MEI FP1 Study Resources Graphs And Inequalities 1-2

MEI Further Pure Mathematics 1 Graphs and Inequalities Section 1: Graphs of rational functions Study Plan Background This section is concerned with sketching graphs of rational functions, i.e. N( x) functions of the form y  , where N(x) and D(x) are polynomials in x and D( x) D(x) is not the zero polynomial. Sketches of graphs can be very useful in solving mathematical problems. You will find that it is often possible to solve an equation or inequality by sketching graphs of the functions involved, or at least to get a rough idea of the number and location of solutions. An accurate graph is not necessary: all you need is a sketch which shows the main features of a function.

Detailed work plan 1. Read pages 74 – 85. This is quite a long section, and goes through the four steps usually involved in sketching a graph. Three further examples are shown in the Notes and Examples. 2. You can explore graphs of rational functions using the Geogebra resource Rational functions. You can also see PowerPoint presentations which show the three examples from the Notes and Examples. 3. Exercise 3A Attempt at least Questions 1, 3, 4, 5, 6, 8, 9, 11, 13 and 14. If you have time you might like to try some of Questions 15-18.

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MEI Further Pure Mathematics 1 Graphs and Inequalities Section 1: Graphs of rational functions Notes and Examples Don’t be discouraged by the amount of reading in the text book for this section! Once you have had some practice in sketching graphs, you will find that it is usually quite straightforward.

You can explore the graphs of rational functions using the Geogebra resource Rational functions. Look in particular at asymptotes and at the behaviour of the graphs for numerically large values of x.

These notes will deal with three further examples. Each of the four steps outlined in the book will be gone through for each of the three examples. You can also look at PowerPoint presentations showing each stage in sketching each of the three curves. The examples are x y (i) ( x  2)( x  3) 4  2x y (ii) 2 x x2  x  4 y (iii) x2  3 Step 1: Find where the graph cuts the axes Find where the graph cuts the y axis by putting x = 0. Find where the graph cuts the x axis by looking for values of x which make the numerator zero.

(i)

(ii)

x ( x  2)( x  3) When x = 0, y = 0. (0, 0) is the only point where the graph cuts either axis. 4  2x y 2 x When x = 0, y = 2 When y = 0, x = 2 The graph cuts the axes at (0, 2) and (2, 0) y

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MEI FP1 Graphs Section 1 Notes and Examples (iii)

x2  x  4 x2  3 When x = 0, y = 43 The numerator is never zero (the discriminant of the quadratic expression is negative), so there are no values of x for which y = 0. So the graph does not cut the x axis. The graph cuts the y axis at (0, 43 ) y

Step 2: Find the vertical asymptotes Find the values of x which make the denominator zero. The text book also shows how to look at the behaviour of the graph near the asymptotes. This is not always necessary, so can be left to a later stage if needed.

(i)

(ii)

(iii)

x ( x  2)( x  3) The vertical asymptotes are x = -2 and x = 3. 4  2x y 2 x The vertical asymptote is x = -2 x2  x  4 y x2  3 There are no vertical asymptotes as the denominator is not zero for any real values of x. y

Step 3: Examine the behaviour as x tends to infinity

All the examples you meet in this section will approach a horizontal asymptote for numerically large values of x. If the degree of the denominator is greater than the degree of the numerator, then the graph will approach the x axis (see Examples (i) and (iii)). In these cases you just need to look at the sign of y for large positive and negative x, to see whether the asymptote is approached from above or below. If the degree of the denominator is equal to the degree of the numerator, then the graph will approach another horizontal asymptote (see Example (ii)). In this case you need to work out the value of y for a large positive and a large negative value of x, to see whether the asymptote is approached from above or below.

