5.00 b. Maximum revenue is $1250. 1 20. 58 cm 21. a. b 6 b. n 4 c. m 1 d. x 6 e. 5 6 (contradiction) f. g 10 22. 3 14 1n 122 16, the number is 40 23. a. x 2, x 2, x 7 b. r 0, r 1, r 4 1 5 c. g 3, g 1, g 1, g 3 24. a. x b. h , h 2 2 3 c. n 13 (2 is extraneous) 25. a. x 3, x 3 b. x 4, x 5 c. x 1 (7 is extraneous) 18. a. 冟x 11冟 b.

CHAPTER 1 Exercises 1.1, pp. 98–102 1. first, second 3. radius, center 5. Answers will vary. in 7. Year D 51, 2, 3, 4, 56 college GPA R 52.75, 3.00, 3.25, 3.50, 3.756 1

2.75

2

3.00

3

3.25

4

3.50

5

3.75

9. D 51, 3, 5, 7, 96; R 52, 4, 6, 8, 106 11. D 54, 1, 2, 36; R 50, 5, 4, 2, 36 13. 15. 10 8 6 4 2 108642 2 4 6 8 10

y

10 8 6 4 2

2 4 6 8 10 x

x

y

108642 2 4 6 8 10

17. D: x 僆⺢ R: y 1

y

15 12 9 6 3

2 4 6 8 10 x

x

y

y

54321 3 6 9 12 15

1 2 3 4 5 x

x

y

6

5

2

0

3

3

0

2, 2

3

8 3

0

1

1

3, 3

2

3

1

3

5, 5

0

1 3

6

3

6

8, 8

2

8

13 3

7

9, 9

3

8

4

15

19. D: 5 x 5 R: 0 y 5 10 8 6 4 2 108642 2 4 6 8 10

x

21. D: x 1 R: y 僆⺢

y

10 8 6 4 2

2 4 6 8 10 x

y

108642 2 4 6 8 10

23. D: x 僆⺢ R: y 僆⺢

y

5 4 3 2 1 108642 1 2 3 4 5

2 4 6 8 10 x

x

y

y

2 4 6 8 10 x

x

y

4

3

10

3, 3

9

2

3

4

5

2, 2

2

1

0

5

4

13, 13

1

0

2

121

2

1, 1

0

1

0.5, 0.5

4

15

0

7

2

3

4

1.25

4

3

1

3

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25. 10, 22, 15, 02

69. 11, 22, r 2 13, x 僆 31 2 13, 1 2 134, y 僆 3 2 213, 2 2 13 4

10

10

71. 14, 02, r 9, x 僆 313, 5 4 , y 僆 39, 9 4

10

73. 1x 52 2 1y 62 2 57, (5, 6), r 157

10

10

10 75. 1x 52 2 1y 22 2 25, 15, 22, r 5

29. 13, 12

10

31. 10.7, 0.32

39. 2 234 41. 10 47. right triangle 49. x2 y2 9 10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

53. 1x 42 2 1y 32 2 4 10 8 6 4 2 108642 2 4 6 8 10

y

108642 2 4 6 8 10

2 4 6 8 10 x

y

2 4 6 8 10 x

61. 1x 72 2 1y 12 2 100 20 16 12 8 4 2016 1284 4 8 12 16 20

y

4 8 12 16 20 x

65. 1x 52 2 1y 42 2 9 10 8 6 4 2 108642 2 4 6 8 10

35. 10, 12

37. 11, 02

51. 1x 52 y 3 2

10 8 6 4 2 108642 2 4 6 8 10

y

2 4 6 8 10 x

77. x2 1y 32 2 14, 10, 32, r 114

2

y

2 4 6 8 10 x

79. 1x 22 2 1y 52 2 11, 12, 52, r 111

55. 1x 72 2 1y 42 2 7 10 8 6 4 2

57. 1x 12 2 1y 22 2 9 10 8 6 4 2

33.

1 1 1 20 , 24 2

43. right triangle 45. not a right triangle

y

108642 2 4 6 8 10 15 12 9 6 3 1512963 3 6 9 12 15

10

27. 10, 02, 10, 42

10 8 6 4 2

108642 2 4 6 8 10

y

2 4 6 8 10 x

81. 1x 72 2 y2 37, 17, 02, r 137

59. 1x 42 2 1y 52 2 12 10 8 6 4 2 108642 2 4 6 8 10

y

83. 1x 32 2 1y 52 2 32, 13, 52, r 4 12 2 4 6 8 10 x

63. 1x 32 2 1y 42 2 41 15 12 9 6 3 1512963 3 6 9 12 15

y

3 6 9 12 15 x

10 8 6 4 2 108642 2 4 6 8 10

y

10 8 6 4 2 108642 2 4 6 8 10

10 8 6 4 2 108642 2 4 6 8 10

10 8 6 4 2 108642 2 4 6 8 10

15 12 9 6 3 1512963 3 6 9 12 15

15 12 9 6 3

2 4 6 8 10 x

y

3 6 9 12 15 x

y

3 6 9 12 15 x

y

2 4 6 8 10 x

y

2 4 6 8 10 x

y

2 4 6 8 10 x

y

3 6 9 12 15 x

y

3 6 9 12 15 x

85. a. y x2 6x 87. b. y x2 1y 32 2 36 89. f. 1x 12 2 1y 22 2 49 91. j. 6x y x2 9 93. a. (3, 183), (5, 241), (7, 299), (9, 357), (11, 415); yes b. $473 y c. 2014 d. (11, 415)

300

200

(3, 183) 0 1 2 3 4 5 6 7 8 9 1011 x

2 4 6 8 10 x

1512963 3 6 9 12 15

1512963 3 6 9 12 15

400

67. (2, 3), r 2, x 僆 30, 4 4, y 僆 3 1, 5 4

15 12 9 6 3

y

95. a. 1x 52 2 1y 122 2 625 b. no 97. Red: 1x 22 2 1y 22 2 4; Blue: 1x 22 2 y2 16; Area blue 12␲ units2

cob19545_saa_001-009.qxd

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Student Answer Appendix 99.

15 12 9 6 3 1512963 3 6 9 12 15

101. 103. 105. c. r2 109.

No, distance between centers is less than sum of radii.

3 6 9 12 15 x

3. negative, downward 5.

x

y

6

10 8

6

3

108642 2 4 6 8 10

4

0

2

3

0

y

11. 0.5 4 13. 4 0.5 0.5 0.5 ✓ 19 3 1 4 2 12 2 4 19 3 4 4 4 15.

19 4 ✓

(5, 2)

108642 2 4 6 8 10

21.

10 8 6 4 (6, 0)2

17.

y

10 8 6 (0, 1) 4 2 (5, 0)

23.

y

108642 2 2 4 6 8 10 x 4 (0, 4) (3, 6) 6 8 10

27.

10 8 6 4 (5, 0) 2

29.

y

108642 2 2 4 6 8 10 x (0, 2) 4 6 (5, 4) 8 10

33. m 1; 12, 42 and 11, 32 4 8

4

1

8

4

2

7

4

10

19.

y

10 8 (2, 8) 6 4 , ( T 0) 2 (0, 3)

(3, 0)

y

10 8 6 4 2 (0, 0) (4, 2) 108642 2 2 4 6 8 10 x 4 6 (2, 1) 8 10

10 8 6 4 2 108642 2 4 6 8 10

y

(2, 0)

y 80

(1, 75)

1 , 0 40 (0, 25) 2

(

8

)

4 40

4

8

x

80

31.

y

10 8 6 4 (4, R) 2

25.

y

10 8 6 4 (0, 4) 2 (3, 0)

(4, 6) (2, 3) (0, 0)

108642 2 2 4 6 8 10 x 4 6 (6, 4) 8 10

2 4 6 8 10 x

4

(4, 2) 4

8

x

4

(1, 8)

8

8 6 4 2

4 2

108642 2 (3, 4) 4 6 8 10

2 4 6 8 10 x

108642 2 2 4 6 8 10 x 4 6 (0, R) 8 10

2 4 6 8 10 x

8

x

8

2 4 6 8 10 x

108642 2 4 6 8 10

2 4 6 8 10 x

(2, 4) (2, 0) 2 4 6 8 10 x

(2, 6)

53. L1: x 2; L2: y 4; point of intersection (2, 4) Justices 55. a. For any two points chosen m 0, indicating 109 8 there has been no increase or decrease in the number 7 6 Nonwhite, nonmale of supreme court justices. b. For any two points 5 4 1 3 chosen m 10 , which indicates that over the last 5 2 decades, one nonwhite or nonfemale justice has been 1 10 20 30 40 50 added to the court every 10 yr. Time t 57. parallel 59. neither 61. parallel 63. right triangle 65. not a right triangle 67. right triangle 69. a. 78.2 yr b. 2015 71. v 1250t 8500 a. $3500 b. 5 yr 73. h 3t 300 a. 273 in. b. 20 months 30 75. Yes they will meet, the two roads are not parallel: 38 12 9.5 . 77. a. $7360 b. 2015 79. a. 21% b. 2016 81. a 6 83. a. 142 b. 83 c. 9 d. 27 2 85. a. 10 15 b. 4 87. a. 1x 32 1x 221x 22 2 b. 1x 242 1x 12 c. 1x 521x 5x 252

1. first 3. range 5. Answers will vary. 7. function 9. Not a function. The Shaq is paired with two heights. 11. Not a function; 4 is paired with 2 and 5. 13. function 15. function 17. Not a function; 2 is paired with 3 and 4. 19. function 21. function 23. Not a function; 0 is paired with 4 and 4. 25. function 27. Not a function; 4 is paired with 1 and 1. 29. function y y 31. 33. 5 7 4 3 2 1 54321 1 2 3 4 5

6 5 4 3 2 1

1 2 3 4 5 x

7654321 1 2 3

1 2 3 x

8 4

x

8

4

(10, 3) 4

4

4

8

8

8

(4, 5)

x

60

Exercises 1.3, pp. 127–131

y

(4, 6) (3, 5) 4

0

108642 2 4 6 8 10

8

(3, 5) 6

35. m 43 ; 17, 12 and 11, 92

y

8

2

10 (4, 10) 8 (2, 7) 6 4 (0, 4) (2, 1) 2

y

108642 2 4 6 8 10

2 4 6 8 10 x

(3, 6) 4 4

41. a. m 125, cost increased $125,000 per 1000 sq ft b. $375,000 43. a. m 22.5, distance increases 22.5 mph b. about 186 mi 45. a. m 23 6 , a person weighs 23 lb more for each additional 6 in. in height b. ⬇3.8 47. In inches: (0, 6) and (576, 18): m 1 48 . The sewer line is 1 in. deeper for each 48 in. in length. y y 49. 51. 10 10

y

y

4 8

10 8 6 (0, 6) 4 (1, 3) 2 (2, 0) 108642 2 4 6 8 10

8

4

no m1 # m2 1

x

2 4 6 8 10 x

3 2 132 92

19 4

yes m1 m2 9.

6 (6, 6) 4 (3, 4) 2 (0, 2) (3, 0)

y

8 4

y 4.8; y 3.6, Answers will vary. Answers will vary. a. center: 16, 22; r 0 (degenerate case) b. center: (1, 4); r 5 1; degenerate case 107. x 74 n 1 is a solution, n 2 is extraneous

1. 0, 0

39. m 4 7 ; 110, 102 and 111, 22

y

(3, 7)

8

Exercises 1.2, pp. 112–116 7.

37. m 15 4 ; 15, 232 and 17, 222

y

SA-5

function 35. w 45; w 僆 3 45, q 2

function 37. 250 6 T 6 450; T 僆 1250, 4502 4

p 僆 1q, 32

)

39.

3 2 1

41.

4 3 2 1

0

43.

4 3 2 1

0

45.

43 21 0 1 2 3 4 5 6 7

0

1

2

3

2

3

4

5

6

m 僆 1q, 5 4

2

3

4

5

6

x 僆 1q, 12 ´ 11, q2

[ 1

)) 1

)

47. 5x| x 26; 3 2, q 2

)

x 僆 12, 52

49. 5x|2 x 16; 32, 1 4

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function, x 僆 34, 5 4 , y 僆 3 2, 3 4 function, x 僆 3 4, q2 , y 僆 3 4, q2 function, x 僆 3 4, 4 4 , y 僆 3 5, 14 function, x 僆 1q, q2 , y 僆 1q, q2 not a function, x 僆 33, 5 4 , y 僆 3 3, 3 4 not a function, x 僆 1q, 34 , y 僆 1q, q2 x 僆 1q, 52 ´ 15, q 2 65. a 僆 3 5 3 , q2 x 僆 1q, 52 ´ 15, 52 ´ 15, q2 v 僆 1q, 3 122 ´ 13 12, 3 122 ´ 13 12, q 2 x 僆 1q, q2 73. n 僆 1q, q2 75. x 僆 1q, q2 x 僆 1q, 22 ´ 12, 52 ´ 15, q 2 x 僆 32, 52 2 ´ 1 52 , q2 81. x 僆 14, q 2 83. x 僆 13, q2

51. 53. 55. 57. 59. 61. 63. 67. 69. 71. 77. 79.

85. x 僆 1 73 , q 2

2000

0

7 1 87. f 162 0, f 12c2 c 3, f 1c 12 c 2 2 89. f 162 132, f 1 32 2 34 , f 12c2 12c2 8c, f 1c 12 3c2 2c 1 1 3 9 91. h132 1, h1 2 3 2 2 , h13a2 , h1a 22 a a2 93. h132 5, h1 2 3 2 5, h13a2 5 if a 6 0 or 5 if a 7 0, h1a 22 5 if a 7 2 or 5 if a 6 2 3 95. g142 8␲, g a b 3␲, g12c2 4␲c, g1c 32 2␲1c 32 2 9 3 97. g142 16␲, ga b ␲, g12c2 4␲c2, g1c 32 1c2 6c 92␲ 2 4 3 99. p152 213, pa b 26, p13a2 26a 3, p1a 12 22a 1 2 14 3 27a2 5 7 3a2 6a 2 101. p152 , p a b , p 13a2 , p1a 12 2 2 5 2 9 9a a 2a 1 103. a. D: 51, 0, 1, 2, 3, 4, 56 b. R: 52,1, 0, 1, 2, 3, 46 c. 1 d. 1 105. a. D: 3 5, 5 4 b. y 僆 3 3, 4 4 c. 2 d. 4 and 0 107. a. D: 3 3, q2 b. y 僆 1q, 44 c. 2 d. 2 and 2 1 109. a. 186.5 lb b. 37 lb 111. A 182 22 1 25 units2 2 113. a. N1g2 2.5g b. g 僆 30, 5 4 ; N 僆 30, 12.5 4 115. a. 30, q ) b. about 2356 units3 c. 800 units3 117. a. c1t2 42.50t 50 b. $156.25 c. 5 hr d. t 僆 3 0, 10.64 ; c 僆 3 0, 500 4 119. a. Yes. Each x is paired with exactly one y. b. 10 P.M. c. 0.9 m d. 7 P.M. and 1 A.M. 121. negative outputs become positive y x

10 8 6 4 2

108642 2 4 6 8 10

f 1 32 2

4. a. E1x2 7.5x 950 b. $1100, $1175, $1250 c. x 僆 30, 75, 5 4 and y 僆 3200, 2000, 200 4 d. 47 snowboards

15 4,

y

y

2 4 6 8 10 x

108642 2 4 6 8 10

125. a. 19 16

b. 12x 321x 82

b. 1

y

10 8 6 4 (6, 4) 2 (3, 0) 108642 2 2 4 6 8 10 x 4 (0, 4) 6 8 10

18 7

D: x 僆 1q, q 2, R: y 僆 1q, 72 D: x 僆 31, 7 4 , R: y 僆 36, 9 4 D: x 僆 1q, 9 4, R: y 僆 3 3, q 2 D: x 僆 1q, q 2, R: y 僆 1q, 9 4

Exercises 1.4, pp. 143–147 1.

7 4

, (0, 3)

3. 2.5


7. y 4 5 x 2

2. m 3. positive, loss is decreasing (profit is increasing); m yes; 1.5 1 , each year Data.com’s loss decreases by 1.5 million.

11. y 5 3 x 5

9. y 2x 7

x

y

x

y

x

y

5

6

5

3

5

2

18 5

2

3

2

10 3 5 3

0

2

0

7

0

5

1

6 5 2 5

1

9

1

20 3

3

13

3

10

3

35 6 x

4: y

35 6 ,

5 5 15. y 3 x 7: 3 , 7

4 21.

10 8 6 4 2 108642 2 4 6 8 10

23.

y

2 4 6 8 10 x

(0, 2) (2, 5) (4, 8)

y

10 8 6 4 (0, 3) (3, 4) 2 (3, 2) 108642 2 4 6 8 10

2 4 6 8 10 x

25. a. 3 b. y 3 c. The coeff. of x is the slope and the 4 4 x 3 constant is the y-intercept. 27. a. 25 b. y 25 x 2 c. The coeff. of x is the slope and the constant is the y-intercept. 29. a. 45 b. y 45 x 3 c. The coeff. of x is the slope and the constant is the y-intercept. 2 2 31. y 2 3 x 2, f 1x2 3 x 2, m 3 , y-intercept (0, 2) 5 5 33. y 5 , x 5, f 1x2 x 5 m 4 4 4 , y-intercept (0, 5) 1 1 1 35. y 3 x, f 1x2 3 x, m 3, y-intercept (0, 0) 3 3 37. y 3 4 x 3, f 1x2 4 x 3, m 4 ,y-intercept (0, 3)

Mid-Chapter Check, p. 132 1.

1. 2. 3. 4.

10 8 6 (0, 5) 4 2 (3, 1) 108642 2 2 4 6 8 10 x 4 (6, 3) 6 8 10

3 123. a. x 僆 1q, 22 ´ 12, q 2 ; x 2y 1 y ; y 僆 1q, 12 ´ 11, q2

127. a. 1x 321x 521x 52 c. 12x 52 14x2 10x 252

Reinforcing Basic Concepts, pp. 132–133

17. y 19.

2 4 6 8 10 x

b. x 僆⺢ x 1y 3; y 僆 33, q2

200 5. x 3; no; input 3 is paired with more than one output. 6. y 2; yes 7. a. 0 b. x 僆 3 3, 5 4 c. 1 and 1 d. y 僆 34, 5 4 8. from x 1 to x 2; steeper line S greater slope 3 9. ¢F ¢p 4 ; For each increase of 4000 pheasants, the fox population increases by 300; 1100 foxes. 10. a. x 僆 53, 2, 1, 0, 1, 2, 3, 46 , y 僆 53,2,1, 0, 1, 2, 3, 46 b. x 僆 3 3, 4 4 , y 僆 3 3, 44 c. x 僆 1q, q 2 , y 僆 1q, q 2

13. y 2x 3: 2, 3

10

x

8 6 y x 4 2

yx

75

39. y 23 x 1

3 2,

45. y 4x 10

41. y 3x 3 47. y

3 2 x

43. y 3x 2

1

49. y 75 2 x 150

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SA-7

Student Answer Appendix 51. y 35 x 4

53. y 23 x 5

y

10 8 6 4 (0, 4) 2 108642 2 4 6 8 10

108642 2 4 6 8 10

2 4 6 8 10 x

59.

y

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

10 8 6 4 2 108642 2 4 6 8 10

55.

y

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

(0, 5)

61.

y

2 4 6 8 10 x

10 8 6 4 2 108642 2 4 6 8 10

y

2 4 6 8 10 x

y

2 4 6 8 10 x

29 63. y 25 x 4 65. y 5 67. y 12 3 x 7 5 x 5 69. y 5 71. perpendicular 73. neither 75. parallel 5 77. a. y 3 b. y 43 x 20 79. a. y 49 x 31 4 x 2 3 9 9 3 1 b. y 4 x 4 81. a. y 2 x 2 b. y 2x 2 y y 83. 85. 10 10 8 6 4 2

8642 2 4 6 8 10

2 4 6 8 10 x

5. Answers will vary. b. positive

y

1600 1400 1200 1000 x 0

108642 2 4 6 8 10

2 4 6 8 10 x

9. a.

4

8

12 16 20 24 28

Year (1980 → 0)

b. linear c. positive

y

150 Congresswomen

125

y 3.1 0.51x 1.82 6 5 1x

c 99. B 101. D 103. m a a. m 3 b , y-intercept b 4 , y-intercept 2 5 (0, 2) b. m 5 , y-intercept (0, 3) c. m 6 , y-intercept (0, 2) d. m 53 , y-intercept (0, 3) 105. a. As the temperature increases 5°C, the velocity of sound waves increases 3 m/s. At a temperature of 0°C, the velocity is 331 m/s. b. 343 m/s c. 50°C 107. a. V 20 3 t 150 b. Every 3 yr the value of the coin increases by $20; the initial value was $150. c. $223.33 d. 15 years, in 2013 e. 3 yr 109. a. N 7t 9 b. Every 1 yr the number of homes with Internet access increases by 7 million. c. 1993 d. 86 million e. 13 yr f. 2010 111. a. P 58,000t 740,000 b. Each year, the prison population increases by 58,000. c. 1,900,000 113. Answers will vary.

2. a.

3. c.

4. b. 5. f.

6. h.

117. a. 9

50 x

5

10 15 20 25 30 35 Year (1970 → 0)

11. a.

b. positive 50 40 30 20 10 0

10 20 30 40 50 x Year (1970 → 0)

13. a. (A) (D) (C) (B) y b. 60 y 60 55 50 45 40 35 30 25

b. 9|x|

55 50 45 40 35 30 25 0 1 2 3 4 5 6 7 8 9 x

1. intersection-of-graphs, Y1, Y2, x-coordinate, intersection 5. Answers will vary. y 10 8 6 4 2

15. 17. 19. 21.

0 1 2 3 4 5 6 7 8 9 x

1 4 6 8 10 x

x3

y 60 55 50 45 40 35 30 25

y 60 55 50 45 40 35 30 25

0 1 2 3 4 5 6 7 8 9 x

0 1 2 3 4 5 6 7 8 9 x

c. positive, c. positive, c. negative, c. negative, d. m ⬇ 3.8; d. m ⬇ 4.2; d. m ⬇ 2.4; d. m ⬇ 4.6 a. linear b. positive c. strong d. m ⬇ 4.2 a. nonlinear b. positive c. NA d. NA a. nonlinear b. negative c. NA d. NA a. b. positive c. f 1x2 2.4x 62.3, y f 152 74.3174,3002, f 1212 112.7 1112,7002 100

(3, 2)

c. strong

y


108642 2 4 6 8 10

75

0

119. 113.10 yd2

3. literal, two 7.

100

25

GDP per capita (1000s)

89. y 2 42 ; For each 5000 additional sales, income rises $6000. 91. y 100 20 1 1x 0.52 ; For every hour of television, a 1 student’s final grade falls 20%. 93. y 10 35 2 1x 2 2 ; Every 2 in. of rainfall increases the number of cattle raised per acre by 35. 95. C 97. A

115. 1. d.

3. linear

1800 Cost (cents)

y

10 8 6 4 2

1. scatterplot 7. a.

2 4 6 8 10 12 x

y 4 38 1x 32

y 5 21x 22 87.


8 6 4 2

108642 2 4 6 8 10

9. x 3 11. x 僆 1q, q2 13. x 9 15. no solutions 17. x 3, answers match 19. x 2 21. x 0 23. no solutions 25. x 僆 1q, q 2 27. x 僆 13, q2 , verified graphically in Exercise 7 29. x 僆 1q, 0 4 31. x 僆 15, q2 P C 33. x 僆 1q, q2 35. no solutions 37. C 39. r 1M 2␲ 2Sn T1P2V2 3V 41. T2 43. h 45. n P1V1 a1 an 4␲r2 21S B2 A 16 C 20 47. P 49. y 51. y x x S B B 9 3 4 53. y x 5 55. a 3; b 2; c 19; x 7 5 16 57. a 6; b 1; c 33; x 3 59. a 7; b 13; c 27; x 2 61. h 17 cm 63. 510 ft 65. 56 in 67. 3084 ft 69. 48; 50 71. 5; 7 73. 11: 30 A.M. 75. 36 min 77. 4 quarts; 50% O.J. 79. 16 lb; $1.80 lb 81. 12 lb 83. 16 lb 85. Answers will very 87. 69 89. x 7 91. a. 12x 3212x 32 b. 1x 32 1x2 3x 92

Officers (1000s)

57.

10 8 6 4 2

90 80 70 60 3 5 7 9 11 13 x Year (1990 → 0)

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23. Using (5, 7.6) and (20, 23.3): y ⬇ 1.05x 2.37; GDP in 2010 will be near 44,370 25. a.

Making Connections, p. 177 1. d

3. a

5. b

7. c 9. f

11. d

13. f

15. a

Summary and Concept Review, pp. 178–182 1. x 僆 57, 4, 0, 3, 56 y 僆 52, 0, 1, 3, 86

2. 40

0

8

3

4

2

5

1

3

0

0

8

y

x y 5 0 4 3 2 221 ⬇ 4.58 0 5 2 221 ⬇ 4.58 4 3 5 0

5 4 3 2 1 54321 1 2 3 4 5

b. positive, larger radius 1 larger area c. perfect correlation d. m 2␲ 27. a. b. linear c. positive y d. y 0.96x 1.55, 63.95 in.

1 2 3 4 5 x

x 僆 35, 5 4 y 僆 30, 54

3.

0

10

10

75.5

Wingspan (in.)

7

10

69.5 63.5 57.5 51.5 51 57 63 69 75 x Height (in.)

4. 65 mi 29. a.

b. linear c. positive d. y 9.55x 70.42; about 271,000 The number of applications, since the line has a greater slope.

y Patents (1000s)

260 220 180

x 3

5 7 9 11 13 Year (1990 → 0)

Percent of men

Percent of women

86

55 45 35 25

82

10 20 30 40 50x

78 74 0

Year (1950 → 0)

54321 1 2 3 4 5

1 2 3 4 5 x

5 108642 2 2 4 6 8 10 x 4 9 6 8 10

10 20 30 40 50 x Year (1950 → 0)

b. women: linear c. positive; b. men: linear c. negative d. yes, |slope| is greater 33. a. 2000

5 9 ,

10 8 6 9 4 2

12. a. a. linear b. y 108.18x 330.20 c. $1736.54 billion; about $2601.98 billion 35. a. r ⬇ 0.9783 b. r ⬇ 0.9783 c. they are almost identical; context, pattern of scatterplot, anticipated growth, etc. 37. No. Except for the endpoints of the domain, one x is mapped to two y’s. AP 39. r Pt

1 3,

1 2 3 4 5 x

y

3 2 4 6 8 10 x

10, 32

b. perpendicular 10 8 6 4 2

y

b.

(1, 1)

108642 2 2 4 6 8 10 x 4 (0, 2) 6 8 10

13

0

108642 2 4 6 8 10

114, 72

10. a. parallel 11. a.

0

y 5 4 3 2 1

8 6 4 2

70 0

7.

y

8. 1x 1.52 2 1y 22 2 6.25 y 9. a. b. 10

31. a. 65

10, 52, 13, 02

10

5 4 3 2 1 54321 1 2 3 4 5

140 100

6.

5. 1 52 , 32

y

10 8 6 (0, 2) 4 2 (3, 0) 108642 2 2 4 6 8 10 x 4 6 8 10

y

10 8 6 4 2 (0, 1) 108642 2 4 6 8 10

b.

10 8 6 4 2

2 4 6 8 10 x

(2, 2)

y

(w, 0)

108642 2 2 4 6 8 10 x 4 (0, 2) 6 8 10

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Student Answer Appendix 13. a. vertical b. horizontal c. neither

5

Practice Test, pp. 183–184

y x5

2y x 5

(0, ) 5 2

1. 2. 5. 6. 7.

(5, 0)

5

5x

y 4 5

14. yes 15. m y-intercept (0, 2); when the rodent population increases by 3000, the hawk population increases by 200. 16. a. x 僆 3 5 b. x 僆 1q, 22 ´ 12, 32 ´ 13, q2 4 , q2 2 3,

2 17. 14; 26 18. It is a function. 9 ; 18a 9a 19. I. a. D 51, 0, 1, 2, 3, 4, 56, R 52, 1, 0, 1, 2, 3, 46 c. 2 II. a. x 僆 1q, q2, y 僆 1q, q 2 b. 1 c. 3 III. a. x 僆 3 3, q2, y 僆 3 4, q 2 b. 1 c. 3 or 3

b. 1

4 20. a. y 4 3 x 4, m 3 , y-intercept (0, 4) b. y 53 x 5, m 53 , y-intercept 10, 52

21. a.

10 8 6 4 2 108642 2 4 6 8 10

22. a.

10 8 6 4 2 108642 2 4 6 8 10

b.

y

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

b.

y

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

4 3 2

y 11 54321 1 2 3 4 5

2 1

x 4 1 2 3 4 5 x

321 1 2 3 4 5 6 7

1 2 3 4 5 6 7 x

20 2 2 9. V 20 10. y 6 3 t 3 , 663 mph 5 x 5 11. a. (7.5, 1.5), b. ⬇ 61.27 mi 12. L1: x 3 L2: y 4 13. a. x 僆 54, 2, 0, 2, 4, 66 y 僆 52, 1, 0, 1, 2, 36 b. x 僆 3 2, 6 4 y 僆 31, 4 4

14. a. 300

y

b. 30

c. W1h2 25 2h

d. Wages are $12.50 per hr.

a2 6a 7 e. h 僆 30, 40 4 ; w 僆 3 0, 500 4 b. 2 a 6a 9 ¢sales 13.5 16. a. ¢time 1 b. sales are increasing at a rate of 13.5 million phones per year c. 2008: about 15 million sales, 2010: about 42 million sales, 2011: about 55.5 million sales 17. a. x 9 b. no solution 18. a. x 僆 3 3, q2 b. x 僆 1q, q2 19. a. b. linear c. positive 30 7 15. a. 2

2 4 6 8 10 x

y

2 4 6 8 10 x

11 23. y 5, x 2; y 5 24. y 3 25. f 1x2 43 x 4 x 4 2 2 26. m 5 , y-intercept (0, 2), y 5 x 2. When the rabbit population increases by 500, the wolf population increases by 200. 27. a. 1y 902 15 b. (14, 0), (0, 105) c. f 1x2 15 2 1x 22 2 x 105 d. f 1202 45, x 12 28. x 6 29. x 僆 1q, q2 30. x 僆 1q, 12 31. h ␲rV 2 32. L P 2 2W 33. x c a b 2 9 2 2 34. y 3 x 2 35. 8 gal 36. 12 8 ␲ ft ⬇ 15.5 ft 37. 23 hr 40 min 100 38. a.

0

P a. x 27 b. x 2 c. C 1 d. W P 2 2L k 30 gal 3. S 177 4. about 5.1 sec a and c are nonfunctions, they do not pass the vertical line test neither y y 8. 12,32; r 4 5 3

0

50

3

20. a. f 1x2 0.91x 10.78

b. f 1502 ⬇ 35

Strengthening Core Skills, pp. 184–185 1. a. 13 , increasing b. y 5 13 1x 02 , y 13 x 5 c. (0, 5), 115, 02

130 2. a. 7 b. y 9 3 , decreasing y 7 3 x 9 c. (0, 9), 1 22 7 , 02

50

7 3 1x

02 ,

b. linear c. positive 39. a. f 1x2 0.35x 56.10 b.

3. a. 12 , increasing b. y 2 12 1x 32 , y 12 x 12 c. 10, 12 2, 11, 02

100

0

130

50 c. strong 40. f 11202 98.1. over 98%

c. strong: r ⬇ 0.974

4. a. 34 , increasing b. y 4 34 1x 52 , y 34 x 14 1 c. 10, 1 4 2, 1 3 , 02

10 8 6 4 2 108642 2 4 6 8 10

10 8 6 4 2 108642 2 4 6 8 10

10 8 6 4 2 108642 2 4 6 8 10

10 8 6 4 2 108642 2 4 6 8 10

y

2 4 6 8 10 x

y

2 4 6 8 10 x

y

2 4 6 8 10 x

y

2 4 6 8 10 x

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5. a. 3 b. y 5 4 , decreasing 7 y 3 4 x 2 c. 10, 72 2, 1 14 3 , 02

6. a. 1 b. y 7 2 , decreasing y 1 2 x 6 c. 10, 62 , 112, 02

3 4 1x

22 ,

10 8 6 4 2 108642 2 4 6 8 10

1 2 1x

22 ,

15 12 9 6 3 1512963 3 6 9 12 15

y

2 4 6 8 10 x

y

3 6 9 12 15 x

c. f 1x2 0: x 僆 1q, 14 ´ 31, q 2; f 1x2 6 0: x 僆 11, 12 d. f 1x2c: x 僆 10, q 2, f 1x2T: x 僆 1q, 02 e. min: (0, 1) 61. a. D: t 僆 [1983, 2009], R: I 僆 [5, 14]; b. I1t2c: t 僆 (1983, 1984) ´ (1986, 1987) ´ (1993, 1994) ´ (1998, 2000) ´ (2005, 2006); I1t2T: t 僆 (1984, 1986) ´ (1989, 1993) ´ (1994, 1998) ´ (2000, 2003) ´ (2006, 2009); I(t) constant: (1987, 1989) ´ (2003, 2005); c. global max: I 14 in 1984, global min: I 5. in 2009 (also an endpoint min); d. greatest increase: (1993, 1994), greatest decrease: (1985, 1986) y 63. zeroes: (8, 0), (4, 0), (0, 0), (4, 0); 8 min: 110, 62,(2, 1), (4, 0); max: (6, 2), (2, 2) 4

8

4

4

8

x

4

CHAPTER 2

8

Exercises 2.1, pp. 196–201 1. linear, bounce 3. increasing 5. Answers will vary. y 7. 9. even 11. even 5

5

5 x

65. no; no; Answers will vary. 67. a. Thorpe; Rosolino b. two times, at 190 sec and 219 sec c. about 29 sec 1219 190 29 sec2 d. Thorpe e. about 2 sec 1223 221 2 sec2 f. 221 sec 3 min 41 sec 69. h1k2 h1k2 1 1 3 1k2 3 4 2 1k3 2 2 1 1 3 2 1k 2 1k3 2 2 1 1 3 2 1k 2 1k3 2 2 ✓ 71. a. 4 12 x2 b. 4 9 x2 73. V 5184␲ cm3, SA 1152␲ cm2

5

Exercises 2.2, pp. 212–217 13.

15. odd

y

17. not odd

10

10

10 x

10

19. neither 21. odd 23. neither 25. x 僆 3 1, 14 ´ 3 3, q2 27. x 僆 1q, 12 ´ 11, 12 ´ 11, q2 29. x 僆 3 2, q2 31. x 僆 1q, 24 33. V1x2c: x 僆 13, 12 ´ 14, 62 ; V1x2T: x 僆 1q, 32 ´ 11, 42 ; constant: none 35. f 1x2c: x 僆 11, 42 ; f 1x2T: x 僆 12, 12 ´ 14, q2 ; constant: x 僆 1q, 22 37. a. p1x2c: x 僆 1q, q2 ; p1x2T: none b. down, up 39. a. f 1x2c: x 僆 13, 02 ´ 13, q 2 ; f 1x2T: x 僆 1q, 32 ´ 10, 32 b. up, up 41. a. x 僆 1q, q2; y 僆 1q, 54 b. x 1, 3 c. H1x2 0: x 僆 3 1, 34 ; H1x2 0: x 僆 1q, 1 4 ´ 3 3, q2 d. H1x2c: x 僆 1q, 22 ; H1x2T: x 僆 12, q 2 e. local max: y 5 at (2, 5) 43. a. x 僆 1q, q2; y 僆 1q, q2 b. x 1, 5 c. g1x2 0: x 僆 3 1, q 2 ; g1x2 0: x 僆 1q, 1 4 ´ 53.56 d. g1x2c: x 僆 1q, 12 ´ 15, q2 ; g1x2T: x 僆 11, 52 e. local max: y 6 at (1, 6); local min: y 0 at (5, 0) 45. a. x 僆 34, q 2; y 僆 1q, 34 b. x 4, 2 c. Y1 0: x 僆 3 4, 2 4 Y1 0: x 僆 32, q 4 d. Y1c: x 僆 14, 22 ; Y1T: x 僆 12, q 2 e. local max: y 3 at (2, 3); endpoint min y 0 at 14, 02 47. a. x 僆 1 ¥ , ¥ 2, y 僆 1 ¥ , ¥ 2 b. x 4 c. p1x2 0: x 僆 3 4, q2; p1x2 0: x 僆 1q, 4 4 d. p1x2c: x 僆 1q, 32 ´ 13, q 2; p1x2T: never decreasing e. local max: none; local min: none 49. max: y ⬇ 1.58 at x ⬇ 0.78; min: y ⬇ 0.47 at x ⬇ 2.55 51. max: y ⬇ 1.54 at x ⬇ 6.21, y ⬇ 3.28 at x ⬇ 2.55; min: y ⬇ 3.28 at x ⬇ 2.55, y ⬇ 1.54 at x ⬇ 6.21 53. max: y ⬇ 3.08 at x 83 ; min: y 0 at x 4 (endpoint) 55. a. x 僆 1q, 3 4 ´ 3 3, q 2 ; y 僆 30, q2 b. 13, 02 , (3, 0) c. f 1x2c: x 僆 13, q2 ; f 1x2T: x 僆 1q, 32 d. even 9y2 36 e. x 57. a. x 僆 3 0, 2604 , y 僆 3 0, 80 4 b. 80 ft B 2 c. 120 ft d. yes e. (0, 120) f. (120, 260) 59. a. x 僆 1q, q2; y 僆 31, q 2 b. (1, 0), (1, 0)

1. stretch, compression 3. (5, 9), upward 5. Answers will vary. 7. a. quadratic; b. up/up, (2, 4), x 2, (4, 0), (0, 0), (0, 0); c. D: x 僆⺢, R: y 僆 3 4, q 2 9. a. quadratic; b. up/up, (1, 4), x 1, (1, 0), (3, 0), (0, 3); c. D: x 僆⺢, R: y 僆 34, q 2 11. a. quadratic; b. up/up, (2, 9), x 2, (1, 0), (5, 0), (0, 5); c. D: x 僆⺢, R: y 僆 39, q 2 13. a. square root; b. up to the right, (4, 22 , (3, 0), (0, 2); c. D: x 僆 34, q 2 , R: y 僆 32, q 2 15. a. square root; b. down to the left, (4, 3), (3, 0), (0, 3); c. D: x 僆 1q, 4 4 , R: y 僆 1q, 3 4 17. a. square root; b. up to the left, (4, 0), (4, 0), (0, 4); c. D: x 僆 1q, 4 4 , R: y 僆 30, q 2 19. a. absolute value; b. up/up, (1, 4), x 1, (3, 0), (1, 0), (0, 2); c. D: x 僆⺢, R: y 僆 3 4, q2 21. a. absolute value; b. down/down, (1, 6), x 1, (4, 0), (2, 0), (0, 4); c. D: x 僆⺢, R: y 僆 1q, 6 4 23. a. absolute value; b. down/down, (0, 6), x 0, (2, 0), (2, 0), (0, 6); c. D: x 僆⺢, R: y 僆 1q, 6 4 25. a. cubic; b. up/down, (1, 0), (1, 0), (0, 1); c. D: x 僆⺢, R: y 僆⺢ 27. a. cubic; b. down/up, (0, 1), (1, 0), (0, 1); c. D: x 僆⺢, R: y 僆⺢ 29. a. cube root; b. down/up, (1, 1), (2, 0), (0, 2); c. D: x 僆⺢, R: y 僆⺢ 31. square root function; y-int (0, 2); x-int (3, 0); initial point (4, 2); up on right; D: x 僆 34, q2, R: y 僆 32, q 2 33. cubic function; y-int (0, 2); x-int (2, 0); inflection point (1, 1); up, down; D: x 僆⺢, R: y 僆⺢ 35. 10

5

10

5 the graph of g is f shifted up 2 units; the graph of h is h shifted down 3 units

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55.

