KT Classroom Functions -Part 1

MATHEMATICS Learner’s Study and Revision Guide for Grade 12 FUNCTIONS ‐ Part 1 PARENT FUNCTIONS

Revision Notes, Exercises and Solution Hints by

Roseinnes Phahle

Examination Questions by the Department of Basic Education 1

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Contents Unit 5 Basic functions

Domain and range

Transformations and characteristics of functions Illustrations of transformations

Combined effect of parameters

A tip on sketching functions

Exercise 5.1

Answers

Parabola

Discriminant

Points of intersection

Examination questions with solution hints and answers More questions from past examination papers

Answers

3 4 5 6 17 18 20 21 22 24 26 28 34 39

TRANSFORMATION of functions is a theme that runs through this Unit and is fundamental to all the functions you study up to Grade 12. Pay very special attention to this Unit and follow it with Functions Parts 2 and 3 in order to gain a complete knowledge of everything about ALL the functions covered up to Grade 12.

How to use this revision and study guide 1. Study the revision notes given at the beginning. The notes are interactive in that in some parts you are required to make a response based on your prior learning of the topic in class or from a textbook.

2. “Warm‐up” exercises follow the notes. Some exercises carry solution HINTS in the answer section. Do not read the answer or hints until you have tried to work out a question and are having difficulty.

3. The notes and exercises are followed by questions from past examination papers.

4. The examination questions are followed by blank spaces or boxes inside a table. Do the working out of the question inside these spaces or boxes.

5. Alongside the blank boxes are HINTS in case you have difficulty solving a part of the question. Do not read the hints until you have tried to work out the question and are having difficulty.

6. What follows next are more questions taken from past examination papers.

7. Answers to the extra past examination questions appear at the end. Some answers carry HINTS and notes to enrich your knowledge.

8. Finally, don’t be a loner. Work through this guide in a team with your classmates.

2

Functions – Part 1

REVISION UNIT 5: PARENT FUNCTIONS Below are all the basic functions you study from Grade 10 to Grade 12. In the boxes shown, illustrate

[

]

their shapes in the domain x ∈ − 360 o ; 360 o in the case of the trigonometric functions and state their special features.

Shape Special features Basic Function f ( x )

y = x

Straight line

or linear

2 Parabolic

y=x or quadratic

Hyperbolic 1 y=

x

x

Exponential y = b ,

b is a constant

Sine or sin

y = sin x

y = cos x

Cosine or

cos

y = tan x Tangent or

tan

In general, these basic functions can be denoted by y = f ( x ) or g ( x ) or using any letters.

Name

We will look upon these basic functions as PARENT FUNCTIONS.

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DOMAIN AND RANGE OF A FUNCTION Domain: The set of values for which a function is defined is called the domain denoted by D f .

Range: The set of values which the function can take is called the range denoted by R f .

Illustrations:

Consider the function

f ( x ) = x 2 − 3x

As can be seen from the sketch or by substituting any real value into f ( x ) , this function is defined for any real value of x . Its domain is thus all real values and we write this as

D f : x ∈ ℜ or D f : x ∈ (− ∞;+∞ )

which means the set of all values of x such that x is a member or element of the set.

However, the function itself consists only of real values that are greater than or equal to ‐4. The range is thus

R f : y ≥ −4 or R f : y ∈ [− 4; ∞ )

Consider the function

f ( x ) =

1

x−2

The function is defined for all real values of x with the exception of x = 2 . Thus the domain is

D f : x ∈ ℜ − {2}

and, from the graph, we see that y is defined for all values except 0. Thus

R f : y ∈ ℜ − {0}

4

Asymptote: The graph neither crosses nor touches the dotted line x = 2. All the graph does is to get closer and closer to it.

The line is called an asymptote.

Functions – Part 1

TRANSFORMATION OF FUNCTIONS The following are the transformations which f ( x ) can undergo and their effects on the graph of f ( x ) : Transformation of f ( x )

Effect of the transformation on f ( x ) a ⋅ f (x ) , assuming a > 0 . Stretches or shrinks the ordinate y by scale factor a according to whether a > 1 or 0 < a < 1 . In other words the graph is vertically

stretched up or down depending to whether a is positive or negative.

Reflects the graph in the x ‐axis. − f (x ) Reflects the graph in the y ‐axis. f (− x ) Translates or shifts the graph by q units up or down according to whether f (x) + q q is positive or negative.