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MEI FP1 Graphs Section 1 Notes and Examples (i)

y

x ( x  2)( x  3)

x 1  0. x2 x For large positive values of x, all three of x, (x + 2) and (x – 3) are positive, so y is positive. For large negative values of x, all three of x, (x + 2) and (x – 3) are negative, so y is negative. So as x → ∞, y → 0 from above and as x → -∞, y → 0 from below. 4  2x y 2 x 2 x As x → ±∞, y →  2 . x For large positive values of x (try x = 100 for example), y is slightly greater than -2. For large negative values of x (try x = -100 for example), y is slightly less than -2. So as x → ∞, y → -2 from above and as x → -∞, y → -2 from below. x2  x  4 y x2  3 x2 As x → ±∞, y → 2  1 . x For large positive values of x (try x = 100 for example), y is slightly greater than 1. For large negative values of x (try x = -100 for example), y is slightly less than 1. So as x → ∞, y → 1 from above and as x → -∞, y → 1 from below. As x → ±∞, y →

(ii)

(iii)

Step 4: Complete the sketch

Put on all the information you have gained from steps 1, 2 and 3. It may then be possible to complete the sketch. If not, look at the behaviour of the graph on either side of the vertical asymptotes (see Example (i)).

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MEI FP1 Graphs Section 1 Notes and Examples (i)

y

x ( x  2)( x  3)

Information from Steps 1-3 is shown on this diagram.

This allows us to complete part of the graph, but there is not enough information to the section between the asymptotes

Look at the sign of y between -2 and 0. x and (x – 3) are both negative, (x + 2) is positive, so y is positive. Look at the sign of y between 0 and 3. x and (x + 2) are both positive, (x – 3) is negative, so y is negative. This further information allows us to complete the sketch.

(ii)

y

4  2x 2 x

Information from Steps 1-3 is shown on this diagram.

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This gives enough information to complete the sketch.

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MEI FP1 Graphs Section 1 Notes and Examples (iii)

y

x2  x  4 x2  3

Information from Steps 1-3 is shown on this diagram.

This gives enough information to complete the sketch. We do not know the exact positions of the turning points, but for a sketch this does not matter.

In the enrichment material in the next section, you will learn a method which will allow you to find the actual positions of the turning points in (iii) without using calculus. This example is shown in the Notes and Examples for section 2. However, this is not part of the syllabus.

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Further Pure Mathematics 1 Graphs and Inequalities Section 1: Graphs of rational functions Crucial points 1. Factorise where possible Make sure that you always factorise both the numerator and the denominator if possible, and if they are not already given in factorised form. If you don’t, you may miss vertical asymptotes or points where the graph cuts the x axis. 2. Put all the information that you have on to your initial sketch Remember when completing the sketch that the graph cannot cross the x axis at any point other than the points which you found in Step 1. 3. Don’t be too hasty in completing the sketch Make sure that there is only one possible way in which you can do it. If there isn’t, then obtain the extra information you need, such as the sign of y near the asymptotes.

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Further Pure Mathematics 1 Graphs and Inequalities Section 1: Graphs of rational functions Exercise In questions 1 to 5: (i) find the co-ordinates of the points where the graph cuts the axes (ii) find any vertical and horizontal asymptotes (iii) state the behaviour of the curve as x → ± ∞ (iv) sketch the curve 1. y = 3 +

2 x

2. y =

x+3 2x − 3

3. y =

2x − 3 ( x + 2 x − 3)

4. y =

x ( x − 3)( x + 1)

5. y =

( x − 5)( x − 1) ( x + 1)( x − 3)

2

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Further Pure Mathematics 1 Graphs and Inequalities Section 1: Graphs of rational functions Solutions to Exercise 1. y = 3 + (i)

(ii)

(iii) (iv)

2

x

=

3x + 2

x

When x = 0, y is undefined. When y = 0, x = − 23 The graph cuts the axes at ( − 23 ,0 ) . There is a vertical asymptote at x = 0 As x → ∞, y → 3 There is a horizontal asymptote at y = 3. As x → ∞, y → 3 from above. As x → -∞, y → 3 from below.

− 23

2. y = (i)

(ii)

(iii)

x +3 2x − 3

When x = 0, y = -1 When y = 0, x = -3 The graph cuts the axes at (0, -1) and (-3, 0). There is a vertical asymptote at x = 23 . As x → ∞, y → 21 There is a horizontal asymptote at y = 21 . As x → ∞, y → 21 from above. As x → -∞, y → 21 from below.