10

10

10

10

10

10

10

10 the graph of q is p vertically stretched; the graph of r is p vertically compressed

the graph of q is p shifted down 5 units; the graph of r is p shifted up 2 units 57. 39.

41.

y

8

8

4

4 8

10

y

4

4

8

8

x

4

4

4

4

8

8

43.

8

x

10

10

10

10

10 the graph of Y2 is Y1 vertically stretched; the graph of Y3 is Y1 vertically compressed y y 59. 61. 10

10

8 6 4 2

8 4 8

4

4

8

108642 2 4 6 8 10

x

4

10 the graph of q is p shifted left 5 units 45.

8

63. g 65. i 67. e 75. left 2, down 1

10

69. j 71. l 73. c 77. left 3, reflected across x-axis, down 2

y

10

10

2 4 6 8 10 x

79. left 3, down 1

y

8

8

4

4

y 8 4

(3, 1) 8

4

4

8

8

x

(2, 1) 4

4

4

4

8

8

(3, 2)

8

x

4

4

8

x

4

8

8

10 81. left 1, down 2

the graph of Y2 is Y1 shifted right 4 units 47.

49.

y

8

51.

y

y

8

8

8

4

4

4

4

4

8

x

4 8

53.

8

4

4

8

x

8

4 4

8

8

8

x

8

y

8

8

8

4

4

4

4

(1, 2)

y

85. left 1, reflected across x-axis, stretched vertically, down 3

y

y 4

4

83. left 3, reflected across x-axis, down 2

4

8

8

4

4

4 8 4 (3, 2)

8

8

x

x

8

4

4 8 4 (1, 3)

x

8

8 4 8

4

4 4 8

8

x

87. left 2, reflected across x-axis, compressed vertically, down 1 y

5 4

8

4

(2, 1)4 8

91. right 3, compressed vertically, up 1

y

y 8

(1, 3) 3

4 8

89. left 1, reflected across x 1, reflected across x-axis, stretched vertically, up 3

4

2 1 4

8

x

987654321 1 2 3 4 5

(3, 1) 1 x

8

4

4 4 8

8

x

cob19545_saa_010-017.qxd

SA-12

11/26/10

b.


c.

y

y

y

8

8

8

4

4

4

4

4

8

8

x

4

4

8

8

x

4

4

4

4

8

8

8

d.

95. a. y

8 4 8

8

x

4

8

x

4

4

4

8

8

x

4

4

4

4

8

8

8

c.

x

8

4 4

8

y

8

4

4

b.

y

8

Page SA-12


93. a.

8

7:50 PM

d. y

y 8

8 4 8

4

4

4

8

8

x

4

4

4

4

8

8

97. f 1x2 1x 22 2 103. a. V(r) 艐 4.2r3 120

8

x

99. p1x2 1.52x 3 101. f 1x2 45 0 x 4 0 b. about 65 units3, V⬇65.4 units3; yes; 3 3 c. r 2 4␲ V

100 80 60 40 20 1

2

r

3

105. a. compressed vertically T(x) b. 2.25 sec 5

107. a. compressed vertically P(v) 1000 b. 216 W 800

4

600

3

400

2

200

1 4

8

12

16

20

24

28 v

20 40 60 80 100 x

1. reverse 3. 7, 7 5. no solution; answers will vary. 7. 54, 66 9. 52, 126 11. 53.35, 0.856 13. 5 87 , 26 15. 512 , 12 6 17. { } 19. 510, 66 21. {3.5, 11.5} 23. 51.6, 1.66 25. 53, 12 6 27. 15, 32 29. { } 7 8 14 3 31. c , d 33. 5 6 35. a1, b 37. c , 0 d 3 3 5 4 39. 1q, 102 ´ 14, q 2 41. 1q, 34 ´ 33, q 2 3 7 7 43. aq, d ´ c , q b 45. aq, d ´ 3 1, q2 3 3 7 7 2 ´ 11, q2 49. 1q, 02 ´ 15, q 2 47. 1q, 15 51. { } 53. 1q, q 2 55. a. x 2 and x 6 b. 1q, 2 4 ´ 36, q2 c. [2, 6] 57. a. x 3 and x 0.2 b. 1q, 34 ´ 30.2, q 2 c. 13, 0.22 59. a. 45 d 51 in. b. d L x d L 61. in feet: [32,500, 37,600]; yes 63. in feet: d 6 210 or d 7 578 65. a. 冟s 37.58冟 3.35 b. [34.23, 40.93] 67. a. 冟s 125冟 23 b. [102, 148] 69. a. 冟d 42.7冟 6 0.03 b. 冟d 73.78冟 6 1.01 c. 冟d 57.150冟 6 0.127 d. 冟d 2171.05冟 6 12.05 e. golf: t ⬇ 0.0014 71. a. x 4 b. 3 43 , 4 4 c. x 0 d. 1q, 35 4 e. { } 73. a. 52, 86 b. 1q, 32 ´ 111, q 2 c. 3 6, 24 d. 52, 106 77. x 僆 1q, 52 2

75. ␳ V2W 2 CA

Mid Chapter Check, pp. 228–229 1. neither 2. max: y ⬇ 11.12 at x ⬇ 0.50, y ⬇ 8.55 at x ⬇ 1.47; min: y ⬇ 7.80 at x ⬇ 0.75 3. increasing on 1q, 0.502 ´ 10.75, 1.472, decreasing on 10.50, 0.752 ´ 11.47, q 2 4. g1x2 2x 4 2 5. a. cubic b. up on the left, down on the right; inflection point: (2, 2); x-int: (4, 0); y-int: (0, 5) c. 1q, q2 ; 1q, q2 d. k 1 6. 10 q(x) is a reflection of p(x) across the x-axis, and r(x) is the same as q(x), but compressed by a factor of 12 10 10

109. a. vertical stretch by a factor of 2 b. 12.5 ft d(t) 14 12 10 8 6 4 2

7. a. 54, 146 1

2

3

111. x 僆 10, 42 ; yes, x 僆 14, q2 ; yes y

c. k 僆 1q, q 2

f(x)

8 4 8

4

g(x) 4

8

10. w 僆 3 8, 264 ; no, yes

1. x 3 or x 7 2. x 僆 3 5, 34

8

3. x 僆 1q, 1 4 ´ 34, q 2


113. Any points in Quadrants III and IV will reflect across the x-axis and move to Quadrants I and II.

5 4321 1 2 3 4 5

b. 566 19 23 b. y 僆 aq, b ´ a , q b 2 2

Reinforcing Basic Concepts, p. 229 x

4

5 4 3 2 1

10 8. a. q 僆 18, 02

9. a. d 僆 1q, 0 4 ´ 3 4, q 2

t

4

b. { }

f(x) f(x) x2 4 1 2 3 4 5 x

5 4 3 2 1 5 4321 1 2 3 4 5

F(x) F(x) 兩x2 4兩 1 2 3 4 5 x

115. P 14x2 16x 142 units f 1x2T : x 僆 1q, 42

1. as x S q, y S 2 3. vertical, y 2 5. Answers will vary. 7. a. as x S q, y S 2 9. a. as x S q, y S 1 as x S q, y S 2 as x S q, y S 1 b. as x S 1 , y S q b. y 1 as x S 1 , y S q c. as x S 2 , y S q as x S 2 , y S q 11. down 1, x 僆 1q, 02 ´ 10, q 2 , y 僆 1q, 12 ´ 11, q2 y

117. f 1x2c: x 僆 14, q 2 ;

x 0

8 4

(1, 0) 8

4

4 8 4 y 1 8

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Student Answer Appendix 13. left 2, x 僆 1q, 22 ´ 12, q 2 , y 僆 1q, 02 ´ 10, q 2

43. g1x2 increases faster a

4 2 7 b a. x 1, 0, and 1, 5 3 b. 11, 02 ´ (0, 1), c. 1q, 12 ´ 11, q2

y

8

x 2 8

4

冢0, q冣 y 0

4

4

8

45. g 1x2 increases faster a

x

4 8

47. 49. 55. 59. 61.

15. right 2, reflected across x-axis, x 僆 1q, 22 ´ 12, q 2 , y 僆 1q, 02 ´ 10, q 2 y

8

冢0, q冣 8

4

x 2

4

4 4

8

y 0

SA-13

63.

x

1 1 7 b a. x 0 and 1, b. (0, 1), c. 11, q 2 3 6 5 2 g1x2 increases faster a 7 b a. x 0 and 1, b. (0, 1), c. 11, q2 4 3 3 0, q2 51. 1q, q 2 53. 1q, q 2 defined: b, c, d; undefined: a 57. defined: a, b, d; undefined: c F is the graph of f shifted left 1 unit and down 2; verified P is the graph of p shifted right 2 units and reflected across the x-axis; verified d2F a. F becomes very small b. y x12 c. m2 km 1

65. a. It decreases; 75, 25, 15 c. as p S 0, D S q D(p)

8

b. It approaches 0.

90

17. left 2, down 1, x 僆 1q, 22 ´ 12, q2 , y 僆 1q, 12 ´ 11, q 2

70 50 30

y

10

8 4

x 2 8

4

10 30 50 70 90 p (1, 0) 4

8

y 1 4 冢0, q冣

67. a. It decreases; 100, 25, 11.1. c. as d S 0, l S q 2200 I(d)

x

b. toward the light source

1800

8

1400

19. right 1, x 僆 1q, 12 ´ 11, q2 , y 僆 10, q2

1000 y

600 200

8

(0, 1) 4 y 0

8

4

4

8

69. a. $20,000, $80,000, $320,000; cost increases dramatically b. Cost ($1000)

x

4

900

8

21. left 2, reflected across x-axis, x 僆 1q, 22 ´ 12, q2 , y 僆 1q, 02

x 1

700 500

y

300 100

8 4

y 0

8

4

4

8

x

冢0, ~冣

4 8

x 2

23. down 2, x 僆 1q, 02 ´ 10, q2 , y 僆 12, q 2

d 5 15 25 35 45 55

Percent 10 30 50 70 90

c. as p S 100, C S q 71. a. 253 ft/sec (about 172 mph) b. approx. 791 ft 73. a. size 11 b. approx. 5 ft, 5 in. 75. a.

y

b. P 32.251 w0.246 c. about 63 days d. about 6.9 kg

75

8 4

冢√q, 0冣 y 2 8 4

冢√q, 0冣 4

8

x

4

2

8

30

x 0

25. left 2, up 1, x 僆 1q, 22 ´ 12, q 2, y 僆 11, q2

y 8 4

y 1

8

4

冢0, @冣 4

4

8

10 x

77. a.

100

8

x 2

27. reciprocal quadratic, S1x2

1

2 1x 12 2 1 2 29. reciprocal function, Q1x2 x1 1 5 33. S 2 31. reciprocal quadratic, v1x2 1x 22 2 35. S q

37. 1, q

39. g1x2 increases faster 13 7 22 a. x 0 and 1, b. 1q, 02 ´ (0, 1), c. 11, q2 41. f 1x2 increases faster 14 7 22 a. x 1, 0, and 1, b. 11, 02 ´ (0, 1), c. 1q, 12 ´ 11, q 2

2500

40,000

10 b. S 1.687a

0.386

c. about 33 species

d. about 37,200 mi2

79. The area is always 1 unit2; The area is always 1x units2

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81. y 2 3 x 5,

4 8

4

25. discontinuity at x 1, redefine f 1x2 3 at x 1; c 3

83. c 2mE

y 8

y 8 4

(3, 3) 4

8

8

x

4

4

4

4

8

8

1 x1 27. f 1x2 e 2 3x 6

Exercises 2.5, pp. 254–258 1. continuous 3. smooth 5. Each piece must be continuous on the corresponding interval, and the function values at the endpoints of each interval must be equal. Answers will vary. x2 6x 10 7. a. f 1x2 e 3 5 2x 2

y 8 4 8

4

4

x

8

4 8

33. a. S1t2 e 35. a.

y 8 4 4

4 x 6 2 x2

x2 2x 3 x 1 x1 x 7 1 31. Graph is discontinuous at x 0; f 1x2 1 for x 7 0; f 1x2 1 for x 6 0.

9. 2, 2, 12 , 0, 2.999, 5 11. 5, 5, 0, 4, 5, 11 13. D: x 僆 3 6, q 2 ; R: y 僆 3 4, q2 p122 4 p102 2

8

x

29. p1x2 e

0x5 b. y 僆 3 1, 114 5 6 x9

15. D: x 僆 3 2, q 2 ; R: y 僆 3 4, q2

8

4

x

8

t2 6t 5

Year (0 S 1950) 5 15 25 35 45 55 65

0t5 b. S1t2 僆 30, 9 4 t 7 5

Percent 7.33 14.13 14.93 22.65 41.55 60.45 79.35

4

b. Each piece gives a slightly different value due to rounding of coefficients in each model. At t 30, we use the “first” piece: P1302 13.08. 0.09h 0 h 1000 200 C(h) 180 37. C1h2 e 160 0.18h 90 h 7 1000 140

y 8 4 8

4

4

x

8

4 8

19. D: x 僆 1q, q 2 ; R: y 僆 3 0, q2

C112002 $126 y

39. C1t2 e

8 4 8

4

4

8

120 100 80 60 40 20

0.75t 0 t 25 1.5t 18.75 t 7 25

60 55 50 45 40 35 30 25 20 15 10 5

x

4

C 1452 $48.75

8

21. D: x 僆 1q, q 2 ; R: y 僆 1q, 32 ´ 13, q 2

5 4 3 2 1

41. S1t2 e

y

54321 1 2 3 4 5

600

1.35t2 31.9t 152 0 t 12 2.5t2 80.6t 950 12 6 t 22

43. c1m2 e

3.3m 7m 111

0 m 30 m 7 30

8 4 4

4 4 8

8

x

$2.11

1400

t 10

$498 billion, $653 billion, $931 billion

y

1000

C(t)

1 2 3 4 5 x

23. discontinuity at x 3, redefine f 1x2 6 at x 3; c 6 8

h 200

20

30

40

450 250 150 50 2

6 10 14 18 22

t (years since 1980)

300 250 200 150 100

50

(12, 340)

350

350

Cost of call in cents

17. D: x 僆 1q, 92 ; R: y 僆 3 2, q 2

Spending (in billions of dollars)

8

(30, 99)

50 10 20 30 40 50 60

Duration of call in min

cob19545_saa_010-017.qxd

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⎞ ⎪ ⎬ ⎪ ⎠

d.

a 6 2 10 8 2 a 6 13 6 13 a 6 20 4 20 a 6 65 2 0 10 20 30 40 50 60 70 a 65 Age in years $38 47. a. C1w 12 17< w 1= 88 b. 0 6 w 13 c. 88¢ d. 173¢ e. 173¢ f. 173¢ g. 190¢ y 5 x 3 4 49. yes; h1x2 • 2x 1 3 6 x 6 2 2 5 x2 Cost of admission

0 2 45. C1a2 5 7 5

SA-15

4

2

2

4

x

e. k 6; A 6s2; 55,303,776 m2 25. a. Distance varies directly as time squared. b. D kt2 c. 500

2 4

51. f(x) has a removable discontinuity at x 2; g(x) has a discontinuity at x 2 53. x 7, x 4 55. y 43 x 2

8

0

Exercises 2.6, pp. 265–269 1. constant 3. directly, height, square 7. d kr 9. F ka 11. y 0.025x x

y

500

12.5

650

16.25

750

18.75

5. Answers will vary. d.

13. w 9.18h; $321.30; the hourly wage; k $9.18/hr 192 15. a. k 192 b. 360 47 , S 47 h 320 c. 330 stairs d. S 331; yes 280 192 Number of stairs

50

s(h) 冢 47 冣h

240

e. k 16; d 16t2; about 3.5 sec; 121 ft 27. F dk2 29. S Lk 31. Y

200 160

12,321 Z

Z2

120 80 40 10 20 30 40 50 60 70 80 90100

Height h (in meters)

17. A kS2 19. P kc2 21. k 0.112; p 0.112q2

q

p

45

33. w

3,072,000,000

226.8

39. V ktr2

55

338.8

41. C 6.75R 2 S

70

548.8

4000

7,500,000

37

9

74

2.25 1

111

23. a. Area varies directly as a side squared. b. A ks2 c. 75,000,000

0

Y

r2

; 48 kg

R

35. l krt 37. A kh1B b2

S

C

120

6

22.5

200

12.5

350

15

8.64 10.5

43. E 0.5mv2; 612.50 J 45. a. cube root family b. answers will vary c. 0.054 or 5.4% d. A 1R 12 3 Amount A

Rate R

1.0

0.000

1.05

0.016

1.10

0.032

1.15

0.048

1.20

0.063

1.25

0.077

cob19545_saa_010-017.qxd

SA-16

11/26/10

7:51 PM

Page SA-16


47. T 48 49. M 16 E; ⬇41.7 kg V ; 32 volunteers 51. D 21.61S; ⬇144.9 ft 53. C 8.5LD; $76.50 55. C ⬇ 14.4

pp 104 2 d1 22 ; 3

16. cube root y 8

about 222 calls

4

(2, 0)

57. a. about 23.39 cm b. about 191% 59. a. M kwh2 1 L1 2 b. 180 lb 61. 6.67 10 7

4

8

x

4 8

17. a. (2, 3)

Making Connections, p. 270 1. g

4

9y2

65. a. x 僆 1q, 42 ´ 14, 42 ´ 14, q2 4x2 b. x 僆 1q, 42 ´ 14, q2 63.

8

3. a 5. d

7. a

9. b

11. h

13. c

54321 1 2 3 4 5

15. f

Summary and Concept Review, pp. 271–275

1. D: x 僆 1q, q2 , R: y 僆 35, q2 , f 1x2c: x 僆 12, q 2 , f 1x2T: x 僆 1q, 22 , f 1x2 7 0: x 僆 1q, 12 ´ 15, q 2 , f 1x2 6 0: x 僆 11, 52 2. D: x 僆 3 3, q2 , R: y 僆 1q, 0 4 , f 1x2c: none, f 1x2T: x 僆 13, q2 , f 1x2 7 0: none, f 1x2 6 0: x 僆 13, q 2 3. D: x 僆 1q, q 2, R: y 僆 1q, q2 , f 1x2c: x 僆 1q, 32 ´ 11, q 2 , f 1x2T: x 僆 13, 12 , f 1x2 7 0: x 僆 15, 12 ´ 14, q2 , f 1x2 6 0: x 僆 1q, 52 ´ 31, 42 4. a. odd b. even c. neither d. odd y 5. zeroes: 16, 02 , (0, 0), 8 (10, 6) (6, 0) (9, 0) (3, 4) 4 min: 13, 82, 8 4 4 8 x 17.5, 22 (7.5, 2) 4 max: 16, 02, 13, 42

5 4 3 2 1

b.

y

(1.5, 5)

(3, 3) (4, 1) (0.5, 1)

5 4 3 2 1

冢4, w冣 (1, 1)

54321 1 2 3 4 5

1 2 3 4 5 x

c.

y

5 4 3 2 1

54321 1

1 2 3 4 5

y

冢1, w冣 1 2 3 4 5 x

(1.5, 0.5)2 3 4 5

18. 54, 106 19. 57, 36 20. 55, 86 21. 54, 16 22. 1q, 62 ´ 12, q2 23. [4, 32] 24. { } 25. { } 26. 1q, q2 27. 32, 6 4 28. 1q, 2 4 ´ 3 10 29. a. 0 r 2.5 0 1.7 3 , q2 b. highest: 4.2 in., lowest: 0.8 in. 30. 31. y y 8

8

4

4

(1, 0) 8

4

4 4

8

8

x

4

4

8

x

(0, 3.25)4

(0, 0.5)

8

8

32. a. ⬇$32,143; $75,000; $175,000; $675,000; cost increases dramatically c. as p S 100, C S q

b.

700

C(p)

600 500 400 300

8

200

6. max: y 0.73 at x 0.48; min: y 0.73 at x 0.48 7. squaring function a. up on left/up on the right; b. x-intercept: 14, 02, 10, 02 ; y-intercept: (0, 0) c. vertex 12, 42 d. x 僆 1q, q 2, y 僆 34, q2 8. square root function a. down on the right; b. x-intercept:(0, 0); y-intercept: (0, 0) c. initial point 11, 22 ; d. x 僆 31, q 2 , y 僆 1q, 2 4 9. cubing function a. down on left/up on the right b. x-intercept(s): 12, 02 ; y-intercept: (0, 2) c. inflection point: (1, 1) d. x 僆 1q, q 2 , y 僆 1q, q 2 10. absolute value function a. down on left/down on the right b. x-intercepts: (1, 0), (3, 0); y-intercept: (0, 1) c. vertex: (1, 2); d. x 僆 1q, q 2 , y 僆 1q, 24 11. cube root function a. up on left, down on right b. x-intercept: (1, 0); y-intercept: (0, 1) c. inflection point: (1, 0) d. x 僆 1q, q2 , y 僆 1q, q2 12. quadratic

13. absolute value y

y

8

8

4

4

(3, 0) 8

4

4

8

x

8

4

4

(2, 5) 4

4

8

8

14. cubic

8

x

8

40

33.

4

3

7

Domain of f (x) is 1q, q 2 ; Domain of g(x) and h(x) is 3 0, q2 .

2 b. 570 km 5 x 4 4 6 x 3 b. R: y 僆 3 2, q2 35. a. f 1x2 • x 1 3 1x 3 1 x 7 3 y 36. D: x 僆 1q, q 2, 8 R: y 僆 1q, 82 ´ 18, q 2 , 4 h(x) discontinuity at x 3; 8 4 4 8 x define h1x2 8 at x 3 34. a. 88.4 hr,

4 8

y

8

4

4

(5, 2) (6, 3) 4

4

4 8

(0, 1)

8

x

8

4

4 4 8

8

x

80 100

5

(9, 4) 8

60

100

37. 4, 4, 4.5, 4.99, 3 13 9, 3 13.5 9 38. D: x 僆 1q, q 2 R: y 僆 34, q2

y

8

p 20

(3, 6)

15. square root y

100

8

4

4 4 8

8

x

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SA-17

Student Answer Appendix 20x x2 39. • 30x 20 2 6 x 4 40x 60 x 7 4 For 5 hr the total cost is $140.

16. a 1, b 4 17. 1617 KWH/year y 18. a. 4, 4, 6.25 b.

Cost

160 120

8

80

4

40 0

2

4

6

8

8

Hours

3 40. k 17.5; y 17.5 1x

(2, 4)

4

4

8

(2, 4) 4

x

8

41. k 0.72; z 0.72v w2

x

y

v

w

z

216

105

196

7

2.88

0.343

12.25

38.75

1.25

17.856

729

157.5

24

0.6

48

19. M kd3 1 P12 2, approx. 2.2788 108

20. 520 lb

Strengthening Core Skills, p. 278 1. k 1 or 1 3 263 62

42. t 160 43. 4.5 sec

Cumulative Review Chapters R–2, pp. 279–280


1 7 1. f 122 23, f a b 2 4

1. a. D: x 僆 3 4, q2 ; R: y 僆 33, q 2 b. f 112 ⬇ 2.2 c. f 1x2 6 0: x 僆 14, 32 ; f 1x2 7 0: x 僆 13, q 2 d. f 1x2c: x 僆 14, q2 ; f 1x2T: none e. f 1x2 31x 4 3 y y 2. 3. 5 5 4 3 2 1 5 4321 1 2 3 4 5

5 4321 1 2 3 4 5

(3, 2)

(2, 3) 8

4

4

8

8

x

4

4 4

8

8

22 8. 1q, 20 3 2 ´ 1 3 , q2 y 11. 5 x 3

9. 34, 02 12. 13.5 sec

y 12 x

y

7 2

11.

8

4

5

4

8

5

x

8

x

5 x

5

13. a. D: x 僆 1q, 84 , R: y 僆 3 4, q 2 b. 5, 3, 3, 1, 2 c. 12, 02 d. f 1x2 6 0: x 僆 12, 22 ; f 1x2 7 0: x 僆 1q, 22 ´ 32, 84 e. min: (0, 4), max: (8, 7) f. f 1x2c: x 僆 10, 82 ; f 1x2T: x 僆 1q, 02 y

8 4 8

4

4

8

x

4 8

15. a.

x7 1x 521x 22

b.

b2 4ac 4a2

17. x2 2

19. center 13, 62, r 3 21. W 31 cm, L 47 cm

5 23. a. x 4 b. x 5, 13, 13 3 ,2 25. P 15 197 units ⬇ 24.8 units. No, it is not a right triangle. 52 1 1972 2 102 27. 3.1

10. x 0.75 4.7

4.7

冢0, s冣 y 0

5

5

13. a. 1q, q2 15. a.

5 x

b. 3 0, q 2

c. 3 0, q 2

3.1 14. VA: x 2; HA: y 1

29.

10

25

10 0.1

10

1.1

10 3 b. S(t) 17.27 t 2.50

c. 3.05 mm

y

4 8

4

4

3 5

8

1 2 3 4 5 x

8

(0, 5) 4

5. x 1

4

4. max: y 8 at x 2; min: y 7 at x ⬇ 5.87 and y 7 at x ⬇ 1.87 5. I. a. square root b. x 僆 3 4, q 2, y 僆 33, q 2 c. 12, 02, 10, 12 d. up on right e. x 僆 12, q2 f. x 僆 3 4, 22 II. a. cubic b. x 僆 1q, q2 , y 僆 1q, q2 c. 12, 02, 10, 12 d. down on left, up on right e. x 僆 12, q2 f. x 僆 1q, 22 III. a. absolute value b. x 僆 1q, q 2 , y 僆 1q, 4 4 c. 11, 02, 13, 02, 10, 22 d. down on left, down on right e. x 僆 11, 32 f. x 僆 1q, 12 ´ 13, q2 IV. a. quadratic b. x 僆 1q, q2 ; y 僆 3 5.5, q2 c. 10, 02, 15, 02, 10, 02 d. up on left, up on right e. x 僆 1q, 02 ´ 15, q 2 f. x 僆 10, 52 y y 6. 7. 8

b.

9.

4 3 2 1 1 2 3 4 5 x

1 3

7. a.

3. 29.45 cm

d. 0.95 sec

cob19545_saa_018-038.qxd

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CHAPTER 3

67. w 23 or w 1 2


69. m

1. 3 2i 3. 2, 3 12 5. (b) is correct. 7. a. 12i b. 7i c. 313 d. 6 12 9. a. 3i 12 b. 5i 12 c. 15i d. 6i 2 13 3 12 11. a. i119 b. i 131 c. i d. i 5 8 13. a. 1 i b. 2 i 13 15. a. 4 2i b. 2 i 12 17. a. 5 0i; a 5, b 0 b. 0 3i; a 0, b 3 12 12 i; a 0, b 19. a. 0 18i; a 0, b 18 b. 0 2 2 21. a. 4 5i 12; a 4, b 5 12 b. 5 3i13; a 5, b 3 13 7 712 7 712 1 110 1 110 i; a , b i; a , b 23. a. b. 4 8 4 8 2 2 2 2 25. a. 19 i b. 2 4i c. 9 10i 13 1 27. a. 3 2i b. 8 c. 2 8i 29. a. 2.7 0.2i b. 15 i 12 1 c. 2 i 31. a. 15 0i b. 16 0i 33. a. 21 35i 8 b. 42 18i 35. a. 12 5i b. 1 5i 37. a. 41 b. 74 39. a. 11 b. 17 41. a. 5 12i b. 7 24i 36

73.

43. a. 21 20i b. 7 6i 12 45. a. 4 5i; 41 b. 3 i12; 11 47. a. 7i; 49 b. 12 23 i; 25 49. no 51. yes 36 53. yes 55. no 57. yes 59. verified 61. a. 1 b. 1 c. i 2 4 21 14 15 10 i b. i d. i 63. a. 0 i b. 0 i 65. a. 7 5 13 13 13 13 2 3 67. a. 1 i b. 1 i 69. a. 113 b. 5 c. 111 4 3 71. A B 10 AB 40 73. 17 5i2 75. 125 5i2 V 77. 1 74 i2

79. a. 1a bi21a bi2 a2 abi abi 1bi2 2 a2 b2 112 a2 b2 ✓ b. 1x 6i2 1x 6i2 c. 1m i 1321m i 132 d. 1n 2i1321n 2i 132 e. 12x 7i212x 7i2 83. 5.6 hr (5 hr 36 min) 85. John

81. 8 6i

Exercises 3.2, pp. 308–313 1. exact, approximate 3. discriminant, 2 7. x 4 or 3

9. x

3 2

or 1

81. 83. 85.

17. x ⬇ 1.14 or 2.64

19. { }

21. m 4

23. y 217; y ⬇ 5.29 25. no real solutions x ⬇ 1.15 29. n 9; n 3

31. w 5 13; w ⬇ 3.27 or w ⬇ 6.73 33. no real solutions

35. m 2 3 12 37. 9; 1x 32 2 7 ; m ⬇ 2.61 or m ⬇ 1.39

89. p 23 1394 6 ; p ⬇ 3.97 or p ⬇ 2.64 91. two rational; factorable 93. two nonreal 95. two rational; factorable 97. two nonreal 99. two irrational 101. one repeated; factorable 103. x 32 12 i

47. p 3 15; p ⬇ 0.76 or p ⬇ 5.24 49.

113 m 3 2 2 ; 5 3 15 n2 2 ;n a. x 12 or x

m ⬇ 0.30 or m ⬇ 3.30

51. ⬇ 5.85 or n ⬇ 0.85 53. 4 b. x 0.5 or x 4 55. a. n 3 or n 3 b. n 3 or n 1.5 2 57. a. p 38 141 b. p ⬇ 1.18 or p ⬇ 0.43 8 59. a. m 72 133 b. m ⬇ 6.37 or m ⬇ 0.63 2 61. x 6 or x 3 63. m 52 65. n 1

15 2 ;

n ⬇ 2.12 or n ⬇ 0.12

13 2 i

105. x 12

107. x 54

111. x 僆 1q, 5 4 ´ 3 1, q 2 115. x 僆 3 17, 17 4

3 17 4 i

109. x 僆 10, 42

113. x 僆 11, 72 2

117. x 僆 3 32

133 2 ,

32

133 2 4

x 僆 1q, 53 4 ´ 31, q 2 121. x 僆 1q, q 2 123. { } x 僆 1q, 52 ´ 15, q 2 127. { } 129. x 僆 1q, q 2 x 僆 1q, q 2 133. 13, 12 135. 1q, 3 2 4 ´ 32, q2 1q, 1.32 ´ 11.3, q 2 139. {2.9} 141. { } 143. 1q, q2 x 僆 1q, 5 4 ´ 35, q 2 147. x 僆 1q, 0 4 ´ 35, q 2 { } 151. a 153. b v 1v2 64h 155. t 157. t 3 1138 sec, t ⬇ 8.87 sec 2 32 2 159. a. P x 120x 2000 b. 10,000 161. t 2.5 sec, 6.5 sec 163. 36 ft, 78 ft 165. 30,000 ovens 167. x ⬇ 13.5, or the year 2008 169. a. 7x2 6x 16 0 b. 6x2 5x 14 0 c. 5x2 x 6 0 3 171. z 2i; z 5i 173. z i; z 2i 4 175. z 1 i; z 13 i 177. a. P 2L 2W, A LW b. P 2␲r, A ␲r2 c. P c h b1 b2, A 12 h1b1 b2 2 d. P a b c, A 12 bh 179. 700 $30 tickets; 200 $20 tickets 119. 125. 131. 137. 145. 149.

Exercises 3.3, pp. 322-326 1. 25 3. 0, f (x) 5. Answers will vary. 2 7. left 2, down 9 9. right 1, reflected across x-axis, up 4 y

y

8

8

4

(5, 0) 8

(0, 3) 4 (1, 0)

(1, 0) 4

4 4

8

x

8

(0, 5)

4

y 16

4

(0, 7) 8 (0.7, 0)

(0.6, 0)

4

4

8

x

8

12

(C, 121 ) 12

(0, 3)

4

8

(74, 258)

y

8

(0, 6) 4

(3, 0) 4

8

x

17. left 67 , reflected across x-axis, stretched vertically, up 121 12

y

4

8

16

8

4

(4.7, 0) 4

8

15. right 74 , stretched vertically, down 25 8

8

(2, 15)

4

(0, 5)

(1, 8) 8

(q, 0)

x

13. right 2, reflected across x-axis, stretched vertically, up 15

8

4

8

8

y

8

(3, 0)

4

11. left 1, stretched vertically, down 8

(2.6, 0)

(1, 4)

4

(2, 9) 8

39. 94 ; 1n 32 2 2 41. 19 ; 1p 13 2 2 43. x 1; x 5 45. p 3 16; p ⬇ 5.45 or p ⬇ 0.55

0.80i

87. a 3 3 12 2 ; a ⬇ 0.88 or a ⬇ 5.12

11. x 1 or 2

15. x ⬇ 1.61 or 3.11

27. x

79.


13. x ⬇ 4.19 or 1.19

1 421 ;

77.

16 3 71. n 32 2 2 i; m ⬇ 1.5 1.12i 123 4 1 w 5 or w 2 75. a 6 6 i; a ⬇ 0.16 p 35 2 16 5 ; p ⬇ 1.58 or p ⬇ 0.38 1 w 10 121 10 ; w ⬇ 0.56 or w ⬇ 0.36 a 34 131 4 i; a ⬇ 0.75 1.39i p 1 3 12 2 i; p ⬇ 1 2.12i 12 w 1 3 3 ; w ⬇ 0.14 or w ⬇ 0.80

x

(3, 0) 8

4

(s, 0) 4

4 8

8

x

cob19545_saa_018-038.qxd

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Student Answer Appendix 19. right 52 , down 17 4

21. left 1, down 7

y 8

4 (0, 2)

4

(4.6, 0)

4

4 4

8

5

8

10. f 1x2 49 1x 32 2 2

y

(2, 2) (4, 0)

(0, 0)

7

(3.6, 0) x

3 x

(1.6, 0)

4

4 4

(e, *)

8

x 2

y

8

(0.4, 0) 8

9.

SA-19

(1, 7)

8

8

x 5

(0, 6)

Reinforcing Basic Concepts, pp. 327-328 23. right 2, reflected across x-axis, up 6

25. left 3, compressed vertically up 52

y

y

8 (2, 6)

8

(0, 2) 4

冢3, e冣

(4.4, 0) 8

4

4

8

8

x

4

4

4

(0.4, 0) 4

4

8

8

27. right 52 , reflected across x-axis, stretched vertically, up 11 2

4

y

(0, 7)

4

(0, 3)

(4.2, 0)

4

x

8

冢e, y冣

(0.8, 0) 8

8

29. right 32 , stretched vertically, down 6

y 8

(0, 7)

4

8

8

x

4

(2.7, 0) 4

4

(0.3, 0) 4

8

8

8

x

冢w, 6冣

31. left 3, compressed vertically, down 19 2 y 8 4

(7.4, 0) 8

(1.4, 0) 4

4 8 4 (0, 5)

x

8

冢3, p冣

33. y 11x 22 2 1 35. y 11x 22 2 4 3 37. y 1x 22 2 3 39. i. x 3 15 ii. x 4 13 2 1 iii. x 4 214 iv. x 2 12 v. t 2.7, t 1.3 vi. t 1.4, t 2.6 41. a. 10, 66,0002; when no cars are produced, there is a loss of $66,000. b. (20, 0), (330, 0); no profit will be made if fewer than 20 or more than 330 cars are produced. c. 175 d. $240,250 43. a. $2 b. $44 c. $8800 d. $23; $44,100 45. 6000; $3200 47. a. h1t2 16t2 240t 544 b. 544 ft; that is when the fuel is exhausted. c. 1344 ft d. 1344 ft e. It is coming back down. f. 1444 ft g. 17 sec 49. a. 14.4 ft b. 41 ft c. 48.02 ft d. 90 ft 51. a. 25 ft b. approx. 3.43 sec c. 67.25 ft 53. a. 2500 ft2, 50 ft 50 ft b. 5000 ft2, 50 ft 100 ft 55. a. approx. 29.5" wide by 18.7" long b. approx. 930 in2 57. a. $1.25 each, $781.25 b. about $0.85 each 59. Answers will vary. 7 1 2 7 61. y ax b 63. m 43 , y-intercept (0, 3) 18 2 2 3 65. g1x2 1 x13

7 5 b 112 ✓ 2 2 a 7# 7 c 112 ✓ 2 2 a 2 3 12 2 3 12 4 b Exercise 2: ✓ 2 2 2 a 14 7 c 2 3 12 # 2 3 12 ✓ 2 2 4 2 a b Exercise 3: 15 213i2 15 2 13i2 10 ✓ a c 15 2 13i2 15 213i2 25 12 37 ✓ a b b b Exercise 4: x1 x2 since radical terms sum to 0, and 2a b ; a 2a a 2 2 2 2 2 c b 2b 4ac b b 4ac 4ac c x1x2 since a b a b 2 2 a 2a 2a a 4a 4a2 4a Exercise 1:

Exercises 3.4, pp. 335-339 1. quadratic, negative 3. average, slope, secant line 5. Answers will vary. 7. a. y 1.45x2 2.59x 4.55 b. 4.1475 c. x 6.446 9. a. y 0.851x2 3.153x 64.428 b. 65.037 c. x 7.623 11. a. y 3.485x2 26.424x 60.505 b. 10.422 c. x 1.746 or 5.836 13. a. y 0.113x2 5.796x 61.583 b. 12.882 c. x 7.368 or 44.019 15. 0 17. 1.4 19. 0.2 21. 1.6 23. 0.1 25. 1.2 27. a. 48 ft/sec b. 32 ft/sec c. 16 ft/sec d. 32 ft/sec 29. a. 0 m/sec b. 0 m/sec c. 4.9 m/sec d. 4.9 m/sec 31. Answers will vary. 500 450 400 350 300 250 200 150 100 50 0

(5, 400) (4, 384) (7, 336) (3, 336) (2, 258)

1 2 3 4 5 6 7 8 9 10

33. Answers will vary. 80

(2.5, 57.725)

70

(3.5, 62.625) (4.5, 57.725)

60 50

(2, 51.6)

40 30

(5, 51.6) (1, 32)

(6, 32)

20 10

Mid-Chapter Check, p. 327

0

1. sum 4, product 13; both yield real numbers

2. i

11 2i2 211 2i2 5 0 1 4i 4i2 2 4i 5 0 1 142 2 5 0 0 0✓ Yes. 7 1 4. x 1 and 3 5. x 2 253 6. x 4 417 7. 1q, 1 4 ´ 3 0.37, q 2 8. a. about 29.6 in. b. about 7.17 ft c. yes; 3 7.17 7 21.5 3.