Translates or shifts the graph by p units to the right. f (x − p ) Translates or shifts the graph by p units to the left. f (x + p )

a ⋅ f (x ± p ) + q

First the graph of f ( x ) is stretched or shrunk by scale factor a and then shifted horizontally by p units and vertically by q units.

CHARACTERISTICS OF FUNCTIONS Helping you describe and sketch the graphs of functions will be the following characteristics: • •

Domain : values of x for which a function is defined

Range: the set of all values of y or f ( x ) which can be obtained from the domain

•

x ‐intercepts obtained by putting y = 0 and this solving the equation f ( x ) = 0

•

y ‐intercept obtained by putting x = 0 and thus evaluating y = f (0)

•

Equations of the asymptotes, if any

•

In the case of trigonometric functions, the amplitude =

1 (maximum value –minimum value) 2

• •

360 o Periods of trigonometric functions, for example, the period of sin kx is . k

The turning points

Algebraic and trigonometric functions: These two types of functions are treated together in what follows in order to stress that transformations work in the same way for both.

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ILLUSTRATIONS OF THE ABOVE TRANSFORMATIONS TRANSFORMATION: f ( x ) + a is f ( x ) shifted upward by a units

o o f (x ) = sinx; x ∈ 0 ;360 f ( x ) = sin x + 0,5;

[

]

Answer the following: Answer the following: Period = Period = Turning points = Turning points = Maximum value = Maximum value = Minimum value = Minimum value = Amplitude = Amplitude = Domain: Domain: Range: Range:

In the space opposite, describe the

transformation illustrated.

TRANSFORMATION: f ( x ) − a is f ( x ) shifted downward by a units

f (x ) = x 2

Domain: Range:

In the space opposite, describe the transformation illustrated.

]

f (x ) = x 2 − 1,5

Answer the following: x ‐intercepts = y ‐intercept = Domain: Range:

6

[

x ∈ 0 o ;360 o

Functions – Part 1

TRANSFORMATION: f ( x ) ± a is f ( x ) shifted upward or downward by a units

The graph of f ( x ) = 2 x is shown below. On the same axes sketch the graph of g ( x ) = 2 x + 3 .

y 9 8 7 6 5 4 3 2 1

x -13

-12

-11

-10

-9

-8

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

8

9

10

11

12

13

-1 -2 -3 -4 -5 -6 -7 -8 -9

Answer the following with respect to f ( x ) : y − intercept is at?

Equation of the asymptote is?

Domain: Range:

Describe the transformation:

The graph of f ( x ) =

1 is shown. On the same x 1 axes sketch the graph of g ( x ) = − 2 x

Answer the following with respect to f ( x ) : Equations of the asymptote are?

Domain: Range:

Answer the following with respect to g ( x ) : Equations of the asymptote are?

Domain: Range:

Describe the transformation:

Answer the following with respect to g ( x ) : y − intercept is at?

Equation of the asymptote is?

Domain: Range:

y 9 8 7 6 5 4 3 2 1

x -13

-12

-11

-10

-9

-8

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

8

9

10

11

12

13

-1 -2 -3 -4 -5 -6 -7 -8 -9

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TRANSFORMATION: f ( x ± a ) is f ( x ) shifted to the left or right by a units

[

]

(

f ( x ) = sin x; x ∈ − 90 o ;270 o

[

)

]

f (x ) = sin x + 30 o ; x ∈ − 90 o ;270 o

y

y

2

2

1

1

x -60

-30

30

60

90

120

150

180

210

x

240

-60

30

-1

-1

-2

-2

Answer the following: Period = Turning points = Maximum value = Minimum value = Amplitude = Domain: Range: x ‐intercepts = y ‐intercept =

In the space opposite, describe the transformation illustrated.

The graph of f ( x ) = x 2 is shown. On the same axes sketch the gaph of g ( x ) = ( x − 3)

Answer the following with respect to f ( x ) : Domain: Range:

Answer the following with respect to g ( x ) : Domain: Range:

Describe the transformation:

2

60

90

120

150

180

210

240

1

2

3

4

5

Answer the following: Period = Turning points = Maximum value = Minimum value = Amplitude = Domain: Range: x ‐intercepts = y ‐intercept =

35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 -5

8

-30

-4

-3

-2

-1

-1 -2

y

x

Functions – Part 1

TRANSFORMATION: f ( x + a ) is f ( x ) shifted to the left by a units

f (x ) = 2 x

f (x ) = 2

( x+3)

Answer the following: Answer the following: 1. y ‐intercept is at? Explain?