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Further Pure Mathematics 1 (iv)

2x − 3 2x − 3 = x + 2 x − 3 ( x + 3)( x − 1) (i) When x = 0, y = 1. When y = 0, x = 23 The graph cuts the axes at (0, 1) and ( 23 , 0 ) . (ii) There are vertical asymptotes at x = -3 and x = 1. As x → ∞, y → 0 There is a horizontal asymptote at y = 0. (iii) As x → ∞, y → 0 from above. As x → -∞, y → 0 from below. (iv)

3. y =

4. y =

2

x

( x − 3)( x + 1) When x = 0, y = 0. The graph cuts the axes at (0, 0). (ii) There are vertical asymptotes at x = 3 and x = -1. As x → ∞, y → 0 There is a horizontal asymptote at y = 0.

(i)

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Further Pure Mathematics 1 (iii) (iv)

As x → ∞, y → 0 from above. As x → -∞, y → 0 from below.

( x − 5 )( x − 1 ) ( x + 1)( x − 3) (i) When x = 0, y = − 53 . When y = 0, x = 5 or 1. The graph cuts the axes at ( 0, − 53 ) , (5, 0) and (1, 0). (ii) There are vertical asymptotes at x = -1 and x = 3. As x → ∞, y → 1 There is a horizontal asymptote at y = 1. (iii) As x → ∞, y → 1 from above. As x → -∞, y → 1 from below. (iv)

5. y =

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MEI Further Pure Mathematics 1 Graphs and Inequalities Section 1: Graphs of rational functions Multiple Choice Test 1) The graph of y =

( x + 1)( x − 2) crosses the coordinate axes at the points (2 x + 1)(1 − x)

(a) (-1, 0), (2, 0), ( − 12 , 0 ) and (1, 0)

(b) (-1, 0), (2, 0) and (0, -2) (d) (1, 0), (-2, 0), ( − 12 , 0 ) and (-1, 0)

(c) (1, 0), (-2, 0) and (0, -2) (e) I don’t know

2) The graph of y =

x(2 x + 1) has the following asymptotes: ( x − 2)( x + 3)

(a) x = 2, x = -3, y = 2 (c) x = 2, x = -3

(b) x = -2, x = 3, y = 2 (d) x = 0, x = 2, x = -3, x = − 12

(e) I don’t know

3) The behaviour of the graph y =

x for numerically large x can be ( x − 3)(2 − x)

described as follows: (a) As x → ∞, y → 0 from above As x → -∞, y → 0 from below (c) As x → ∞, y → -1 from below As x → -∞, y → -1 from above (e) I don’t know

(b) As x → ∞, y → 0 from below As x → -∞, y → 0 from above (d) As x → ∞, y → -1 from above As x → -∞, y → -1 from below

Questions 4 and 5 refer to the following functions: x2 + 1 x f ( x) = 4 A B f ( x) = 2 x −1 x 2 x +1 1 C f ( x) = D f ( x) = x +1 ( x + 2)( x − 2)

4) Which of these functions are even? (a) A, C and D (c) A and D (e) I don’t know

(b) A only (d) B and D

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MEI FP1 Graphs Section 1 MC test 5) Which of these functions are odd? (a) A and C (c) B and D (e) I don’t know

(b) B and C (d) B only

Questions 6, 7 and 8 refer to the four graphs P, Q, R and S shown below.

P

Q

R

S

6) One of the graphs above is y = (a) P (c) R (e) I don’t know

(b) Q (d) S

7) One of the graphs above is y = (a) P (c) R (e) I don’t know

1 . This is graph ( x + 2)( x − 3)

x . This is graph ( x + 2)( x − 3) (b) Q (d) S

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MEI FP1 Graphs Section 1 MC test 8) One of the graphs above is y =

x2 . This is graph ( x + 2)( x − 3)

(a) P (c) R (e) I don’t know

(b) Q (d) S

9)

This is a sketch of the graph 2 x−3 2 (c) y = ( x − 3) 2 (e) I don’t know

2 ( x + 3) 2 2 (d) y = x+3

(a) y =

(b) y =

10)