2

1

2

3

4

5

6

7

8

¢F 35. ¢y 37. ¢y 39. ¢m 9.8 ¢x 39 ¢x 1 ⬇ 41. ¢A 43. a. 14,570 ft b. 14,174 ft c. 198 ft/sec 37.70 ¢r d. h1102 13,340 ft; h1122 12,624 ft; average rate of change 358 ft/sec, almost twice as fast 45. a. R 43.07t2 976.53t 126.8 b. 4598 c. t ⬇ 8.26 days (early in the ninth day) d. about 5408 participants 47. a. T 5.92n2 83.13n 349.86 b. about 82 sec c. 16 tourists d. about 58 sec with seven tourists

cob19545_saa_018-038.qxd

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50 49. a. ¢weight ¢time 1 , positive, 50 g are gained each week b. 25th to 29th: ¢w 50 ¢w 250 ¢t 1 ; 32nd to 36th: ¢t 1 ; the weight gain is five time greater in the later weeks. 51. a. 7 b. 7 y c. They are the same. 10 8 (2, 8) d. Slopes are equal. 6 4

(1, 1) 2 54321 2 4 6 (2, 8) 8 10

(1, 1) 1 2 3 4 5 x

53. a. 176 ft b. 320 ft c. 144 ft/sec d. 144 ft/sec; The arrow is going down. 55. a. 17.89 ft/sec; 25.30 ft/sec b. 30.98 ft/sec; 35.78 ft/sec c. Between 5 and 10. d. 1.482 ft/sec, 0.96 ft/sec 57. a. about 3569 in3 b. about 3800 in3. c. ¢r ¢t 0.74 in./sec when 3 t 僆 3 0, 1 4 and 0.04 in./sec when t 僆 3 18, 194 ; ¢V ¢r ⬇ 64.745 in /in. when t 僆 3 0, 1 4 and 22.602 when t 僆 3 18, 194 . Answers will vary. 59. 1 12 , q 2 |x| x 6 1 61. a. 1q, 12 ´ 12, q 2 b. { } 63. g1x2 e 1 x 1 2 1x 32


1. 1f g21x2 , A 傽 B 3. intersection, g(x) 5. Answers will vary. 7. a. x 僆⺢ b. f 122 g122 13 9. a. h1x2 x2 6x 3 b. h122 13 c. they are identical 11. a. x 僆 3 3, q2 b. h1x2 1x 3 2x3 54 c. h142 75, 2 is not in the domain of h. 13. a. x 僆 35, 3 4 b. r1x2 1x 5 13 x c. r122 17 1, 4 is not in the domain of r. 15. a. x 僆 3 4, q2 b. h1x2 1x 412x 32 c. h142 0, h1212 225 17. a. x 僆 31, 7 4 b. r 1x2 1x2 6x 7 c. 15 is not in the domain of r, r 132 4 19. a. x 僆 1q, 42 ´ 14, q2 b. h1x2 x 4, x 4 21. a. x 僆 1q, 42 ´ 14, q2 b. h1x2 x2 2, x 4 23. a. x 僆 1q, 12 ´ 11, q 2 b. h1x2 x2 6x, x 1 x1 25. a. x 僆 1q, 52 ´ 15, q2 b. h1x2 ,x5 x5 2x 3 27. a. x 僆 1q, 22 b. r 1x2 12 x 15 c. 6 is not in the domain of r, r 162 2 x5 29. a. x 僆 15, q2 b. r1x2 1x 5 c. r162 1, 6 is not in the domain of r. x2 36 13 c. r 162 0, r162 0 , qb b. r 1x2 2 12x 13 2x 4 33. a. h1x2 b. x 僆 1q, 32 ´ 13, q2 c. x 2, x 0 x3 35. sum: 3x 1, x 僆 1q, q 2 ; difference: x 5, x 僆 1q, q2 ; product: 2x2 x 6, x 僆 1q, q2 ; 2x 3 quotient: , x 僆 1q, 22 ´ 12, q 2 x2 37. sum: x2 3x 5, x 僆 1q, q 2 ; difference: x2 3x 9, x 僆 1q, q2 ; product: 3x3 2x2 21x 14, x 僆 1q, q 2 ; 31. a. x 僆 a

quotient:

x2 7 2 2 , x 僆 aq, b ´ a , qb 3x 2 3 3

39. sum: x2 3x 4, x 僆 1q, q 2 ; difference: x2 x 2, x 僆 1q, q2 ; product: x3 x2 5x 3, x 僆 1q, q 2 ; quotient: x 3, x 僆 1q, 12 ´ 11, q2 41. sum: 3x 1 1x 3, x 僆 3 3, q2 ; difference: 3x 1 1x 3, x 僆 3 3, q2 ; product: 13x 12 1x 3, x 僆 3 3, q2 ; 3x 1 quotient: , x 僆 13, q2 1x 3

43. sum: 2x2 1x 1, x 僆 31, q2 ; difference: 2x2 1x 1, x 僆 3 1, q2 ; product: 2x2 1x 1, x 僆 31, q2 ; 2x2 , x 僆 11, q2 quotient: 1x 1 7x 11 , x 僆 1q, 22 ´ 12, 32 ´ 13, q 2 ; 45. sum: 1x 32 1x 22 3x 19 , x 僆 1q, 22 ´ 12, 32 ´ 13, q2 ; difference: 1x 32 1x 22 10 , x 僆 1q, 22 ´ 12, 32 ´ 13, q2 ; product: 1x 321x 22 2x 4 , x 僆 1q, 22 ´ 12, 32 ´ 13, q2 quotient: 51x 32 47. a. 6000 b. 3000 c. 8000 d. C192 T192 ; 4000 49. a. $1 billion b. $5 billion c. 2003, 2007, 2010 d. t 僆 12000, 20032 ´ 12007, 20102 e. t 僆 12003, 20072 f. R152 C152 ; $4 billion 1 51. a. 4 b. 0 c. 2 d. 3 e. f. 6 g. 3 h. 1 i. 1 3 2 j. undefined 53. h1x2 3 x 4 55. h1x2 4x x2 57. a. 1q, q 2 b. 1q, 22 ´ 12, q 2 59. a. 31, q 2 b. 11, q2 61. a. 3 2, 34 b. 3 2, 32 63. a. 3 4, q 2 b. 34, q2 65. a. 1q, q 2 b. 1q, 42 ´ 14, q 2 67. a. 34, 3 4 b. 34, 32 69. a. V 400 400e0.08t b. f 1t2 400, g1t2 400e0.08t, V112 ⬇ 400 369 31 ft/sec, V122 ⬇ 400 341 59 ft/sec, V1202 ⬇ 400 81 319 ft/sec c. 400 ft/sec 71. a. P1x2 12,000x 108,000 b. nine boats must be sold 73. a. P1n2 11.45n 0.1n2 b. $123 c. $327 d. C11152 7 R11152 75. km 115 77. Anywhere between km 115 and km 199. Answers will vary. 79. a. 4 sec b. about 494 ft 81. a. 1995 to 1996; 1999 to 2004 b. 30; 1995 c. 20 seats; 1997 d. The total number of seats in the senate (50); the number of additional seats held by the majority 83. 67 a. x 23 , where Y1 Y2; 67 b. x 4, where Y1 0; 68 a. x 0, where Y1 Y2; 68 b. x 1.2 or 3, the zeroes of Y1 and Y2. 85.

x

148 9

87. a. 6x 5y 13 b. d 261

Exercises 3.6, pp. 365–369 1. composition 3. domain, g(x) 5. Answers will vary. 7. 0, 0, 4a2 10a 14, a2 9a 9. a. h1x2 12x 2 b. H1x2 21x 3 5 c. D of h(x): x 僆 3 1, q 2 ; D of H(x): x 僆 33, q 2 11. a. h1x2 13x 1 b. H1x2 3 1x 3 4 c. D of h(x): x 僆 3 13 , q 2 ; D of H(x): x 僆 33, q 2 13. a. h1x2 x2 x 2 b. H1x2 x2 3x 2 c. D of h(x): x 僆 1q, q2 ; D of H(x): x 僆 1q, q2 15. a. h1x2 x2 7x 8 b. H1x2 x2 x 1 c. D of h(x): x 僆 1q, q 2 ; D of H(x): x 僆 1q, q 2 17. a. h1x2 冟3x 1冟 5 b. H1x2 3冟x冟 16 c. D of h(x): x 僆 1q, q 2 ; D of H(x): x 僆 1q, q2 19. h132 6, h1 222 ⬇ 9.071 h1 12 2 2.75, h152 50 21. h132 1, h1 222 ⬇ 145.91 h1 12 2 64, h152 ERR; x 5 is not in the domain of g 23. h132 2, h1 222 5 h1 12 2 5.5, h152 ERR; g(5) is not in the domain of f

25. a. 1 f ⴰ g2 1x2 : For g(x) to be defined, x 0. 2g1x2 5 For f 3g1x2 4 , g1x2 3 so x . g1x2 3 3 domain: all real numbers except x 3 and x

5 3

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Student Answer Appendix b. 1g ⴰ f 21x2 : For f (x) to be defined, x 3. 5 , f 1x2 0 so x 0. For g3 f 1x2 4 f 1x2 domain: all real numbers except x 3 and x 0 10 5x 15 ; 1g ⴰ f 21x2 c. 1 f ⴰ g21x2 ; the domain of a 5 3x 2x composition cannot always be determined from the composed form 27. a. 1 f ⴰ g21x2 : For g(x) to be defined, x 5. 4 , g1x2 0 and g(x) is never zero For f 3 g1x2 4 g1x2 domain: {x | x 5} b. 1g ⴰ f 21x2 : For f(x) to be defined, x 0. 4 1 , f 1x2 5 so x . For g3 f 1x2 4 f 1x2 5 5 4 domain: all real numbers except x 0 and x 5 x c. 1 f ⴰ g21x2 4x 20; 1g ⴰ f 21x2 ; the domain of a 4 5x composition cannot always be determined from the composed form 29. a. 41 b. 41 31. g1x2 1x 2 1, f 1x2 x3 5 33. p1x2 21x 42 2 3, q1x2 12x 72 2 1 35. a. 2 b. 2 c. 2 d. 1 e. 1 f. 1 g. 3 h. ⬇0.5 37. a. f 1x2: x 1, g1x2: x 2 c. h1x2: x 僆 1q, 22 ´ 31, q 2 39. 2 41. 2x h 43. 2x 2 h 45. x1x2 h2 ¢g ¢g ¢g 2x 2 h b. 3.9 c. 3.01 47. a. ¢x ¢x ¢x y d. The rates of change have opposite sign, with the 7 6 secant line to the left being slightly more steep. 5

654321 1 2 3

¢g d.

¢x

1 2 3 4 x

¢g 12.61 c. ⬇ 0.49 ¢x ¢x Both lines have a positive slope, but the line at x 2 is much steeper.

3x2 3xh h2 y

20 16 12 8 4 54321 4 8 12 16 20

¢g

b.

1 2 3 4 5 x

2x h ; x2 1x h2 2 ¢j 3 0.50, 0.51 4: ⬇ 15.5; ¢x ¢j 3 1.50, 1.51 4 : ⬇ 0.6; ¢x Answers will vary. ¢g 3x 2 3xh h2; 53. ¢x ¢g 3 2.01, 2.00 4 : ⬇ 12.1; ¢x ¢g 3 0.40, 0.41 4: ⬇ 0.5; ¢x Answers will vary. 55. a. f 3g1x2 4 1x 22 2 41x 22 3 x2 4x 4 4x 8 3 x2 1✓ b. verified 57. h1x2 x 2.5; 10.5 59. a. 4160 b. 45,344 c. M1x2 453.44x; yes 61. a. 6 ft b. 36␲ ft2 c. A1t2 9␲t 2 ; yes 63. a. L102 500 lions and H15002 400 hyenas b. H3 L1x2 4 400 0.0075x, 1H ⴰ L2116,0002 520 hyenas c. prior to an increase of 30,000 ¢j

51.

¢x

c.

48 42 36 30 24 18 12 6 6 12

b.

y

¢d ⬇ 0.04 ¢h As height increases you can see farther, but the sight distance increases at a slower rate.

x 100 200 300 400 500

¢d ¢d ⬇ 15, June: ⬇ 3, 5 times faster ¢t ¢t ¢d b. t 6.75, late June c. decreasing a 6 0b; 5000 units/month ¢t 69. Answers will vary. 1 ¢y 1 ¢y 71. a. For y : ⬇ 3.91; For y 2 : ⬇ 15.53 x ¢x x ¢x 1 ¢y b. Less—decrease is more gradual; For y : ⬇ 1.54; x ¢x 1 ¢y For y 2 : ⬇ 3.83 x ¢x y y y 73. a. b. c. 67. a. March:

8

8

f(x)

4 8

4

4

4

8

8

x

4

8 4

g(x) 4

8

8

x

h(x)

4

4

4

4

4

8

8

8

8

75. y 32 x

Making Connections, p. 370 1. b

3. h

5. d

7. a

9. g

11. f

13. c

15. a


4 3 2 1

49. a.

¢d ⬇ 0.2 ¢h

65. a.

1. 6i 12 2. 24i 13 3. 2 i 12 4. 3i12 5. i 6. 21 20i 7. 2 i 8. 5 7i 9. 13 10. 20 12i 11. 15i2 2 9 34 15i2 2 9 34 2 25i 9 34 25i2 9 34 25 9 34✓ 25 9 34✓ 12. 12 i 152 2 412 i 152 9 0 4 4i 15 5i2 8 4i 15 9 0 5 152 0 ✓ 12 i 152 2 412 i 152 9 0 4 4i 15 5i2 8 4i 15 9 0 5 152 0✓ 13. a. x 5 or x 2 b. x 5 or x 5 c. x 53 or x 3 d. x 2 or x 2 or x 3 14. a. x 3 b. x 2 15 c. x i 15 d. x 5 15. a. x 3 or x 5 b. x 8 or x 2 c. x 1 110 2 ; x ⬇ 2.58 or x ⬇ 0.58 d. x 2 or x 13 16. a. x 2 i 15; x ⬇ 2 2.24i b. x 32 12 c. x 32 12 i 2 ; x ⬇ 2.21 or x ⬇ 0.79 17. a. 1q, 22 ´ 13, q 2 b. 31, 1 4 c. 1q, q 2 18. a. 34, 1 4 b. 1q, 52 ´ 14, q 2 c. {2} 19. a. 1.3 sec b. 4.7 sec c. 6 sec 20. a. 0.8 sec b. 3.2 sec c. 5 sec y y 21. 22. 10 8 6 4 2 (0, 0)

8 4

(5, 0)

(6, 0)

8

108642 2 2 4 6 8 10 x 4 6 8 10 (3, 9)

23.

(3, 0) 4

8

(0, 3) 4

4

(2, 1) 4

4

(0.3, 0) 8

x

8

4

(2.7, 0) 4

4

4

8

8

(0, 5)

x

y

8

8

8

8

24.

y

4

(4, 1)4

8

冢w, 6冣

x

x

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25. a. 0 ft b. 108 ft 26. $3.75; 3000 27. a.

14. 12, 02, y 2x2 4x 15. 16. a. 1300

d. 144 ft, t 3 sec

c. 2.25 sec 400

0

0

15

50

a. 40 ft, 48 ft

; quadratic

b. A1x2 2.144x2 1.010x 25.847, about $48.2 billion c. about $460.2 billion d. year 16 S 2016 28. a. 2000

c. 14 sec

14

quadratic

0 b. y ⬇ 6.68x2 3.48x 176.30 1300

0 0

b. 49 ft

14

80

0 200

, quadratic

b. Fd 1x2 0.35x 0.3x, 1242 units c. about 82 mph 2.95 11.93 ¢N ¢N ⬇ 29. a. , 295 stores per year b. for 2000 to 2002: , ¢t 1 ¢t 1 1193 stores per year; 295 4 ⬇ 1180 ✓ c. for 2002 to 2004: ¢N 13.42 ¢N 13.20 ⬇ ⬇ , 1342 stores per year; for 2006 to 2008: , 1320 ¢t 1 ¢t 1 3 3 stores per year, very close 30. a. 20 ft b. approx. 18.251 ft c. approx. 1.749 ft3/sec d. approx. 0.149 ft3/sec e. t ⬇ 22.4 sec 31. a2 7a 2 32. 147 33. x 僆 1q, 23 2 ´ 1 23 , q 2 34. a. 4 b. 6 c. 1 d. 14 5 35. a. P1x2 84.95n 10.002n2 20n 30,0002 0.002n2 64.95n 30,000 b. $3700 c. $344,750 d. 456 36. 4x2 8x 3 37. 99 38. x; x 39. f 1x2 1x 1; g1x2 3x 2 1 40. f 1x2 x2 3x 10; g1x2 x3 41. A1t2 ␲12t 32 2 42. a. 0 b. 7 c. 7 d. 2 e. 4 43. 2x 1 h; 3.01 2

c. about 396,000; about 4,264,000 17. a. 4750 books per year b. approx. 11,267 books per year c. 2400 books per year and 18,100 books per year 18. 3x 1; x 僆 3 13 , q 2 19. a. No, new company and sales should be growing b. 15 for [4, 5]; 19 for [5, 6] c. ¢S ¢t 4t 3 2h. For small h, sales volume is approximately units 37,000 unit units in month 10, 69,000 in month 18, and 93,000 in month 24 1 mo 1 mo 1 mo 20. a. V1t2 43 ␲1 1t2 3 b. 36␲ in3

Strengthening Core Skills, pp. 378–379 1.

(1 2兹2, 0)

12. a. x 3

13 3

b. x 1 3i

13. a. f 1x2 1x 52 2 9 20 16 12 8 4

b. g1x2 12 1x 42 2 8

y

y (5, 9) (8, 0)

42 4 2 4 6 8 10 12 14 16 x 8 (2, 0) 12 16 (0, 16) 20

(4, 8)

18 16 (0, 16) 14 12 10 8 6 4 2

108642

2 4 6 8 10 x

x e (5, 9)

4

4

4

8

x

8

4

4

4

4 4

x3

4

b. i

(5 2兹2, 0)

(5 2兹2, 0) 8

4

4

x

8

4

x

8

(5, 8) x5

4

(3, 2) 4

y

8

(6, 11)

8

x

8

8

4.

y (0, 11)12

8

4 4

(2, 7) (0, 7) (1, 8) 8

3.

y 8 (0, 9)

(1 2兹2, 0) 冢e, %冣

8

1. a. 10, 11 b. 1q, 12 ´ 11, q2 c. 1q, 73 4 ´ 3 3, q 2 d. { } 2 2 2. x 5i 3. x 1 i 23 4. x 23 , x 6 5. a. t 5 (May) 15 3 i

2.

y 8


b. t 9 (Sept.) c. July; $3000 more 6. a. 43 7. a. 1 b. i 13 c. 1 8. 32 32 i 9. 34 10. 12 3i2 2 412 3i2 13 0 4 12i 9 8 12i 13 0 13 13 0 00✓ 11. a. x 5 12 b. x 54 17 2 4 i

x 1

8

8

12

(0, 17) 16

5.

6.

y

(6, 21) (3, 3) 8

50 40 30 20 (0, 21) 10

410 20 30

x 3 40 50

4

y (0, 8)

x

8

冢r, 8冣

8 4

8

(10, 17)

4

冢#, ≥冣 4

8

x

4 8

x#


3. a. 1x 12 1x2 x 12 b. 1x 321x 221x 22 1009 11 5. all reals 7. verified 9. y x ; 39 min, driving time 60 60 increases 11 min every 60 days 11. Month 9 13. a. f (x) b. g(x) R1R2 R2

1. R R1

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5

y

5

5 x

5

17. X 63 19. a. f 142 1, g122 4, 1 f ⴰ g2122 1 b. g142 0, f 182 4, 1g ⴰ f 2182 0 g 4 c. 1 fg2 102 122142 8, 1 f 2102 2 2 d. 1 f g2112 3 5 2, 1g f 2192 2 2 0 21. 2x2 9.2x 14.5 0 and 1.2x2 5.52x 8.7 23. y 3x2 1.5x 7 25. x 僆 1q, 0.22 ´ 10.2, 0.22 ´ 10.2, q2

SA-23

59. P1x2 1x 221x 2321x 232, P1x2 x3 2x2 3x 6 61. P1x2 1x 521x 22321x 2232, P1x2 x3 5x2 12x 60 63. P1x2 1x 121x 221x 2102 1x 2102, P1x2 x4 x3 12x2 10x 20 65. P1x2 1x 221x 321x 42 67. p1x2 1x 32 2 1x 321x 12 69. f 1x2 21x 32 2 1x 221x 52 71. p1x2 1x 321x 32 2 73. p1x2 1x 22 3 75. p1x2 1x 32 1x 32 3 77. p1x2 1x 321x 32 2 1x 42 2 79. 4-in. squares; 16 in. 10 in. 4 in. 81. a. week 10, 22.5 thousand b. one week before closing, 36 thousand c. week 9 83. a. 198 ft3 b. 2 ft c. about 7 ft 85. k 10 87. k 3 89. The theorems also apply to complex zeroes of polynomials. 91. S3 36; S5 225 93. Yes, John wins. 95. G1t2 1400t 5000


CHAPTER 4 Exercises 4.1, pp. 390–394 1. synthetic, zero 3. P(c), remainder 5. Answers will vary. 7. x3 5x2 4x 23 1x 221x2 3x 102 3 9. 2x3 5x2 4x 17 1x 3212x2 x 72 4 11. x3 8x2 11x 20 1x 521x2 3x 42 0 2x2 5x 3 0 13. a. 12x 12 x3 x3 b. 2x2 5x 3 1x 3212x 12 0 x3 3x2 14x 8 0 15. a. 1x2 5x 42 x2 x2 b. x3 3x2 14x 8 1x 221x2 5x 42 0 x3 5x2 4x 23 3 1x2 3x 102 x2 x2 b. x3 5x2 4x 23 1x 221x2 3x 102 3 17. a.

2x3 5x2 11x 17 13 12x2 3x 12 x4 x4 b. 2x3 5x2 11x 17 1x 4212x2 3x 12 13 21. x3 5x2 7 1x 121x2 4x 42 11 23. x3 13x 12 1x 421x2 4x 32 0 25. 3x3 8x 12 1x 1213x2 3x 52 7 27. n3 27 1n 321n2 3n 92 0 29. x4 3x3 16x 8 1x 221x3 5x2 10x 42 0 19. a.

31.

2x3 7x2 x 26 x2 3 x4 5x2 4x 7

12x 72

7x 5

x2 3 4x 3

35. 41. 47. 51.

1x 42 2 x2 1 x 1 a. 30 b. 12 37. a. 2 b. 22 39. a. 1 b. 3 a. 31 b. 0 43. a. 10 b. 0 45. a. yes b. yes a. no b. yes 49. a. yes b. yes 53. 2 1 0 7 6 3 1 2 5 6 2 4 6 3 3 6 1 2 3 0 1 1 2 0

55.

2 3

33.

9 18 4 8 6 16 8 9 24 12 0

2

57. P1x2 1x 221x 321x 52, P1x2 x3 4x2 11x 30

1. coefficients 3. a bi 5. b; 4 is not a factor of 6 7. P1x2 1x 221x 221x 3i21x 3i2 x 2, x 2, x 3i, x 3i 9. Q1x2 1x 22 1x 221x 2i2 1x 2i2 x 2, x 2, x 2i, x 2i 11. P1x2 1x 121x 121x 12 x 1, x 1, x 1 13. Q1x2 1x 52 1x 521x 52 x 5, x 5, x 5 15. 1x 52 3 1x 92 2; x 5, multiplicity 3; x 9, multiplicity 2 17. 1x 72 2 1x 22 2 1x 72; x 7, multiplicity 2; x 2, multiplicity 2; x 7, multiplicity 1 19. P1x2 x3 3x2 4x 12 21. P1x2 x4 x3 x2 x 2 23. P1x2 x4 6x3 13x2 24x 36 25. P1x2 x4 2x2 8x 5 27. P1x2 x4 4x3 27 29. a. yes b. yes 31. 34.7, 4.6 4 , 3 2.3, 2.24 , [0.9, 1] 3 5 1 15 3 5 33. 51, 15, 3, 5, 14 , 15 4 , 4 , 4 , 2 , 2 , 2 , 2 6 1 15 3 5 35. 51, 15, 3, 5, 2 , 2 , 2 , 2 6 1 7 2 7 1 7 28 4 37. 51, 28, 2, 14, 4, 7, 16 , 14 3 , 3 , 3 , 3 , 6 , 2 , 2 , 3 , 3 6 1 1 1 1 1 3 3 3 3 3 39. 51, 3, 32 , 2 , 16 , 4 , 8 , 32 , 2 , 16 , 4 , 8 6 41. 1x 421x 121x 32, x 4, 1, 3 43. 1x 321x 221x 52, x 3, 2, 5 45. 1x 321x 121x 42, x 3, 1, 4 47. 1x 221x 321x 52, x 2, 3, 5 49. 1x 421x 121x 22 1x 32, x 4, 1, 2, 3 51. 1x 721x 221x 12 1x 32, x 7, 2, 1, 3 53. 12x 3212x 12 1x 12; x 32 , 12 , 1 55. 12x 32 2 1x 12; x 32 , 1 57. 1x 221x 1212x 52; x 2, 1, 52 59. 1x 1212x 12 1x 152 1x 152; x 1, 12 , 15, 15 61. 1x 2213x 22 1x 2i21x 2i2; x 2, 23 , 2i, 2i 63. x 1, 2, 3, 3 65. x 2, 1, 2 67. x 2, 3 2 3 2 ,4 69. x 3, 1, 53 71. x 1, 2, 3, i17 73. x 2, 23 , 1, i 13 75. x 1, 2, 4, 2 77. x 3, 4, 12 79. x 1, 32 , i 13 81. x 12 , 1, 2, i 13 83. a. possible roots: 51, 8, 2, 46; b. neither 1 nor 1 is a root; c. 3 or 1 positive roots, 1 negative root; d. roots must lie between 2 and 2 85. a. possible roots: 51, 26; b. 1 is a root; c. 2 or 0 positive roots, 3 or 1 negative roots; d. roots must lie between 3 and 2 87. a. possible roots: 51, 12, 2, 6, 3, 46; b. x 1 and x 1 are roots; c. 4, 2, or 0 positive roots, 1 negative root; d. roots must lie between 1 and 4 89. a. possible roots: 51, 20, 2, 10, 4, 5, 12 , 52 6; b. x 1 is a root; c. 1 positive root, 1 negative root; d. roots must lie between 2 and 1 91. 1x 4212x 32 12x 32; x 4, 32 , 32 93. 12x 1213x 22 1x 122; x 12 , 23 , 12 13 95. 1x 221x 122 14x2 32; x 2, 12, 13 2 i, 2 i 97. a. 5 b. 13 c. 2 99. yes 101. yes 103. a. 4 cm 4 cm 4 cm b. 5 cm 5 cm 5 cm 105. length 10 in., width 5 in., height 3 in.

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107. 1994, 1998, 2002, about 5 yr 109. a. 8.97 m, 11.29 m, 12.05 m, 12.94 m b. 9.7 m, 3.7 111. a. yes b. no c. about 14.88 113A. a. 1x 5i21x 5i2 b. 1x 3i21x 3i2 c. 1x i17 21x i 172 113B. a. x 17, 17 b. x 213, 2 13 c. x 312, 3 12 115. a. C1z2 1z 4i21z 321z 22 b. C1z2 1z 9i21z 421z 12 c. C1z2 1z 3i21z 1 2i21z 1 2i2 d. C1z2 1z i21z 2 5i21z 2 5i2 e. C1z2 1z 6i21z 1 i 1321z 1 i 132 f. C1z2 1z 4i21z 3 i 1221z 3 i 122 g. C1z2 1z 2 i21z 3i21z i2 h. C1z2 1z 2 3i21z 5i21z 2i2 117. a. w 150 ft, l 300 b. A 15,000 ft2 119. r 1x2 2 1x 4 2

73.

54321 8 16 24 32 40

1. zero, m 3. bounce, flatter 5. Answers will vary. 7. polynomial, degree 3 9. not a polynomial, sharp turns 11. polynomial, degree 2 13. up/down 15. down/down 17. down/up; 10, 22 19. down/down; 10, 62 21. up/down; 10, 62 23. a. even b. 3 odd, 1 even, 3 odd c. f 1x2 1x 32 1x 12 2 1x 32, deg 4 d. x 僆⺢, y 僆 39, q2 25. a. even b. 3 odd, 1 odd, 2 odd, 4 odd c. f 1x2 1x 321x 121x 221x 42, deg 4 d. x 僆⺢, y 僆 1q, 25 4 27. a. odd b. 1 even, 3 odd c. f 1x2 1x 12 2 1x 32, deg 3 d. x 僆⺢, y 僆⺢ 29. degree 6; up/up; 10, 122 31. degree 5; up/down; 10, 242 33. degree 6; up/up; 10, 1922 35. degree 5; up/down; 10, 22 37. b 39. e 41. c y y y 43. 45. 47. 10 10 10

108642 2 4 6 8 10

49.

10 8 6 4 2 108642 2 4 6 8 10

55.

10 8 6 4 2 108642 2 4 6 8 10

61.

10 8 6 4 2 108642 2 4 6 8 10

67.

10 8 6 4 2 54321 2 4 6 8 10

8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

y

51.

108642 2 4 6 8 10

2 4 6 8 10 x

y

57.

63.

1 2 3 4 5 x

20 16 12 8 4 54321 4 8 12 16 20

2 4 6 8 10 x

y

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

y

10 8 6 4 2

69.

200 160 120 80 40 54321 40 80 120 160 200

75.

y

1 2 3 4 5 x

30 24 18 12 6 54321 6 12 18 24 30

y

1 2 3 4 5 x

h1x2 1x 421x 232 1x 232 1x i 232 1x i232 f 1x2 21x 52 21x 2221x 2221x 2321x 232 P1x2 16 1x 421x 121x 32, P1x2 16 1x3 13x 122 a. 132 2 112 2 122 2 142 2 122 2 21132 9 1 4 16 4 26 30 30 ✓ b. 1x 321x 121x 22 1x 42 x4 2x3 13x2 14x 24 ✓ 85. a. 280 vehicles above average, 216 vehicles below average, 154 vehicles above average b. 6:00 A.M. 1t 02, 10:00 A.M. 1t 42 , 3:00 P.M. 1t 92 , 6:00 P.M. 1t 122 c. max: about 300 vehicles above average at 7:30 A.M.; 300 v(t) 240 min: about 220 vehicles below average at 12 noon 180 77. 79. 81. 83.


8 6 4 2

40 32 24 16 8

120 60 60 120 180 240 300

87. a. 3 89. a.

b. 5

2 4 6 8 10 12 14 16 t

c. B1x2 14 x1x 42 1x 92, $80,000 700

0

12

8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

y

53.

108642 4 8 12 16 20

2 4 6 8 10 x

y

59.

65.

1 2 3 4 5 x

20 16 12 8 4 54321 4 8 12 16 20

1 2 3 4 5 x

y

20 16 12 8 4 108642 4 8 12 16 20

2 4 6 8 10 x

y

20 16 12 8 4

71.

40 32 24 16 8 54321 8 16 24 32 40

100

2 4 6 8 10 x

, quartic;

y

2 4 6 8 10 x

y

2 4 6 8 10 x

b. t ⬇ 1.7 (7:42 A.M.), 227 vehicles; t ⬇ 9.9 (3:54 P.M.), 551 vehicles c. t ⬇ 7.93 (1:56 P.M.) and t ⬇ 11.27 (5:16 P.M.) 91. a. 150

y

12

0 1 2 3 4 5 x

20 y

1 2 3 4 5 x

, quartic;

cob19545_saa_018-038.qxd

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SA-25

Student Answer Appendix b. morning: t ⬇ 1.72 (11:43 A.M.); evening: t ⬇ 9.11, (7:07 P.M.) c. t ⬇ 4.98 (2:59 P.M.), about 33 customers d. t ⬇ 7.92 (5:55 P.M.) and t ⬇ 10.02 (8:01 P.M.) 93. a. f 1x2 S q, f 1x2 S q b. g1x2 S q, g1x2 S q; x4 0 for all x 95. c 18 97. verified 99. verified, x 1 2i 101. yes

31. (0, 0) cross, (3, 0) cross 33. 14, 02 cross, (0, 4) 35. (0, 0) cross, (3, 0) bounce 37.

10 8 6 4 2 108642 2 4 6 8 10

Mid Chapter Check, p. 428

1. a. x3 8x2 7x 14 1x2 6x 521x 22 4 x3 8x2 7x 14 4 b. x2 6x 5 x2 x2 2. f 1x2 12x 321x 12 1x 121x 22 3. f 122 7 4. f 1x2 x3 2x 4 5. g122 8 and g132 5 have opposite signs 6. f 1x2 1x 221x 121x 221x 42 7. x 2, x 1, x 1 3i y y 8. 9. 10 15

108642 3 6 9 12 15

108642 2 4 6 8 10

2 4 6 8 10 x

10 8 6 4 2 108642 2 4 6 8 10

49.

8 6 4 2

12 9 6 3

43.

10 8 6 4 2

2 4 6 8 10 x

10. a. degree 4; three turning points b. 2 sec c. A1t2 1t 12 2 1t 321t 52 , A1t2 t 4 10t3 32t2 38t 15 A122 3; altitude is 300 ft above hard-deck, A142 9; altitude is 900 ft below hard-deck

108642 2 4 6 8 10

55. f 1x2 59.

Exercise 1. 1.532

Exercise 2. 2.152, 1.765

Exercises 4.4, pp. 440–445 1. as x S q, y S 2 3. denominator, numerator 5. about x 98 7. x 3, x 僆 1q, 32 ´ 13, q2 9. x 3, x 3, x 僆 1q, 32 ´ 13, 32 ´ 13, q2 5 5 11. x 5 2 , x 1, x 僆 1q, 2 2 ´ 12 , 12 ´ 11, q 2 13. No V.A., x 僆 1q, q2

63. a.

8

5

51.

y

10 8 6 4 2

47.

y

53.

y

2 4 6 8 10 x

57. f 1x2

x2 4

v(x)

2 4 6 8 10 x

61.

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

108642 2 4 6 8 10

2 4 6 8 10 x

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

108642 2 4 6 8 10

2 4 6 8 10 x

41.

y

10 8 6 4 2

10 8 6 4 2 108642 2 4 6 8 10

y

2 4 6 8 10 x

y

2 4 6 8 10 x

y

2 4 6 8 10 x

9 x2 12 10 8 6 4 2

108642 2 4 6 8

g(x)

2 4 6 8 10 x

Population density approaches zero far from town. b. 10 mi, 20 mi c. 4.5 mi, 704 people per square mi

Population (100s)

6 4 2

15. x 3, yes; x 2, yes

17. x 3, no 19. x 2, yes; x 2, no 21. y 0, crosses at 1 32 , 02 23. y 4, crosses at 121 25. y 3, does not cross 4 , 42 27. q1x2 0, r 1x2 8x directly; the graph will cross the horizontal asymptote at x 0.

45.

y

1x 22 1x 32

108642 2 4 6 8 10

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

1x 42 1x 12

10 8 6 4 2

Reinforcing Basic Concepts, p. 429

39.

y

5

15 25 35 45

Distance (mi)

65. a. 5 hr; about 0.28 b. 0.019, 0.005; As the number of hours increases, the rate of change decreases. c. h S q, C S 0; horizontal asymptote 67. a. $20,000, $80,000, $320,000; cost increases dramatically b. c. as p S 100 , C S q Cost ($1000) 900 700 500 300 100

7

7

Percent 10 30 50 70 90

69. a. 2; 10 50

W(t) (1, 46)

40

b. 10; 20

c. On average, 6 words will be remembered for life.

30

5

29. q1x2 2, r 1x2 8x 8; the graph will cross the horizontal asymptote at x 1. 10

20 10 10 20 30 40 50 t

71. a. b. 35%; 62.5%; 160 gal c. 160 gal; 200 gal d. 70%; 75%

y 0.9 0.7

10

10

0.5 0.3 0.1 70 140 210 280 350 x

10

73. a. $225; $175 b. 2000 heaters c. 4000 heaters d. The horizontal asymptote at y 125 means the average cost approaches $125 as monthly production gets very large. Due to limitations on production (maximum of 5000 heaters) the average cost will never fall below A150002 135.

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75. a. 5 b. 18 c. The horizontal asymptote at y 95 means her average grade will approach 95 as the number of tests taken increases; no d. 6 77. a. 16.0, 28.7, 65.8, 277.8 b. 12.7, 37.1, 212.0 c. a. 22.4, 40.2, 92.1, 388.9 b. 17.8, 51.9, 296.8; answers will vary. 4 1 3 79. y 81. 16 x 3 , 4 ; 13x 16214x 32 0 3 3

31.


37.

1. slant 3. two

108642 2 4 6 8 10

10 8 6 4 2

x 2

x2 2x 3 9. G1x2 • x 1 4

108642 2 (2, 4) 4 6 8 10

10 8 6 4 2

x 1 x 1

3x 2x 2x 3 11. H1x2 μ 3 2

2 4 6 8 10 x

43.

y

yx

(4, 0)

x3 8 P1x2 • x2 13. 12

10 8 6 4 2

x

108642 2 4 6 8 10

x2

10 8 6 4 (2, 0) 2

(4, 0)

冢w, w冣

41. 10

x1 8 6 4 (2, 0) 2

108642 2 4 6 8 10

(0, 0.2)

y x3

yx1

108642 2 4 6 8 10

47.

2 4 6 8 10 x

y

yx x1 (2, 0)

2 4 6 8 10 x

(0, 0)

y

8 7 6 (0, 4) 5 4 3 2 y x2 1 1

yx (1, 0) 2 4 6 8 10 x

54321 1 2

1 2 3 4 5 x

51. 119.1

16 12 8 4

6 4 2 4 8 12 16

2 4 6 8 10 x

x3

y

(0, d)

108642 2 4 6 8 10

x1

y

(3.2, 0)

(2, 0)

y

45.

y

49.

y

10 8 6 4 (2, 0) 2

x3

108642 2 2 4 6 8 10 x 4 (0, 4) 6 8 10

2 4 6 8 10 x

10 8 6 4 2

2

35.

10 8 yx1 6 4 (2, 0) 2 (2, 0)

108642 2 2 4 6 8 10 x 4 6 (0, 0) 8 10

108642 2 2 4 6 8 10 x 4 (1, 4) 6 8 10

3 2 3 x 2

39.

y

108642 2 4 6 8 10

y 14 x

2

(4.8, 0)

6 4 (0, 1) 2 yx1

y

y yx1

108642 2 2 4 6 8 10 x 4 (1, 0) 6 8 10

2 4 6 8 10 x

10

x 2

10 8 6 4 (2, 0) 2

x 1 8


x2 4 7. F1x2 • x 2 4

33.

y

10 8 yx5 6 4 (0.8, 0) 2 (1, 0)

(0, 2) (3.2, 0) 2

4

6 x

53. a. a 5, y 3a 15 b. 60.5

c. 10

4x2 53x 250 ; x 0, y 4x 53 55. a. A1x2 x b. cost: $307, $372, $445; Avg. cost: $307, $186, $148.33

y 16

x2

12

c. 8, $116.25 d.

(2, 12)

8

300

4 8

x3 7x 6 15. Q1x2 • x 1 4

19.