1. y ‐intercept is at?

2. Equation of the asymptote is?

2. Equation of the asymptote is?

3. Domain?

3. Domain?

4. Range?

4. Range?

Describe the transformation in the space opposite:

TRANSFORMATION: f ( x − a ) is f ( x ) shifted to the right by a units

1 f (x ) = x

Answer the following: What are the equations of the asymptote?

f (x ) =

1 x−2

Answer the following: 1. What are the equations of the asymptote?

2. What has changed, and how?

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TRANSFORMATION: f ( x + a ) is f ( x ) shifted to the left by a units

[

]

(

f ( x ) = tan x; x ∈ − 360 o ;360 o

)

f ( x ) = tan x + 45o

Answer the following: 1. The x − intercepts are at?

2. The asymptotes are at?

Answer the following: 1. The x − intercepts are at?

2. The asymptotes are at?

3. Describe the change.

TRANSFORMATION: af ( x ) stretches f ( x ) upwards or downwards by scale factor ± a units

o o f ( x ) = sin x; x ∈ 0 ;360 f ( x ) = 2 sin x; x ∈ 0 o ;360 o

[

]

Answer the following: Period = Turning points = Maximum value = Minimum value = Amplitude = Domain: Range: x ‐intercepts = y ‐intercept =

Describe the transformation in the space opposite:

[

Answer the following: Period = Turning points = Maximum value = Minimum value = Amplitude = Domain: Range: x ‐intercepts = y ‐intercept =

10

]

Functions – Part 1

TRANSFORMATION: af ( x ) stretches f ( x ) upwards or downwards by scale factor ± a units

f (x ) = x 2

f ( x ) = −2x 2

6

4

2

0 0

-5

0

5

10

-2

In the space opposite, describe the transformation illustrated.

TRANSFORMATION: af ( x ) stretches f ( x ) upwards or downwards by scale factor ± a units

f (x ) = 2 x

Answer the following: 1. y ‐intercept is at?

2. Equation of the asymptote is?

3. Domain?

4. Range?

f ( x ) = 3.2 x

Answer the following: 1. y ‐intercept is at?

2. Equation of the asymptote is?

3. Domain?

4. Range?

Describe the transformation:

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TRANSFORMATION: af ( x ) stretches f ( x ) upwards or downwards by scale factor ± a units

f (x ) =

f (x ) = −

1 x

3 x

Answer the following questions: Answer the following questions: Write down the equations of the asymptotes: Write down the equations of the asymptotes:

Domain: Domain: Range: Range:

Describe the transformation:

TRANSFORMATION: f ( x ) + a is f ( x ) shifted upward by a units

f (x ) = 2 x

f (x ) = 2 x + 3

6

6

5

5

4

4

3

3

2

2

1

1

0 0

-5

0

5

0

10

-5

-1

Answer the following: y − intercept is at?

Equation of the asymptote is?

Domain: Range:

Answer the following: y − intercept is at?

Equation of the asymptote is?

Domain: Range:

Describe the transformation:

12

0

-1

5

Functions – Part 1

TRANSFORMATION: f ( x ) + a is f ( x ) shifted upward by a units

f (x ) =

f (x ) =

1

x

1 + 1,5 x

Answer the following: Answer the following: What are the equations of the asymptote?

What are the equations of the asymptote?

x ‐intercept = Domain: Range:

Domain:

Range:

Describe the transformation:

TRANSFORMATION: f (ax) is f ( x ) shrinked by scale factor 1/ a units

[

]

[

f ( x ) = sin x; x ∈ 0 o ;360 o

]

f ( x ) = sin 3x; x ∈ 0 o ;360 o

Answer the following: The x − intercepts are at?

Period = Amplitude = Domain: Range:

Answer the following: The x − intercepts are at?