This is a sketch of the graph x −1 2− x 1− x (c) y = 2− x (e) I don’t know (a) y =

x −1 2+ x 1− x (d) y = 2+ x

(b) y =

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MEI Further Pure Mathematics 1 Graphs and Inequalities Section 2: Inequalities Study Plan Background In this section you will learn a variety of different techniques for solving inequalities. Inequalities are very important: it is often necessary to find out for what values of x a function f(x) is greater than or less than a particular constant, or perhaps greater than or less than the value of another function g(x). For example, f(x) might describe how far a comet is from Earth at time x, and we might want to know whether the comet will ever get within a distance of 15 million km from Earth. The graph sketching skills introduced in section 1 will be useful in this section. Many inequalities can be solved by graphical methods; for others algebraic techniques may be more useful, and sometimes graphical and algebraic methods are combined. This section also includes some work on the range of values taken by a function. This is enrichment material and is therefore optional, but if you choose to work through this section it should help to deepen your understanding of the graphs of rational functions.

Detailed work plan 1. Read pages 87 – 91. There is a summary of the different methods that can be used in the Notes and Examples. 2. Exercise 3B Attempt Questions 1, 2, 3, 4, 6(i),(ii),(iv),(vii),(viii), 7 and 8. 3. The final section is enrichment material. If you have time, read pages 93 – 94. There is a further example in the Notes and Examples. If you would like to, attempt Questions 1, 2 and 3 of Exercise 3C.

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MEI Further Pure Mathematics 1 Graphs and Inequalities Section 2: Inequalities Notes and Examples These notes contain subsections on  Inequalities  The range of values taken by a function (enrichment work)

Inequalities The main difficulty with solving inequalities is deciding which of the bewildering number of possible methods to use in any given situation! In fact it comes down to personal preference. Some people like to use graphs as much as possible, others prefer to use algebraic methods. Here is a summary of the most commonly used methods. (Of course any of ≥, ≤, > or < could be involved). To solve inequalities of the form f(x) ≥ 0   

Sketch the graph of y = f(x) and look for the values of x for which the graph is (in this case) above or on the x axis (see Example 3.3 Solution 1). Find the critical points of f(x) and consider the sign of f(x) in each region (see Example 3.3 Solution 2, Example 3.4 Solution 2, Example 3.5 Solution 1). Multiply through by the square of the denominator (since this is always positive) and sketch the resulting polynomial, remembering to exclude any points for which the denominator is zero (see Example 3.4 Solution 3, Example 3.5 Solution 2).

To solve inequalities of the form f(x) ≥ g(x) 



Sketch the graphs of y = f(x) and y = g(x), and look for the values of x for which the graph of y = f(x) meets or is above the graph of y = g(x). This will involve solving the equation f(x) = g(x) (See Example 3.4 Solution 1). Rearrange the inequality into the form h(x) ≥ 0 and use any of the methods above (See Example 3.4 Solutions 2 and 3, and Example 3.5 Solutions 1 and 2).

All of these methods are covered in the examples in the text book. Read through these examples carefully. When you come to Exercise 3B, in the

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MEI FP1 Graphs Section 2 Notes and Examples early questions you are asked to sketch graphs, and so it then makes sense to use the graphs to solve the inequalities. However, in Question 6 onwards, you are left to make your own choice of method. If you have time, try doing some of the questions by several different methods. The worked solutions on the website show more than one method in many cases, but of course you may choose a different method. The main pitfall when solving inequalities is dealing with points at which the function is undefined. Look at Example 3.4 in the textbook. Solution 1 is a graphical method. From the graph it is easy to see that x = 0 is undefined, as there is an asymptote here. So although the inequality that is being solved involves ≥, the solution turns out to be -3 ≤ x < 0 or x ≥ 1, so that x = 0 is not included in the solution. In Solution 2 the method of critical points is used. Here, you must consider each of the critical points themselves and decide whether they are included in the solution. Critical points which make the numerator zero (x = -3 and x = 1) are included, those which make the denominator zero (x = 0) are not. In Solution 3 the inequality is multiplied through by x². This can only be done if x ≠ 0, so x = 0 must be excluded from the solution. This is normally only an issue with inequalities involving ≥ or ≤. Those involving > or < are simpler, since the critical points will not be involved in the solution in any case.