10 8 6 4 (2, 0) 2

x 1

108642 2 4 6 8 10

25. 10 8 6 4 (1, 0) 2 108642 2 4 6 8 10

x2 2x 3 2 2

x 3 x1

10 8 6 4 (兹3, 0) 2

yx (2, 0) 2 4 6 8 10 x

y yx2

27. 10 8 6 4 2

108642 2 4 6 8 10

8

x

y

0

0

(1, 2)

108642 2 4 6 8 10

2 4 6 8 10 x

10 8 6 4 2

(兹3, 0)

108642 2 4 6 8 10

29.

y yx3 (1, 0) 2 4 6 8 10 x

10 8 6 4 2

(2, 0)

108642 2 4 6 8 10

2x3 48 x x c. S(x) is asymptotic to y 2x2. d. x ⬇ 2 ft 3.5 in.; y ⬇ 2 ft 3.5 in. 2x 55 59. a. A1x, y2 xy; R1x, y2 1x 2.52 1y 22 b. y ; x 2.5 2 2x 55x A1x2 c. A(x) is asymptotic to y 2x 60 x 2.5 V 2V d. x ⬇ 11.16 in.; y ⬇ 8.93 in. 61. a. h b. S 2␲r2 r ␲r2

57. a. S1x, y2 2x 4xy; V1x, y2 x2y b. y 2

y

10 8 6 (3, 2) 4 2

23.

y

20

1 2 3 4 5 x

108642 2 2 4 6 8 10 x 4 6 8 y x 10

(2, 0) 2 4 6 8 10 x

54321 2 (1, 4) 4 6 8 10

4

x 3, x 1

21.

y

10 8 6 4 2

x 1

x3 3x2 x 3 17. R1x2 μ

4 2

y yx

2 4 6 8 10 x

c. S

2␲r3 2V r

65. S

67. y 34 x 4, m 34 , 10, 42

yx (1, 0)

2

; S1x2

d. r ⬇ 5.76 cm, h ⬇ 11.51 cm; S ⬇ 625.13 cm2

63. Answers will vary. y

12

␲r3 2V ; r ⬇ 3.1 in., h ⬇ 3 in. r

2 69. a. P 30 cm, b. CD 60 13 cm, c. 30 cm , 4320 2 2 d. A 750 169 cm , and A 169 cm

2 4 6 8 10 x

Exercises 4.6, pp. 465–469 1. vertical, multiplicity 3. empty 5. Answers will vary. 7. x 僆 13, 52 9. x 僆 34, q 2 ´ 516 11. x 僆 1q, 2 4 ´ 526 ´ 34, q 2 13. x 僆 12 13, 2 132

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SA-27

Student Answer Appendix 15. x 僆 1q, 3 4 ´ 516 17. x 僆 13, 12 ´ 12, q 2 19. x 僆 1q, 32 ´ 11, 12 ´ 13, q2 21. x 僆 1q, 22 ´ 12, 12 ´ 13, q2 23. x 僆 31, 1 4 ´ 536 25. x 僆 1q, 22 ´ 12, 32 27. x 僆 1q, 2 4 ´ 11, 12 ´ 33, q 2 29. x 僆 3 3, 22 31. x 僆 1q, 22 ´ 12, 12 33. x 僆 1q, 22 ´ 3 2, 32 35. x 僆 1q, 52 ´ 10, 12 ´ 12, q2 37. x 僆 14, 2 4 ´ 11, 2 4 ´ 13, q 2 39. x 僆 17, 32 ´ 12, q2 41. x 僆 1q, 2 4 ´ 10, 22 43. x 僆 1q, 172 ´ 12, 12 ´ 17, q 2 45. x 僆 13, 7 47. x 僆 12, q2 49. x 僆 11, q 2 4 4 ´ 12, q2 51. 1q, 32 ´ 13, q2 53. x 僆 1q, 3 4 ´ 35, q 2 55. x 僆 3 3, 0 4 ´ 33, q2 57. d 59. b 61. a. verified 1 b. D 41p 34 21p 32 2; p 3, q 2; p 3 4 ,q 4

c. 1q, 32 ´ 13, 3 d. verified 4 2 63. d1x2 k1x3 192x 10242 a. x 僆 15, 8 4 b. 320k units c. x 僆 3 0, 32 d. 2 ft 65. a. verified b. horizontal: r2 20, as r1 increases, r2 decreases to maintain R 40; vertical: r1 20, as r1 decreases, r2 increases to maintain R 40 c. r1 僆 120, 402 67. R1t2 0.01t 2 0.1t 30 a. 3 0°, 30°2 b. 120°, q 2 c. 150°, q2 69. a. n 4 b. n 9 c. 13 x2 71. a. yes, x2 0 b. yes, 2 0 x 1 x1x 22 73. x1x 221x 12 2 7 0; 7 0 1x 12 2 75. R1x2 6 0 for x 僆 12, 82 ´ 18, 142 77.

3 2 1 54321 1 2 3 4 5 6 7

F1x2 e

y

f 1x2 6

x 4 x 4

19. degree 5; up/down; 10, 42 y 21. 22. 10

20. degree 4; up/up; 10, 82 y 23. 20

8 6 4 2 54321 2 4 6 8 10

16 12 8 4 54321 4 8 12 16 20

1 2 3 4 5 x

8 6 4 2

9. d

11. a

13. g

108642 2 4 6 8 10

29. V1x2

108642 2 4 6 8 10

2 4 6 8 10 x

2 4 6 8 10 x

x2 x 12

; V102 2 x2 x 6 30. a. y 15; as 冟x冟 S q A1x2 S 15. As production increases, average cost decreases and approaches 15. b. x 7 2000 31. removable discontinuity at 12, 52 ;

79. x 僆 3 3, 9 4

1. q1x2 x2 6x 7; r 8 2. q1x2 x 1; r 3x 4 3. 7 2 13 6 9 14 14 7 7 14 2 1 1 2 0 Since r 0, 7 is a root and x 7 is a factor. 4. x3 4x 5 1x 221x2 2x2 5 5. 1x 421x 121x 32 6. h1x2 1x 121x 421x2 2x 22 7. 12 4 8 3 1 2 5 1 4 10 2 0 Since r 0, 12 is a root and 1x 12 2 is a factor. 8. 3i 1 2 9 18 3i 9 6i 18 1 2 3i 6i 0 Since r 0, 3i is a zero 9. 7 1 9 13 10 7 14 7 1 2 1 3 h172 3 10. P1x2 x3 x2 5x 5 11. C1x2 x4 2x3 5x2 8x 4 12. a. C102 350 customers b. more at 2 P.M., 170 c. busier at 1 P.M., 760 7 710 13. 51, 10, 2, 5, 12 , 52 , 14 , 54 6 14. x 12 , 2, 52 15. P1x2 12x 321x 421x 12 16. only possibilities are 1, 3, none give a remainder of zero 17. [1, 2], [4, 5]; verified 18. one sign change for g1x2 S 1 positive zero; three sign changes for g1x2 S 3 or 1 negative zeroes; 1 positive, 3 negative, 0 complex, or 1 positive, 1 negative, 2 complex; 1 positive, 1 negative, 2 complex, verified

y

10 8 6 4 2 108642 2 4 6 8 10

33.

20 16 12 8 4

x 1

108642 4 8 12 16 20

35. a.

34.

y

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

2 4 6 8 10 x

10 8 6 4 2

x 1

108642 2 4 6 8 10

15. c


1 2 3 4 5 x

8 6 4 2

x2 3x 4 32. H1x2 • x1 5 7. h

y

24. a. even b. x 2, odd; x 1, even; x 1, odd c. deg 6: P1x2 1x 221x 12 2 1x 12 3 25. a. 5x冟x 僆⺢; x 1, 46 b. HA: y 1; VA: x 1, x 4 c. V102 94 (y-intercept); x 3, 3 (x-intercepts) d. V112 43 26. No—even multiplicity; yes—odd multiplicity y y 27. 28. 10 10

1 2 3 4 5x

5. a

54321 3 6 9 12 15

1 2 3 4 5 x

Making Connections, p. 470 1. e 3. b

15 12 9 6 3

y

2 4 6 8 10 x

y

2 4 6 8 10 x

y 10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

b. about 2450 favors c. about $2.90 ea. 36. factored form 1x 421x 121x 22 7 0 Neg

Pos

4

37.

0

1

Pos 2

1x 52 1x 22 x2 3x 10 0 x2 x2

Neg

Pos 2

38.

Neg

0

1x 221x 12 x1x 22

Pos

Neg 2

1

Pos

outputs are positive or zero for x 僆 32, 22 ´ 35, q 2

5

0

Neg 2

outputs are positive for x 僆 14, 12 ´ 12, q 2

Pos 0

Neg 1

Pos 2

outputs are negative or zero for x 僆 32, 02 ´ 31, 22

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19.

1. a. f 1x2 1x 52 2 9 b. g1x2 12 1x 42 2 8 y

20 16 12 8 4

y

18 16 (0, 16) 14 12 10 8 6 4 2

(5, 9) (8, 0) (4, 8)

42 4 2 4 6 8 10 12 14 16 x 8 (2, 0) 12 16 (0, 16) 20

108642

54321 2 4 6 8 10

2 4 6 8 10 x

2. 12, 02, y 2x2 4x 3. a. 40 ft, 48 ft b. 49 ft c. 14 sec 2 14x 3 4. x 5 2 5. x2 2x 9 x 2 x 2x 1 0 15 10 24 6. 3 1 3 9 18 24 1 3 6 8 0 r0✓ 7. 1 8. P1x2 x3 2x2 9x 18 9. Q1x2 1x 22 2 1x 12 2 1x 12, 2 mult 2, 1 mult 2, 1 mult 1 10. a. 1, 18, 2, 9, 3, 6 b. 1 positive zero, 3 or 1 negative zeroes; 2 or 0 complex zeroes c. C1x2 1x 221x 12 1x 3i21x 3i2 11. a. 2002, 2004, 2008 b. 4 yr c. deficit of $7.5 million y y y 12. 13. 14. 20 10 500 16 12 8 4

300 100 108642 100

108642 4 8 12 16 20

2 4 6 8 10 x

300 500

8 6 4 2

40 30 20 10 108642 10 20 30 40 50

108642 2 4 6 8 10

2 4 6 8 10 x

15. a. removal of 100% of the contaminants dramatic increase c. 88% y y 16. a. b. 50 10

2 4 6 8 10 x

b. $500,000; $3,000,000;

108642 2 4 6 8 10

d. 0 i 25. a. Y6


1. second, one 3. 111, 22, 15, 02, 11, 22, 119, 42 5. False, answers will vary. 7. one-to-one 9. one-to-one 11. not one-to-one, fails horizontal line test: x 3, x 0.5 and x 2 are paired with y 0 13. not a function 15. one-to-one 17. not oneto-one, y 1 is paired with x 6 and x 8 19. one-to-one 21. not one-to-one; h1x2 6 3, corresponds to two x-values 23. one-to-one 25. not one-to-one; y 3 corresponds to more than one x-value 27. f 1 1x2 5 11, 22, 14, 12, 15, 02, 19, 22, 115, 526 29. v1 1x2 5 13, 42, 12, 32, 11, 02, 10, 52, 11, 122, 12, 212, 13, 3226 5 x3 x 35. f 1 1x2 31. f 1 1x2 x 5 33. p1 1x2 4 4 37. t1x2 x3 4 39. x 僆⺢, y 僆⺢; f 1 1x2 x3 2, x 僆⺢, y 僆⺢; verified 3 41. x 僆⺢, y 僆⺢; f 1 1x2 2x 1, x 僆⺢, y 僆⺢; verified

b. x 僆 1q, 42 ´ 10, 22

51. 53. 57. 61.

3 b. h 1 55; no c. 28.6%, 29.6% d. ⬇11.7 hr e. 4 hr, 43.7% f. The amount of the chemical in the bloodstream becomes neglible. x2 x 6 20. V1x2 2 ; V102 2 x 2x 3

67. 73. 75.

Strengthening Core Skills, pp. 476–477

Exercise 1. x 僆 1q, 3 4 Exercise 2. x 僆 12, 12 ´ 12, q2 Exercise 3. x 僆 1q, 42 ´ 11, 32 Exercise 4. x 僆 32, q 2 Exercise 5. x 僆 1q, 22 ´ 12, q 2 Exercise 6. x 僆 33, 1 4 ´ 3 3, q 2

77. 79.

3. a. 1x 121x x 12

7. verified 9. y

increases 11 min every 60 days 15.

5

5

17. X 63

y

5 x

5

13. f

1

1x2

x 1, y 1; verified

1x2 1x 5, x 0, y 5 8 3, x 7 0, y 7 3 a. x 7 3, y 7 0 b. v 1x2 Ax 1 a. x 4, y 2 b. p 1x2 1x 2 4, x 2, y 4 1 f ⴰ g2 1x2 x, 1g ⴰ f 21x2 x 55. 1 f ⴰ g21x2 x, 1g ⴰ f 21x2 x 1 f ⴰ g2 1x2 x, 1g ⴰ f 21x2 x 59. 1 f ⴰ g21x2 x, 1g ⴰ f 21x2 x x5 f 1 1x2 63. f 1 1x2 2x 5 65. f 1 1x2 2x 6 3 x3 1 3 3 f 1 1x2 2 x 3 69. f 1 1x2 71. f 1 1x2 2 2x 1 2 2 x2 2 2 , D: x 0, R: y D: x , R: y 0; f 1 1x2 3 3 3 x2 3, D: x 0, R: y 3 D: x 3, R: y 0; p1 1x2 4 1 D: x 0, R: y 3; v 1x2 2x 3, D: x 3, R: y 0 y D: x 僆 3 0, q 2 , R: y 僆 3 2, q2 ; 5 4 D: x 僆 32, q 2 , R: y 僆 30, q 2 3 1

1

54321 1 2 3 4 5

b. 1x 321x 22 1x 22

1009 11 ; 39 min, driving time x 60 60

11. month 9

x x,

2 1

Cumulative Review Chapters R–4 pp. 477–478

5. all reals

b. Y7

CHAPTER 5

8 12 16 20x

2

59 2 i c. 27.63 14.59i 41 41 c. Y4 d. Y3 e. Y5 f. Y2

23. a. 349.36 131.38i b.

49.

0.4

R1R2 1. R R1 R2

1 2 3 4 5 x

47. a. x 5, y 0 b. f

2 4 6 8 10 x

0.8

4

21. 87.91, 80.09, 1.99

45. x 1, y 1; f 1 1x2 1

17. 800 18. a. x 僆 1q, 3 4 ´ 3 1, 4 4 19. a. 1.2 y

4 0

y

43. x 2, y 0; f 1 1x2 8x 2, x 0, y 2; verified

8 6 4 2 2 4 6 8 10 x

10 8 6 4 2

x3 3 2

81.

5 4 3 2 1 54321 1 2 3 4 5

1 2 3 4 5 x

y

1 2 3 4 5 x

D: x 僆 10, q 2 , R: y 僆 1q, q2 ; D: x 僆 1q, q 2 , R: y 僆 10, q2

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5 4 3 2 1 54321 1 2 3 4 5

D: x 僆 1q, 44 , R: y 僆 1q, 4 4 ; D: x 僆 1q, 44 , R: y 僆 1q, 4 4

y

(2, 9) 108

y0

(2, 10) 10

108642 2 4 6 8 10

108642 2 4 6 8 10

25. e

27. a

2 4 6 8 10 x

29. b

y

7.6

10 8 (0, 9) 6 4 2 (2, 1) 108642 2 4 6 8 10

2 4 6 8 10 x

31. 2.718282 33. 7.389056 y 37. 39. 10

35. 4.113250 4 3 2 1

8

5

(1, 5.4) 6 4

87. a. h1 1x2 1 x x b. (0, 0), 11, 12 2 , and 12, 23 2 are on the graph of h; (0, 0), 1 12 , 12 , and 1 23 , 22 are on the graph of h1 c. verified d. 3.1

4.7

y

8 6 4 2 (0, 2)

y1

2 4 6 8 10 x

23. y 1 13 2 x; right 2

y0

7.6

y

6 4 2 (0, 1)

1 2 3 4 5 x

85. a. f 1 1x2 x 2 1 b. 13, 52 , (0, 1), and (1, 3) are on the graph of f; 15, 32 , (1, 0), and (3, 1) are on the graph of f 1 c. verified d. 5

21. y 1 13 2 x; up 1

19. y 3x; reflect across y-axis

(2.3, 0) 2 108642 2 (3, 1) 4 6 8 10

654321 1 2 3 4 5 6

2 4 6 8 10 x

41.

y

y

10 8 6 (0, 6.4) 4 2 (2, 0)

(0, 1) 1 2 3 4 x

108642 2 4 6 8 10

(2, 5.4)

2 4 6 8 10 x

43. 3 45. 32 47. 13 49. 4 51. 3 53. 3 55. 2 57. 2 59. 2 61. 3 63. x ⬇ 2.8 65. x ⬇ 3.2 67. a. 1732, 3000, 5196, 9000 b. yes c. as t S q, P S q d. 50,000 P

4.7

40,000 30,000 20,000 10,000

3.1

1

89. a. 31.5 cm b. The result is 80 cm. It gives the distance of the projector from the screen. 91. a. 63.5°F b. f 1 1x2 2 7 1x 592; independent: temperature, dependent: altitude c. 22,000 ft 1x 93. a. 144 ft b. f 1 1x2 , independent: distance fallen, dependent: 4 time fallen c. 7 sec 3x 95. a. 28,260 ft3 b. f 1 1x2 3 , independent: volume, dependent: B␲ height c. 9 ft 97. Answers will vary. 99. a. P 2l 2w b. A ␲r2 c. V ␲r2h d. V 13 ␲r2h e. C 2␲r f. A 12 bh g. A 12 1b1 b2 2h h. V 43 ␲r3 i. a2 b2 c2 101. ⬇0.472, ⬇0.365; rate of change is greater in [1, 2] due to shape of the graph.

Exercises 5.2, pp. 499–503 1. bx, b, b, x 3. a, 1 5. False; for 冟b冟 6 1 and x2 7 x1, bx2 6 bx1, so 1 1 function is decreasing 7. 16, 2, 8, 11.036 9. 1, 64 , 4 , 64 11. 13.

y0

y

10 8 6 4 2

10

(0, 1)

108642 2 4 6 8 10

2 4 6 8 10 x

y0

increasing

15. y 3x; up 2 10 8 6 (1, 5) 4 (0, 3) 2

108642 2 4 6 8 10

6 4 2 (0, 1) 108642 2 4 6 8 10

2 4 6 8 10 x

decreasing

2 4 6 8 10 x

(1, 9)10

y0

(3, 1)

y

8 6 4 2

108642 2 4 6 8 10

3 4 days

2 4 6 8 10 x

5t

a. $100,000 b. 3 yr 71. a. ⬇ $86,806 b. 3 yr a. $40 million b. 7 yr 75. no, they will have to wait about 5 min 32% transparent 79. 17% transparent 81. ⬇$25,526 a. 8 g b. 48 min 85. 51 87. 75 89. 9.5 10 7; answers will vary 7 91. 5; ; 2a2 3a; 2a2 4ah 2h2 3a 3h 9 93. a. no solution b. 55, 66 69. 73. 77. 83.

Exercises 5.3, pp. 512–516 1. logb x, b, b, greater 1

3. 11, 02 , 0

5. 5; answers will vary 7. 23 8

1 3

9. 7 11. 9 1 13. 8 2 15. 21 2 17. 72 49 2 19. 10 100 21. e4 ⬇ 54.598 23. log464 3 25. log319 2 1 7

0

27. 0 ln 1

1 33. log100 2 1 1 3 3 35. log48 2 37. log48 2 39. 1 41. 2 43. 1 45. 47. 2 2 49. 2 51. 1.6990 53. 0.4700 55. 5.4161 57. 0.7841 59. shift up 3 61. shift right 2, up 3 10 8 6 4 2 108642 2 4 6 8 10

29. log13 27 3

y

(1, 3) 2 4 6 8 10 x

63. shift left 1

17. y 3x; left 3 y

y2

y

(2, 9) 8

(2, 9)

2

10 8 6 4 2

31. log 1000 3

10 8 6 4 2 108642 2 4 6 8 10

y

(3, 3) 2 4 6 8 10 x

65. shift left 1, reflect across x-axis y

108642 2 2 4 6 8 10 x 4 6 (0, 0) 8 10

y

10 8 6 4 2 (0, 0) 108642 2 2 4 6 8 10 x 4 6 8 10

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67. II 69. VI 71. V 73. x 僆 1q, 12 ´ 13, q 2 75. x 僆 1 32 , q 2 77. x 僆 13, 32 79. pH ⬇ 4.1; acid 81. a. ⬇4.7 b. ⬇7.9 83. about 398 times 85. about 3.2 times 87. a. ⬇2.4 b. ⬇1.2 89. a. 20 dB b. 120 dB 91. about 501 times 93. about 3162 times 95. 6194 m 97. a. about 5434 m b. 4000 m 99. a. 2225 items b. 2732 items c. $117,000 101. a. about 58.6 cfm b. about 1605 ft2 103. a. 95% b. 67% c. 39% 105. ⬇4.3; acid 107. Answers will vary. a. 0 dB b. 90 dB c. 15 dB d. 120 dB e. 100 f. 140 dB 109. a. 2 b. 3 c. 5 3 2 2 y 111. D: x 僆⺢, R: y 僆⺢ 10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

113. x 僆 1q, 52; f 1x2 1x 521x 42 2 x3 3x2 24x 80

Exercises 5.4, pp. 523–526 1. e 3. extraneous 5. Answers will vary; Yes, 1.5663025 1.5663025 7. x ⬇ 29.964 9. x ⬇ 1.778 11. x ⬇ 2.200 13. x ⬇ 1.260 65 15. x ln 2, x ⬇ 4.7881 17. x log1782 5, x ⬇ 3.1079 4 8 ln 2.32 19. x , x ⬇ 1.1221 21. x e3 4, x ⬇ 10.3919 0.75 e0.4 5 23. x 5 101.25, x ⬇ 12.7828 25. x , x ⬇ 1.7541 2 27. ln12x2 14x2 35. lna

29. log1x2 12

x5 b 37. ln1x 22 x

31. log34

43. 1x 22 log 8 45. 12x 12 ln 5 47. 51. 3 log a log b

53. ln x 14 ln y

x b x1

41. log5 1x 22

39. log242 1 2

33. log a

log 22 49. 4 log5 3

55. 2 ln x ln y

ln 60 ; 2.104076884 57. 12 3 log1x 22 log x 4 59. ln 7 log 1.73205 ln 152 ; 3.121512475 63. ; 0.499999576 61. ln 5 log 3 65.

log 0.125

;3

log 0.5 log1x2 ; f 152 ⬇ 1.4650; f 1152 ⬇ 2.4650; f 1452 ⬇ 3.4650; 67. f 1x2 log132 outputs increase by 1; f 133 # 52 ⬇ 4.4650 log1x2 ; h122 ⬇ 0.3155; h142 ⬇ 0.6309; h182 ⬇ 0.9464; 69. h1x2 log192 outputs are multiples of 0.3155; h124 2 ⬇ 410.31552 ⬇ 1.2619 71. verified 73. a. N AXm b. ⬇ 3500 people 75. no, pH ⬇ 7.32 77. no, pH ⬇ 5.9 and the soil must be treated further 79. 600601 81. zeroes at x 3 and x 2; HA: y 1, VA x 1, x 1 83. x 1 or x 9

Mid-Chapter Check, p. 526 5

1. a. 23 log279 b. 54 log81243 2. a. 83 32 b. 12960.25 6 3. a. x 5 b. b 54 4. a. x 3 b. b 5 5. a. $71,191.41 b. 6 yr 6. F1x2 4 # 5x3 2 7. f 1 1x2 1x 12 2 3, D: x 僆 3 1, q2 ; R: y 僆 3 3, q 2 ; verified 8. a. 4 log381, verified 2 b. 4 ⬇ ln 54.598, verified 9. a. 273 9, verified b. e1.4 ⬇ 4.0552, verified 10. ⬇7.9 times more intense

Reinforcing Basic Concepts, p. 527 Exercise 1. Answers will vary. b. ln1x2 42 c. logx x 3

Exercise 2. a. log1x2 3x2

Exercises 5.5, pp. 535–538 1. variable, constant 3. uniqueness, one, one 5. False; answers will vary. 7. x 32 9. x 6.4 11. x 20, 5 is extraneous 13. x 2, 52 is extraneous 15. x 0 17. x 52 19. x 23 e2 63 21. x 32 23. x 19 25. x ; x ⬇ 6.1790 9 9 3 27. x 2; 9 is extraneous 29. x 3e 12 ; x ⬇ 59.7566 31. no solution 33. x 2 13; 2 13 is extraneous ln 128,967 ln 231 35. x ; x ⬇ 2.7968 37. x ; x ⬇ 2.4371 ln 7 3 ln 5 ln 2 39. x ; x ⬇ 1.7095 41. x ⬇ 4.815, x ⬇ 102.084 ln 3 ln 2 43. x ⬇ 2.013, x ⬇ 3.608 45. x ⬇ 46.210 C p 1 b ln a a 47. t , t ⬇ 55.45 49. about 3.2 cmHg k 51. a. 30 fish b. about 37 months 53. about 50.2 min 55. $15,641 57. a. 6 hr b. 18.0% 59. Mf 52.76 tons 61. a. 26 planes b. 9 days 63. x 1.609438 65. a. b. y 2 ln1x 32 y 2x1 x 2y1 x 2 ln1 y 32 x ln x 1y 12 ln 2 ln1y 32 2 x ln x y1 e2 y 3 ln 2 x ln x 1y y e2 3 ln 2 x 67. a. y ex ln 2 eln 2 2x; x y 2 1 ln y x ln 2, eln y ex ln 2 1 y ex ln 2 b. y bx, ln y x ln b, eln y ex ln b, y exr for r ln b 69. a. d b. e c. b d. f e. a f. c 71. a. x 僆 3 32 , q 2, y 僆 3 0, q 2 b. x 僆 1q, q 2, y 僆 3 3, q2 73. 13.5 tons

Exercises 5.6, pp. 547–552 1. Compound 3. Q0ert 5. Answers will vary. 7. $4896 9. 250% 11. $2152.47 13. 5.25 yr 15. 80% 17. 4 yr 19. 16 yr 21. $7561.33 23. about 5 yr 25. 7.5 yr 27. no 29. a. no b. 9.12% 31. 7.9 yr 33. 7.5 yr 35. a. no b. 9.4% 37. a. no b. approx 13,609 euros 39. No; $234,612.02 41. about 7 yr 43. 22 yr 45. a. no b. $298.31 Ap nt A A 1b 47. a. t b. p 49. a. r n a pr 1 rt Bp Q1t2 A b ln a b ln a p Q0 Q1t2 b. t 51. a. Q0 rt b. t r r e n ln a1 b n 53. $709.74 55. a. 5.78% b. 91.67 hr 57. 0.65 g 59. about 816 yr 61. about 12.4% 63. $17,027,502.21 65. 7.2% 3 67. a. f 1x2 x3, f 1x2 x, f 1x2 1x, f 1x2 1 x, f 1x2 1x 1 b. f 1x2 冟x冟, f 1x2 x2, f 1x2 2 c. f 1x2 x, f 1x2 x3, f 1x2 1x, x 1 1 3 f 1x2 1 x d. f 1x2 , f 1x2 2 x x 69. P1x2 x4 4x3 6x2 4x 15

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1. scatterplot, context, situation 3. beyond 5. (1) clear out old data, (2) enter new data, (3) display the data, (4) calculate the regression equation, (5) display and use the results; Answers will vary. 7. e 9. a 11. d 13. linear 15. exponential 17. logistic 19. exponential 21. As time increases, the amount of radioactive 0.9 material decreases but will never truly reach 0 or 0.7 become negative. Exponential with b 6 1 and k 7 0 0.5 x is the best choice. y ⬇ 11.04220.5626 0.3

Year (1970 → 0)

49. exponential, y ⬇ 103.83 11.05952 y a. 220 b. The 22nd note, or F# 210 c. frequency doubles, yes 190

2

3

4

5

Grams

170 150 130 110

900

2

4

6

8

10 12 x

Note number

700

51. exponential, y ⬇ 8.02 11.05642 x $41.59/mo, $54.72/mo

500 300

y 44

Monthly charge

Sales

23. Sales will increase rapidly, then level off as the market is saturated with ads and advertising becomes less effective, possibly modeled by a logarithmic function. y ⬇ 120.4938 217.2705 ln1x2

6

Frequency

x

Time (hours)

1

100 5 15 25 35 Cost

c. y ⬇

b. about 1750

y 2000 1750

1 10.2e

Cumulative cases

28 20 12

4

1000 750

20

28

y

9

50

70

90 x

Profit

3

250 30

logistic

55. logistic, y ⬇

c. 34.8 lb

1 32.280e0.336x

15

;

y 150

Logistic

125 100 75 50 25 6

10

14

18

Year (1990 S 0)

x

Time (days)

39. logarithmic, y ⬇ 78.8 10.3 ln x y a. 51,000 b. 1977

7

9

11

x

15

175 Exponential

2

5 90

5

Month

222.133

25

70

3

27

35

50

1

21

45

30

36 x

6 3

500

27. 4.95 29. 6.25 31. 5.75 33. 6.84 35. a. about 19 boards b. about 15 days 37. logarithmic, y ⬇ 27.4 13.5 ln x y a. 9.2 lb b. 29 days

10

12

Year (1980 S 0)

53. quadratic, y ⬇ 1.18x2 10.99x 4.60; month 8

1250

Days after outbreak

Offices (1000s)

36

4

0.11x

1500

10

Weight (pounds)

1719

Subscriptions (millions)

25. a.

y 50 45 40 35 30 25 20 15 10 5 5 10 15 20 25 30 35 40 x

0.1

0

Percent of U.S. Population

47. linear, P1t2 ⬇ 0.51t 22.51, 2005: 40.4%, 2010: 43.0%, 2015: 45.5%


57.

x

about 55 million, about 184 million, about 214 million; 2014

power regression, a. y ⬇ x0.665, 9.5 AU; b. 84.8 yr

7.5

c. 29,900

90 70 50

0

30

15

x

10 0

20

40

60

80 100

Year (1900 S 0)

c. 2013

0

450

59.

350

4000

150 50 6

10

14

18

Year (1980 S 0)

x

3000 2000 1000

43. quadratic y ⬇ 0.576x2 8.879x 394 y a. 360 million b. about 513 million 480 c. from 1984 to 1990

20

40

60

80

100

x

Predators

61. a.

linear, W ⬇ 1.24L 15.83, 32.5 lb, 35.3 in.

W 25

440

Weight (lb)

Chicken production (millions)

5000

250

2

400 360 320 2

6

10

14

18

Year (1980 S 0)

x

250 150 50 5

7

10

9

Year (1990 S 0)

11

x

20

24

28

32

Length (in.)

b. 329.2 million

c. 2010

61. b.

L

logarithmic, C1a2 ⬇ 37.9694 3.4229 ln 1a2, about 49.3 cm, about 34 mo

C(a) Circumference (cm)

350

3

15

16

450

1

20

5

45. linear, y ⬇ 6.555x 165.308 y a. 224.3 million Debit cards (millions)

a. power regression, y ⬇ 58,555.89x1.056; b. about 295 rodents c. about 17 predators

y

Rodents

Number of farms (1000s)

41. exponential, y ⬇ 346.7910.942 x y a. 155,100 b. 54,200

52 48 44 40 36 5

10

15

20

Age (months)

25

a

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63. D: x 僆 1q, 22 ´ 12, 12 ´ 11, 52 ´ 15, q2, x 65. max: 10.4, 1.82 min: 13.5, 3.52, 12.3, 1.42

51.

1 2

f 1x2c: x 僆 13.5, 0.42 ´ 12.3, q 2

f 1x2T: x 僆 1q, 3.52 ´ 10.4, 2.32

20

0

35

Making Connections, p. 565 1. a

3. e

5. c

7. e

9. b

11. g

13. c

15. d 8

Summary and Concept Review, pp. 566–570 1. no

2. no

3. yes

4. f

1

x2 1x2 3

5. f

1

1x2 1 x 2

6. f 1x2 x 1; x 0 7. f 1x2: D: x 僆 3 4, q 2, R: y 僆 3 0, q2; f 1 1x2: D: x 僆 30, q 2, R: y 僆 34, q2 8. f 1x2: D: x 僆 1q, q2, R: y 僆 1q, q 2; f1 1x2: D: 1q, q 2, R: y 僆 1q, q2 9. f 1x2: D: x 僆 1q, q2, R: y 僆 10, q2; f 1 1x2: D: x 僆 10, q 2, R: y 僆 1q, q2 2 10. a. $3.05 b. f 1 1t2 t 0.15 , f 1 13.052 7 c. 12 days y y y 11. 12. 13. 10 10 10 1

2

8 6 4 2 108642 2 4 6 8 10

8 6 4 2

y3

108642 y 1 2 4 6 8 10

2 4 6 8 10 x

108642 2 4 6 8 10

2 4 6 8 10 x

108642 2 4 6 8 10

8 6 4 2

30. x 僆 132 , q 2 33. a. x e32 34. a. x

108642 2 4 6 8 10

2 4 6 8 10 x

2 4 6 8 10 x

b. x 102.38 c. x ln 9.8 b. x

log 19 0.2

108642 2 4 6 8 10

1 3

Practice Test, p. 571 x1

2 4 6 8 10 x

32. a. 4.79 b. 107.3I0 1 d. x log 7 2

, x ⬇ 6.3938

c. 12x 12ln 5 d. 13x 22ln 10

ln p ln q

c. 53 log x 43 log y 52 log x 32 log y d. log 4 53 log p 43 log q 32 log p log q log 45 log 128 ln 124 38. a. ⬇ 2.215 b. ⬇ 4.417 c. ⬇ 6.954 log 6 log 3 ln 2 ln 0.42 7 d. ⬇ 0.539 39. x ln ln 2 , x ⬇ 2.8074 ln 5 5 40. x ln 41. 1 2 ln 3 , x ⬇ 0.9530 ln 3 1, 0 ⬇ x .4650 42. x e2 1, x ⬇ 6.3891 44. x 4.25 45. a. 17.77% 47. 18.5% 50. 55.0%

43. x 5; 2 is extraneous b. 23.98 days

0 8.0

b. ⬇ 6.8 microns 1 1.0e0.7t c. 14 1.962 7.84; 5.56 hr (about 5 hr 34 min)

10 , x ⬇ 333.3333 d. x e2.75, x ⬇ 0.0639 3 3 35. a. ln 42 b. log930 c. ln 1 xx d. log1x2 x2 12 b. 2 log74

8

a. logistic, g1t2

c. x

37. a. ln x 14 ln y b.

0

y 2

3

36. a. 2 log59

10

2 4 6 8 10 x

8 6 4 2

31. x 僆 1q, 02 ´ 16, q 2

ln 4 , x ⬇ 2.7726 0.5

52.

8 6 4 2

1 14. 2 15. 2 16. 52 17. 12.1 yr 18. 32 9 19. 53 125 20. e3.7612 ⬇ 43 21. log525 2 22. ln 0.7788 ⬇ 0.25 23. log381 4 24. 5 25. 1 26. 12 y y y 27. 28. 29. x 3 10 10 10 8 6 4 2

a. logarithmic, y ⬇ 12.772 1.595 ln x b. 16.9 mi/gal c. the year 2011

1. 34 81 2. log255 12 3. 52 logbx 3 logby logbz m 1n3 5 4. logb 5. x 10 6. x 7. 2.68 8. 1.24 3 1p y y 9. 10. 10 10 8 6 4 2 108642 2 4 6 8 10

8 6 4 2

y3

108642 2 4 6 8 10

2 4 6 8 10 x

x2

2 4 6 8 10 x

11. a. 4.19 b. 0.81 12. f is a parabola (hence not one-to-one), x 僆⺢, y 僆 33, q2 ; vertex is at 12, 32 , so restricted domain could be x 僆 3 2, q 2 to create a one-to-one function; f 1 1x2 1x 3 2, x 僆 3 3, q 2, y 僆 32, q 2 . 13. x 1 lnln 89 14. x 1, x 5 is extraneous 3 , x ⬇ 5.0857 15. ⬇5 yr 16. a. P kM␣ b. P ⬇ 15.3 lb 1M 202 17. 19.1 months 18. a. no b. $54.09 19. a. 10.2 lb b. 34 weeks 39.1156 ; 0.89 sec 20. logistic; y 1 314.6617e5.9483x 40

2

0

46. 38.6 cmHg

48. Almost, she needs $42.15 more. 49. a. no

b. $268.93 5

Strengthening Core Skills, p. 573 Exercise 1. about 126 times hotter

Exercise 2. about 4.2 hb

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Cumulative Review Chapters R–5, p. 574

2 1. x 2 7i 3. 14 5i2 814 5i2 41 0 9 40i 32 40i 41 0 0 0 ✓ 5. f 1g1x22 x; g1 f 1x22 x; Since 1 f ⴰ g21x2 1g ⴰ f 2 1x2, they are inverse functions. 7. a. T1t2 455t 2645 11991 S year 12 ¢T 455 b. , triple births increase by 455 each year ¢t 1 c. T162 5375 sets of triplets, T1172 10,380 sets of triplets y 9. D: x 僆 3 10, q 2, R: y 僆 39, q 2 10 8 h1x2c: x 僆 12, 02 ´ 13, q 2 h1x2T: x 僆 10, 32 6

21.

10 8 6 4 2 108642 2 4 (1, 5) 6 8 10

y

Y1

SA-33

5 5X 16 X , Y2 2 3 10

2 4 6 8 10 x

10

10

4 2

108642 2 4 6 8 10

10

2 4 6 8 10 x

2V 11. x 3, x 2 1multiplicity 22; x 4 13. 2␲a b 5x 3 1 y 15. a. f 1 1x2 b. c. 1 f 1x2 2 x f 10 2 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

17. x 5, x 6 is an extraneous root 19. a. about 88 hp for sport wagon, about 81 hp for minivan b. ⬇3294 rpm c. minivan, 208 hp at 5800 rpm 21. x ⬇ 5.064 23. x ⬇ 0.649, x ⬇ 4.967 25. x ⬇ 2.013, x ⬇ 3.608

CHAPTER 6 Exercises 6.1, pp. 586–591

23. 14, 12 25. 13, 52 27. second equation, y, 14, 32 29. second equation, x, 110, 12 31. second equation, x, 1 52 , 74 2 33. 12, 52 2 35. 12, 12 37. 13, 12 39. 12, 32 41. 1 11 2 , 22 43. 12, 32 45. 13, 42 47. 16, 122 49. (2, 8); consistent/ independent 51. ; inconsistent 53. 5 1x, y2 |6x y 226; consistent/ dependent 55. (4, 1); consistent/independent 57. 13, 42; 4 consistent/independent 59. 1 1 61. 1 mph, 2 , 3 2; consistent/independent 4 mph 63. 2318 adult tickets; 1482 child tickets 65. premium: $3.97, regular: $3.87 67. nursing student $6500; science major $3500 69. 150 quarters, 75 dimes 71. a. 3 mph b. 5 mph 73. a. 3.6 ft/sec b. 4.4 ft/sec 75. a. 100 lawns/mo b. $11,500/mo 77. a. 1.6 billion bu, 3 billion bu, yes b. 2.7 billion bu, 2.25 billion bu, yes c. $6.65, 2.43 billion bu 79. about 227 boards at $410 a piece 81. about 90,000,000 gal at $3.04 gal 83. 1776; 1865 85. Tahiti: 402 mi2, Tonga: 290 mi2 87. y 2x 3 89. $50,000 91. 1x 521x 221x 12 13x 12 y 93. x 僆 32, 8 4 25 20 15 10 5

1. inconsistent 3. consistent, independent 5. Multiply the first equation by 6 and the second equation by 10. 7. y 74 x 6, 9. y x 2 11. x 3y 3 y 4 3 x 5 13. y x 2, x 3y 3 15. yes

108642 5 10 15 20 25

2 4 6 8 10 x


17.