Period = Amplitude = Domain: Range:

Describe the transformation:

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TRANSFORMATION: − f ( x ) is f ( x ) flipped upside down and reflected about the x ‐axis

o o f ( x ) = cos x; x ∈ − 180 ;360 f ( x ) = − cos x; x ∈ − 180 o ;360 o

[

]

[

]

Answer the following: Answer the following: Period = Period = Turning points = Turning points = Maximum value = Maximum value = Minimum value = Minimum value = Amplitude = Amplitude = Domain: Domain: Range: Range:

Describe the transformation:

TRANSFORMATION: − f ( x ) is f ( x ) flipped upside down and reflected about the x ‐axis

f (x ) = x 2

Answer the following: Domain: Range:

.

f (x ) = − x 2

Answer the following: Domain: Range:

Describe the transformation:

14

Functions – Part 1

TRANSFORMATION: − f ( x ) is f ( x ) flipped upside down and reflected about the x ‐axis

The graph of f ( x ) = 2 x is shown below. O the same axes sketch in the graph of g ( x ) = −2 x .

y 9 8 7 6 5 4 3 2 1

x -13

-12

-11

-10

-9

-8

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

8

9

10

11

12

13

-1 -2 -3 -4 -5 -6 -7 -8

-9

The graph of f ( x ) =

1 1 is shown below. On the same axes sketch in the graph of g ( x ) = − . x x

y 9 8 7 6 5 4 3 2 1

x -13

-12

-11

-10

-9

-8

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

8

9

10

11

12

13

-1 -2 -3 -4 -5 -6 -7 -8

The graph of f ( x ) = tan x is shown below. On the same axes sketch the graph of g ( x ) = − tan x .

-9

y 9 8 7 6 5 4 3 2 1

x -330

-300

-270

-240

-210

-180

-150

-120

-90

-60

-30

30

60

90

120

150

180

210

240

270

300

330

-1 -2 -3 -4 -5 -6 -7 -8

-9

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TRANSFORMATION: f (− x ) is the reflection of f ( x ) about the y ‐axis

f (x ) = 2 x

f (x ) = 2 − x

y

y 9

9

8

8

7

7

6

6

5

5

4

4

3

3 2

2

1

1

x

x -13

-12

-11

-10

-9

-8

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

8

9

10

11

12

-13

13

-12

-11

-10

-9

-8

-7

-6

-5

-4

-3

-2

-1

1

-1

-1

-2

-2

-3

-3

-4

-4

-5

-5

-6

-6

-7

-7

-8

-8

-9

-9

2

3

4

5

6

7

8

9

10

Answer the following: y − intercept is at?

Equation of the asymptote is?

Describe the transformation:

Answer the following: y − intercept is at?

Equation of the asymptote is?

The graph of f ( x ) =

1 1 is shown below. On the same axes sketch the graph of g ( x ) = − . x x

y 8 7 6 5 4 3 2 1

x -8

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

8

-1 -2 -3 -4 -5 -6 -7

Answer the following with respect to f ( x ) : What are the equations of the asymptote?

-8

Answer the following with respect to g ( x ) : What are the equations of the asymptote?

Describe the transformation:

16

11

12

13

Functions – Part 1

COMBINED EFFECT OF THE PARAMETERS a, p and q We now look at how f ( x ) is transformed by a. f ( x ± p ) ± q .

In a step by step way, this is how you would describe the transformation of f ( x ) : STEP 1: You start off with f ( x ) STEP 2: f ( x ) → f (x ± p ) ; that is a horizontal shift of f ( x ) to the left or right by ± p units STEP 3: f ( x ± p ) → a. f ( x ± p ) ; that is f ( x ± p ) is vertically stretched upwards or downwards by scale factor ± a STEP 4: a. f ( x ± p ) → a. f ( x ± p ) ± q ; that is a. f ( x ± p ) is vertically shifted by ± q units In a single sentence, we can describe the above transformation as:

f (x ) is shifted or translated either to the left or right by p units, stretched by a scale factor a and then shifted or translated upwards or downwards by q units.

Whether the translations are to the left or right, upward or downward will depend on the signs of the parameters.

Trigonometric functions: The functions sin kx , cos kx and tan kx transform the periods of sin x and

cos x by

360 o 180o and tan x by . k k

SUMMARY The table below summarises the effect of the parameters a , p and q on f ( x ) and shows a striking similarity between transformations in algebra, trigonometry and geometry :

Basic

function a ⋅ f ( x ) f (x) + q f (x − p ) f (x ) Effect of Multiplies the y by a Adds q to y Adds p to x a , p (x; y ) → (x; ay ) (x; y ) → (x; y + q ) (x; y ) → (x + p; y ) and q

PARENT FUNCTIONS

a ⋅ f (x − p ) + q

Combines all the effects.

(x, y ) → (x + p; ay + q )

Do you see now why we said the basic functions can be looked upon as parent functions? Can you also see the relationship between the transformation of functions and transformation geometry?