The range of values taken by a function Note that this section is enrichment material and therefore optional. This section shows you a method for finding the positions of turning points without using calculus. It follows on from the ideas introduced in the subsection “How many turning points” within the “Graphs of rational functions” section in the text book. In the Notes and Examples for section 1, three examples of sketching a curve x2  x  4 were given. The third example was the graph y  , which turned out x2  3 to look like this:

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MEI FP1 Graphs Section 2 Notes and Examples The following example shows how to find the turning points of this graph.

Example 1 (i) (ii)

x2  x  4 Find the set of possible values of . x2  3 x2  x  4 Hence sketch the graph of y  , giving the coordinates of the turning x2  3 points.

Solution (i)

x2  x  4 x2  3 (x² + 3)y = x² + x + 4 x²y + 3y – x² – x – 4 = 0 (y – 1)x² – x + 3y – 4 = 0 y

For x to be real, the discriminant of this quadratic equation must be greater than or equal to zero. (-1)² – 4(y – 1)(3y – 4) ≥ 0 1 – 4(3y² – 7y + 4) ≥ 0 1 – 12y² + 28y – 16 ≥ 0 12y² – 28y + 15 ≤ 0 (2y – 3)(6y – 5) ≤ 0 To solve this inequality, sketch the graph f(y) = (2y – 3)(6y – 5)

5 6

From the graph, So (ii)

5 6

 y

3 2

3 2

5 x2  x  4 3   6 x2  3 2

The minimum possible value of y is 56 . Substituting this value of y into the quadratic equation found in (i): (y – 1)x² – x + 3y – 4 = 0

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MEI FP1 Graphs Section 2 Notes and Examples

 56  1 x 2  x  52  4  0  16 x 2  x  32  0 x2  6x  9  0 ( x  3) 2  0 x  3 So the minimum point is  3, 56  .

The maximum possible value of y is 32 . Substituting this value of y into the quadratic equation found in (i): (y – 1)x² – x + 3y – 4 = 0  32  1 x 2  x  92  4  0 1 2

x 2  x  12  0

x2  2x  1  0 ( x  1) 2  0 x 1 So the maximum point is 1, 32 

We can now use the earlier sketch of the graph, and add the coordinates of the turning points.

1, 32   3, 56 

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Further Pure Mathematics 1 Graphs and Inequalities Section 2: Inequalities Crucial points 1. Be careful to use the correct inequality sign in your answers If the question involves < or >, the solution set should involve < or >. However, be careful when the question involves ≤ or ≥, as sometimes the answer may involve < or > (see point 2 below). 2. Watch out for points where the function is undefined For example, a function is undefined at values of x which give a vertical asymptote. These points must not be included in your solution set. 3. Remember the rules for manipulating inequalities Remember that you cannot multiply through by a quantity which may be positive or negative, such as (x – 3). However, you could multiply through by a quantity such as (x – 3)² which is always positive – but if you do, remember that in this case x = 3 must be excluded from the solution set. 4. Watch out for points where a function becomes zero but does not change sign If you want to solve an inequality of the form f(x) ≥ 0, an isolated point where the graph of y = f(x) touches the x axis must be included. For example, to solve the inequality (x + 1)²(x – 2) ≥ 0, the graph below shows that x ≥ 2 is part of the solution, but x = -1 is also part of the solution, since y = 0 when x = -1. y = (x + 1)²(x – 2)

y

-1

2

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x

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Further Pure Mathematics 1 Graphs and Inequalities Section 2: Inequalities Exercise 1. With the help of sketch graphs find values for x which satisfy 1 2 (i) (ii) 3 − x < >1 x−3 x 2. Sketch the curve y = 0≤

x −1 ≤ 2 for x. x+2

x −1 and use this to help solve the inequality x+2

3. Use any appropriate method to solve the following inequalities: x −1 ( x − 1)( x − 2) (i) (ii) >1 >1 ( x − 3)( x − 4) x−3 1 1 1 1 (iii) (iv) < < 2 x −1 x +1 x x+2 Enrichment questions 4. (i) (ii)

5. (i) (ii) (iii)

2 − 3x 3 + 3x + x 2 Sketch the curve showing the co-ordinates of the points of intersection with the axes and any turning points.