19.

1. triple 3. equivalent, systems 5. z 5 7. Answers will vary. 9. Answers will vary. 11. yes, no; R2, R3 13. (5, 7, 4) 15. (2, 4, 3) 17. (4, 0, 3) 19. (5, 12, 13) 21. no solution, inconsistent 23. 5 1x, y, z2 |x 僆⺢, y 2 x, z 2 x6; 1p, 2 p, 2 p2 , other solutions possible 5 2 5 2 25. e1x, y, z2| x z , y z 2, z 僆⺢ f ; a p , p 2, pb, 3 3 3 3 other solutions possible 27. 1p, 2p, p 12 29. 1p 9, p 4, p2 1 3 31. e 1x, y, z2 ` x y 2z 6 f 33. a1, , 2b 2 2 35. 1p 17, p 4, p2 37. (12, 6, 4) 39. 11, 5, 62 1 1 41. 11, 2, 32 43. a , , 2b 45. (5 cm, 3 cm, 4 cm) 5 2 47. $80,000 at 4%; $90,000 at 5%; $110,000 at 7% 49. World War II, 1945; Korean, 1953; Vietnam, 1973 51. Declaration of Independence, 1776; 13th Amendment, 1865; Civil Rights Act, 1964 53. 1 L 20% solution; 3 L 30% solution; 6 L 45% solution 55. saturated: 1.2 g, monounsaturated: 1.0 g, polyunsaturated: 0.6 g 57. h(t) 5t2 30t 1;

yes

10 8 6 4 2 108642 2 4 6 8 10

y

Y1

12 3X , Y2 X 9 2 10

2 4 6 8 10 x

(6, 3)

a. 46 ft

10

10

b. 17.2 ft

59. 2

61. 3 1, 32 4

63. x 1

Mid-Chapter Check, p. 603

10

1. (1, 1) consistent 2. (5, 3) consistent 3. 20 oz 4. No; R2, R3 5. 2R1 R2 6. (1, 2, 3) 7. (1, 2, 3) 8. 1p, p 5, p 42 9. Morphy: 13, Mozart: 8, Pascal: 16 10. prelude: 2.75 min, storm: 2.5 min, sunrise: 2.5 min, finale: 3.25 min

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Reinforcing Basic Concepts, pp. 603–604

7.

y

Exercise 1. Premium: $4.17/gal, 15.3R 35.7P 211.14 Regular: $4.07/gal e P R 0.10 Exercise 2. Verified 9.

1. region , solutions 3. Answers will vary. y 5. a. 3 or 4 not possible

x

y

y

x

y

y

x

x

x

5 4 3 2 1

y

1 2 3 4 5 x

13. 14, 32, 13, 42 15. (2, 5), 14, 72 17. 13, 42 , 14, 32 , (3, 4), 14, 32 19. 14i, 16 24i2, 14i, 16 24i2 21. 14, 32, 14, 32 12i 16, 72 12i 16, 72 23. 12 i, 2 i2, 12 i, 2 i2 25. 18, 12, 17, 42 27. 15, log 5 52 29. 13, ln 9 12, 14, ln 16 12 31. (0, 10), (ln 6, 45) 33. 13, 12 , (2, 1024) 35. 13, 212 , 11, 12 , (2, 4) 37. 12, 42 , (6, 4) 39. (3, 5), 13, 52 41. 12.43, 2.812, (2, 1) 43. (0.72, 2.19), (2, 3), (4, 3), (5.28, 2.19) y y 45. 47. 10 10

108642 2 4 6 8 10

x

8 6 4 2

y

y

20 16 12 8 4

y

x

x

2016 1284 4 8 12 16 20

x

x

51.

y

5 4 3 2 1 54321 1 2 3 4 5

4 8 12 16 20 x

65. $1.83; $3, 90,000 gal e

e. 3 or 4 solutions not possible

y

1 2 3 4 5 x

y

y

10P2 6D 144 67. Answers will vary. 8P2 8P 4D 12

69. 18 in. by 18 in. by 77 in. x

f.

2 4 6 8 10 x

53. h ⬇ 45.8 ft: h 40 ft: h 30 ft 55. The company breaks even if either 18,400 or 48,200 cars are sold. x2y 2000 57. e 2 ; approx. (12.4, 13) or (20, 5); The pool will likely x 4xy 800 have the dimensions 20 ft by 20 ft by 5 ft. 59. 8.5 m 10 m 61. 5 km 9 km 63. 8 ft 8 ft 25 ft

y

y

108642 2 4 6 8 10

2 4 6 8 10 x

49. no solution d.

parabola, parabola; 11, 22 , (2, 1)

8 6 4 2

y

2 4 6 8 10 x

2 4 6 8 10 x

54321 1 2 3 4 5

x

y

11.

parabola, line; 11, 52, 12, 22

y

circle, absolute value; 16, 82 , (8, 6)

y

10 8 6 4 2 108642 2 4 6 8 10

y

x

c.

108642 2 4 6 8 10

x


b. 3 or 4 not possible

10 8 6 4 2

x

71. a. x 2

2 110 3

b. x 0, x 8 c. x 2 73. a. m 400 1 , the copier depreciates by $400 a year b. y 400x 4500 c. $1700 d. 9.5 yr

x

Exercises 6.4, pp. 625–629 y

y

x

y

x

1. half, planes 3. solution 5. The feasible region may be bordered by three or more oblique lines, with two of them intersecting outside and away from the feasible region. 7. No, No, No, No 9. No, Yes, Yes, No x

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10 8 6 4 2 108642 2 4 6 8 10

y

13.

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

15.

45.

y

2 4 6 8 10 x

0

9.4

(0, 0) is in the solution region, and 3(0) 2(0) 8 is true.

10

10

0 (3, 2) yx 1 47. e xy 7 3

yx 1 49. • x y 6 3 y0 51. (5, 3) 53. (12, 11) 55. 26 at (2, 2) 57. 264 at (4, 3) 59. 5 6 H 6 10 61. 100

10

10 17.

9.3

J (in thousands)

11.

(0, 0) is in the solution region, and 4(0) 5(0) 20 is true.

10

80 60 40 20 20

10

10 23.

8 6 4 2 108642 2 4 6 8 10

27.

10 8 6 4 2 108642 2 4 6 8 10

33.

10 8 6 4 2 108642 2 4 6 8 10

39.

10 8 6 4 2 108642 2 4 6 8 10

43.

108642 2 4 6 8 10

2 4 6 8 10 x

y

29.

35.

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

y

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

y

10 8 6 4 2

41.

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

25.

y

31.

37.

2 4 6 8 10 x

60

80 100

5 4 3 2 1 54321 1 2 3 4 5

y

J A 50,000 J 20,000 A 25,000 300 acres of corn; 200 acres of soybeans 240 sheet metal screws; 480 wood screws 65 traditionals, 30 Double-T’s 3 buses from company X; 4 buses from company Y y a. the region is a square 10 8 b. Max is 35.1 at (3, 3) 6 (3, 3) 4 c. optimal solutions occur at vertices 2

63. 65. 67. 69. 71.

2 4 6 8 10 x

y

108642 2 4 6 8 10

2 4 6 8 10 x

73. x 5, i 13

75. 324

Making Connections, p. 629 2 4 6 8 10 x

1. c

3. g

5. d

7. a 9. g

11. e 13. f

y

1.

y 3x 2y 4

10 8 6 4 x 3y 8 2

5

10 8 6 4 2

(4, 4)

108642 2 4 6 8 10

1 2 3 4 5 x

2.

2 4 6 8 10 x

54321 2

( 8

1 2 3 4 5 x

)

10

4. no solution; inconsistent

y

2x y 2 4 3 2 1

2 4 6 8 10 x

54321 1 2

x 2y 4 3

1 2 3 4 5 x

(U, T)

4 5

9.4

0

y x 0.3y 1.4

0.2x 0.5y 4 6 q, 3 1.4

5. 15, 12 ; consistent 6. (7, 2); consistent 7. 13, 12 ; consistent 8. 1 114 , 1 9. Willis Tower is 1450 ft; Hancock Building is 6 2 ; consistent 1127 ft. 10. $1.20 11. (0, 3, 2) 12. (1, 1, 1) 13. no solution, inconsistent 14. 3 aces, 4 face cards, 5 numbered cards 15. 1530 quarters, 1180 dimes, 710 nickels 16. circle, line, (4, 3), 13, 42 17. parabola, line, 13, 22 18. parabola, circle, 1 23, 22, 1 23, 22, 1i 22, 22, 1i 22, 32 19. circle, parabola, i 110 10 11, 32, 11, 32, 1 i 1310 , 10 3 2, 1 3 , 3 2

(2, 4)

15. b


3.

y

9.3

0

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

y

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

y

40

A (in thousands)

10

19. No, No, No, Yes y 21. 10

SA-35

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20.

21.

y 6 5 4 3 2 1 654321 1 2 3 4 5 6

654321 1 2 3 4 5 6

1 2 3 4 5 6 x

parabola, circle 22.

108642 2 4 6 8 10

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

24.

3.1

1 2 3 4 5 6 x

4.7

4.7

note the open circle showing noninclusion at 10, 32 ; circle, parabola

23.

y

10 8 6 4 2

17.

y 6 5 4 3 2 1

3.1

y

18. a. 2 4 6 8 10 x

9.3

10

b. equilibrium is achieved when approx 150,000 plugs are sold at a price near $1.18

10

19.

9.3 25.

y 10 8 6 4 2

(3, 4)

6 4 2 2 4 6

2 4 6 8 10 x

Maximum of 270 occurs at both (0, 6) and (3, 4) 2 x 6, and all points on this on the line y 3

10 3x 2y 12 8 x 4y 10 6 4 2 (2, 3) 108642 2 4 6 8 10

(2, 3)

2. 1 5 , 5

2 4

2

3. 13, 22

4. 12, 1, 42

2 4 6 8 10 x

8 6 4 2

13. 14. 15. 16.

Strengthening Core Skills, pp. 634–635

1.

5. 51x, y, z2|x 2z 1, y 5z 6, z 僆⺢6 6. a 5, b 2 7. 21.59 cm by 35.56 cm 8. Tahiti, 402 mi2; Tonga, 290 mi2 9. Corn 25¢, Beans 20¢, Peas 29¢ 10. $15,000 at 7%, $8000 at 5%, $7000 at 9% y 11. 12. Maximum of P 400 at (8, 0) 10

108642 2 4 6 8 10

x2 y2 7 1 • x2 y2 6 4 x 7 0, y 6 0

1 2 3 4 5 x


Practice Test, pp. 632–633 y

54321 1 2 3 4 5

Exercise 1. 11, 42, elimination

line in between these endpoints.

26. 50 cows, 425 chickens

1.

20. Answers may vary. Possible solution:

y

5 4 3 2 1

2 4 6 8 10 x

30 plain; 20 deluxe 11 27, 1 17 2, 11 17, 1 17 2 1 13, 12, 1 13, 12, 1i 121, 52, 1i 121, 52 15 ft, 20 ft

y

10 8 6 4 (3, 0) 2 (0, 2) 108642 2 4 6 8 10

2 4 6 8 10 x

3.

10 8 6 4 2 108642 2 4 6 8 10

y

5.

(3, 1)

(1, 0) (4, 0)

2 4 6 8 10 x

45 36 27 18 9

108642 9 18 27 36 45

y

(3, 0) 2 4 6 8 10 x

7. a. D: x 僆 1q, q 2 b. R: y 僆 1q, 4 4 c. f 1x2c: x 僆 1q, 12 f 1x2T: x 僆 11, q2 d. n/a e. max: 11, 42 f. f 1x2 7 0: x 僆 14, 22 f 1x2 6 0: x 僆 1q, 42 ´ 12, q 2

¢y 7 ¢x 4 9. x 9, i 12 11. 3x2 5 13. 1 15. 1x 221x 221x 3 3222 1x 3 3 222 g.

17. x 僆 1q, 11 2 4 ´ 13, q 2 19. ⬇9.7 yr y 21. 23. 80 arrows, 25 balls, 15 bats 10 8 6 4 (3, 0) 2 108642 2 4 6 8 10

x 2

(3, 0) 2 4 6 8 10 x

y 1 (0, 2.25)

x2

25. x 5, y ⬇ 4.7

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SA-37


27. a.

3 b. y 1 x 8, verified

10

15

15

10 29. x ⬇ 39.74

CHAPTER 7 Exercises 7.1, pp. 646–650 1. square 3. 2; 3; 1 5. Multiply R1 by 2 and add that result to R2. This sum will be the new R2. 7. 3 2, 5.8 9. 4 3, 1 1 2 1 1 0 1 3 § ; diagonal entries 1, 0, 1 11. £ 1 2 1 1 3 x 2y z 0 x 4y 5 1 13. e S 13, 2 15. • y 2z 2 S 111, 4, 32 2 y 12 z3 x 3y 4z 29 y 32 z 21 2 S 14, 15, 32 z3 1 3 3 2 3 23 12 15 § 23. £ 0 21. £ 0 2 1 0 4 0 17. •

19. c 1 3 10

1 0

6 28

1 3 13

2 d 6

8 6 § 34

2R1 R2 S R2 27. 5R1 R2 S R2 4R1 R3 S R3 3R1 R3 S R3 29. (20, 10) 31. (1, 6, 9) 33. (1, 1, 2) 35. (1, 1, 1) 37. (2, 1, 3) 1 1 39. (10, 6, 8) 41. a , 1, b 43. 11, 3 2 , 22 2 4 45. linear dependence (p 4, 2p 8, p) 47. coincident dependence {(x, y, z)|3x 4y 2z 2} 49. inconsistent, no solution 51. linear dependence, 1p, 13 p, 4 53 p2 53. (2, 1, 1) 55. (3, 0, 2) 57. (1, 2, 3, 1) 59. 28 units2 61. LA to STL, 1600 mi; STL to CIN, 310 mi; CIN to NY, 570 mi 63. Moe 90, Larry 45, Curly 30 65. 15 m, 36 m, 39 m 67. $2000 at 5%; $3000 at 7%; $5000 at 9% 69. x 84°; y 25° 71. x3 x 10; x3 x 6; x4 2x3 8x 16; x2 2x 4 73. C 30,000 in the year 2011 (t ⬇ 6.39) 25.

Exercises 7.2, pp. 658–662 1. aij; bij 3. scalar 5. Answers will vary. 7. 2 2, a12 3, a21 5 9. 2 3, a12 3, a23 6, a22 5 11. 3 3, a12 1, a23 1, a31 5 13. true 15. conditional, c 2, a 4, b 3 10 0 d 19. different orders, sum not possible 17. c 0 10 13 12 14 8 263 19 32 8 1 5 21 21. c 85 23. £ 4 2 8 § 35 d 16 16 7 31 9 8 4 2 25. c 31. c

20 25

1 0 d 0 1 79 37. c 50 1 0 43. £ 0 1 0 0

5 2

1 2 0 2 § 29. £ 0 1 4 3 6 12 24 90 d 33. matrix mult. not possible 35. c 6 15 57 30 42 18 60 1 0 d 39. c d 41. c d 19 12 42 36 0 1 1 3 0 0 4 4 4 3 57 1 3 1 0 § 45. c 19 d 47. £ 2 8 8 § 1 5 19 57 1 11 1 1 4 16 16

15 d 10

27. £ 0 2

1 7 2 3 2

0 1 § 6

1.75 2.5 4 28 4 d 51. c d 53. verified 7.5 13 8 17 3 55. verified 57. P 21.448 cm; A 27.7269 cm2 T S T S 59. a. S 3820 1960 S 4220 2960 V D £ 2460 1240 § M D £ 2960 3240 § P 1540 920 P 1640 820 b. 3900 more by Minsk c. d. 8361.6 5116.8 3972.8 2038.4 V £ 2558.4 1289.6 § £ 5636.8 4659.2 § 1601.6 956.8 3307.2 1809.6 4388.8 3078.4 M £ 3078.4 3369.6 § 1705.6 852.8 61. 322,000 19,000 23,500 14,000 4 ; Total Profit N: $22,000, S: $19,000, E: $23,500, W: $14,000 63. a. $108.20 b. $101 First row, total cost for science from each 100 101 119 restaurant; second row, total cost for math c. c d 108.2 107 129.5 from each restaurant. 49. c

65. c

32.4 10.3 21.3 d a. 10 b. 20 29.9 9.6 19.5 c. p13 gives the approximate number of females expected to join the writing club 0 2n1 2n1 n 67. £ 2 1 1 2n 1 § 69. 11, 1, 22 71. ⬇4.39 2n1 0 2n1

Mid-Chapter Check, pp. 662–663

1. 3 3, 0.9 2. 2 4, 0 3. 12, 32 4. 12, 0, 52 5. 1 p 3, 2p 8, p2 13 17 4 6 5 5 6. a. c d b. c d c. c d 35 9 12 2 5 15 0.8 0.5 2.2 3 1.5 6 1 0 0 7. a. £ 0.1 0.8 1 § b. £ 1.5 c. £ 0 1 0 § 0 3 § 2.1 0.3 1.9 6 1.5 6 0 0 1 24 4 17 32 13 26 18 24 8. a. c b. c c. £ 0 d d 5 § 5 0 5 2 10 4 16 39 4 33 d. c d 9. used: $80, new: $125 2 29 4375 110 10. c d , p11: total rebates paid to individuals, p21: total rebates 2400 59 paid to business, p12: free AAA years given to individuals, p22: free AAA years given to business

Reinforcing Basic Concepts, p. 663 Exercise 1: p32 Exercise 2: 1st row of A with 3rd column of B 2nd row of A with 2nd column of B Exercise 3: 3A 4 S 3 1; 3 B 4 S 1 3 3 A 4 S 3 2; 3B 4 S 2 3 3A 4 S 3 3; 3 B 4 S 3 3 3 A 4 S 3 n; 3B 4 S n 3; n 僆⺞ Exercise 4: only FE P

Exercises 7.3, pp. 673–678 1. diagonal; zeroes 3. AB; BA; I; A1 5. Answers will vary. 1

9 7. verified 9. verified 11. verified 13. verified 15. c 1 9

2 9 5 d 18

cob19545_saa_018-038.qxd

SA-38

5 17. c 2

25.

29.

31.

37. 45. 57.

9 80 1 £ 80 1 20

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1.5 d 0.5 31 400 41 400 1 100

19. verified 21. verified 23. 27 400 3 400 § 17 100

27. c

2 5

2 39 £ 13 4 39

1 13

0 2 13

10 39 1 3 § 19 39

3 x 9 dc d c d 7 y 8

1 2 1 x 1 £1 0 1 § £y§ £3§ 2 1 1 z 3 2 1 4 5 w 3 2 5 1 3 x 4 ≥ ¥ ≥ ¥ ≥ ¥ 33. (4, 5) 35. (12, 12) 3 1 6 1 y 1 1 4 5 1 z 9 1 15 , 13 2 39. 11.5, 0.5, 1.52 41. (3, 2, 5) 43. 11, 0.5, 1.5, 0.52 1, yes 47. 0, no 49. 1 51. singular matrix 53. 34 55. 7 singular matrix 59. det1A2 5; (1, 6, 9) 61. det1A2 0 1

13 63. A1 c 2 13

5 13 3 d 13

verified 65. singular

67. 31 behemoth, 52 gargantuan, 78 mammoth, 30 jumbo 69. Jumpin’ Jack Flash: 3.75 min Tumbling Dice: 3.75 min You Can’t Always Get: 7.5 min Wild Horses: 5.75 min 71. 30 of clock A; 20 of clock B; 40 of clock C; 12 of clock D 73. p1 72.25°, p2 74.75°, p3 80.25°, p4 82.75° 75. y x2 4x 5 77. y x3 2x2 9x 10 79. $450 in the CD, $350 in the MM 81. $1500 in retirement fund, $1500 in mutual fund, $1800 in stock fund 83. 2 oz Food I, 1 oz Food II, 4 oz Food III 85. Answers will vary. 87. x 6, x 3, x 2 9 1 89. x 僆 aq, d ´ c , qb 2 2

Exercises 7.4, pp. 689–692 1. a11a22 a21a12 3. constant 5. Answers will vary. 2 5 7 5 2 7 7. D ` ` ; Dx ` ` ; Dy ` ` 9. 15, 92 3 4 1 4 3 1 26 25 , b 13. no solution 11. a 3 3 4 1 2 5 1 2 2 1 † Dx † 8 2 1 † 15. a. D † 3 1 5 3 3 5 3 4 5 2 4 1 5 Dy † 3 8 1 † Dz † 3 2 8† 1 3 3 1 5 3 b. 冟D冟 22, solutions possible c. 冟D冟 0, Cramer’s rule cannot be used: coefficients R1 R2 R3 3 5 1 b 21. 10, 1, 2, 32 17. (1, 2, 1) 19. a , , 4 3 3 A A B B C 23. 25. x3 x2 x1 x2 x3 A B C A B C 2 27. 29. x x3 x1 x x2 x A Bx C Dx E 5 4 2 2 31. 33. 2 x1 x x 1 x 2 1x 22 4 3 7 2 1 35. 37. 2x 5 x3 x x1 x1 3 1 4 5 4 2 39. 41. x x1 2x 1x 12 2 x 2x 4 5 x1 1 3x 2 2 43. 45. 2 x2 x x 3 1x 12 2 3 2 1 47. 49. 320 32␲ ⬇ 420.5 in2 x1 x3 1x 32 3

51. 8 cm2 53. 27 ft2 55. 19 m3 57. yes 59. no 61. yes, yes, yes 15,000x 25,000y 2900 63. e ; 6%, 8% 25,000x 15,000y 2700 65. F1 F2 6013 ⬇ 103.9 lb 67. a. h 10,000 16t2 b. 10,000 ft c. 3600 ft d. 25 sec 69. A 195 units2 y 71. 73. 32x1 342x; x 1.25 10 8 6 4 2 54321 2 4 6 8 10

1 2 3 4 5 x

Exercises 7.5, pp. 697–699 1. 214.5 ft2 of skin, 231.0 ft2 of wood veneer, 516 tension rods, and 498 ft of hoop 3. 955 ft2 of skin, 1021.5 ft2 of wood veneer, 2180 tension rods, and 2129.5 ft of hoop 5. 92,250 gal gasoline, 595,000 lb corn, 227,500 oz yeast, and 134,750 gal water 7. 5 Silver, 9 Gold, and 2 Platinum 9. one bundle of first class 9.25 measures of grain; one bundle of second class 4.25 measures of grain; one bundle of third class 2.75 measures of grain 11. Answers will vary. 13. Answers will vary. 15. Answers will vary. 17. Answers will vary. 19. Answers will vary. 21. Answers will vary. 23. Answers will vary. 25. Answers will vary. 27. Answers will vary.

Making Connections, p. 699 1. G

3. D

5. A

7. B

9. CD

11. E

13. D

15. H


1. Answers will vary. 2. 12, 42 3. (1, 6, 9) 4. 165 p 1, 25 p 1, p2 5. (2, 7, 1, 8) 7.25 5.25 6.75 6.75 d 7. c d 8. not possible 6. c 0.875 2.875 1.125 1.125 1 0 4 2 6 1 0 d 10. c d 11. £ 5.5 1 1 § 9. c 1 7 0 1 10 2.9 7 3 6 4 8 12 0 3 1 § 13. not possible 14. £ 2 4 4 § 12. £ 4.5 2 3.1 3 16 0.4 20 15.5 6.4 17 17 2 § 16. D 17. It’s an identity matrix. 15. £ 9 18.5 20.8 13 18. It’s the inverse of B. 19. E 20. It is the inverse of F. 21. Matrix multiplication is not generally commutative. 22. 18,62 19 25 37 36 31 , b 25. 11, 1, 22 26. a , , b 23. 12, 0,32 24. a 35 14 19 19 19 91 5 2x 1 units2 28. 2 27. 2 x2 x 3 29. Year 1: fir 2670 ft2, jequitba 2752 ft2, teak 1090 ft2; Year 2: fir 3570 ft2, jequitba 3668 ft2, teak 1456 ft2; Year 3: fir 4470 ft2, jequitba 4584 ft2, teak 1822 ft2; Year 4: fir 5370 ft2, jequitba 5500 ft2, teak 2188 ft2; Year 5: fir 6270 ft2, jequitba 6416 ft2, teak 2554 ft2; Total: fir 22,350 ft2, jequitba 22,920 ft2, teak 9110 ft2 30. PIE ARE SQUARE

Practice Test, pp. 702–703 1 16 1 1. a3, b 2. a3p , p, 2p 3b 3. a2, 1, b 2 3 3 6 5 1.2 1.2 3 1 2 d b. c d c. c d d. c 4. a. c 8 9 1.2 2 3 5 2.5

1 d 1.5

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e. 2

0 5. a. £ 0.5 0.2

0.1 0.6 0.8

0 0 § 0.9

0.3 b. £ 0.06 0.18

0.06 0.06 0.24

0.12 0 § 0.48

40 10 0.31 0.13 0.08 0 17 17 40 17 c. £ 0.01 0.05 0.02 § d. £ 17 10 10 e. 500 17 § 35 30 5 0.39 0.52 0.02 17 17 6. (1, 6, 0), (1, 1, 1), (3, 4, 2), answers vary as (2p 1, 5p 6, p) 97 18 2 b 10. (1, 6, 9) 7. a2, b 8. 13, 2, 32 9. a , 3 34 17 6 11. x 1, y 1, z 2 12. B £ 13 § 13. (1, 4), (2, 1), (4, 1) 11 14. 5 mi2 15. r 2, s 1 16. Dr. Brown owes $31,000; Dr. Stamper owes $124,000 17. 7.5 hr, 15.5 hr 18. 11 one-day, 6 two-day, 3 five-day 19. federal program: $200,000; municipal bonds: $1,300,000; 1 3x 2 bank loan: $300,000 20. 2 x3 x 3x 9

Exercises 8.2, pp. 725–729 1. c2 | a2 b2 | 7. x2 y2 49

3. 2a, 2b


9.3

14.1

14.1

9.3 (4.2, 5.6), (4.2, 5.6) 9. 1x 52 2 y2 3 3.1

Strengthening Core Skills, pp. 704–705 9.4

0

Exercise 1: (1, 4, 1)

Cumulative Review Chapters R–7, pp. 705–706 1. a. 8 i b. 16 30i c. 3 2i d. i 7 4 x 7. y 5 5 y 11. 10

9. a. 18

8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

13.

b. 1

c. 7

4 3 2 1 1 987654321 1 2 3 4 5 6

3. x

3 2

5. x

1 3

d. 3 is not in the domain y

15. 1q, 12 ´ 15, q2

3.1

16, 122, 16, 122 11. 1x 12 2 1y 52 2 25 12.4

1 x

17. f 1x2c: x 僆 10, 22 ´ 14, q2 , f 1x2T: x 僆 1q, 02 ´ 12, 42 , 5 25 constant: none 19. 2y 21. x3 4x2 2x 20 2 212y 12 5 b 25. condominium: $400,000; delivery truck: $60,000; 23. a3, 2 business: $140,000 27. a. 144 b. 1148 29. 1q, 4 3 4 ´ 3 2, q 2

CHAPTER 8

9.4

9.4

0 (5, 8), (5, 2)

13. 1x 62 2 1y 52 2 9 center: (6, 5), r 3 10

15. 1x 22 2 1y 52 2 25 center: 12, 52, r 5

y


10

y

r3

1. geometry, algebra 3. perpendicular 5. point, intersecting 7. 12, 22; verified 9. 12, 22; verified 11. 1 132 , 92; verified 13. 1x 22 2 1y 22 2 52 15. 1x 22 2 1y 22 2 52 13 2 25 2 17. ax b 1y 92 2 a b 19. a. d 13; B, C, E, G 2 2 8 15 b. Answers will vary. 21. Verified, d 23. a. B, C, E 5 1 b. Answers will vary. 25. Verified 27. y x2 16 29. 4x2 3y2 48 31. Verified, verified 33. 3x2 y2 3 35. Verified 37. about 12 yr y 39. f 1x2 1x 221x 12 2 1x 32 16 8 4321 8 16

(6, 5) 10

10

10 x

r5

10 x

(2, 5) 10

10

17. 1x 32 y 14 center: (3, 0), r 114 2

10

2

19.

10

y

y 10

r 艐 3.7 10

10 x

10

10 x

(3, 0) 10

21.

10

23.

y

10

y

1 2 3 4 x

10

10 x

10

10

10 x

10

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y2 x2 1, (0, 0), a 4, b 2 16 4 b. (4, 0), (4, 0), (0, 2), (0, 2) d. 3.1

25. a.

43. c.

1x 32 2 25

y

10

10

1y 22 2 10

1

y (3, 2 √10) (3, 2)

10

10 x

10

10 x (8, 2

(2, 2) 10

4.7

10

3.1 (2.4, 0.8), (2.4, 0.8), (2.4, 0.8), (2.4, 0.8) y2 x2 27. a. 1, (0, 0), a 3, b 4 9 16 b. (0, 4), (0, 4), 13, 02 , (3, 0) c. d. 3.1

(3, 2 √10)

D: x 僆 32, 8 4, R: y 僆 32 110, 2 1104 45. k 20 47. k 20 49. a. (2, 1) b. (3, 1) and (7, 1) c. (2 121, 1) and (2 121, 1) y d. (2, 3) and (2, 1) e.

4.7

10

10

10 x

y

10

10

10

10 x

51. a. (4, 3) b. (4, 2) and (4, 8) y and (8, 3) e.

c. (4, 0) and (4, 6)

d. (0, 3)

10

10

4.7

4.7

10

10 x

10

53. a. (2, 2) b. (5, 2) and (1, 2) c. (2 13, 2) and (2 13, 2) d. (2, 2 16 ) and y (2, 2 16 ) e.

3.1 (2.4, 0.8), (2.4, 0.8), (2.4, 0.8), (2.4, 0.8), y2 x2 29. a. 1, (0, 0), a 15, b 12 5 2 b. (15, 0), ( 15, 0), (0, 12), (0, 12) c.

10

10

y

10 x

5 10 5

y2

2

5 x

55.

x 1 36 20

a.

y 10

5

31. ellipse

10

33. circle

y

10

y

10

10 x

10 10

10

10 x

10 x

r≈3 10

35. ellipse

10

37. x2

y

1y 32 2

10

10 x

9.4

y

10

5 x

(1, 3)

16 10

(2, 3) (6, 1) 10

4

(0, 5)

D: x 僆 31, 1 4, R: y 僆 35, 1 4 1

41.

1x 32 2 4

y

10

(2, 1)

1y 52 2 10

1

c. L 6.6 57.

1x 32 9

d. verified 2

1y 22 2 25

1

a.

y 3

2

y

8 x

(3, 5 √10)

(2, 1) 10 x

(2, 1) 10

6.2

(1, 3)

(0, 3) 5

1y 12 2

9.4

(0, 1) 5

39.

6.2

1

4 5

1x 22 2

b.

10

D: x 僆 3 6, 2 4 , R: y 僆 3 1, 3 4

10

10 x

(1, 5) 10

(5, 5) (3, 5) (3, 5 √10)

D: x 僆 3 1, 5 4, R: y 僆 3 5 110, 5 1104

7

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SA-41

Student Answer Appendix b.

7.

5

9.

y

10

10

(0, 0) (⫺3, 0)

(⫺2, 0)

(3, 0)

⫺10

13.

10

(7, 0)

⫺10

10 x

10 x

⫺10

15.

y

(0, 0)

(2, 0)

⫺10

10 x

12

y

10

(⫺7, 0)

⫺10

⫺6

11.

y (0, 0)

⫺10 10

17.

y

y

10

(⫺6, 0) (0, 0)

(0, 3) (0, 0)

(6, 0)

⫺7

⫺10

c. L ⫽ 3.6 d. verified y2 x2 ⫹ ⫽ 1, 1⫾ 17, 02 59. 16 9 2 1x ⫹ 32 1y ⫹ 12 2 ⫹ ⫽ 1, 1⫺3, ⫺1 ⫾ 2132 61. 4 16 63. A ⫽ 12␲ units2 65. 27 ⬇ 2.65 ft, 2.25 ft y2 y2 x2 x2 ⫹ 2 ⫽ 1; 6.4 ft 71. ⫹ ⫽1 67. 8.9 ft, 17.9 ft 69. 2 6.25 2.25 15 8 2 y x2 ⫹ ⫽ 1 75. a ⬇ 142 million miles, b ⬇ 141 million 73. 2 36 135.252 2 miles, orbit time ⬇ 686 days 77. about 66,697 mph 79. 90,000 ␲ yd2 81. a. 45 ft b. 106 ft c. 25 ft 83. ellipse, since squared terms are positive and A ⫽ B; 61x ⫹ 32 2 ⫹ 31y ⫺ 42 2 ⫽ 0, the constant term becomes zero; the graph is the single point 1⫺3, 42 log 20 85. ⬇ 2.73 log 3 y 87. a. x 僆 10, 62 b. ⫺2 0 x ⫺ 3 0 ⫹ 10 7 4 10 0x ⫺ 30 6 3 (0, 4) (6, 4) x ⫺ 3 7 ⫺3 or x ⫺ 3 6 3 x 7 0 or x 6 6 ⫺8 12 x 0 6 x 6 6

⫺10

19.

10

3.

y 8 4

8

12

x

⫺4

⫺4

4

8

12

16 x

⫺8

⫺12

⫺12

4.

8 7 6 5 4 3 2 1 ⫺6⫺5⫺4⫺3⫺2⫺1 ⫺1 ⫺2

5.

y

7 6 5 4 3 2 1 ⫺7⫺6⫺5⫺4⫺3⫺2⫺1 ⫺1 ⫺2 ⫺3

1 2 3 4 x

y

9.

2

y x ⫹ ⫽1 25 194

2

10.

(0, ⫺3)

251x ⫺ 22 2

⫺10

23. 1⫺4, ⫺22, 12, ⫺22, y ⫽ ⫺2, 1⫺1, ⫺22, x ⫽ ⫺1 25. 14, 12, 14, ⫺32, x ⫽ 4, 14, ⫺12, y ⫽ ⫺1 y y 27. 29. 31. 10 (9, ⫺2)

6.

1 2 3 x

2 1 ⫺2⫺1 ⫺1 ⫺2 ⫺3 ⫺4 ⫺5 ⫺6 ⫺7 ⫺8

10 x (0, ⫺1) ⫺10

33.

10

10

(0, ⫺3)

⫺10

35.

y

(3, ⫺2)

6

10 x

(⫺5, ⫺1)

冢⫺5, ⫺1 ⫺兹7冣

⫺10

37.

y

10

(1, 2) (6, ⫺1) ⫺10

⫺10

⫺10

10

10 x

(1, ⫺8)

(2, ⫺1)

⫺10

⫺14

41.

y

10

(⫺3, 0)

43.

y

10

⫺10

(⫺√6, 0)

(0, 0) ⫺10

10 x

冢4 ⫹兹5, 3冣 y (0, 0)

(0, 2)

(3, 0)

y (4, 3)

冢4 ⫺兹5, 3冣

10 x (1, ⫺3)

10 x

(⫺2, ⫺1)

45.

1x ⫹ 52 2 9 10

y

⫺10

10 x

(√6, 0) 10 x

⫺10

⫺

1y ⫹ 22 2 36

⫺10

⫽1

y

9.3

(⫺8, ⫺2)

1 2 3 4 5 6 7 8 x

2

⫺10

10 x

(⫺2, ⫺2)

⫺18.8

⫺10

(⫺5, ⫺2)

9.4

⫺15.3 (0, ⫺10), (0, 6), (⫺10, ⫺10), (⫺10, 6)

47.

1y ⫺ 32 2 1

⫺ 14

1x ⫹ 22 2 4

⫽1

y

14.4

(⫺2, 4) (⫺2, 3)

91y ⫺ 32 2

⫺14

⫽1 2 4 2 2 Exercise 2: 281x ⫺ 12 ⫹ 481y ⫹ 22 ⫽ 1 25 25

(⫺2, 2) ⫺6

Exercises 8.3, pp. 740–744 1. transverse 3. perpendicular, transverse, center

⫺10

x

(⫺3, ⫺2)

⫺10

y x ⫹ ⫽1 16 4

⫹

冢⫺5, ⫺1 ⫹兹7冣

(0, 1) ⫺10

y

10

10

Reinforcing Basic Concepts, p. 730 Exercise 1:

10 x

(0, ⫺6)

⫺10

1 2 3 4 5 6 7 x

7. y ⫽ 1 x2 12 1x ⫹ 32 2 1y ⫺ 12 2 8. a. ⫹ ⫽ 1; D: x 僆 3⫺5, ⫺1 4, R: y 僆 3⫺3, 5 4 4 16 b. 1x ⫺ 32 2 ⫹ 1y ⫺ 22 2 ⫽ 16; D: x 僆 3 ⫺1, 7 4 , R: y 僆 3⫺2, 6 4 2

(0, 0)

(0, ⫺2)

⫺3⫺2⫺1 ⫺1 ⫺2 ⫺3 ⫺4 ⫺5

⫺4

⫺8

⫺10

(0, 6)

y

5 4 3 2 1

10 x (0, ⫺2√3)

y

⫺10

10 x

(0, 0)

4 4

10

(0, 0) ⫺10

Mid-Chapter Check, pp. 729–730

⫺4

⫺10

21.

y

⫺10

2.

(0, 0)

(0, 3)

39.

y

(0, 2√3) ⫺10

10 x

(0, ⫺3)

⫺10

1.

⫺10

10 x

5 Answers will vary.

6 x

⫺20.8

⫺10.4 (2.8, 5.6), (2.8, 0.4), (⫺6.8, 0.4), (⫺6.8, 5.6)

16.8

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49. circle; A B 51. circle; A B 53. hyperbola; A, B opposite signs 55. hyperbola; A, B opposite signs 57. circle; A B 59. ellipse; A B 61. 8, 2a 8, 2b 6 63. 12, 2a 16, 2b 12 1y 42 2 1x 32 2 65. 1 a. (3, 4) b. (0, 4) and (6, 4) 9 4 c. 13 113, 42 and 13 113, 42 d. 2a 6, 2b 4 y e. 10

10

7. x 僆 1q, q 2, y 僆 34, q 2

9. x 僆 1q, q2, y 僆 1q, 18 4

y

10

y (2, 18)

20

(0, 10) 10

(1, 0)

(3, 0)

(1, 0) (0, 3)

(5, 0)

3

10 x

7 x

(1, 4) 20

10

11. x 僆 1q, q 2, y 僆 310.125, q2

13. x 僆 3 4, q2, y 僆 1q, q2 y 8

y

10 x

4

12

1y 32

2

2

x 1 a. (0, 3) b. (0, 7), (0, 1) 16 4 c. 10, 3 2 152, 10, 3 2 152 d. 2a 4, 2b 8 y e. y 2x 3, y 2x 3 14 67.