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A TIP ON SKETCHING FUNCTIONS When you sketch functions, always start with a sketch of the parent function shown by the dotted curves in the previous illustrations and those that follow. Then apply transformations to the sketch in a step by step manner.

ILLUSTRATIONS Example 1: Sketch the graph of y = ( x − 1) − 9 2

y 9 8 7

Step 2 6 5 4

Step 1

3

Step 3

2 1

x -13

-12

-11

-10

-9

-8

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

8

9

10

11

12

13

-1 -2 -3 -4 -5 -6 -7 -8 -9

Can you interpret the above graphs in terms of the transformations undergone by y = x 2 ?

Example 2: Sketch the graph of y = −2.3 x +5 − 4 y 9 8

Step 4

7 6

Step 2

5 4 3

Step 1

2 1

x -13

-12

-11

-10

-9

-8

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

8

9

10

11

12

13

-1

Step 3

-2 -3 -4 -5 -6 -7 -8 -9

Can you interpret the above graphs in terms of the transformations undergone by y = 3 x ?

18

Functions – Part 1

(

)

Example 3: Sketch the graph of y = 2 + 3 sin( x − 30 o ) for values of x ∈ − 90 o ; 270 o y 7

6

5

Step 4

4

3

Step 3

2

Step 2

1

Step 1 x -90

-80

-70

-60

-50

-40

-30

-20

-10

10

20

30

40

50

60

70

80

90

100 110

120

130 140

150

160 170

180

190 200

210

220 230

240

250 260

270

-1

-2

-3

-4

Can you interpret the above graphs in terms of the transformations undergone by y = sin x ?

Example: Sketch the graph of y =

3 + 6 x−4 y 9 8 7

Step 4

Step 2

6 5 4 3

Step 3

2 1

Step 1

-13

-12

-11

-10

-9

-8

-7

-6

-5

-4

-3

-2

-1

1

x 2

3

4

5

6

7

8

9

10

11

12

13

-1 -2 -3 -4 -5 -6 -7 -8 -9

Can you interpret the above graphs in terms of the transformations undergone by y =

1 ? x

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EXERCISE 5.1 1. Find the equation of the curve obtained when the graph of y = x 2 − 2 x − 3 is made to undergo the following transformations and the coordinates of the turning points in each case: a. shifted by ‐2 units parallel to the y ‐axis

b. translated by +5 units parallel to the x ‐axis

c. stretched vertically by a scale factor of 3, followed by a translation of +1 in the direction of the y ‐axis

d. reflected in the x ‐axis

e. reflected in the y ‐axis 2. Find the equation of the curve obtained when the graph of y = cos x is made to under the following transformations: a. stretched by ‐3 units in the direction of the y ‐axis

b. shifted by +3 units in the direction of the y ‐axis

c. translated by 30 o in the negative direction of the x ‐axis

d. period reduced to 120 o

e. shifted by + 45 o along the x ‐axis, followed by stretching by scale factor 2 and a downward translation by 3 units 3. State the transformation(s) that must be applied to the graph of y = x 2 to obtain the graph of:

y = (x − 4) 2

a.

y = x2 + 3

b.

y = −x2

c.

d.

y = 3( x − 2 ) − 5 2

20

Functions – Part 1

ANSWERS EXERCISE 5.1 In this question you are required to transform a function and thereafter to find the coordinates of its turning points.

The best and quickest way to proceed is first, by completing the square, find the coordinates of the turning point of the function you are asked to transform. Then apply the transformations to this turning point which is (1; ‐4).

Of course, you can go about the long way of completing the square on each of the transformed functions to find the coordinates of the turning points.

(

)

(

)

(

)

1. a. y = x 2 − 2 x − 3 − 2 = x 2 − 2 x − 5

Turning point at (1; ‐6)

2 b. y = ( x − 5) − 2( x − 5) − 3 = x 2 − 12 x + 32

Turning point at (6; ‐4)

c. y = 3 x 2 − 2 x − 3 + 1 = 3 x 2 − 6 x − 8

Turning point at (1; ‐11)

d. y = − x 2 − 2 x − 3 = − x 2 + 2 x + 3

Turning point at (1; 4)

2 e. y = (− x ) − 2(− x ) − 3 = x 2 + 2 x − 3

Turning point at (‐1; ‐4)

2. a. y = −3 cos x b. y = cos x + 3

(

)

c. y = cos x + 30 o d. y = cos 3 x

(

)

e. y = 2 cos x − 45 o − 3

3. a. Shift by 4 units to the right parallel to the x ‐axis.

b. Shift upwards by 3 units parallel to the y ‐axis.

c. Reflection in the x ‐axis.

d. Shift by 2 units to the right parallel to the x ‐axis; followed by stretching by scale factor +3; then a shift downwards by 5 units parallel to the y ‐axis.