Find the range of values that y can take for real x if y =

4x 1 + x2 Find the points at which the curve cuts the axes, and any asymptotes. Sketch the curve.

Find the range of values that y can take for real x given that y =

6. Show that for real x, 0
1 x −3

1 crosses the y-axis at ( 0, − 31 ) . x −3 There is a vertical asymptote at x = 3 As x → ∞, y → 0 from above As x → -∞, y → 0 from below

The graph of y =

y=

1 x −3

y=1

The graphs cross where

1 = 1 ⇒ 1 = x −3 x −3 ⇒x =4

From the graph, the solution of the inequality is 3 < x < 4. (ii) 3 − x
1 x −3

2

x

y=

2

x

y=3–x

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Further Pure Mathematics 1 The graphs cross where 3 − x =

2

x

⇒ 3x − x 2 = 2 ⇒ x 2 − 3x + 2 = 0 ⇒ ( x − 1)( x − 2) = 0 ⇒ x = 1 or x = 2

From the graph, the solutions to the inequality 3 − x < are 0 < x < 1 and x > 3.

2.

y=

2

x

x −1 x +2

The graph crosses the axes at (1, 0) and (0, − 21 ). There is a vertical asymptote at x = -2 As x → ∞, y → 1 from below As x → -∞, y → 1 from above

y=

x −1 x +2

y=2 y=0

x −1 crosses the graph y = 0 at x = 1 x +2 x −1 The graph y = crosses the graph y = 2 at the point for which x +2 x −1 = 2 ⇒ x − 1 = 2x + 4 x +2 ⇒ x = −5 x −1 ≤ 2 are x ≤ −5 and From the graph, the solutions of the inequality 0 ≤ x +2 x ≥ 1. The graph y =

3. (i)

x −1 >1 x −3

Multiplying both sides by ( x − 3)2 :

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Further Pure Mathematics 1 ( x − 1)( x − 3) > ( x − 3)2 ( x − 1)( x − 3) − ( x − 3)2 > 0 ( x − 3)(( x − 1) − ( x − 3)) > 0 2( x − 3) > 0

x >3 ( x − 1)( x − 2) >1 ( x − 3)( x − 4) ( x − 1)( x − 2) −1 >0 ( x − 3)( x − 4) ( x − 1)( x − 2) − ( x − 3)( x − 4) >0 ( x − 3)( x − 4) x 2 − 3 x + 2 − ( x 2 − 7 x + 12) >0 ( x − 3)( x − 4) 4 x − 10 >0 ( x − 3)( x − 4) The critical points are x = 52 , x = 3 and x = 4.

(ii)

x< -

4x – 10 X–3 X–4 4 x − 10 ( x − 3)( x − 4)

5 2

5 2

The solutions of the inequality are (iii)

<x 4.

1 1 < x −1 x +1

y=

1 x +1

y=

1 x −1

From the graph, the solution of the inequality is -1 < x < 1.

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1 1 < x −1 x +1

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Further Pure Mathematics 1 (iv)

1

x

2


2. 2 − 3x 3 + 3x + x 2 3 y + 3 xy + yx 2 = 2 − 3 x

4. (i) y =

yx 2 + (3 y + 3)x + 3 y − 2 = 0 For this quadratic equation to have real solutions, (3 y + 3)2 − 4 y(3 y − 2) ≥ 0 9 y 2 + 18 y + 9 − 12 y 2 + 8 y ≥ 0 3 y 2 − 26 y − 9 ≤ 0 (3 y + 1)( y − 9) ≤ 0 − 31 ≤ y ≤ 9

f(y)

− 31

9

y

For real x, y can take values such that − 31 ≤ y ≤ 9 . (ii)

When x = 0, y =

2 3

When y = 0, x = 23 There are no vertical asymptotes as the denominator is never zero. As x → ∞ , y → 0 from below As x → −∞ , y → 0 from above

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Further Pure Mathematics 1 Turning points:

When y = − 31 , − 31 x 2 + 2 x − 3 = 0

x 2 − 6x + 9 = 0 ( x − 3)2 = 0

x =3 When y = 9, 9 x 2 + 30 x + 25 = 0 (3 x + 5 )2 = 0

x = − 53

Turning points are ( 3, − 31 ) and ( − 53 , 9) .