10

1x 32 2

(1, 0)

12 6

6

12

4

4 8 (3, 0) 4 (0, 1)

x

15. x 僆 1q, 164 , y 僆 1q, q 2

17. x 僆 1q, 0 4, y 僆 1q, q2

y

y

8

(0, 7)

4

(16, 3)

16 8

6

8

8

(16, 0)

(7, 0)

16

4 (0, 4)

16 8

x

8

4 (0, 1)

4

8

8

19. x 僆 39, q 2, y 僆 1q, q 2

y

10

4 (0, 6)

(0, 2) (4, 0)

(0, 0) 4

4

8

10

x

10 x

(0, 2)

8

10 x

10

23. x 僆 1q, 0 4, y 僆 1q, q2 10

10

25. x 僆 36.25, q2, y 僆 1q, q2

y

(0, 2) (6, 0) (0, 1) 10

10

10 x

10 x

(1, 0) (0, 3)

(6.25, 0.5) 10

27. x 僆 321, q 2, y 僆 1q, q 2 (21, 5)

10

29. x 僆 1q, 11 4, y 僆 1q, q2

15 (0, 5 兹21)

25

9.4

冢

(3, 0)

y

2 2 3 2x

9

89. a.

33. x 僆 32, q2, y 僆 1q, q2

y

10

y

(2, 3)

(2, 3) 10

x2 b2, as x S q, y S 2

Ba 1 4

1y 22 2 0

冣

(0, 7)

b2

1x 42

91. yes, 3

10

冢

15

31. x 僆 1q, q 2, y 僆 33, q2

83. 12 microns

2

15 x (11, 2) 4 兹22 0, 2

(4, 0)

15 x

10

b2

b x2 x B a2 a

b. 1x 22 2

15

10 x

(11, 0)

y2 x2 85. 1, about (24.1, 60) or 124.1, 602 225 2275 87. y

冣

15

25 x

15

81. 40 yd

y

15

4 兹22 0, 2

(0, 5 兹21)

6.2

y

10

y

9.4

x

(9, 3) 8

4

y2 y2 1x 22 2 x2 71. 1 73. 1 36 28 9 9 2 2 y 1y 12 2 1x 22 2 x 75. 1, 1 113, 02 77. 1, 4 by 2 15 4 9 4 5 2 2 79. a. y 3 2x 9 b. x 僆 1q, 3 4 ´ 3 3, q 2 c. 6.2

16

21. x 僆 34, q2, y 僆 1q, q2

y

8

10

x

8

6 (0, 7)

1 a. 13,22 b. 11, 22 and 15, 22 4 12 c. 11, 22 and 17, 22 d. 2a 4, 2b 4 13 y e. 10 69.

(3.5, 0)

8

12 (1.25, 10.125)

10 x

1y 22 2

(0, 3)

(4, 1)

6

10

1y 42 1 5

10

35. x 僆 31, q2, y 僆 1q, q2 10

2

37.

10

(0, 2) 10

(1, 3)

0

y

y

10 x

y 2

(0, 0)

20

20 x

(19, 0)

10

93. 42 solid, 46 liquid 10

Exercises 8.4, pp. 751–754 1. horizontal, right, a 0

39.

10

41.

y

10

y6

3. ( p, 0), x p 5. Answers will vary.

y x1

3

(1, 0)

(0, 0) 20 x

10 (0, 6)

10

冢0, 2 冣

(0, 0) 20

43.

y

10

y 10

3 2

10 x

(0, 0)

10

10 x

10

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10

47.

y

49.

y

10

10

y

89.

y

5

x2 9

x 2

5

冢 2, 0冣

9

( 2 , 0)

10

15

10 x

y 2

(0, 0) 10

51.

10

10

53.

y

55.

20 (0, 6)

(4, 2)

(5, 6) 25

10

25 x

57.

10

(1, 3) 10

y 4

y 8

10

10

10 x

(7, 2)

20

15

x3

10 x

(4, 3)

10

59.

y

1

2

4

5 x

91. 6 in.; (13.5, 0) 93. ⬇14.97 ft, (0, 41.75) 95. y2 5x or x2 5y, 1.25 cm 97. 1x 22 2 12 1y 82; p 18 ; 12, 82 99. y 2x2 5x 6 101. symmetric to the y-axis, f 1x2 f 1x2; symmetric to the origin, f 1x2 f 1x2

Making Connections, p. 754 1. d

y (5, 5)

3

(4, 2.5) y

x 5

(7, 4)

1

15 x

10

y

10

(4, 2.5)

, 冢 25 64 0冣

(4, 0)

(0, 0)

10

10 x

x

(4, 2)

25 64

3. f 5. g

7. c

9. h

11. e

13. b

15. g

x7


(2, 3) 10

10 x

10 x

(6, 5) 10

15

61. x2 8y 63. y2 16x 65. x2 20y 67. 1y 22 2 121x 22 69. 1x 42 2 121y 72 71. 1x 32 2 81y 22 73. y2 81x 12; vertex (1, 0); focus (1, 0) 75. 1y 22 2 81x 22 ; directrix: x 0; endpoints (4, 6) and (4, 2) 77. 14, 32, 14, 32, 14, 32, 14, 32 6.2

1. verified (segments are perpendicular and equal length) 2. x2 1y 12 2 34 3. yes 4. verified y y 5. 6. 7. 10 10

10

8.

9.

y

10

y

10 x

5

10

y2

2

10.

10

10

5 x

9.4

10 x

10

5

y

10

10 x

10

5

9.4

10

10 x

10

x 1 25 9 6.2

6.2

79. 13, 52, 13, 52, 10, 42

6.2

9.4

9.4

9.4

6.2

81. 15, 52, 15, 52, 15, 52, 15, 52 9.3

9.4

6.2 one possibility is (3, 2.4), (3, 2.4), (3, 2.4), (3, 2.4) y2 y2 x2 x2 11. a. 1 b. 1 169 25 144 400 1x 22 2 1y 12 2 y 12. 1 13. 25 4 8 5 4

y

(2 兹21, 1) 3 (3, 1)

14.1

2 1

321 1 2 3 4 5

14.1 14.

20

4

(2 兹21, 1) (2, 1)

8

9.3

85. 11.863, 11.8082, 17.863, 1.8082

8

x

8

1 2 3 4 5 6 7 x

center: (0, 0) vertices: 0, 5)

15.

y

15

16.

y

9

4 20 12 4 4

4 4

(7, 1)

12

83. (4, 4), (6.187, 1.571) 87. 16 units2

4

12

20 x

12 20

center: (2, 3) vertices: (2, 1), (2, 7)

15

9

3 3

y

12 (0, 1)

3 4

20

4 3

9

15 x

9 15

center: (2, 1) vertices: (5, 1), (1, 1)

20 12 4 4

4

12

20 x

12 20

center: (0, 1) vertices: (0, 2), (0, 4)

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17.

9.

y

15

(1, 6)

9

(6, 1) 5

25

10

35 x

15

11.

y2 x2 1 9 16

10

(2, 3)

6.2 15, 5.32, 15, 5.32, 15, 5.32, 15, 5.32, y2 y2 x2 x2 19. a. 1 b. 1 225 64 16 9 2 2 1x 52 1y 22 y 1 21. 20. 10 9 4

6.2

9.4

10

10 x

6.2

10

16 3

(2, 3), (2, 3), (2, 3), (2, 3) 17. 154.89 million miles; 128.41 million miles 18. y 1x 12 2 4; D: x 僆 1q, q2, R: y 僆 34, q 2; focus: 11, 15 4 2 19. 1x 12 2 1y 12 2 25; D: x 僆 34, 6 4, R: y 僆 34, 6 4 1x 32 2 y2 1; D: x 僆 36, 0 4, R: y 僆 36, 6 4; 20. 9 36 foci: 13, 313 2, 13, 3 13 2

10

22.

23.

y

10

10

y y5

(0, 2)

(6, 0)

(0, 0) 10

10

10 x

(6.25, 0.5)

(0, 3)

15

10

(0, 5) (10, 5)

25. circle, parabola; (3, 2), (3, 2)

y

(4, 2)

(0, 2)

10 x

(10, 5)

10

10

Strengthening Core Skills, p. 759 1.

(8, 2) 15 x

y 2

(4, 0)

1x 22 2 2 1 22 5 2

a

10

3.

Practice Test, pp. 757–758 1. c 2. d 3. a y 5. 10

6.

1x 32 1 72 2

2 1 23 2

b

2

1y 32 2

1y 12 1 65 2

(4, 3)

(6, 3)

1x 12 2 27 2 1 5 14 2

5 27 14 ,

1y 22 2 23 1 5 12 2

b

5 23 12

2

1; a 72 , b 65

2

1

2

10

(3, 2.2) (6.5, 1)

y

(0.5, 1) 10 x

(3, 0.2) 10

10 x

10

2.

10

F2 (2 兹15, 3) y (2, 3)

1

a

2 3

C (2, 3) F1 (2 兹15, 3)

10

10 x

10

22 5 , 2

4. b

r3

10

9.4

(0, 2)

14 x 2

y2 x2 1 16 12

(0, 2)

(8, 2)

y 3 x

15

(4, 0)

(5, 2) (2, 2)

3冣

15 x

10 16 , b, 12, 02 b. 12, 22, 12, 22, 12, 22, 12, 22 3 3 14. 15 ft, 20 ft 15. 1x 22 2 1y 52 2 8 16.

6

7 , 4

冢

(1, 3)

13. a. a

9.4

4 3

y

15

10 x

10

2

15

x 5

(2, 3)

y 3x

12.

y

9 4

10

9.4

10 x (1 兹13, 3)

C (1, 3)

10

x

6.2

24.

10

(1 兹13, 3)

10

y

3

y 3 2 (x 1)

y

V1 (3, 3) V2 (1, 3)

10 x

(1, 0)

center: (6, 1) vertices: (2, 1), (10, 1)

10

10

F2 (1, 3 兹5)

15

9

18.

3

y 3 2 (x 1)

C (1, 3) F1 (1, 3 兹5)

3 25 15 5 3

10.

y

10

4.

1x 32 1 4 15 3 2

2

2

1y 12 1 92 2

2

(3, 1)

2

10

9 5 1; a 4 1 3 ⬇ 3, b 2 (艐6, 1)

y

(艐0, 1)

10

7. (6, 4)

10

y

8.

10

r5

(0, 4) (3 兹13, 4)

(3 兹13, 4) 10

(3, 4) 10

10 x 2 y 4 3 (x 3)

10 x

y

10

10

10 x

(5, 2) 10

Cumulative Review Chapters R–8, p. 760 1. x 2, x 2i 3. x 7; x 4 is extraneous ln 7 7. x 2 9. x 5; x 2 is extraneous ln 3

5. x 3 2i

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10

13.

y

y

10

(0, 5)

(3, 1) 10

10 x

10 x

10

15.

10

10

17.

y

x 3 0, fl

10

10

y

(5, 2)

x3

( )

19.

x 1 y

h(x)

f(x) (0, 0)

10

10

10 x

10

10 x

10 x

r3

(2, 0) 10

arithmetic; d ␲ 19. 2, 5, 8, 11 21. 7, 5, 3, 1 6 0.3, 0.33, 0.36, 0.39 25. 32 , 2, 52 , 3 27. 34 , 58 , 12 , 38 2, 5, 8, 11 a1 2, d 5, an 5n 3, a6 27, a10 47, a12 57 a1 5.10, d 0.15, an 0.15n 4.95, a6 5.85, a10 6.45, a12 6.75 35. a1 32 , d 34 , an 34 n 34 , 33 39 a6 21 4 , a10 4 , a12 4 37. 61 39. 1 41. 2.425 43. 9 45. 43 47. 21 49. 26 51. a. appears linear b. d 0.75 c. an 0.75n 0.75 17. 23. 29. 31. 33.

(2, 3) 10

7

10

10

21. a. x 僆 1q, q 2 b. y 僆 1q, 44 c. f 1x2c: x 僆 1q, 12, f 1x2T: x 僆 11, q 2 d. none e. max: (1, 4) f. f 1x2 7 0: x 僆 14, 22 g. f 1x2 6 0: x 僆 1q, 42 ´ 12, q2 23. (3, 4), (3, 4), 13, 42, 13, 42 25. 313 L 27. x 6, x 2 2 ln 2 ln 3 ln 12 29. x ⬇ 1.387 ax or b ln 2 ln 3 ln 6

CHAPTER 9 Exercises 9.1, pp. 770–772 1. pattern, order 3. recursive 5. formula defining the sequence uses the preceding term(s); answers will vary. 7. 1, 3, 5, 7; a8 15; a12 23 9. 0, 9, 24, 45; a8 189; a12 429 11. 1, 2, 3, 4; a8 8; a12 12 1 2 3 4 8 12 1 1 1 1 1 1 13. , , , ; a8 ; a12 15. , , , ; a8 ; a12 2 3 4 5 9 13 2 4 8 16 256 4096 1 1 1 1 1 17. 1, , , ; a8 ; a12 2 3 4 8 12 1 1 1 1 1 1 19. , , , ; a8 ; a12 2 6 12 20 72 156 21. 2, 4, 8, 16; a8 256; a12 4096 1 1 1 23. a. 79 b. {1, 2, 7, 14, 23} 25. a. 15 b. {1, 1 2 , 3, 4 , 5} 1 1 1 1 625 7776 27. a. 32 b. {2, 1, 2 , 4 , 8 } 29. a. approx. 2.6 b. {2, 49 , 64 27 , 256 , 3125 } 1 1 1 1 1 31. a. 36 b. {13 , 10 , 21 , 36 , 55 } 33. 2, 7, 32, 157, 782 35. 1, 4, 19, 364, 132,499 37. 64, 32, 16, 8, 4 1 1 1 39. 336 41. 36 43. 28 45. 21 , 13 , 14 , 15 47. 13 , 120 , 15,120 , 3,991,680 9 32 137 49. 1, 2, 2 , 3 51. 15 53. 64 55. 60 57. 2 1 4 7 10 59. 1 5 15 29 47 95 61. 1 2 132 4 152 6 172 4 63. 0.5 2 4.5 8 15 65. 6 8 10 12 14 50 1 1 1 1 1 1 27 67. a b a b 3 8 15 24 35 48 112 5

69. a.

5

兺 4n

73.

兺 12k 12 q

兺 5n

b.

n1

71. a.

n1

5

3

兺 1n 32

75.

n1

k1 7

n

2

兺3

兺

77.

n n

n3 2

n1

SA-45

q

b.

79. 35

兺

0

7

2 53. a. appears nonlinear b. no common difference 472 55. d 3, a1 1 57. d 0.375, a1 0.65 59. d 115 126 , a1 63 61. 1275

63. 601.25

65. 534

1

2 k1 k

81. 100

83. 35, verified 85. $7.25, $7.75, $8.25, $8.75, $9.25; $17,760 87. an 600010.82 n1; 6000, 4800, 3840, 3072, 2457.60, 1966.08 89. ⬇2690 91. approaches 1 n

93.

兺 ca ca j

j1

1

ca2 ca3

c1a1 a2 a3 n

c

兺a

p

p

can1 can

an1 an 2

j

j1

1 1 or 0.25 97. 3x 81 ;x4 4 99. 1 15, 22, 1 15, 22, 1 15, 22, 1 15, 22 95.

Exercises 9.2, pp. 779–782 1. common, difference 3.

n1a1 an 2

, nth 5. Answers will vary. 2 7. arithmetic; d 3 9. arithmetic; d 2.5 11. not arithmetic; all prime 1 13. arithmetic; d 24 15. not arithmetic; an n2

67. 75. 81. 85.

82.5 69. 74.04 71. 21022 73. S6 21; S75 2850 at 11 P.M. 77. 5.5 in.; 54.25 in. 79. a7 220; a12 2520; yes a. linear function b. quadratic 83. t ⬇ 3.6 f 1x2 49x 972; 1364

Exercises 9.3, pp. 791–795 1. multiplying 3. a1rn1 5. Answers will vary. 7. r 2 9. r 2 11. not geometric; an n2 1 13. r 0.1 15. not geometric; ratio of terms decreases by 1 17. r 25 240 19. r 12 21. r 4x 23. not geometric; an n! 3 25. 5, 10, 20, 40 27. 6, 3, 3 29. 4, 4 13, 12, 1213 2 ,4

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31. 0.1, 0.01, 0.001, 0.0001 1 n1 3 1 25 33. an 24a b ; a7 35. an 152 n1; a4 2 8 20 4 37. an 21 222 n1 1 222 n1; a7 16 1 1 39. a1 27 , r 3, an 27 132 n1, a6 9, a10 729, a12 6561 1 1 41. a1 729, r 13 , an 7291 13 2 n1, a6 3, a10 27 , a12 243

43. a1 12 , r 12, an 12 1 122 n1, a6 2 12, a10 812, a12 16 12 45. a1 0.2, r 0.4, an 0.210.42 n1 a6 0.002048, a10 0.0000524288, a12 0.000008388608 47. 5 49. 11 51. 9 53. 8 55. 13 57. 9 59. a. appears exponential b. r 0.2 c. an 131.2510.22 n1 200

8

0

3 n1 119. a0 462; an 277.2 a b , a5 ⬇ 35.9 in3, 7 strokes 5 121. a0 8, an 610.752 n1, 2 days, 8 days 123. a0 50; an 100122 n1, a10 51,200 bacterium, 12 half-hours (6 hr) 4 n1 125. a0 2 m; an 1.6 a b , a7 ⬇ 0.42 m, 5 total distance a0 2Sq 18 m 127. about 67,109 in. This is almost 1.06 mi. 129. 40,000 1750x vs. 40,000(1.04)x; 6 yr 131. a. For an arithmetic sequence, the difference d ak ak1 must be constant. For ak a1rk1 and ak1 a1rk2, we have d log1a1rk1 2 log1a1rk2 2 log1a1 2 log1rk1 2 3log1a1 2 log1rk2 2 4 log1a1 2 1k 12log1r2 log1a1 2 1k 22log1r2 log 1r2 3 1k 12 1k 22 4 log1r2 ✓ ak b. For a geometric sequence, the ratio r must be constant. ak1 For ak a1 1k 12d and ak1 a1 1k 22d we have r

50 61. 63. 69. 77. 81.

a. appears nonexponential b. no common ratio 32 r 23 , a1 729 65. r 32 , a1 243 67. r 32 , a1 256 81 10,920 71. 3872 73. 2059 75. 728 27 ⬇ 143.41 8 257.375 85 79. ⬇1.60 8 10.625 1364

10a1 1k12d

10a1 1k22d 10a1101k12d 10a1101k22d 101k12d

101k22d 101k12d 1k22d 10d 31k12 1k224 10d✓ 133. x 0 135. p1502 ⬇ 2562.1, p1752 ⬇ 3615.6, p11002 ⬇ 4035.1, p11502 ⬇ 4189.1


83.

31,525 2187

85.

387 512

87. 521 89. 3367 91. 14 15 1 2 93. 27 95. no 97. 125 25 1296 7 3 99. 12 101. 4 103. 13 105. 32 107. No finite sum exists. 109. 1296 111. an 2410.82 n1, a7 ⬇ 6.3 ft, Sq 120 ft 113. a0 46,000; an 36,80010.82 n1, a4 $18,841.60, 10 yr 115. a0 160; an 155.210.972 n1, a8 ⬇ 125.4 gpm, 10 mo 117. a0 277; an 283.3711.0232 n1, a10 ⬇ 347.7 million

1. finite, universally 3. induction, hypothesis 5. Answers will vary. 7. an 10n 6 a4 10142 6 40 6 34; a5 10152 6 50 6 44; ak 10k 6; ak1 101k 12 6 10k 10 6 10k 4 9. an n a4 4; a 5 5; ak k; ak1 k 1 11. an 2n1 a4 241 23 8; a5 251 24 16; ak 2k1; ak1 2k11 2k 13. Sn n15n 12 S4 415142 12 4120 12 41192 76; S5 515152 12 5125 12 51242 120; Sk k15k 12; Sk1 1k 12151k 12 12 1k 1215k 5 12 1k 1215k 42 n1n 12 15. Sn 2 414 12 4152 S4 10; 2 2 515 12 5162 S5 15; 2 2 k1k 12 Sk ; 2 1k 121k 22 1k 121k 1 12 Sk1 2 2

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Student Answer Appendix 17. Sn 2n 1 S4 24 1 16 1 15; S5 25 1 32 1 31; Sk 2k 1; Sk1 2k1 1 19. an 10n 6; Sn n15n 12 S4 415142 12 4120 12 41192 76; a5 10152 6 50 6 44; S5 515152 12 5125 12 51242 120; S4 a5 S5 76 44 120 120 120 Verified n1n 12 21. an n; Sn 2 414 12 4152 S4 10; 2 2 a5 5 ; 515 12 5162 S5 15; 2 2 S4 a5 S5 10 5 15 15 15 Verified 23. an 2n1; Sn 2n 1 S4 24 1 16 1 15; a5 251 24 16; S5 25 1 32 1 31; S4 a5 S5 15 16 31 31 31 Verified 25. a. an n3; Sn 11 2 3 4 p n2 2 S1 12 13 S5 11 2 3 4 52 2 152 225 1 8 27 64 125 225 S9 11 2 p 92 2 452 2025 1 8 p 729 2025 n1n 12 2 n2 1n 12 2 b. c d 2 4 27. 1. Show Sn is true for n 1. S1 111 12 1122 2 Verified 2. Assume Sk is true: 2 4 6 8 10 p 2k k1k 12 and use it to show the truth of Sk1 follows. That is: 2 4 6 p 2k 21k 12 1k 121k 1 12 Sk ak1 Sk1 Working with the left hand side: 2 4 6 p 2k 21k 12 k1k 12 21k 12 1k 121k 22 Sk1 Since the truth of Sk1 follows from Sk, the formula is true for all n. 29. 1. Show Sn is true for n 1. 511211 12 5122 S1 5 2 2 Verified 2. Assume Sk is true: 5k1k 12 5 10 15 p 5k 2 and use it to show the truth of Sk1 follows. That is:

SA-47

51k 121k 1 12 5 10 15 p 5k 51k 12 2 Sk ak1 Sk1 Working with the left hand side: 5 10 15 p 5k 51k 12 5k1k 12 51k 12 2 5k1k 12 101k 12 2 1k 1215k 102 2 51k 121k 22 2 Sk1 Since the truth of Sk1 follows from Sk, the formula is true for all n. 31. 1. Show Sn is true for n 1. S1 112112 32 5 Verified 2. Assume Sk is true: 5 9 13 17 p 14k 12 k12k 32 and use it to show the truth of Sk1 follows. That is: 5 9 13 17 p 14k 12 341k 12 1 4 1k 12121k 12 32 Sk ak1 Sk1 Working with the left hand side: 5 9 13 17 p 14k 12 3 41k 12 14 k12k 32 4k 5 2k2 3k 4k 5 2k2 7k 5 1k 1212k 52 Sk1 Since the truth of Sk1 follows from Sk, the formula is true for all n. 33. 1. Show Sn is true for n 1. S1

3131 12 2

313 12 2

3122 2

3

Verified 2. Assume Sk is true: 3 9 27 p 3k

313k 12

2 and use it to show the truth of Sk1 follows. That is: 3 9 27 p 3k 3k1

313k1 12

2 Sk ak1 Sk1 Working with the left hand side: 3 9 27 p 3k 3k1 313k 12 3k1 2 k 313 12 213k1 2 2 3k1 3 213k1 2 2 313k1 2 3 2 313k1 12 2 Sk1 Since the truth of Sk1 follows from Sk, the formula is true for all n. 35. 1. Show Sn is true for n 1. S1 211 2 22 2 4 2 2 Verified 2. Assume Sk is true: 2 4 8 p 2k 2k1 2 and use it to show the truth of Sk1 follows. That is: 2 4 8 p 2k 2k1 2k2 2 Sk ak1 Sk1

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Working with the left hand side: 2 4 8 p 2k 2k1 2k1 2 2k1 212k1 2 2 2k2 2 Sk1 Since the truth of Sk1 follows from Sk, the formula is true for all n. 37. 1. Show Sn is true for n 1. 1 1 1 S1 2112 1 21 3 Verified 2. Assume Sk is true: 1 1 1 1 k p 3 15 35 12k 1212k 12 2k 1 and use it to show the truth of Sk1 follows. That is: 1 1 1 1 p 3 15 35 12k 1212k 12

1 k1 121k 12 12121k 12 12 21k 12 1 Sk ak1 Sk1 Working with the left hand side: 1 1 1 1 1 p 3 15 35 12k 1212k 12 12k 12 12k 32 k 1 2k 1 12k 1212k 32 k12k 32 1 12k 1212k 32 2k2 3k 1 12k 1212k 32 12k 121k 12 12k 1212k 32 k1 2k 3 Sk1 Since the truth of Sk1 follows from Sk, the formula is true for all n. Show Pn is true for n 1. P1: 31 2112 1 321 33 Verified Assume Pk: 3k 2k 1 is true and use it to show the truth of Pk1 follows. That is: 3k1 21k 12 1. Working with the left hand side: 3k1 313k 2 312k 12 6k 3 Since k is a positive integer, 6k 3 2k 3 Showing Pk1: 3k1 2k 3 Since the truth of Pk1 follows from Pk, the statement is true for all n. Show Pn is true for n 1. P1: 3 # 411 41 1 3 # 40 4 1 3#13 33 Verified Assume Pk: 3 # 4k1 4k 1 is true and use it to show the truth of Pk1 follows. That is: 3 # 4k11 4k1 1. Working with the left hand side: 3 # 4k 3 # 414k1 2 4 # 314k1 2 414k 12 4k1 4

39. 1.

2.

41. 1.

2.

Since k is a positive integer, 4k1 4 4k1 1 Showing Pk1: 3 # 4k 4k1 1 Since the truth of Pk1 follows from Pk, the statement is true for all n. 43. n2 7n is divisible by 2 1. Show Pn is true for n 1. Pn: n2 7n 2m P1: 112 2 7112 2m 1 7 2m 6 2m Verified 2. Assume Pk: k2 7k 2m for m 僆⺪ and use it to show the truth of Pk1 follows. That is: 1k 12 2 71k 12 2p for p 僆⺪. Working with the left hand side: 1k 12 2 71k 12 k2 2k 1 7k 7 k2 7k 2k 6 2m 2k 6 21m k 32 is divisible by 2. Since the truth of Pk1 follows from Pk, the statement is true for all n. 45. n3 3n2 2n is divisible by 3 1. Show Pn is true for n 1. Pn: n3 3n2 2n 3m P1: 112 3 3112 2 2112 3m 1 3 2 3m 6 3m 2m Verified 2. Assume Pk: k3 3k2 2k 3m for m 僆⺪ and use it to show the truth of Pk1 follows. That is: 1k 12 3 31k 12 2 21k 12 3p for p 僆⺪. Working with the left hand side: 1k 12 3 31k 12 2 21k 12 k3 3k2 3k 1 31k2 2k 12 2k 2 k3 3k2 2k 31k2 2k 12 3k 3 k3 3k2 2k 31k2 2k 12 31k 12 3m 31k2 2k 12 31k 12 is divisible by 3. Since the truth of Pk1 follows from Pk, the statement is true for all n. 47. 6n 1 is divisible by 5 1. Show Pn is true for n 1. Pn: 6n 1 5m P1: 61 1 5m 6 1 5m 5 5m 1m Verified 2. Assume Pk: 6k 1 5m or 6k 5m 1 for m 僆⺪ and use it to show the truth of Pk1 follows. That is: 6k1 1 5p for p 僆⺪. Working with the left hand side: 6k 1 616k 2 1 615m 12 1 30m 6 1 30m 5 516m 12 is divisible by 5. Since the truth of Pk1 follows from Pk, the statement is true for all n. 49. verified 1 1 3 3 8 7 51. A B c d; A B c d ; 2A 3B c d; 7 4 1 2 6 7 6 7 5 3 0.3 0.1 AB c d ; BA c d ; B1 c d 10 0 5 11 0.4 0.2 2 2 53. 1x 42 1y 32 25

Mid-Chapter Check, p. 803 1. 3, 10, 17, a9 59

2. 4, 7, 12, a9 84

3. 1, 3, 5, a9 17

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6

4. 360

5.

兺 13k 22

6. d

7. e

8. a

1. one 3. 1a 12b2 2 5 5. Answers will vary. 7. x5 5x4y 10x3y2 10x2y3 5xy4 y5 9. 16x4 96x3 216x2 216x 81 11. 41 38i 13. 35 15. 10 17. 1140 19. 9880 21. 1 23. 1 25. c5 5c4d 10c3d2 10c2d3 5cd4 d5 27. a6 6a5b 15a4b2 20a3b3 15a2b4 6ab5 b6 29. 16x4 96x3 216x2 216x 81 20 2 31. 11 2i 33. x9 18x8y 144x7y2 35. v24 6v22w 33 2v w 4 3 2 2 10 37. 35x y 39. 1792p 41. 264x y 43. ⬇0.25 45. a. ⬇17.8% b. ⬇23.0% 47. a. ⬇0.89% b. ⬇7.0% c. ⬇99.0% d. ⬇61.0% 49. 2n1, 2048 y f 132 1 51. 10

9. b

k1

10. c 11. a. a1 2, d 3, an 3n 1 b. a1 32 , d 34 , an 34 n 34 12. n 25, S25 950 13. n 16, S16 128

14. S10 5

14,762 15. S10 16. a. a1 2, r 3, an 2132 n1 27 b. a1 12 , r 12 , an 1 12 2 n 17. n 8, S8 1640 18. 343 27 6 19. 1785 20. ⬇4.5 ft; ⬇127.9 ft

Reinforcing Basic Concepts, pp. 803–804 Exercise 1: $71,500

Exercises 9.5, pp. 811–816 1. experiment, well-defined 7. a. 16 possible

t

3. N

8 6 4 2


108642 2 4 6 8 10

Begin

W

W

X

X

Y

Z

W

X

Y

Y

Z

W

X

53. g1x2 7 0: x 僆 12, 02 ´ 13, q 2

Z

Y

Z

W

X

Y

8 6 4 2 2 4 6 8 10 x

Exercises 9.6, pp. 823–829 1. n1E2 3. 0, 1, 1, 0 5. Answers will vary. 7. S 5HH, HT, TH, TT6, 14 9. S {coach of Patriots, Cougars, 1 Angels, Sharks, Eagles, Stars}, 16 11. P1E2 49 13. a. 13 b. 41 1 c. 21 d. 26 15. P1E1 2 18 , P1E2 2 58 , P1E3 2 34 17. a. 43 b. 1 c. 41 d. 12 19. 43 21. 67 23. 0.991 1 11 60 25. a. 12 b. 12 c. 98 d. 56 27. 10 29. 143 31. b, about 12% 21 7 33. a. 0.3651 b. 0.3651 c. 0.3969 35. 0.9 37. 24 39. 0.59 2 7 2 9 41. a. 61 b. 36 c. 19 d. 94 43. a. 25 b. 50 c. 0 d. 25 e. 1 5 1 2 8 3 1 45. 34 47. 11 49. a. b. c. d. e. f. 15 18 9 9 4 36 12 1 51. 14 ; 256 ; answers will vary. 53. a. 0.33 b. 0.67 c. 1 d. 0 1 9 e. 0.67 f. 0.08 55. a. 21 b. 12 c. 81 57. a. 16 b. 41 c. 16 5 3 3 1 9 2 d. 16 59. a. 26 b. 26 c. 13 d. 26 e. 13 f. 11 61. a. 81 26 5 1 3 47 2 3 9 11 b. 16 c. 16 63. a. 100 b. 25 c. 100 d. 50 e. 100 65. a. 429 8 1 1 b. 2145 67. 3360 69. 1,048,576 ; answers will vary; 20 heads in a row. 71. (9, 1, 1) y 73. x 僆 3 1, 02 ´ 3 1, q 2 5 4 3 2 1

54321 1 2 3 4 5

1 2 3 4 5 x

10 8 6 4 (2, 0) 2

Z

b. WW, WX, WY, WZ, XW, XX, XY, XZ, YW, YX, YY, YZ, ZW, ZX, ZY, ZZ 9. 32 11. 15,625 13. 2,704,000 15. a. 59,049 b. 15,120 17. 360 if double veggies are not allowed, 432 if double veggies are allowed. 19. a. 120 b. 625 c. 12 21. 24 23. 4 25. 120 27. 6 29. 720 31. 3024 33. 40,320 35. 6; 3 37. 90 39. 336 41. a. 720 b. 120 c. 24 43. 360 45. 60 47. 60 49. 120 51. 30 53. 60, BANANA 55. 126 57. 56 59. 1 61. verified 63. verified 65. 495 67. 364 69. 252 71. 8! 40,320 73. 8 nPr 3 336 75. 20 nCr 5 15,504 77. 8 nCr 4 70 79. a. ⬇1.2% b. ⬇0.83% 81. 7776 83. 324 85. 800 87. 6,272,000,000 89. 518,400 91. 357,696 93. 6720 95. 8 97. 10,080 99. 5040 101. 2880 103. 5005 105. 720 107. 52,650, no 109. 6 · (6 nCr 3 2) 2 · 20 148 111. an 1 1n 124; 137; 2415 y 113. 10

108642 2 4 6 8 10

2 4 6 8 10 x

108642 2 4 6 8 10

y

(0, 0) (3, 0) 2 4 6 8 10 x

Making Connections, p. 837 1. c

3. b

5. d

7. b

9. d

11. a 13. e 15. g

Summary and Concept Review, pp. 838–842 5 11 1. 1, 6, 11, 16; a10 46 2. 1, 35 , 25 , 17 ; a10 101 4 3. an n ; a6 1296 4. an 17 1n 12132; a6 2 5. 255 6. 112 7. 140 8. 35 9. 2, 6, 12, 20, 30 10. 12 , 34 , 54 , 94 , 17 256 4 7

11.

兺 1i

2

3i 22; 210

12. a. about 134 hawks b. 8 yr

i1

an 2 31n 12; 119 14. an 3 1221n 12; 65 740 16. 1335 17. 630 18. 11.25 19. 875 20. 7.55 m 3645 22. 32 23. 2401 24. 10.75 25. 6560 26. 819 512 does not exist 28. 50 29. 4 30. 63,050 31. does not exist 32. 5 9 6561 2 n1 33. a0 121,500, a1 81,000, an 81,000 a b , a7 ⬇ 7111 ft3 3

13. 15. 21. 27.

34. a0 1225, a1 ⬇ 1311, an 131111.072 n1, a15 ⬇ 3380, S15 ⬇ 32,944 111 12 35. (1) Show Sn is true for n 1: S1 1✓ 2 k1k 12 (2) Assume Sk is true: 1 2 3 p k 2 Use it to show the truth of Sk1: 1 2 3 p k 1k 12

1k 121k 1 12

left hand side: 1 2 3 p k 1k 12

k1k 12

1k 12 2 1k 121k 22

2

k1k 12 21k 12

2

36. (1) Show Sn is true for n 1: S1

2 111 12 3 2112 14 6

(2) Assume Sk is true: 1 4 9 p k2

1✓

k1k 1212k 12

6 Use it to show the truth of Sk1: 1k 121k 1 12 321k 12 14 1 4 9 p k2 1k 12 2 6

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left hand side: 1 ⫹ 4 ⫹ 9 ⫹ p ⫹ k2 ⫹ 1k ⫹ 12 2 ⫽

k1k ⫹ 1212k ⫹ 12

⫹

61k ⫹ 12 2

⫽

1k ⫹ 12 3 12k2 ⫹ k ⫹ 6k ⫹ 6 4

6 6 6 1k ⫹ 1212k2 ⫹ 7k ⫹ 62 1k ⫹ 121k ⫹ 2212k ⫹ 32 ⫽ 6 6 (1) Show Pn is true for n ⫽ 1: P1: 41 ⱖ 3112 ⫹ 1✓ (2) Assume Pk is true: 4k ⱖ 3k ⫹ 1 Use it to show the truth of Pk⫹1: 4k⫹1 ⱖ 31k ⫹ 12 ⫹ 1 ⫽ 3k ⫹ 4 left hand side: 4k⫹1 ⫽ 414k 2 ⱖ 413k ⫹ 12 ⫽ 12k ⫹ 4 Since k is a positive integer, 12k ⫹ 4 ⱖ 3k ⫹ 4 showing 4k⫹1 ⱖ 3k ⫹ 4 (1) Show Pn is true for n ⫽ 1: P1: 6 # 71⫺1 ⱕ 71 ⫺ 1✓ (2) Assume Pk is true: 6 # 7k⫺1 ⱕ 7k ⫺ 1 Use it to show the truth of Pk⫹1: 6 # 7k ⱕ 7k⫹1 ⫺ 1 left hand side: 6 # 7k ⫽ 7 # 6 # 7k⫺1 ⱕ 7 # 7k ⫺ 1 ⱕ 7k⫹1 ⫺ 7 Since k is a positive integer, 7k⫹1 ⫺ 7 ⱕ 7k⫹1 ⫺ 1. (1) Show Pn is true for n ⫽ 1: P1: 31 ⫺ 1 ⫽ 2 or 2112 ✓ (2) Assume Pk is true: 3k ⫺ 1 ⫽ 2p or 3k ⫽ 2p ⫹ 1 for p 僆⺪ Use it to show the truth of Pk⫹1: 3 k⫹1 ⫺ 1 ⫽ 2q for q 僆⺪ left hand side: 3k⫹1 ⫺ 1 ⫽ 3 # 3k ⫺ 1 ⫽ 3 # 12p ⫹ 12 ⫺ 1 ⫽ 3 # 2p ⫹ 3 ⫺ 1 ⫽ 3 # 2p ⫹ 2 ⫽ 213p ⫹ 12 ⫽ 2q is divisible by 2 six ways Begin ⫽

37.

38.

39.

40.