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THE PARABOLA Consider a quadratic equation:

ax 2 + b x + c = 0

The formula for solving this quadratic equation is

The solutions of the equation are called ROOTS. There is a relationship between the roots of an equation, its factors and the x ‐intercepts when the equation is graphed by putting

y = ax 2 + bx + c

The shape the graph is called a PARABOLA. Note first that a comparison of the above two equations implies that y = 0 which is the equation of the

x ‐axis. A solution of the quadratic equation is therefore given by the points of intersection of the parabola and the x ‐axis. It is these points that are called the x ‐intercepts and which form the roots of the equation.

Note also that if the roots of the equation

ax 2 + b x + c = 0

are

x = x 1 and x = x 2

which we can write as

x − x 1 = 0 and x − x 2 = 0

ax 2 + bx + c = a(x − x 1 )( x − x 2 )

then

‐ all of which has demonstrated a relationship between the roots of an equation, its factors and the x ‐ intercepts.

Let’s extend the idea that the graphical solution of an equation is given by the intersection of the x ‐axis which is a line with equation y = 0 and the graph of the equation and apply this idea to the graphical solution of 22

Functions – Part 1

ax 2 + bx + c = k , where k is any constant.

The solutions or roots will now be given by the points of intersection of the graphs of

y = ax 2 + bx + c

y =k

and

Illustrations x 2 + 2x − 3 = 0

x 2 + 2x − 3 = 2

x 2 + 2x − 5 = 0

We have drawn the graphs of y = x 2 + 2 x − 3 and y = 2 The two graphs intersect at x = −3,5 and x = 1,5 which must be roots of x 2 + 2 x − 3 = 2 x 2 + 2 x − 3 − 2 = 0 x 2 + 2 x − 5 = 0 Use the quadratic formula in the space below to verify the solutions obtained from the graph:

x =

Instead of drawing two graphs as in the previous column, we have drawn a single graph that of

y = x 2 + 2 x − 5

in order to solve

x 2 + 2 x − 3 = 2

Can you read the solutions off the graph above and do they agree with the solutions in the previous column?

x = x =

The x ‐intercepts are x = −3 and x = 1 which are the roots of the above equation.

Factorizing

x 2 + 2 x − 3 = 0 = (x + 3)( x − 1)

Preparation for the Mathematics examination brought to you by Kagiso Trust

The turning points of a parabola The parent graph to the parabola is y = x 2 with a turning point at the origin (0; 0). Let’s apply the following series of transformations to the parent graph.

STEP 1: Shift the graph to the right by p units so that y = ( x − p ) .

2

The origin is thus shifted to be at ( p; 0) .

STEP 2: Stretch the graph by a scale factor of a so that y = a ( x − p ) .

2

The origin in STEP 1 remains unchanged.

STEP 3: Shift he graph upwards by q units so that y = a ( x − p ) + q ..

2

The origin now shifts upwards by q units to be at ( p; q )

Example: In the space below, use the method of completing the square to verify that the coordinates of

the turning point of y = 3 x 2 + 12 x + 7 are (− 2; - 5) . A sketch is shown to verify the coordinates.

19

y

18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 -13

-12

-11

-10

-9

-8

-7

-6

-5

-4

-3

-2

-1

-1

x 1

2

3

4

5

6

7

8

9

10

11

12

13

-2 -3 -4 -5 -6 -7 -8 -9 -10

The DISCRIMINANT: b 2 − 4ac Below are shown the graphs of the quadratic equation y = ax 2 + bx + c : 1.

y = x 2 − 2x − 8

2.

y = x 2 − 2 x + 1

3.

y = x 2 − 2x + 2 24

-11

Functions – Part 1

In each case calculate b 2 − 4ac and comment on the relation of the graph to the x ‐axis.

y = x 2 − 2x − 8

y

19 18

17 16 15

b − 4ac =

14

2

13 12 11 10 9

Comment:

8 7 6 5 4 3 2 1 -13

-12

-11

-10

-9

-8

-7

-6

-5

-4

-3

-2

-1

x 1

-1

2

3

4

5

6

7

8

9

10

11

12

13

-2 -3 -4 -5 -6 -7 -8 -9

-10 -11

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

y

19 18

17 16 15

b 2 − 4ac =

14 13 12 11 10

Comment:

9 8 7 6 5 4 3 2 1 -13

-12

-11

-10

-9

-8

-7

-6

-5

-4

-3

-2

-1

x 1

-1

2

3

4

5

6

7

8

9

10

11

12

13

-2 -3 -4 -5 -6 -7 -8 -9

y = x 2 − 2x + 2

-10 -11

19

y

18

17 16 15

b 2 − 4ac =

14 13 12 11

Comment:

10 9 8 7 6 5 4 3 2 1 -13

-12

-11

-10

-9

-8

-7

-6

-5

-4

-3

-2

-1

-1

x 1

2

3

4

5

6

7

8

9

10

11

12

13

-2 -3 -4 -5 -6 -7 -8 -9

-10 -11

What can you conclude from the above about the roots of a quadratic equation? Write your conclusions in the spaces below and describe the graph in each case.

b 2 − 4ac > 0

b 2 − 4 ac = 0

b 2 − 4 ac < 0

You can see that b 2 − 4ac tells you whether the roots are real and distinct, or real and coincident or there are no real roots (in other words, at Grade 12 we would say “no solution”).

Preparation for the Mathematics examination brought to you by Kagiso Trust

POINTS OF INTERSECTION To determine the coordinates of the points of intersection of two graphs, simply solve their equations simultaneously. This idea was graphically illustrated on page 69 ‐ see the middle column on that page.

Example: The graphs of y = x 2 − x − 5 and y = 2 x + 5 are shown below. Verify by solving the two equations simultaneously that the coordinates of A are (5; 15) and those of B are (‐2; 1).

19

y

18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 -13

-12

-11

-10

-9

-8

-7

-6

-5

-4

-3

-2

-1

-1

x 1

2

3

4

5

6

7

8

9

10

11

12

13

-2 -3 -4 -5 -6 -7 -8 -9 -10 -11

Solution:

x 2 − x − 5 = 2x + 5

x 2 − x − 2x − 5 − 5 = 0

x 2 − 3 x − 10 = 0

(x + 2)(x − 5) = 0

so that

x = −2

and

Substituting into

y = 2x + 5

y = 2( −2) + 5 = 1

x = 5

and

Hence A(5;15) and B(- 2;1) .

26

y = 2(5) + 5 = 15

Functions – Part 1

Example: Determine the equations of the graphs of the hyperbola and parabola shown in the sketch below. y 9

8

7

6

5

4

3

2

xA(2;1)

1

x -2

-1

1

2

3

4

5

-1

-2

-3

Solution:

By inspection of the asymptotes of the graph:

y=

so that

y=

The equation of a hyperbola is of the form:

a +q x− p

p = 3 and q = 3

a + 3 x−3 a + 3 = −a + 3 2−3

Substituting the coordinates of the point A:

1=

a = 3 −1 = 2

Hence the equation of the hyperbola is

y=

2 + 3 x−3

The x ‐intercepts of the parabola are 1 and ‐2.

The equation of the parabola is therefore

y = a(x − 1)(x + 2)

The y ‐intercept the parabola makes is at (0; 6) so that

6 = a(0 − 1)(0 + 2) = a(− 1)(2) = −2

∴

a = −3

Hence the equation of the parabola is

y = −3( x − 1)( x + 2) = −3x 2 − 3x + 6

Preparation for the Mathematics examination brought to you by Kagiso Trust

PAPER 1 QUESTION 4

DoE/ADDITIONAL EXEMPLAR 2008

28

Functions – Part 1

PAPER 1 QUESTION 4

Number Hints and answers 4.1 A and B are the x ‐intercepts made by the graph of f . At these points f ( x ) or y is zero.

DoE/ADDITIONAL EXEMPLAR 2008 Work out the solutions in the boxes below

Replace f ( x ) by zero and solve the resulting equation.

Answers: A(‐3;0) and B(1;0)

4.2

4.3

The given equation g ( x ) = a.3 x + q means that the parent exponential curve has been translated vertically upwards by q units.

To find the value of q you will need to find the y ‐intercept of the graph of f at the point C. Work out the y ‐intercept and replace q by its value.

Now to work out the value of a . You need a point with known coordinates on the graph of g . You can straight away tell the coordinates of D because D is the turning point of the graph of f . What are the coordinates of D? Substitute them and the value of q in the equation of g to find the value of a .