( − 53 , 9)

y

2 3 2 3

5. (i)

( 3, − 31 )

x

4x 1 + x2 y + yx 2 = 4 x

y=

yx 2 − 4 x + y = 0 For this quadratic equation to have real solutions: 16 − 4 y 2 ≥ 0

y2 ≤ 4 −2 ≤ y ≤ 2 For real x, y can take values such that −2 ≤ y ≤ 2 . (ii) When x = 0, y = 0 The only point at which the curve cuts the axes is the origin. There are no vertical asymptotes as the denominator is never zero As x → ±∞ , y → 0 so y = 0 is a horizontal asymptote.

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Further Pure Mathematics 1 (iii) When y = 2, 2 x 2 − 4 x + 2 = 0

x2 − 2x + 1 = 0 ( x − 1)2 = 0

x =1 When y = -2, −2 x 2 − 4 x − 2 = 0 x2 + 2x + 1 = 0 ( x + 1)2 = 0

x = −1 so the turning points are (1, 2) and (-1, -2).

y

(1, 2)

x

(-1, -2)

6.

y=

4

2 + 2x + x2 2 y + 2 yx + yx 2 = 4

yx 2 + 2 yx + 2 y − 4 = 0 For this quadratic equation to have real solutions: (2 y )2 − 4 y(2 y − 4) ≥ 0 4 y 2 − 8 y 2 + 16 y ≥ 0 −4 y + 16 y ≥ 0 2

f(y)

y 2 − 4y ≤ 0 y( y − 4) ≤ 0 From graph, 0 ≤ y ≤ 4

4 y

But from the original equation, y cannot be zero 4 ≤ 4. so 0 < 2 + 2x + x2 When x = 0, y = 2 There are no values of x for which y = 0 There are no vertical asymptotes As x → ±∞ , y → 0 , so y = 0 is a horizontal asymptote

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Further Pure Mathematics 1 When y = 4, 4 x 2 + 8 x + 4 = 0

x2 + 2x + 1 = 0 ( x + 1)2 = 0

x = −1 so there is a turning point at (-1, 4)

y (-1, 4)

2

x

7.

x +2 ( x + 3)( x − 1) 2 yx + 2 yx − 3 y = x + 2

y=

yx 2 + (2 y − 1)x − 3 y − 2 = 0 For this quadratic equation to have real solutions, (2 y − 1)2 − 4 y( −3 y − 2) ≥ 0 4 y 2 − 4 y + 1 + 12 y 2 + 8 y ≥ 0 16 y 2 + 4 y + 1 ≥ 0 (4 y + 21 )2 − 41 + 1 ≥ 0 (4 y + 1)2 + 43 ≥ 0 This is true for all values of y, so for real x the function can take all real values.

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Further Pure Mathematics 1 Graphs and Inequalities Section 2: Inequalities Multiple Choice Test 1) The diagram shows the graphs of y = x3 + 2 x 2 − 1 (shown in red) and 2 y = (shown in blue). x

The solution of the inequality x 3 + 2 x 2 − 1 ≤ (a) x ≤ -2 or 0 < x ≤ 1 (c) -2 ≤ x < 0 or x ≥ 1 (e) I don’t know

2 is x

(b) -2 ≤ x ≤ 1 (d) x ≤ -1 or 0 < x < 2

2) The diagram shows the graph of y =

x−3 (the vertical asymptotes are ( x + 1)( x − 2)

shown in blue).