B

A

51k ⫹ 12 2 ⫺ 1k ⫹ 12 5k2 ⫺ k ⫹ 51k ⫹ 12 ⫺ 3 ⫽ 2 2 101k ⫹ 12 ⫺ 6 1k ⫹ 12 351k ⫹ 12 ⫺ 14 5k2 ⫺ k ⫹ ⫽ 2 2 2 2 1k ⫹ 12 15k ⫹ 42 5k ⫹ 9k ⫹ 4 ⫽ 2 2 15k ⫹ 421k ⫹ 12 15k ⫹ 42 1k ⫹ 12 ⫽ ✓ 2 2 1⫺1 1 10. For n ⫽ 1: 2 # 3 ⱕ3 ⫺1 2112 ⱕ 2✓ Assume: 2 # 3k⫺1 ⱕ 3k ⫺ 1 Prove: 2 # 31k⫹12 ⫺1 ⱕ 3k⫹1 ⫺ 1 2 # 31k⫹12 ⫺1 ⫽ 2 # 31k⫺12 ⫹1 ⫽ 2 # 31k⫺12 # 3 ⱕ 13k ⫺ 12 # 3 13k ⫺ 12 # 3 ⫽ 3k⫹1 ⫺ 3 ⱕ 3k⫹1 ⫺ 1✓ 11. a. Begin Prove:

B

A

C

B

C

A

C

A

B

C

B

C

A

B

A

b. ABC, ACB, BAC, BCA, CAB, CBA 12. 302,400 13. 13 14. a. 720 b. 120 c. 20 15. 900,900 16. 302,400 17. a. x4 ⫺ 8x3y ⫹ 24x2y2 ⫺ 32xy3 ⫹ 16y4 b. ⫺4 18. a. x10 ⫹ 1012 x9 ⫹ 90x8 b. a8 ⫺ 16a7b3 ⫹ 112a6b6 5 5 7 19. 0.989 20. a. 14 b. 12 c. 31 d. 12 e. 12 f. 14 g. 12 h. 0 21. a. 0.08 b. 0.92 c. 1 d. 0 e. 0.95 f. 0.03 22. a. 0.1875 59 b. 0.589 c. 0.4015 d. 0.2945 e. 0.4110 f. 0.2055 23. a. 100 53 13 47 b. 100 c. 100 d. 100 24. a. 0.8075 b. 0.0075 c. 0.9925 25. a. about 27.9% b. about 97.6%

C

Strengthening Core Skills, pp. 845–846 B

C

A

C

A

B

Exercise 1. C

B

C

A

B

A

41. 720; 1000 42. 24 43. 220 44. a. 5040 b. 840 c. 35 45. a. 720 b. 120 c. 24 46. 3360 47. a. 220 b. 1320 4 3 7 48. 13 49. 13 50. 65 51. 24 52. 175 53. a. 0.608 b. 0.392 396 c. 1 d. 0 e. 0.928 f. 0.178 54. a. 21 b. 56 55. a. x4 ⫺ 4x3y ⫹ 6x2y2 ⫺ 4xy3 ⫹ y4 b. 41 ⫺ 38i 56. a. a8 ⫹ 8 13a7 ⫹ 84a6 ⫹ 168 13a5 b. 78,125a7 ⫹ 218,750a6b ⫹ 262,500a5b2 ⫹ 175,000a4b3 57. a. 280x4y3 b. ⫺64,064a5b9 58. a. about 93.3% b. about 62.4%

Practice Test, pp. 842–844 1. c. 3. b. d. 5. 7. 9.

a. 12 , 45 , 1, 87 ; a8 ⫽ 16 b. 6, 12, 20, 30; a8 ⫽ 90 11 311 3, 212, 17, 16; a8 ⫽ 12 2. a. 165 b. 420 c. ⫺2343 d. 7 512 a. a1 ⫽ 7, d ⫽ ⫺3, an ⫽ 10 ⫺ 3n a1 ⫽ ⫺8, d ⫽ 2, an ⫽ 2n ⫺ 10 c. a1 ⫽ 4, r ⫽ ⫺2, an ⫽ 41⫺22 n⫺1 a1 ⫽ 10, r ⫽ 25 , an ⫽ 101 25 2 n⫺1 4. a. 199 b. 9 c. 4 d. 6 a. 1712 b. 2183 c. 2188 d. 12 6. a. ⬇8.82 ft b. ⬇72.4 ft $6756.57 8. $22,185.27 ak ⫽ 5k ⫺ 3, ak⫹1 ⫽ 51k ⫹ 12 ⫺ 3, 51k ⫹ 12 2 ⫺ 1k ⫹ 12 5k2 ⫺ k Sk ⫽ , Sk⫹1 ⫽ ; 2 2 2 5112 ⫺ 1 For n ⫽ 1: S1 ⫽ ⫽ 2✓ 2 3

Assume: Sk ⫽

5k2 ⫺ k is true, 2

Exercise 2. Exercise 3. Exercise 4.

4C1

# 13C5 ⫺ 40 52C5

4 # 13C3 # 39C2 52C5

4

# 13C4 # 39C1 52C5

4#

⬇ 0.001 970

⬇ 0.326 170 ⬇ 0.042 917

10C5 ⬇ 0.000 388 52C5

Cumulative Review Chapters R–9, pp. 846–847 1. a. 23 cards are assembled each hour. b. y ⫽ 23x ⫺ 155 c. 184 cards d. ⬇ 6:45 A.M. ⫺5 ⫾ 1109 kVW 3VW 3. x ⫽ ; x ⬇ 0.91; x ⬇ ⫺2.57 5. Y ⫽ ;Y⫽ 6 X 2X 7. verified; reflections across y ⫽ x ⫺1 9. a. 4x ⫹ 2h ⫺ 3 b. 1x ⫹ h ⫺ 22 1x ⫺ 22 y 11. 15 12 9 6 3 ⫺10⫺8⫺6⫺4⫺2 ⫺3 (⫺2, 0) ⫺6 ⫺9 ⫺12 ⫺15

(0, 8) y⫽2 2 4 6 8 10 x

(2, 0)

x ⫽ ⫺1 x ⫽ 1

13. a. x3 ⫽ 125

b. e5 ⫽ 2x ⫺ 1 15. a. x ⬇ 3.19 b. x ⫽ 334 1x ⫹ 32 2 1y ⫺ 32 2 ⫹ ⫽ 1; (⫺3, 3); (⫺7, 3), (1, 3); 17. (5, 10, 15) 19. 16 4 (⫺3 ⫺ 2 13, 3), (⫺3 ⫹ 2 13, 3)

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Student Answer Appendix 21. 1333 23. a. ⬇7.0% b. ⬇91.9% c. ⬇98.9% 12 d. a b 10.042 12 10.962 0; virtually nil 0 9a ⫹ 3b ⫹ c ⫽ 56 25. • 25a ⫹ 5b ⫹ c ⫽ 126; y ⫽ ⫺7x2 ⫹ 91x ⫺ 154 a. mid-June 81a ⫹ 9b ⫹ c ⫽ 98 b. ⬇142 c. 98 d. November to February 27. 20

SA-51

71. a. positive b. negative c. negative d. negative 11 73. ⫺ 75. ⫺2 77. 92 ⫽ 81 is closest 79. 7 81. ⫺2.185 83. 413 6 ⫺11 5 1 7 85. ⫺29 12 or ⫺212 87. 0 89. ⫺5 91. ⫺10 93. ⫺8 95. ⫺4 97. 12 99. 64 101. 4489.70 103. 3 105. D ⬇ 4.3 cm 107. 32°F 109. 179°F 111. about 68.13 cm 113. Tsu Ch’ung-chih: 355 113 115. negative

APPENDIX A.II pp. A-22–A-27 10

0

, (3.4044, 1.4539)

0 29. 0.2794

CHAPTER APPENDIX APPENDIX A.I pp. A-10–A-13 1. proper; subset; element 3. positive; negative; 7; ⫺7; principal 5. Order of operations requires multiplication before addition. 7. a. { 1, 2, 3, 4, 5} b. { } 9. True 11. True 13. True 15. 1.3 ⫺1

0

1

2

3

4

0

1

2

3

4

1. composite 3. space 5. Answers will vary. 7. A ⫽ 2 280 mm 2 or 22.8 cm2 9. A ⬇ 950 mm2 11. P ⫽ 18 cm 13. P ⫽ 24 cm 15. C ⬇ 15.7 cm A ⫽ 20 cm2 A ⫽ 24 cm2 17. c ⫽ 45 cm 19. P ⫽ 21␲d2 15␲ ⬇ 47.1 in. A ⫽ 486 cm2 P ⫽ 108 cm bh ␲r2 21. A ⫽ ⫹ LW 23. A ⫽ 2 4 A ⫽ 54 ft2 A ⫽ 81␲ ⬇ 254.5 in2 25. P ⫽ 2L ⫹ ␲d P ⫽ 240 ⫹ 80␲ ⬇ 491.3 ft; A ⫽ LW ⫹ ␲r2 A ⫽ 9600 ⫹ 1600␲ ⬇ 14,626.5 ft2 27. 504 in2 29. W ⫽ 12 cm 31. a. l

w

17. 2.5 ⫺1

19. a. 40,000 acres b. ten 3’s (0.333 333 333 3, may vary by calculator model) 21. ⬇2.65 ⫺2

⫺1

0

1

2

3

2

3

4

5

W L b. Total Area ⫽ area of larger rectangle plus area of smaller rectangle c. A ⫽ LW ⫹ Iw 33. a.

s

23. ⬇1.73 0

1

25. a. i. { 8, 7, 6} ii. { 8, 7, 6} iii. 1⫺1, 8, 7, 62 iv. 5⫺1, 8, 0.75, 92, 5.6, 7, 35, 66 v. { } vi. 5⫺1, 8, 0.75, 92, 5.6, 7, 35, 66 b. 5⫺1, 35, 0.75, 92, 5.6, 6, 7, 86 c. ᎑ 5.6 E t 0.75 ⫺2 ⫺1 0 1 2 3 4 5 6 7 8

27. a. i. 5149, 2, 6, 46 ii. 5149, 2, 6, 0, 46 iii. 5⫺5, 149, 2, ⫺3, 6, ⫺1, 0, 46 iv. 5⫺5, 149, 2, ⫺3, 6, ⫺1, 0, 46 v. 513, ␲6 vi. 5⫺5, 149, 2, ⫺3, 6, ⫺1, 13, 0, 4, ␲6 b. 5⫺5, ⫺3, ⫺1, 0, 13, 2, ␲, 4, 6, 1496 c. 兹3 p 兹49 ⫺6⫺5 ⫺4 ⫺3 ⫺2 ⫺1 0 1 2 3 4 5 6 7

29. 31. 33. 35. 41. 43. 45. 55. 63. 69.

False; not all real numbers are irrational. False; not all rational numbers are integers. False; 125 ⫽ 5 is not irrational. c; IV 37. a; VI 39. d; III Let a represent Kylie’s age: a ⱖ 6 years. Let n represent the number of incorrect words: n ⱕ 2 incorrect. 3 xⱖ 47. x 6 2 49. 5 ⱕ w 6 32 51. 2.75 53. ⫺4 2 1 3 57. 59. 冟⫺7.5 ⫺ 2.5冟, 冟2.5 ⫺ 1⫺7.5冟, 10 61. ⫺8, 2 2 4 negative 65. ⫺n 67. undefined, since 12 ⫼ 0 ⫽ k implies k # 0 ⫽ 12 undefined, since 7 ⫼ 0 ⫽ k implies k # 0 ⫽ 7

H

h s W L

b. Total volume ⫽ volume of rectangular solid plus volume of pyramid c. V ⫽ LWH ⫹ LWh or V ⫽ LWH ⫹ s2h 35. V ⫽ LWH ⫹ ␲r2h V ⫽ 200 ⫹ 43.75␲ ⬇ 337.4 in3 37. 156.25 gal 39. 12.5 gal 41. 7108.4 yd 43. 661.5 kg 45. K: LW C: S1 ⫹ S2 ⫹ S3 J: ␲r2 1 A: 4S I: LWH D: BH 2 H: S1 ⫹ S2 ⫹ S3 ⫹ S4 E: 2␲r G: S2 h1b1 ⫹ b2 2 B: 2L ⫹ 2W F: S3 L: 2 47. 32 ft 49. 192 ft3 51. 23.91 m 53. 91.44 cm ⫻ 76.2 cm ⫻ 121.92 cm 55. one large at about 5.8¢/in2 57. about 654.5 gal 59. a. 35.3 m2 b. 2.1 m3 c. about $280.88 61. a. 216.5 ft2 b. about 7 c. about $61.07 63. 5 in. ⫻ 3 in. 65. 2500␲ ⬇ 7854 ft3 about 58,752 gal 67. d. 35 ⫹ xy ⫺ 7y


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Index A Absolute maximum, 196 Absolute value of complex numbers, 291, 408 explanation of, 81, 220, 257 on graphing calculator, 221, 224, 225, A–6 multiplicative property of, 222 of real numbers, A–5 – A–6 Absolute value equations distance and, 231 explanation of, 275 graphically solving, 226 on graphing calculator, 221 method to solve, 220–222, 275 property of, 220 Absolute value functions, 204, 225, 257 Absolute value inequalities applications involving, 226–227 distance and, 231 explanation of, 275 graphically solving, 226 on graphing calculators, 226 solutions to “greater than,” 224–225 solutions to “less than,” 223–224 Accumulated value, 540, 543 Addition associative property of, 5, 81 commutative property of, 5, 81 of complex numbers, 284–285 distributive property of multiplication over, 6–7, 285, 768 of matrices, 652–653, 700 of polynomials, 17–18 of radical expressions, 70 of rational expressions, 56–58 Addition method. See Elimination Addition of ordinates, 351 Additive identity, 5–6, 81 Additive inverse, 6, 81 Additive property of equality, 25 of inequality, 27–28 Algebra of functions, 340–346, 373 fundamental theorem of, 394–397, 399 of matrices, 650–658 Algebraic expressions evaluation of, 3–4 explanation of, 2 method to simplify, 7 translating English phrases into, 3 Algebraic fractions. See Rational expressions Algebraic method explanation of, 482 to find inverse functions, 483–485 Algebraic terms, 2 Algorithm, division, 383 Allometric studies, 241

Amortization, 542–545 Analytical geometry algebraic tools used in, 708–709 conic section characteristics and, 710–712 distance between point and line and, 709 explanation of, 708, 755 Angles, right, A–14 Annuities, 542–545 Applications. See also Applications index of absolute value inequalities, 226–227 of algebra of functions, 345–346 of analytic parabolas, 750 of binomial theorem, 834–835 of combinations, 810–811 of composition, 362–363 of data analysis, 170 of difference quotient, 362–364 of direct variation, 262–264 of exponential and logarithmic functions, 492, 498–499, 508–512, 534–535, 539–547 of foci of conic sections, 724–725, 739–740 of functions, 189 of inequalities, 32, 464–465 of inverse functions, 487 of inverse variation, 265 of joint variation, 266 of linear equations, 31–32, 113–114, 142–144 of linear inequalities, 615 of linear programming, 622–625 of matrices, 645–646, 672–673 of piecewise-defined functions, 254–255 of polynomial functions, 405, 421–422 of power functions, 239–242 problem-solving guide for, 157–160 of quadratic equations, 306–307 of quadratic inequalities, 307–308 of rate of change, 332–334, 362–364 of rational functions, 239–242, 439, 451–454 of regression, 556–557 of remainder theorem, 390 of sequences, 768–769, 779, 789–791 of series, 790–791 of system of linear equations in three variables, 598–599 of system of linear equations in two variables, 582–586 of system of linear inequalities, 619–620 Area. See also Surface area explanation of, 32, A–14 formulas for, 33, 37, 455, 491, 627, 660, 727, A–14 – A–16 of triangle, 655, 688 Arithmetic sequences applications of, 779

explanation of, 773, 839 finding nth partial sum in, 777–778 finding nth term of, 774–777 graphs of, 776 identification of, 773–774 Associated minor matrices, 669–670 Associations linear/nonlinear, 166–167 positive and negative, 165 Associative properties, 5, 81 Asymptotes of central hyperbola, A–33 to graph rational functions, 235–236 horizontal, 234, 240, 433–435 oblique or nonlinear, 448–450 vertical, 235, 240, 430–433 Asymptotic behavior, 234, 733 Augmented matrices explanation of, 700 matrix inverses and, 704–705 of system of equations, 638–640, A–30 triangularizating, 640–642 Auxiliary lines, A–15 Average distance, 95 Average rate of change applications of, 333, 334, 362–364 calculation of, 334 difference quotient and, 359–361 explanation of, 332, 373 Average rate of change formula, 361 Axis of symmetry, 205

B Back-substitution, 26, 578 Base, A–7, A–21 Base-b, to solve exponential equations, 530 Base-e exponential functions, 495–496 Binomial coefficients, 831–832 Binomial conjugates, 19–20 Binomial cubes, 291 Binomial expansion, 830, 831, 833, 834 Binomial experiment, 834 Binomial factors, 40–41 Binomial powers, 829–831 Binomial probability, 834–836 Binomials. See also Polynomials expansion of, 830, 831, 833, 834 explanation of, 16, 829 multiplication of, 18–19, 285 Binomial squares, 20–21, 43, 759, 830 Binomial theorem applications of, 834–835 binomial expansion and, 834 explanation of, 829, 833, 842 Bisection, 429 Body mass index formula, 37 Boundary lines horizontal, 123–124, 134

I-1


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Index

Boundary lines—Cont. linear inequalities and, 615–616 vertical, 122–123, 134 Boundary point, 122 Bounded region, 621 Branches, of hyperbola, 731 Break-even analysis, 584, 610

C Calculators. See Graphing calculators Capacity, A–18 – A–19 Cardano, Girolomo, 291 Carrying capacity, 555 Center of circle, 96–98 of hyperbola, 732, 733 Centimeters of mercury, 510 Central circle, 96, 718 Central ellipse, 718 Central hyperbola, 731–732, A–33 Central rectangle, 733 Change in x, 107 Change in y, 107 Change-of-base formula, 522–523 Circles center of, 96–98 central, 96, 718 circumference of, 176, 261, 709, A–15 equation of, 96–97, 715–716, 755 explanation of, 33, 96, 710, 755 on graphing calculator, 99–100, 715 graphs of, 95–99, 715–716 perimeter and area formulas for, 33, A–14 properties of, 714 radius of, 96–98, 103, 491, 710, 714 standard form of, 96 Circumference, 176, 261, 709, A–15 Clark’s rule, A–12 Coefficient matrices, 638–639, 700 Coefficients explanation of, 2 leading, 17, 41 solving linear equations with fractional, 26 Coincident dependence, 598 Collinear points test, 688 Combinations explanation of, 809–810 on graphing calculator, 810 stating probability using, 820 Combined variation, 266 Common difference, for sequence, 773 Common logarithms explanation of, 504 method to find, 505–506 Common ratio, 782 Commutative properties, 5, 81 Complementary events, 818–819 Complements, 818–820 Completing the square applications of, 730–731, 759 explanation of, 104, 297 to graph circles, 715, 716 to graph ellipses, 719–720, 722–723 to graph hyperbolas, 735–736, 738 to graph parabolas, 746 to solve quadratic equations, 297–299 to solve quadratic functions, 313–315

Complex conjugates theorem division and, 288 explanation of, 286, 396 product of, 286 proof of, A–34 – A–35 Complex numbers absolute value of, 291, 408 addition and subtraction of, 284–285 division of, 288 explanation of, 283, 370–371 on graphing calculator, 285, 286 identifying and simplifying, 282–284 multiplication of, 285–287 raised to power, 831 square root of, 408 standard form of, 283–284 Complex polynomial functions, 394 Complex polynomials, 312 Complex zeroes, 475–476 Composite figures explanation of, A–15 – A–16 volume of, A–17 – A–18 Composite functions, 356–357 Composition applications, 362–363 Composition of functions explanation of, 352–354, 374 on graphing calculator, 355, 358–359 method to find, 354–356 notation for, 353 numerical and graphical view of, 357–359 Compounded interest formula, 541 Compound fractions, 58–59 Compound inequalities explanation of, 29 method to solve, 30–31 Compound interest explanation of, 540–542, 569 on graphing calculator, 542 Compressions, 210–211 Conditional equations, 27 Cones explanation of, 710 surface area of, 80 volume of, 35 Conical shells, 52 Conic sections. See also Circles; Ellipses; Hyperbolas; Parabolas; specific conic sections applications of foci of, 724–725, 739–740 characteristics of, 710–712 equations of, 736–737, A–32 – A33 explanation of, 708 graphs of, 200 nonlinear systems and, 749–750 Conjugate axis, 732 Conjugates binomial, 19–20 complex, 286, 288, 396 Consecutive integers, 158 Constant of variation, 261 Constant terms, 2 Constraints, 620 Continuous functions explanation of, 137 piecewise and, 247, 251 Continuous graphs, 91

Continuously compounded interest, 541–542, 545, 569 Contradictions, 27 Convenient values method, 685–687 Convex region, 621 Coordinate, A–2 Coordinate grid, 89 Correlation, strong and week, 167–169 Cost-based pricing, 584 Counting techniques combinations, 809–811 distinguishable permutations, 807–808 fundamental principle of, 806–807 listing and tree diagrams, 804–806 nondistinguishable permutations, 809 review of, 840–841 Cramer’s Rule, 680–682, 701, 704 Cube root function, 205 Cube roots, 66, A–8. See also Radical expressions Cubes binomial, 291 explanation of, 35, A–21 perfect, 44–45, 394, A–7 sum of cubes of n natural numbers, 793 sum or difference of two perfect, 44–45 volume of, A–17 Cubic units, A–16 Cubing function, 205 Cylinders, 52, 162, 176, 311, 349, 455 Cylindrical shells, 52

D Data analysis, applications of, 170 Decay rate, 545–547 Decimal notation, 15–16 Decimals, A–2, A–3 Decision variables, 622 Decomposition of composite function, 356–357 for rational expressions, 682, 685–688 of rational terms, 2 Decomposition template, 682–685 Degenerate cases, 104 Degree, of polynomials, 16 Delta (⌬), 107 Denominator, rationalizing the, 71 Dependent system of equations matrices and, 644–645 in three variables, 596–598 in two variables, 581–582 Dependent variable, 88 Depreciation, 498 Descartes’ rule of signs, 402–403 Descriptive variables, 2 Determinants to find area of triangle, 688 of general matrices, A–31 of singular matrices, 668–672 to solve systems, 679–682 3 ⫻ 3, 670–671 2 ⫻ 2, 701 Diagonal entries, of matrices, 638 Difference of two squares, factoring, 43 Difference quotient applications of, 362–364


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Index average rate of change and, 359–361 explanation of, 337, 359, 374 Directrix of ellipse, 711, 712 of hyperbola, 712 of parabola, 710, 711, 746–749 Direct variation applications of, 262–264 explanation of, 261, 277 Discontinuities asymptotic, 431 removable, 446–447 Discontinuous functions, piecewise-defined, 252, 253 Discriminant of cubic equation, 467 explanation of, 301 of quadratic formula, 301–302 use of, 312 Diseconomies of scale, 610 Distance absolute value and, 231 average, 95 perpendicular, 601, 709 Distance formula, 95–96, 708, 755 Distinguishable permutations, 807–808 Distributive property of multiplication over addition, 6–7, 285, 768 Division of complex numbers, 288 factor theorem and, 387–389 on graphing calculator, A–7 long, 382 with nonlinear divisor, 385 of polynomials, 383, 384 of radical expressions, 71–72 of rational expressions, 56 remainder theorem and, 386–387, 390 synthetic, 383–384, 470, A–28 – A–29 with zero, 384–385, A–6 Division algorithm, 383 Divisor, 385 Domain of functions, 123–125, 128 implied, 124–125 of logarithmic functions, 507–508 of piecewise-defined functions, 248–249 of power function, 237–238 of rational functions, 431 of relations, 88, 134–135 solving quadratic inequality to determine, 305–306 Dominant term, of polynomial function, 413

E e, 495–496 Eccentricity, 758–759 Economies of scale, 610 Electrical resistance, 10 Elementary row operations, 640–641 Elements of set, A–1 Elimination explanation of, 579, 580, 634–635 Gaussian, 40, 642, A–30 Gauss-Jordan, 642–645, A–30 to solve system of linear equations in three variables, 593–596

to solve system of linear equations in two variables, 579–581 to solve system of nonlinear equations, 607–608 Ellipses applications using characteristics of, 724–725 area of, 727 central, 718 characteristics of, 711–712, 724–725 definition of, 720, 721 equation of, 716–717, 720–721, 723–724, 755, A–32 focal chord, 723 foci of, 720–725 graphs of, 717–724 horizontal, 717–718 perimeter of, 727 with rational/irrational values, 759 vertical, 718–719 Elongation, 758–759 Empty set, A–1 End-behavior explanation of, 195 of polynomial graphs, 412–415, 419 of rational functions, 233, 234 Endpoint maximum, 196 Endpoints, 122, A–4 Equality additive property of, 25 of matrices, 650–651 multiplicative property of, 25 power property of, 72, 74 square root property of, 295–297 Equations. See also Linear equations; System of linear equations; System of linear equations in three variables; System of linear equations in two variables; System of nonlinear equations; specific types of equations absolute value, 220–222, 226, 231 of circle, 96–97, 715–716, 755 conditional, 27 of conic sections, 736–737 of ellipse, 716–717, 720–721, 723–724, 755, A–32 equivalent, 25 explanation of, 25, 150 exponential, 496–499, 503–505, 517–519, 527, 531–533, 569 families of, 25 in function form, 137, 138 of functions, 236, 249, 253–254, 316, 438 of hyperbolas, 731–736, 756, A–32 – A–33 intersection-of-graphs method to solve, 150–153, 183, 519 of line, 138–141 literal, 155–156 logarithmic, 517–519, 527–530, 569 logistic, 533–534, 572 matrix, 663, 667–668 of parabola, 746–749, 756 parametric, 467 of piecewise-defined function, 249, 253–254

polynomial, 46–48, 312 present value, 540 quadratic, 46–48, 292–302, 371 of quadratic function, 316 radical, 72–76 rational, 59–60 of rational function, 236, 438 regression, 171–172, 552, 556–557, 570 relations stated as, 89 roots of, 25 of semi-hyperbola, 742 variation, 261, 263, 265 written information translated into, 31–32 Equivalent equations, 25 Equivalent system of equations, 579, 593–594 Euler, Leonhard, 37 Euler’s polyhedron formula, 37 Even functions, 190–191 Events complementary, 818–819 explanation of, 816 mutually exclusive, 822 nonexclusive, 821–823 probability of, 817, 818 Experiments, 805 Exponential decay, 545–547 Exponential equations explanation of, 569 logarithmic form and, 503–505 method to solve, 517–519, 531–533 uniqueness property to solve, 496–499 Exponential form, 504–505, 519, A–7 Exponential functions. See also One-to-one functions applications of, 492, 498–499 base-e, 495–496 evaluation of, 492–493 explanation of, 492, 566–567 on graphing calculator, 496–499 graphs of, 493–495 natural, 496 Exponential growth, 495, 545–546 Exponential notation, 11, A–7 Exponential properties multiplying terms using, 11 simplifying expressions using, 14 summary of, 15 Exponential regression model, 553–554 Exponents explanation of, 11, A–7 power property of, 11–12, 81 product property of, 11, 12, 81 quotient property of, 13, 81 rational, 67–68 zero and negative, 13, 81 Extraneous roots, 60, 527 Extrapolation, 556–557, 570 Extreme values explanation of, 316 quadratic functions and, 316–321

F Factorial formulas, 814 Factorial notation, 765 Factorials, 764–765

I-3


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Index

Factoring chart of methods of, 46 difference of two squares, 43 explanation of, 39, 46, 82 by grouping, 40–41 nested, 53 perfect square trinomials, 43–44 polynomial equations, 46–49 quadratic forms, 45–46 quadratic polynomials, 41–43 sum or difference of two perfect cubes, 44–45 trial-and-error method for, 42–43 Factors greatest common, 39–40 Factor theorem explanation of, 387, 471 to find factors of polynomials, 387–389 finding zeroes using, 389 proof of, A–34 Families of curves, 711 of equations, 25 of functions, 204, 236 of polynomials, 16 Feasible region, 621 Finite sequences, 762, 838 Finite series, 765–766 Focal chord of ellipse, 723 explanation of, 723 of hyperbolas, 738, 743 of parabola, 748, 753 Foci applications of, 724–725 of ellipse, 711, 720–725 explanation of, 710 of hyperbola, 712, 737–740 of parabola, 710, 711, 746–749 Foci formula, 723, 738 Focus-directrix form of equation of parabola, 746–749 F-O-I-L method, 19, 41, 285, 399 Folium of Descartes, 467 Formulas area, 33, 37, 455, 491, 627, 660, 727, A–14 – A–16 average rate of change, 361 binomial coefficients, 831–832 binomial cubes, 291 binomial probability, 836 body mass index, 37 change-of-base, 522–523 Clark’s rule, A–12 compounded interest, 541 distance, 95–96, 708, 755 Euler’s polyhedron, 37 explanation of, 155 exponential growth, 495, 545 factorial, 814 foci, 723, 738 inverse of matrices, 704 lift capacity, 37 midpoint, 95, 708, 755 perimeter, 33, 660, 727, A–14 – A–16 perpendicular distance from point to line, 601, 709, 713, 755

Pick’s theorem, 132 pitch diameter, A–12 population density, 381, 441 Pythagorean Theorem, 75–76, 83 radius, 490 required interest rate, 269 right parabolic segment, 752 simple interest, 539 slope, 107, 108, 110, 138 Stirling’s formula, 814 sum of cubes of first n natural numbers, 793 sum of first n natural numbers, 781 sum of squares of first n natural numbers, 781 surface area of cylinder, 162 surface area of rectangular box with square ends, 323 velocity, 79, 80, 236 vertex, 315–316 vertex/intercept, 323 volume, 34, 218, 491, 627 Fractions compound, 58–59 partial, 682–688, 702 Function families, 204 Function form, of linear equations, 137, 138 Function notation, 125–127 Functions absolute value, 204, 225, 257 algebra of, 340–346, 373 applications of, 189, 345–346 base, 378 composite, 356–357 composition of, 352–359, 374 constant, 137 continuous, 247 cube root, 205 cubing, 205 domain and range of, 122–125, 128 evaluation of, 126–127 even, 190–191 explanation of, 119–120, 181–182 exponential, 492–499, 566–567 on graphing calculator, 127, 244–245 graphs of, 120–122, 127–128, 190–197, 204–214, 273, 343–345, 485–487 identification of, 120 identity, 137, 204 increasing or decreasing, 194–195 inverse, 481–487, 527, 566 linear, 138, 169–171 logarithmic, 504–508, 567 maximum and minimum value of, 196, 621–622 nonlinear, 331–332 objective, 620 odd, 191–192 one-on-one, 480–484, 486, 566 piecewise-defined, 247–255, 276 polynomial, 192–193, 394–405, 411–420 positive and negative, 192–193 power, 236–242, 275, 296 products and quotients of, 341–343 quadratic, 293–294, 313–321, 372 range of, 123–124 rational, 232–236, 239–242, 275, 430–439, 473

as relations, 119–122 root, 236, 238–239 sand dune, 257 smooth, 247 square root, 204 squaring, 204 step, 254–255 sums and differences of, 340–341 vertical line test for, 120–122 zeroes of, 192–194 Fundamental principle of counting (FPC), 806–807, 820 Fundamental properties of logarithms explanation of, 517 to solve equations, 516–519 Fundamental property of rational expressions, 53–54 Fundamental theorem of algebra, 394–397, 399

G Galileo Galilei, 236, 363 Gauss, Carl Friedrich, 394, 642 Gaussian elimination, 640, 642, A–30 Gauss-Jordan elimination, 642–645, A–30 General form, of equation of circle, 98 General functions, transformations of, 211–214 General linear equations, 147 Geometric sequences applications of, 789–791 explanation of, 782–783, 839 finding nth partial sum of, 787–788 finding nth term of, 783–787 sum of infinite, 788–789 Geometric series, 782 Geometry analytical, 708–712 perimeter and area formulas, A–14 – A–16 plane, 708–709 unit conversion factors, A–18 – A–21 verifying theorem from basic, 708 volume, A–16 – A–18 Global maximum, 196 Goodness of fit, 167, 376–377 Graphical solutions intersect method for, 150–152 for linear inequalities, 154–155 x-intercept/zeroes method for, 152–154 Graphing calculator features intersect, 150–151 QuadReg command on, 329, 330 repeat graph, 758 split screen viewing, 399 TABLE feature, 92–93, 402, 511–512, 544 window size, 603–604 Graphing calculators absolute value on, 221, 224, 225, A–6 asymptotes on, 240 circles on, 99–100, 715 combinations on, 810 complex numbers on, 285, 286 composition of functions on, 355, 358–359 compound interest on, 542 division on, A–7 elongation and eccentricity on, 758–759 evaluating expressions on, 4


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Page I-5

Index exponential functions on, 496–499 functions on, 127, 344–345 hyperbolas on, 735, 736, 739 imaginary and complex numbers on, 285 intersection-of-graphs method on, 160, 519, 545–546 inverse functions on, 484–487 linear equations on, 114 linear inequalities on, 617, 619 linear programming on, 633–634 logarithms on, 505, 506, 508, 511–512, 520, 528–535 logistic equations on, 533, 534 matrices on, 643–645, 653, 656–658, 665, 667–669, 672–673 maximums and minimums on, 196–197, 317, 319–321 order of operations on, A–9 parallel lines on, 140 partial sums on, 766–767, 769 permutations on, 808 piecewise-defined functions on, 250 polynomial inequalities on, 460 power functions on, 237 probability on, 821, 823, 835 quadratic equations on, 298, 299, 301, 302 rational functions on, 234, 447, 452–454 regression on, 171–173, 178, 242, 376–377, 552, 554–557 relations on, 92–94 repeating decimals on, A–2 sequences on, 763–765, 775–776, 785–786, 789–790 summation on, 767, 769 system of linear equations on, 577, 578, 580–586, 599 system of nonlinear equations on, 605–608 transformations on, 213, 265 translations on, 206–208 variation on, 264, 279 x-intercept on, 153–154, 293–294, 296 zeroes method on, 154 Graphs of circles, 95–99 continuous, 91 of ellipses, 717–724 end-behavior of, 195, 413 of exponential functions, 493–498 of functions, 120–122, 127–128, 190–197, 204–214, 273, 343–345 of hyperbolas, 731–735, 737–740 of linear equations, 105–114, 136–142 of lines, 138–139 of logarithmic functions, 506–507 one-dimensional, 591 of parabolas, 91, 92, 193, 745–749 of piecewise-defined functions, 248–254 of polynomial functions, 411–420, 472 of power functions, 237–239 quadratic, 378 of quadratic functions, 313–321 of rational functions, 232–236, 435–438, 446–450, 472–473, 485–487 of reflections, 208–210 of relations, 89–94 of semicircles, 92 of sequences, 776, 786

to solve inequalities, 193, 194 symmetry and, 190–192 of system of linear equations, 577 of system of nonlinear equations, 604–605 transformations of, 211–214 of translations, 206, 207 two-dimension, 591 of variation, 262–265 vertically stretching/compressing basic, 210–211 Gravity, 236, 244, 269, 271, 363 Greater than symbol, A–4 Greatest common factors (GCF), 39–40 Grid lines, 89 Grouping, factoring by, 40–41 Grouping symbols, 341 Growth rate, 545

H Half planes, 615–616 Horizontal asymptotes, 234, 433–434 Horizontal boundary lines, 123–124, 134 Horizontal change, 107 Horizontal hyperbolas, 732 Horizontal lines, 109–110 Horizontal line test, 480, 486 Horizontal parabolas, 92, 745–747 Horizontal reflections, 209–210 Horizontal translations, 207–208 Hyperbolas applying properties of, 739, 740 branches of, 731 central, 731–732, A–33 directrix of, 712 equation of, 731–736, 756, A–32 – A–33 equation of semi-, 742 explanation of, 731, 737–738, 756 focal chord of, 738, 743 foci of, 712, 737–740 on graphing calculator, 735, 736, 739 graphs of, 731–735, 737–740 horizontal, 732 with rational/irrational values, 759 standard form of equation of, 733–735 vertical, 733, 735–736

I Identities additive, 5–6, 81 explanation of, 27 multiplicative, 5–6, 81 Identity function, 137, 204 Imaginary numbers calculations with, 285, 291 explanation of, 282 on graphing calculator, 285, 286 historical background of, 291 Imaginary unit (i), 282, 287 Implied domain, 124–125 Inclusion, of endpoint, 122 Inconsistent system of equations matrices and, 644–645 in three variables, 596–598 in two variables, 581, 582 Independent variable, 88 Index, A–8 Index of summation, 766, 838

I-5

Induction. See Mathematical induction Induction hypothesis, 800 Inequalities. See also Linear inequalities absolute value, 223–227, 231, 275 additive property of, 27–28 applications of, 32, 464–465 compound, 29–31 graphs to solve, 193, 194 interval tests to solve, 461–464 joint, 31 linear, 27–29, 154–155 multiplicative property of, 27–28 polynomial, 459–462, 474 push principle to solve, 476–477 quadratic, 303–308, 371 rational, 461–464, 474 writing mathematical models using, A–4 zeroes and, 475–476 Inequality symbols, A–4 Infinite geometric series, 788–789 Infinite sequences, 762, 838 Infinity symbol, 123 Input values, 3 Integers, 81, A–2 Intercept method explanation of, 106, 136, 137 to graph hyperbolas, 713–732 Interest compound, 540–541, 569 continuously compounded, 541–542, 545, 569 simple, 539, 569 Interest rate, 269, 539, 540 Intermediate value theorem (IVT), 397–399 Interpolation, 556–557, 570 Intersection, 29 Intersection-of-graphs method on graphing calculator, 160, 519, 545–546 to solve equations, 150–154, 183, 519, 534 to solve linear inequalities, 154–155 use of, 160, 544 Interval notation, 122, 123 Intervals where function is increasing or decreasing, 194–195 where function is positive or negative, 192–194 Interval test method explanation of, 304, 305 to solve function inequalities, 461–464 to solve quadratic inequalities, 304–305 Inverse additive, 6, 81 of functions, 482, 566 graphs of function and its, 485–487 matrix, 665–667, 696–697, 704–705 multiplicative, 6, 81 Inverse functions algebraic method to find, 482–485 applications of, 487 explanation of, 481–482, 566 on graphing calculator, 484–487 graphs of, 485–487 use of, 527 Inverse variation, 264–265, 277 Irrational numbers, 81, A–3


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Index

J Joint inequality, 31 Joint variation explanation of, 266, 277 on graphing calculator, 279

L Latus rectum, 753 Leading coefficients, 17, 41 Least common denominator (LCD), 26, 56–58 Less than symbol, A–4 Lift capacity formula, 37 Like terms, 7 Limited value, 788 Linear association, 166 Linear depreciation, 143 Linear equations. See also Equations applications of, 31–32, 113–114, 142–144 explanation of, 105, 181 forms of, 186–187 in function form, 137, 138 general, 147 on graphing calculators, 114 graphs of, 105–106, 136–142 horizontal and vertical lines and, 109–111 intercept/intercept form of, 147 intercept method to graph, 106, 136, 137 methods to solve, 82, 136 in one variable, 25–26, 82 parallel and perpendicular lines and, 111–113 in point-slope form, 141–142 properties of equality to solve, 25–26 in slope-intercept form, 137, 138 slope of line and rates of change and, 106–109 standard form of, 46 in two variables, 105, 106 Linear factorization theorem explanation of, 395, 396 proof of, 8, A–35 Linear function models, 169–171, 184 Linear functions, 138 Linear inequalities. See also Inequalities applications of, 615 explanation of, 27 on graphing calculator, 617, 619 intersection-of-graphs method to solve, 154–155 method to solve, 27–29 system of, 618–620 in two variables, 615–616, 618 Linear programming applications of, 622–625 explanation of, 620–621, 631–632 on graphing calculators, 633–634 solutions to problems in, 621–622 Linear regression, 171–173 Linear systems. See System of linear equations; System of linear equations in three variables; System of linear equations in two variables Line of best fit, 171–173 Lines auxiliary, A–15

equations of, 138–141 horizontal, 109–110 parallel, 111–112 perpendicular, 112–113 slope-intercept form and graph of, 138–139 slope of, 106–108 vertical, 109–110 Literal equations, 155–156 Local maximum, 196 Logarithmic equations explanation of, 569 forms of, 532 on graphing calculator, 532–533 method to solve, 517–519, 527–530 system of, 608 Logarithmic form, 504–505, 519 Logarithmic functions domain of, 507–508 explanation of, 504, 567 graphs of, 506–507 Logarithmic regression model, 554–555 Logarithmic scales, 508–509 Logarithms applications for, 508–512, 534–535 base-10, 522 base-e, 522 change-of-base formula and, 522–523 common, 504–506 explanation of, 567 fundamental properties of, 516–519 on graphing calculator, 505, 506, 508, 511–512, 520, 528–535 natural, 504–506 product, quotient, and power properties of, 9, 519–521, 527, 568, A–36 uniqueness property of, 528–529 Logistic equations explanation of, 533 on graphing calculator, 533, 534 investigation of, 572 method to solve, 533–534 regression models and, 555–556 Logistic growth, 533–534, 536 Logistic growth model, 555 Long division, 382 Lorentz transformation, 52 Lower bound, 404