Answer: g (x ) = 3.3 x + 3

dy of f and put it equal to zero to solve Find

dx

for x .

Then use the x value to get the y coordinate.

Answer:

⎛ 3 15 ⎞ ; ⎟ ⎝ 2 4⎠ f (x ) − k means f (x ) is shifted vertically downwards by k units. So for what values of k Point ⎜ −

4.4

will the function be negative always?

Answer: k > 4

Preparation for the Mathematics examination brought to you by Kagiso Trust

PAPER 1 QUESTION 6

DoE/ADDITIONAL EXEMPLAR 2008

30

Functions – Part 1

PAPER 1 QUESTION 6 Number 6.1

Hints and answers The range is defined by all the values which the function y or

f (x ) can take.

Answer:

DoE/ADDITIONAL EXEMPLAR 2008

Work out the solutions in the boxes below There is no calculation to be worked out.

Simply look at the graph and say which values of y the graph can take and which it cannot.

y ∈ (− ∞;5) U (5; ∞ )

or

6.2

6.3

y ∈ R − {5} Put f equals to g and solve the equation for x .

Having found x , find y .

Now write down the coordinates of the point of intersection.

Answer: (0;2) and (‐2;8) You must give verbal expression to the transformation of the parent graph f implied by the mathematical expression

h( x ) =

3 + 5 . x +1

Answers: Reflection about the asymptote x = −1 or Reflection about the asymptote y = 5 or Reflection about the x ‐axis and translated up by 10 units.

There is no calculation to be worked out.

But make sure you understand the answers given here and that you yourself could write any of these descriptions without looking at the answers. The answers are expressed in different ways but they all mean the same transformation.

Preparation for the Mathematics examination brought to you by Kagiso Trust

PAPER 1 QUESTION 4

DoE/NOVEMBER 2008

32

Functions – Part 1

PAPER 1 QUESTION 4

Number

Hints and answers

4.1

All you need do to determine the values of p and q is to look at the positions of the asymptotes.

Having determined the values of p and q , use the point T(5; 3) to find the value of a .

Answers: You write the values of p and q . a = 1 This is a question whose answer is simply worked out but requires quite a somewhat long explanation.

What is the parent function of the function f given in this question?

The parent function has two lines of symmetry along which it can be reflected.

What are the equations of those lines?

What is the translation of the parent function to f ?

The lines of symmetry of the parent function will undergo a similar translation to become y = − x + c .

Apply the translation to y = − x and that will give you a value of c .

Answer: c = 6

4.2

DoE/NOVEMBER 2008

Work out the solutions in the boxes below

Preparation for the Mathematics examination brought to you by Kagiso Trust

MORE QUESTIONS FROM PAST EXAMINATION PAPERS Exemplar 2008

34

Functions – Part 1

Preparatory Examination 2008

Preparation for the Mathematics examination brought to you by Kagiso Trust

Feb – March 2009

36

Functions – Part 1

November 2009 (Unused paper)

DIAGRAM SHEET 1

Preparation for the Mathematics examination brought to you by Kagiso Trust

Feb – March 2010

DIAGRAM SHEET 1

38

Functions – Part 1

ANSWERS November 2009 (Unused) 7.1 x ‐intercept: (‐3; 0) y ‐intercept: (0; 3)

Exemplar 2008 x = 3 7.1 y = −1 x ‐intercept: (5; 0) 7.2

7.2

5 y ‐intercept: (0; − ) 3 7.3

7.3 Horizontal asymptote: y = 1 Vertical asymptote: x = −1 7.4 Sketch:

Sketch:

Preparatory Examination 2008 a = −2 6.1 6.2 y = 1 x = 2

6.3 Sketch:

7.5 x ≤ −3 or x > −1 Or x ∈ (− ∞; − 3] ∪ (− 1; ∞ )

Feb/March 2010 5.1 Vertical asymptote: x = 3 Horizontal asymptote: y = 1 5.2 x ‐intercept: (1; 0) y ‐intercept: (0; 1/3) 5.3 Sketch:

Feb/March 2009 6.1 y =

2 2 x +1 2 + x +1 x + 3 = +1 = + =

x +1 x +1 x +1 x +1 x +1

2 + 2 x −1

⎛ ⎝

9⎞ 4⎠

6.2 Turning point at ⎜1;− ⎟ 6.3 y = 2 x = 2

6.4 h( x ) = −( x − 1) + 2

9

4