The solution of the inequality (a) -1 ≤ x ≤ 2 or x > 3 (c) -1 < x < 2 or x > 3 (e) I don’t know

x−3 ≥ 0 is ( x + 1)( x − 2)

(b) -1 ≤ x ≤ 2 or x ≥ 3 (d) -1 < x < 2 or x ≥ 3

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Further Pure Mathematics 1 (1 + x) 2 3) The diagram shows the graph of y = . (3 + x)(1 − x)

(1 + x ) 2 The solution of the inequality ≤ 0 is (3 + x)(1 − x)

(a) x ≤ -3, x = -1 or x ≥ 1 (c) x < -3 or x > 1 (e) I don’t know

(b) -1 < x < 3 (d) x < -3, x = -1 or x > 1

4) The set of values of x for which (2x – 1)(x + 4)(x – 5) < 0 is (a) x < −4 or 12 < x < 5 (c) x < −5 or − 12 < x < 4 a. I don’t know

(b) −4 < x < 12 or x > 5 (d) −5 < x < − 12 or x > 4

5) The set of values of x for which (x + 3)(x + 1)²(x – 2) ≥ 0 is (a) −3 ≤ x ≤ −1 or x ≥ 2 (c) −3 ≤ x ≤ 2 (e) I don’t know

(b) x ≤ −3, x = −1 or x ≥ 2 (d) x ≤ −3 or − 1 ≤ x ≤ 2

6) The set of values of x for which (a) x < −3, −1 ≤ x ≤ 0 or x > 2 (c) x < −3 or 0 ≤ x < 2 (e) I don’t know

x( x + 1) ≤ 0 is (2 − x)( x + 3)

(b) −3 < x ≤ −1 or 0 ≤ x < 2 (d) −3 < x ≤ 1 or x > 2

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Further Pure Mathematics 1 7) The set of values of x for which

2+ x > 0 is 3 − 2x

(a) −2 < x < 32 (c) − 32 < x < 2 (e) I don’t know

(b) x < −2 or x > 32 (d) x < − 32 or x > 2

8) The set of values of x for which

x 2 + 6 x − 49 < 1 is x +1

(a) x < -10 or -1 < x < 5 (c) -10 < x < 5 (e) I don’t know

(b) -10 < x < -1 or x > 5 (d) x < -10 or x > 5

9) The set of values of x for which

x 1+ x is < x + 2 3− x

(a) -2 < x < 3 (c) -3 < x < 2 (e) I don’t know

(b) x < -2 or x > 3 (d) x < -3 or x > 2

10) The set of values of x for which 2 x + 1 ≥ (a) − 12 ≤ x ≤ 81 or x > 2 (c) x ≤ − 12 or 81 ≤ x < 2 (e) I don’t know

1 − 8x is x−2

(b) x ≤ −3 or 12 ≤ x < 2 (d) −3 ≤ x ≤ 12 or x > 2

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Further Pure Mathematics 1 Graphs and inequalities Chapter assessment 1. A curve has equation y =

x( x + 2) . 4x2 − 9

(i) Write down the coordinates of the points where the curve cuts the coordinate axes. [2] (ii) Write down the equations of the three asymptotes.

[3]

(iii)Describe the behaviour of the curve for large positive and large negative values of x, justifying your description.

[3]

(iv) Sketch the curve.

[4]

(v) Solve the inequality

x( x + 2) 1 < . 4x2 − 9 5

2. A curve has equation y =

[5]

x2 − 9 . x2 + 1

(i) Write down the coordinates of the points where the curve cuts the coordinate axes. [3] (ii) Explain why the curve has no vertical asymptotes, and give the equation of the horizontal asymptote. [2] (iii)Describe the behaviour of the curve for large positive and large negative values of x, justifying your description.

[3]

(iv) Sketch the curve.

[4]

x2 − 9 (v) The equation 2 = k has no real solutions. x +1 What can you say about the value of k?

[4]

3. Solve the following inequalities. (i) ( x + 1)( x + 3) < 5 x + 9 5x + 9 x +1

[5]

1 1 ≤ 2x + 3 x − 2

[5]

(ii) x + 3 < (iii)

[3]

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Further Pure Mathematics 1 4 , giving the equations of the ( x − 2)( x + 2) asymptote(s) and the coordinates of the point(s) where the graph cuts the coordinate axes. [4]

4. (i) Sketch the graph of y =

(ii) Write down the equation of the line of symmetry of the curve, and hence find the coordinates of the local maximum point. [2] 4 = k has no real solutions. ( x − 2)( x + 2) What can you say about the value of k?

(iii)The equation

(iv) Solve the inequality

4 1 ≥ . ( x − 2)( x + 2) 3

[4] [4]

Total 60 marks

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