M Mapping notation, 88, 89, 354 Market equilibrium, 585–586 Mathematical induction applied to sums, 796–798 explanation of, 796, 840 general principle of, 799–800 to prove statement, 798–799 Mathematical models, 2–3, A–4 Matrices addition and subtraction of, 652–653, 700 applications of, 645–646, 692–697, 701, 702 associated minor, 669–670 augmented, 638–642, 704–705, A–30 coefficient, 638–639 determinants and, 668–672, A–31 equality of, 650–651

explanation of, 638, 700 on graphing calculator, 643–645, 665, 667–669, 672, 673 identity, 663–665 inconsistent and dependent systems and, 644–645 inverse of, 665–667, 696–697, 704–705 multiplication and, 653–658, 663, 700 in reduced row-echelon form, 394, 642 in reduced row-echelon form and, A–30 singular, 668–672 to solve system of equations, 640–643 square, 638, 664 in triangularized form, 640, 678 Matrix equations applications of, 672–673 explanation of, 663, 701 on graphing calculator, 672–673 to solve systems of equations, 667–668, 701 Matrix multiplication explanation of, 654–656 on graphing calculator, 656–658 identity matrices and, 663–665 product matrix and, 663 properties of, 657 Matrix of constraints, 638 Maximum values explanation of, 196 of functions, 196, 621–622 on graphing calculator, 196–197, 317, 319–321 Members of set, A–1 Message encryption, 694–697 Metric units, A–19 – A–20 Midinterval points, 418, 420 Midpoint formula, 95, 708, 755 Midpoint of line segment, 94–95 Minimum values explanation of, 196, 621 of functions, 196 on graphing calculator, 196–197, 317, 319, 320 Minors, 669–670 Mixture problems, 159–160, 582–583 Modeling mathematical, 2–3 step function for, 255 system of linear equations in two variables, 582–586 Monomials, 16, 18. See also Polynomials Multiplication associative property of, 5, 81 commutative property of, 5, 81 of complex numbers, 285–287 matrix, 653–658, 663–665, 700 of polynomials, 18–21 of radical expressions, 70–72 of rational expressions, 55 scalar, 653–654 using exponential properties, 11 Multiplicative identity, 5–6, 81 Multiplicative inverse, 6, 81 Multiplicative property of absolute value, 222 of equality, 25 of inequality, 27–28


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Index Multiplicity vertical asymptotes and, 432–433 zeroes of, 396 Mutually exclusive events, 822

N Nappe, 710 Natural exponential functions, 496 Natural logarithms explanation of, 504 method to find, 505–506 Natural numbers, 81, 781, A–1 Negative association, 165 Negative exponents, 13, 15, 81 Negative numbers, A–2 Negative slope, 108 Nested factoring, 53 Newton’s law of cooling, 498–499 Nondistinguishable permutations, 809 Nonexclusive events, 821–823 Nonlinear association, 166 Nonlinear asymptotes, 448–450 Nonlinear divisor, 385 Nonlinear functions, 331–332 Nonlinear systems applications of, 610–611 conic sections and, 749–750 of equations, 604–608, 631 of inequalities, 609–610, 631 Nonrepeating and nonterminating decimals, A–3 Notation/symbols for composition of functions, 353 delta, 107 explanation of, A–1 exponential, 11, A–7 factorial, 765 function, 125–127 grouping, 341 inequality, A–4 infinity, 123 intersection, 29 interval, 122, 123 mapping, 88, 89, 354 proper subset of, A–1 radical, A–8 rate of change, 495 scientific, 15–16 set, 81, 122, A–1, A–2 subscript, 796 summation or sigma, 766, 767, 838 union, 29 nth term of arithmetic sequence, 774–777 explanation of, 762 of geometric sequence, 783–787 Null set, A–1 Number line, 122, A–2 Numbers. See also Integers complex, 282–288, 291, 370–371, 408 imaginary, 282, 285, 286, 291 irrational, 81, A–3 natural, 81, 781, A–1 negative, A–2 positive, A–2

rational, 53, 54, 81, A–2 real, 5–7, 81, A–3, A–4 sets of, 81, A–1 whole, 81, A–1 – A–2

O Objective functions, 620 Objective variables, 155–156, 622 Oblique asymptotes, 448–450 Odd functions, 191–192 One-dimensional graphs, 591 One-to-one functions. See also Exponential functions explanation of, 480, 486, 566 identification of, 480–481 inverse of, 482, 566 restricting domain to create, 483–484 Ordered pairs, 88–91 Ordered triples, 591–593 Order of operations, A–8 – A–9 Order property of real numbers, A–4 Origin, symmetry to, 191–192 Output values, 3

P Parabolas applications of analytic, 750 definition of, 747 explanation of, 91, 710–711, 744–745, 756–757 focal chord of, 748, 753 focus-directrix form of equation of, 746–749, 756 graphs of, 91, 92, 193, 745–749 horizontal, 92, 745–747 right parabolic segment of, 752 vertex of, 196, 205, 316 vertical, 91, 745, 747 with vertical axis, 745–746 Parallel lines equations for, 139–141 explanation of, 111 slope of, 111, 112, 139 Parametric equations, 467 Parent functions, transformations of, 212 Parent graph, 205 Pareto’s law, 525 Partial fractions explanation of, 682 rational expressions and, 682–688, 701 Partial sums of arithmetic sequence, 777–778 explanation of, 766 of geometric sequence, 787–788 on graphing calculator, 766–767, 769 of series, 766 Pascal’s triangle, 829–831 Perfect cubes, 44–45, 394, A–7 Perfect squares, 43, A–7 Perfect square trinomials explanation of, 20 factoring, 43–44 Perimeter explanation of, 32, A–14 formulas for, 33, 660, 727, A–14 – A–16 Permutations distinguishable, 807–808

I-7

on graphing calculator, 808 nondistinguishable, 809 Perpendicular distance, 601, 709, 713, 755 Perpendicular lines equations for, 139–141 explanation of, 112 slope of, 112, 113, 139 Pick’s theorem, 132 Piecewise-defined functions applications of, 254–255 continuous, 247, 251 discontinuous, 252, 253 domain of, 248–249 equation of, 249, 253–254 explanation of, 247, 276 on graphing calculator, 250 graphs of, 249–251 Pitch diameter, A–12 Placeholder substitution, 45 Plane, 591, A–14 Plane geometry, relationships from, 708–709 Point of inflection, 124 Point-slope form explanation of, 141 to find function model, 143–144 linear equations in, 141–142 Pointwise defined relations, 88, 89 Poiseuille’s law, 52 Polygons, 37, A–14 Polyhedron formula, 37 Polynomial equations with complex coefficients, 312 explanation of, 46–48 Polynomial form converting between standard form and, 759 of equation of circle, 715 of equation of ellipse, 718 of equation of hyperbola, 735 Polynomial functions applications of, 405 complex, 394 dominant term of, 413 graphs of, 411–420 zeroes of, 192–193, 394–405, 471 Polynomial graphs end-behavior of, 412–415, 419 guidelines for, 419–420 identification of, 411–412 turning points and, 411 zeroes of multiplicity and, 415–418 Polynomial inequalities explanation of, 459, 474 on graphing calculator, 476 interval tests to solve, 461–462 method to solve, 459–461 push principle to solve, 476 Polynomial modeling, 421–422 Polynomials. See also Binomials; Monomials; Trinomials addition and subtraction of, 17–18 applications of, 421–422 complex, 312 degree of, 16 division of, 383, 384 evaluation of, 386–387 explanation of, 16, 82


I-8

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Index

Polynomials—Cont. factoring, 39–49 factors of, 387–389, 395 families of, 16 identifying and classifying, 16–17 multiplication of, 18–21 prime, 42, 46 quadratic, 41–43 quartic, 425 rational zeroes of, 400–402 real, 397–399 Population density, 381, 441 Positive association, 165 Positive numbers, A–2 Positive slope, 108 Power functions applications of, 239–242 explanation of, 236, 275, 496 on graphing calculators, 237 graphs of, 236–239 transformations of, 238, 239 variation and, 280 Power property of equality, 72, 74 of exponents, 11, 12, 15 of logarithms, 520–521, 527, 568, A–36 Present value equation, 540 Pressure, 10 Prime polynomials, 42, 46 Principal square roots, 282, 408, A–8 Prisms, A–21 Probability binomial, 834–836 defining events and, 816 elementary, 817 explanation of, 816, 841 on graphing calculator, 821, 823, 835 of mutually exclusive events, 822 of nonexclusive events, 821–823 properties of, 818–820 quick-counting and, 820–821 Problem-solving guide, 157–160 Product property of exponents, 11, 12, 81 of logarithms, 520–521, 527, 568, A–36 of radicals, 68–69, 292 Product to power property, 12, 15, 81 Projectile height, 311, 318–319, 324 Projectile velocity, 333 Proof by induction, 796 Proper subset of whole numbers, A–1 Property of negative exponents, 13 Push principle, 476–477 Pythagorean theorem, 75–76, 83

Q Quadrants, 89 Quadratic equations applications of, 306–307 checking solutions to, 327–328 completing the square to solve, 297–299 explanation of, 46–47, 292, 371 on graphing calculator, 298, 299, 301, 302 quadratic formula to solve, 300, 301 square root property of equality and, 295–297 standard form of, 46

zero product property and, 47–48 Quadratic factors, 396 Quadratic forms explanation of, 45–46 u-substitution to factor expressions in, 45–46, 49, 75 Quadratic formula discriminant of, 301–302 explanation of, 299–300 to solve quadratic equations, 300, 301 Quadratic functions completing the square to graph, 313–315 equation of, 316 explanation of, 293–295, 372 extreme values and, 316–321 finding equation of, 316 vertex formula to graph, 315–316 zeroes of, 293–294 Quadratic graphs, 378 Quadratic inequalities applications of, 30–308 domain and, 305–306 explanation of, 303, 371 interval test method for, 304–305 method to solve, 303–306 Quadratic models, 328–331, 372 Quadratic polynomials, 41–43 Quadratic regression, 331–332, 372 Quadrilaterals, A–14 Quartic polynomials, 425 Quick-counting techniques, 820–821 Quotient property of exponents, 13, 15, 81 of logarithms, 520–521, 527, 568, A–36 of radicals, 69 Quotients difference, 337, 359–364, 374 to power property, 12, 15

R Radical equations explanation of, 72–73 methods to solve, 73–75, 83 Radical expressions. See also Cube roots; Square roots addition and subtraction of, 70 evaluation of, 65 explanation of, 65 methods to simplify, 65–66, 68–69 multiplication and division of, 70–72 rational exponents and, 67–68 Radicals explanation of, 65, 83 product property of, 68–69, 292 quotient property of, 69 Radical symbol, A–8 Radicand, 65, 66, A–8 Radioactive elements, 546 Radius of circle, 96–98, 103, 491, 710, 714 of sphere, 490 Range of functions, 123–124 of relations, 88, 134–135 Rate of change applications of, 332 average, 332–334, 359–361, 373

difference quotient and, 359–361 explanation of, 181 nonlinear functions and, 331–332 notation for, 495 rational functions and, 445 slope as, 106–107, 109, 136, 142 Rational equations, 59–60 Rational exponents explanation of, 67 power property of equality and, 74 radical expressions and, 67–68 simplifying expressions with, 68 solving equations with, 74, 83 Rational expressions addition and subtraction of, 56–58 decomposition for, 682, 685–688 explanation of, 53, 82 fundamental property of, 53–54 multiplication and division of, 55–56 partial fractions and, 682–688 in simplest form, 53–55 simplifying compound fractions and, 58–59 Rational functions applications of, 239–241, 439, 451–454 domain of, 431 end-behavior of, 233, 234 equations of, 236, 438 explanation of, 232, 275, 430, 473 on graphing calculator, 234, 452–454 graphs of, 235–236, 435–438, 446–450, 472–473 horizontal asymptotes of, 234, 433–435 with oblique or nonlinear asymptotes, 448–450 reciprocal function and, 232, 233, 235 removable discontinuities and, 446–447 vertical asymptotes of, 235, 430–433 writing equation of, 236 Rational inequalities analysis to solve, 462–463 explanation of, 474 interval test method to solve, 461–464 Rationalizing the denominator, 71 Rational numbers, 53–54, 81, A–2 Rational zeroes theorem, 399–402, 429 Raw data, 165 Real numbers absolute number of, A–5 – A–6 explanation of, 81, A–3 properties of, 5–7, 81, A–4 Real polynomials, 397–399 Reciprocal function, 232, 235 Reciprocal quadratic function, 235 Reciprocal square function, 233 Rectangles central, 733 explanation of, 33, A–14 perimeter and area formulas for, 33, 660, A–14 Rectangular coordinate system, 89 Rectangular solid, 35, 601, A–17 Recursive sequences, 764–765 Reduced row-echelon form, 394, 642, A–30 Reference intensity, 509 Reflections horizontal, 209–210 vertical, 208–209


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Page I-9

Index Regression applications of, 556–557 explanation of, 165 forms of, 552–553 on graphing calculator, 171–173, 178, 242, 376–377, 552, 554–557 quadratic, 331–332, 372 Regression equations applications of, 556–557 explanation of, 171–172, 552, 570 Regression line, 171 Regression models exponential, 553–554 logarithmic, 554–555 logistic equations and, 555–556 selection of, 178, 328, 570 Regular polygon, area of, 37 Relations domain and range of, 88, 134–135 explanation of, 88, 180 functions as, 119–122 on graphing calculator, 92–94 graphs of, 89–94 methods to represent, 88, 89 pointwise defined, 88, 89 Relative maximum, 196 Remainder theorem application of, 390 to evaluate polynomials, 386–387 explanation of, 386, 470, 471 proof of, A–34 Removable discontinuities explanation of, 446 rational functions and, 446–447 Repeated zeros, 475–476 Repeating and nonterminating decimals, A–2 Required interest rate, 269 Residuals, 376–377, 552 Right angles, A–14 Right circular cones, 35, A–17 Right circular cylinders, 35, A–17 Right parabolic segment, 752 Right prisms, A–21 Right pyramids, 35 Right square pyramid, A–17 Right triangles, 75 Root functions explanation of, 236 transformations of, 238–239 Roots cube, 66, A–8 of equation, 25 extraneous, 60, 527 square, 65–66, 282, 408, A–8 Root tests for quartic polynomials, 425 Row-echelon form, 642 Row operations, elementary, 640–641 Run, 107

S Sample outcome, 805 Sample space, 805, 816, 817 Sand dune function, 257 Scalar multiplication, 653–654 Scatterplots explanation of, 165

linear/nonlinear association and, 166–167 positive/negative association and, 165–166 Scientific notation, 15–16, 81 Secant lines, 140, 146 Semicircles, graph of, 92 Semimajor axis, 717 Semiminor axis, 717 Sequences. See also Series; specific types of sequences applications of, 768–769, 779, 789–791 arithmetic, 773–779, 839 explanation of, 762, 838 finding terms of, 762–764 finite, 762 geometric, 782–791, 839 on graphing calculator, 763–765, 775–776, 785–786, 789–790 graphs of, 776, 786 graphs of arithmetic, 776 infinite, 762 recursive, 764–765 Series. See also Sequences explanation of, 762, 765, 838 finite, 765–766 geometric, 782 properties of, 769 Set notation, 81, 122, A–1, A–2 Sets of numbers, 81, A–1 – A–3 Sigma notation, 766, 767, 838 Similar triangles, A–20 Simple interest, 539, 569 Simplest form radical expressions in, 71 rational expressions in, 53–55 Singular matrices determinants and, 669–672 explanation of, 669 Sinking fund, 544 Slope of line, 106–108 positive and negative, 108 as rate of change, 106–107, 109, 136, 142 Slope formula, 107, 108, 110, 138 Slope-intercept form explanation of, 137 graph of line and, 138–139 linear equations in, 137, 138 Smooth functions, 247 Solution region, 616 Solution sets, 27 Spheres, 35, 218, A–17 Spherical shells, 52 Square matrices, 638, 664 Square root function, 204 Square root property of equality, 295–297 Square roots. See also Radical expressions of complex numbers, 408 principal, 282, 408, A–8 simplification of, 65–66 Squares. See also Completing the square binomial, 20–21, 43, 759, 830 difference of two, 43 explanation of, 33 factoring difference of two, 43 perfect, 43, A–7 perimeter and area formulas for, 33, A–16 trinomial, 730

I-9

Square systems, 597 Square units, A–14 Squaring function, 204 Standard form of complex numbers, 283–284 converting between polynomial form and, 759 of equation of circle, 96, 716 of equation of ellipse, 717 of equation of hyperbola, 733–734 of linear equations, 46 of polynomial expressions, 17 of quadratic equations, 46 of system of equations, 579 Statistics, 816 Step functions, 254–255 Stevens, Stanley, 571 Stevens’ law, 571 Stirling’s formula, 814 Stretches, vertical, 210–211 Subscripted variables, A–14 Subscript notation, 796 Substitution applying power property after, 12 back, 578 to check complex root, 286, 287 to check solutions, 26 explanation of, 578 placeholder, 45 to solve system of equations, 578–580 Subtraction of complex numbers, 284–285 of matrices, 652–653, 700 of polynomials, 17–18 of radical expressions, 70 of rational expressions, 56–58 Summation applications of, 803–804 explanation of, 766 index of, 766, 838 properties of, 767–768 Summation notation, 766, 767 Surface area. See also Area of cone, 80 of cylinder, 52, 311, 349, 455 of rectangular box, 323 Symbols. See Notation/symbols Symmetry axis of, 205 to origin, 191–192 polynomial graphs and, 419 y-axis, 190–191 Synthetic division, 383–384, 470, A–28 – A–29 System of linear equations applications of, 645–646 augmented matrix of, 638–640 determinants and Cramer’s Rule to solve, 679–682 equivalent, 579, 593–594 explanation of, 576 on graphing calculators, 643–644 inconsistent and dependent, 581–582, 596–598, 644–645 matrices to solve, 640–643, 663–673, 701, A–30 square, 597 verifying solutions to system of, 576


I-10

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Page I-10

Index

System of linear equations in three variables applications of, 598–599 coincident dependent, 592 elimination to solve, 593–596, 634–635 explanation of, 591–592, 630–631 inconsistent and dependent, 596–598 linearly dependent, 592 solutions to, 592–593 System of linear equations in two variables applications of, 582–586 elimination to solve, 579–581, 634–635 explanation of, 576, 630 on graphing calculator, 577, 578, 580–586 graphs to solve, 577 inconsistent and dependent, 581–582 modeling and, 582–586 substitution to solve, 578–579 System of linear inequalities applications of, 619–620 explanation of, 615 on graphing calculator, 619 method to solve, 618–620 System of logarithmic equations, 608 System of nonlinear equations conic sections and, 749–750 elimination to solve, 607–608 explanation of, 604, 631 on graphing calculator, 605–608 graphs of, 604–605 possible solutions for, 604–605 substitution to solve, 605–607 System of nonlinear inequalities, 609–610, 631 System of two equations in two variables, 576

T Temperature conversions, 132 Terminating decimals, A–2 Toolbox functions direct variation and, 261–264 explanation of, 204–206, 274 horizontal translations and, 207–208 vertical translations and, 206–207 Transcendental functions, 503 Transformations of general functions, 211–214 to graph exponential functions, 495 on graphing calculator, 213, 265 to graph logarithmic functions, 506 Lorentz, 52 of parent graph, 205 of power functions, 238, 239 of reciprocal functions, 235 review of, 274 of root functions, 238 Translations on graphing calculator, 206–208 horizontal, 207–208 vertical, 206–207 Transverse axis, 732 Trapezoids, 33–34, A–14

Tree diagrams, 804–805 Trial-and-error method, for factoring, 42–43 Trials, 805 Triangles explanation of, 33 Pascal’s, 829–831 perimeter and area formulas for, 33, 455, 688, A–14 right, 75 similar, A–20 Triangularizating augmented matrix, 640–642, 678 Trigonometric graphs, 200 Trinomials. See also Polynomials explanation of, 16 factoring, 41–43 perfect square, 20, 43–44 in quadratic form, 45 Trinomial squares, 730 Two-dimension graphs, 591

U Unbounded region, 621 Uniform motion, 158–159, 583–584 Union, 29 Uniqueness property to solve exponential equations, 496–499 to solve logarithms, 528–529 Unique solutions, 592 Unit conversion factors, A–18 – A–21 Upper and lower bounds property, 402, 404 Upper bound, 404 U.S. Customary Units, A–19 u-substitution, 45–46, 49, 75

V Variables dependent, 88 descriptive, 2 independent, 88 object, 155–156 subscripted, A–14 Variation constant of, 261 direct, 261–264, 277 on graphing calculators, 264, 279 graphs of, 262–265 inverse, 264–265, 277 joint or combined, 266, 277 power functions and, 280 Variation equations, 261, 263, 265 Velocity, 333, 363, 364 Velocity formula, 79, 80, 236 Vertex of cone, 710 of hyperbola, 732 of parabola, 196, 205, 316 Vertex formula, 315–316 Vertex/intercept formula, 323 Vertical asymptotes, 235, 430–433 Vertical axis, 591 Vertical boundary lines, 122, 134

Vertical change, 106–107 Vertical format, 3 Vertical hyperbolas, 732, 735–736 Vertical lines, 109–110 Vertical line test, 120–122 Vertical parabolas, 91, 745, 747 Vertical reflections, 208–209 Vertical shifts. See Vertical translations Vertical stretches, 210–211 Vertical translations, 206–207 Volume of box, 52 of cone, 35, 491 of cylinder, 176 of cylindrical shell, 52 explanation of, 34, A–16, A–18 – A–19 formulas for, 34, 491, A–17 method to compute, 35 of open box, 392 of right prism, A–21 of sphere, 218 of triangular pyramid, 691

W Whole numbers, 81, A–1 – A–2 Written information translated into equations, 31–32 translated into mathematical model, 2–3

X x-intercepts explanation of, 106 on graphing calculators, 153–154, 293–294, 296 of quadratic functions, 293–294 x-intercept/zeroes method, 152–154 xy-plane, 89

Y y-axis, 190–191 y-intercepts, 106

Z Zeroes approximation of real, 429 complex, 475–476 division with, 384–385, A–6 factor theorem to find, 389 of functions, 192–194 intermediate value theorem to find, 398 of multiplicity, 396, 415–416, 475 of polynomial functions, 192–193, 394–405, 415–416, 471 of quadratic functions, 293–294 quotient of, 384–385, A–6 rational, 400–402, 429 repeated, 475–476 Zeroes/x-intercept method, 152–154, 298 Zero exponent property, 13, 15 Zero exponents, 13, 81 Zero product property, 47–48

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Photo Credits Chapter R

Chapter 4

Chapter 7

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Chapter 1

Chapter 5

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Chapter 3

Chapter 6

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Appendices II

Chapter 2

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Appendices I

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C-1


cob19545_es.indd Page Sec1:2 28/01/11 3:41 PM s-60user

▼


Special Constants

▼

␲ ⬇ 3.1416 ▼

e ⬇ 2.7183

12 ⬇ 1.4142

Special Products 1x a21x b2 x2 1a b2x ab

1a b21a b2 a2 b2

1a b2 2 a2 2ab b2

1a b2 2 a2 2ab b2

1a b2 a 3a b 3ab b 3

2

2

1a b2 a 3a b 3ab b

3

3

3

2

2

3

Special Factorizations x2 1a b2x ab 1x a21x b2

a2 b2 1a b21a b2

a2 2ab b2 1a b2 2

a2 2ab b2 1a b2 2

a b 1a b21a ab b 2 3

▼

Distance between P1 and P2

13 ⬇ 1.7321

3

2

a b 1a b21a ab b 2

2

3

3

2

2

▼

Formulas from Plane Geometry: P S perimeter, C S circumference, A S area Rectangle

Square

w

P 2l 2w

2

A bh

A

b

Triangle

h 1a b2 2

C

A

b

▼

1 bh 2

h

a

A ␲r2

b

C 2␲r ␲d

Right Parabolic Segment 4 A ab 3

a b

C ⬇ ␲ 221a2 b2 2

▼ b

▼

H

V LWH S 21LW LH WH2

L

W

Cube

Right Circular Cylinder

V s3

V ␲r2h

S 6s2

Right Circular Cone

Right Square Pyramid

Sphere

1 V ␲r2h 3

1 V b2h 3

4 V ␲r3 3

S ␲r 1r s2

h

S b2 b2b2 4h2 r

ISBN: 0-07-351954-5 Author: John W. Coburn Title: College Algebra, 3e

Front endsheets Color: 5 Pages: 2, 3

h

S 4␲r2 b

y y1 m1x x1 2

y mx b, where b y1 mx1

Parallel Lines

Perpendicular Lines

Slopes Are Equal: m1 m2

Slopes Have a Product of 1: m1m2 1

Intersecting Lines

Dependent (Coincident) Lines

Slopes Are Unequal: m1 m2

Slopes and y-Intercepts Are Equal: m1 m2, b1 b2

Logarithms and Logarithmic Properties y logb x 3 b y x

logb b 1

logb bx x

blogb x x logb a

logb 1 0 logc x

M b logb M logb N N

logb x logb c

logb MP P # logb M

Applications of Exponentials and Logarithms A S amount accumulated

P S initial deposit, P S periodic payment

n S compounding periods/year

r S interest rate per year

r R S interest rate per time period a b n

t S time in years

Interest Compounded n Times per Year

Interest Compounded Continuously

r nt A P a1 b n

A Pert

Accumulated Value of an Annuity

Payments Required to Accumulate Amount A P

AR 11 R2 nt 1

Sequences and Series: Arithmetic Sequences

Geometric Sequences

a1, a2 a1 d, a3 a1 2d, ... , an a1 1n 12d

a1, a2 a1r, a3 a1r2, ... , an a1r n1

Sn

S 2␲r 1r h2

s

Slope-Intercept Form (slope m, y-intercept b)

a1 S 1st term, an S nth term, Sn S sum of n terms, d S common difference, r S common ratio

Formulas from Solid Geometry: S S surface area, V S volume Rectangular Solid

Point-Slope Form

P A 冤11 R2 nt 1冥 R

a

¢y y2 y1 x2 x 1 ¢x

Equation of Line Containing P1 and P2

b

r

m


logb 1MN 2 logb M logb N

Circle

c

a2 b2 c2

Ellipse

A

Pythagorean Theorem

A B C 180°

a

Triangle

h

Right Triangle

B

A ␲ab

a

Trapezoid

h

s

P ns a A P 2

As

A lw

Sum of angles

Regular Polygon s

P 4s

l

Parallelogram

Slope of Line Containing P1 and P2

d 21x2 x1 2 2 1y2 y1 2 2

3

▼

Formulas from Analytical Geometry: P1 S (x1, y1), P2 S (x2, y2)

h

Sn

r r

▼

a1 a1rn 1r a1 Sq ; 冟r冟 6 1 1r

n 1a1 an 2 2

Sn

n 冤2a1 1n 12d冥 2

Binomial Theorem n n n 1a b2 n a b anb0 a b an1b1 a b an2b2 0 1 2 n! n1n 121n 22

# # # 132122112;

###

a

0! 1

n n b a 1bn1 a b a0bn n1 n n n! a b k k!1n k2!


▼


Special Constants

▼

␲ ⬇ 3.1416 ▼

e ⬇ 2.7183

12 ⬇ 1.4142

Special Products 1x a21x b2 x2 1a b2x ab

1a b21a b2 a2 b2

1a b2 2 a2 2ab b2

1a b2 2 a2 2ab b2

1a b2 a 3a b 3ab b 3

2

2

1a b2 a 3a b 3ab b

3

3

3

2

2

3

Special Factorizations x2 1a b2x ab 1x a21x b2

a2 b2 1a b21a b2

a2 2ab b2 1a b2 2

a2 2ab b2 1a b2 2

a b 1a b21a ab b 2 3

▼

Distance between P1 and P2

13 ⬇ 1.7321

3

2

a b 1a b21a ab b 2

2

3

3

2

2

▼

Formulas from Plane Geometry: P S perimeter, C S circumference, A S area Rectangle

Square

w

P 2l 2w

2

A bh

A

b

Triangle

h 1a b2 2

C

A

b

▼

1 bh 2

h

a

A ␲r2

b

C 2␲r ␲d

Right Parabolic Segment 4 A ab 3

a b

C ⬇ ␲ 221a2 b2 2

▼ b

▼

H

V LWH S 21LW LH WH2

L

W

Cube

Right Circular Cylinder

V s3

V ␲r2h

S 6s2

Right Circular Cone

Right Square Pyramid

Sphere

1 V ␲r2h 3

1 V b2h 3

4 V ␲r3 3

S ␲r 1r s2

h

S b2 b2b2 4h2 r



h

S 4␲r2 b

y y1 m1x x1 2

y mx b, where b y1 mx1

Parallel Lines

Perpendicular Lines

Slopes Are Equal: m1 m2

Slopes Have a Product of 1: m1m2 1

Intersecting Lines

Dependent (Coincident) Lines

Slopes Are Unequal: m1 m2

Slopes and y-Intercepts Are Equal: m1 m2, b1 b2

Logarithms and Logarithmic Properties y logb x 3 b y x

logb b 1

logb bx x

blogb x x logb a

logb 1 0 logc x

M b logb M logb N N

logb x logb c

logb MP P # logb M

Applications of Exponentials and Logarithms A S amount accumulated

P S initial deposit, P S periodic payment

n S compounding periods/year

r S interest rate per year

r R S interest rate per time period a b n

t S time in years

Interest Compounded n Times per Year

Interest Compounded Continuously

r nt A P a1 b n

A Pert

Accumulated Value of an Annuity

Payments Required to Accumulate Amount A P

AR 11 R2 nt 1

Sequences and Series: Arithmetic Sequences

Geometric Sequences

a1, a2 a1 d, a3 a1 2d, ... , an a1 1n 12d

a1, a2 a1r, a3 a1r2, ... , an a1r n1

Sn

S 2␲r 1r h2

s

Slope-Intercept Form (slope m, y-intercept b)

a1 S 1st term, an S nth term, Sn S sum of n terms, d S common difference, r S common ratio

Formulas from Solid Geometry: S S surface area, V S volume Rectangular Solid

Point-Slope Form

P A 冤11 R2 nt 1冥 R

a

¢y y2 y1 x2 x 1 ¢x


b

r

m


logb 1MN 2 logb M logb N

Circle

c

a2 b2 c2

Ellipse

A

Pythagorean Theorem

A B C 180°

a

Triangle

h

Right Triangle

B

A ␲ab

a

Trapezoid

h

s

P ns a A P 2

As

A lw

Sum of angles

Regular Polygon s

P 4s

l

Parallelogram

Slope of Line Containing P1 and P2

d 21x2 x1 2 2 1y2 y1 2 2

3

▼

Formulas from Analytical Geometry: P1 S (x1, y1), P2 S (x2, y2)

h

Sn

r r

▼

a1 a1rn 1r a1 Sq ; 冟r冟 6 1 1r

n 1a1 an 2 2

Sn

n 冤2a1 1n 12d冥 2

Binomial Theorem n n n 1a b2 n a b anb0 a b an1b1 a b an2b2 0 1 2 n! n1n 121n 22

# # # 132122112;

###

a

0! 1

n n b a 1bn1 a b a0bn n1 n n n! a b k k!1n k2!


▼


The Toolbox and Other Functions

▼

linear

linear

y

y

identity

constant

y

y mx b

y

Fundamental Counting Principle: Given an experiment with two tasks completed in sequence, if the first can be completed in m ways and the second in n ways, the experiment can be completed in m ⴛ n ways.

yb

yx

Permutations—Order Is a Consideration: (Al, Bo, Ray) and (Ray, Bo, Al) finish the race in a different order. n! . The permutations of r objects selected from a set of n (unique) objects is given by nPr ⴝ (n ⴚ r)! Combinations—Order Is Not a Consideration: (Al, Bo, Ray) and (Ray, Bo, Al) form the same committee. n! . The combinations of r objects selected from a set of n (unique) objects is given by nCr ⴝ r!(n ⴚ r)! Basic Probability: Given S is a sample space of equally likely events and E is an event defined relative to S. n(E) , where n1E2 and n1S2 represent the number of elements in each. The probability of E is P(E) ⴝ n(S) For any event E1: 0 P1E1 2 1 and P1E1 2 P1~E1 2 1.

(0, b)

(0, b)

x

x y mx b

m 0, b 0

absolute value

x

m 0, b 0

m 1, b 0

squaring

cubing

y

y

x

square root y x3

x

y 兹x x

ceiling function

Probability of E1 and E2

Probability of E1 or E2

P1E1 傽 E2 2 P1E1 2P1E2 2 1E1, E2 independent2

P1E1 ´ E2 2 P1E1 2 P1E2 2 P1E1 傽 E2 2

x

floor function

y y x

y

m 0, b 0 y

y x2

y x

cube root

x

y

Quick-Counting and Probability

▼

reciprocal

y y x

Conic Sections y

y y

y

circle with center at (h, k)

1 x

3

y 兹x

r x

x

x

1 2

x

k

logarithmic

y

y y

1 x2

logistic

y bx (b 0)

(a, 0)

x

1

x

(0, 1 c a )

y a f 1x h2 k vertical reflections vertical stretches/compressions

▼



Average Rate of Change of f(x)

f(x2) ⴚ f(x1) ⌬y ⴝ ⌬x x2 ⴚ x1


k (h, k)

1

y2 b2

1

x

If a b, the ellipse is oriented vertically.

(y k)2 b2

h

x2 a2

y2 b2

c2 a2 b2

p0

(0, p) x

x

If term containing y leads, the hyperbola is oriented vertically. 1

y

1 y p

(a, 0) (c, 0)

(c, 0)

x2 4py vertical parabola focus (0, p) directrix y p y

( p, 0) (x h)2 a2

(a, 0)

For linear function models, the average rate of change on the interval 3x1, x2 4 is constant, and given by the slope formula: ¢y y2 y1 . The average rate of change for other function models is nonconstant. By writing the slope formula in function form x2 x1 ¢x using y1 f 1x1 2 and y2 f 1x2 2, we can compute the average rate of change of other functions on this interval:


hyperbola with center at (h, k)

central hyperbola

S

S

y f 1x2

S

Transformation of Given Function

(y k)2 b2

c2 |a2 b2|

Transformations of Basic Graphs Given Function

(a, 0)

x2 a2

x y

▼

(c, 0)

(0, b)

1

x

(c, 0)

h

c 1 aebx

(h a, k)

(0, b)

x2 y2 r2

yc y

x

h

ellipse with center at (h, k), a b

(x h)2 a2

(h, k b)

central ellipse

(x, y)

(0, 0)

y

y y logb x (b 0)

(h, k)

(x h)2 (y k)2 r2 r

exponential

(h a, k)

(h, k)

central circle

reciprocal square

k

(h, k b)

p0

x p y2 4px horizontal parabola focus ( p, 0) directrix x p

x


▼


The Toolbox and Other Functions

▼

linear

linear

y

y

identity

constant

y

y mx b

y

Fundamental Counting Principle: Given an experiment with two tasks completed in sequence, if the first can be completed in m ways and the second in n ways, the experiment can be completed in m ⴛ n ways.

yb

yx

Permutations—Order Is a Consideration: (Al, Bo, Ray) and (Ray, Bo, Al) finish the race in a different order. n! . The permutations of r objects selected from a set of n (unique) objects is given by nPr ⴝ (n ⴚ r)! Combinations—Order Is Not a Consideration: (Al, Bo, Ray) and (Ray, Bo, Al) form the same committee. n! . The combinations of r objects selected from a set of n (unique) objects is given by nCr ⴝ r!(n ⴚ r)! Basic Probability: Given S is a sample space of equally likely events and E is an event defined relative to S. n(E) , where n1E2 and n1S2 represent the number of elements in each. The probability of E is P(E) ⴝ n(S) For any event E1: 0 P1E1 2 1 and P1E1 2 P1~E1 2 1.

(0, b)

(0, b)

x

x y mx b

m 0, b 0

absolute value

x

m 0, b 0

m 1, b 0

squaring

cubing

y

y

x

square root y x3

x

y 兹x x

ceiling function

Probability of E1 and E2

Probability of E1 or E2

P1E1 傽 E2 2 P1E1 2P1E2 2 1E1, E2 independent2

P1E1 ´ E2 2 P1E1 2 P1E2 2 P1E1 傽 E2 2

x

floor function

y y x

y

m 0, b 0 y

y x2

y x

cube root

x

y

Quick-Counting and Probability

▼

reciprocal

y y x

Conic Sections y

y y

y

circle with center at (h, k)

1 x

3

y 兹x

r x

x

x

1 2

x

k

logarithmic

y

y y

1 x2

logistic

y bx (b 0)

(a, 0)

x

1

x

(0, 1 c a )

y a f 1x h2 k vertical reflections vertical stretches/compressions

▼



Average Rate of Change of f(x)

f(x2) ⴚ f(x1) ⌬y ⴝ ⌬x x2 ⴚ x1


k (h, k)

1

y2 b2

1

x

If a b, the ellipse is oriented vertically.

(y k)2 b2

h

x2 a2

y2 b2

c2 a2 b2

p0

(0, p) x

x

If term containing y leads, the hyperbola is oriented vertically. 1

y

1 y p

(a, 0) (c, 0)

(c, 0)

x2 4py vertical parabola focus (0, p) directrix y p y

( p, 0) (x h)2 a2

(a, 0)

For linear function models, the average rate of change on the interval 3x1, x2 4 is constant, and given by the slope formula: ¢y y2 y1 . The average rate of change for other function models is nonconstant. By writing the slope formula in function form x2 x1 ¢x using y1 f 1x1 2 and y2 f 1x2 2, we can compute the average rate of change of other functions on this interval:


hyperbola with center at (h, k)

central hyperbola

S

S

y f 1x2

S

Transformation of Given Function

(y k)2 b2

c2 |a2 b2|

Transformations of Basic Graphs Given Function

(a, 0)

x2 a2

x y

▼

(c, 0)

(0, b)

1

x

(c, 0)

h

c 1 aebx

(h a, k)

(0, b)

x2 y2 r2

yc y

x

h

ellipse with center at (h, k), a b

(x h)2 a2

(h, k b)

central ellipse

(x, y)

(0, 0)

y

y y logb x (b 0)

(h, k)

(x h)2 (y k)2 r2 r

exponential

(h a, k)

(h, k)

central circle

reciprocal square

k

(h, k b)

p0

x p y2 4px horizontal parabola focus ( p, 0) directrix x p

x

College Algebra: Graphs & Models

College Algebra: Graphs & Models, 3rd Edition

Algebra & Trigonometry: Graphs & Models

Precalculus: Graphs & Models

College Algebra

College Algebra

College algebra

College Algebra

College Algebra

College algebra

College algebra

College Algebra

Precalculus: Graphs & Models, 3rd Edition

Concurrency, Graphs and Models

Elementary and Intermediate Algebra: Graphs and Models, 4th Edition

College Algebra, 8th Edition

College Algebra and Trigonometry

College Algebra, 5th Edition

College Algebra, 9th Edition

College Algebra, 8th Edition

College algebra, (Fourth Edition)

College Algebra, (Ninth Edition)

College Algebra, 9th Edition

College Algebra Demystified

College Algebra Demystified

College Algebra Demystified

College Algebra Demystified

Schaum's Outline of College Algebra

College Algebra: A Graphing Approach

Schaum's Easy Outline: College Algebra

Schaum's Outline of College Algebra

College Algebra: Graphs & Models