Ch02 Functions and Transformations

2 Functions and transformations AREAS OF STUDY • The behaviour of functions of a single real variable, including key fe...

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2 Functions and transformations AREAS OF STUDY

• The behaviour of functions of a single real variable, including key features of their graphs such as axis intercepts, stationary points and points of inflection, domain (including maximal domain) and range, asymptotic behaviour, symmetry and the algebra of functions, including composition of functions. The behaviour of these functions is linked to applications in practical situations. • Identification of key features of graphs of: – power functions y = xn for n ∈ N and n = −1, −2, 1 and transformations of these to the 2 form y = a(x + b)n + c where a, b and c ∈ R – the modulus function y = | x | • Transformations from y = f (x) to y = Af (n(x + b)) + c (where A, n, b and c ∈ R and f is one of the functions specified above) and the relation between the original function and the graph of the transformed

• •



• •

2A 2B 2C 2D 2E 2F 2G 2H 2I

Transformations and the parabola The cubic function in power form The power function (the hyperbola) The power function (the truncus) The square root function in power form The absolute value function Transformations with matrices Sum, difference and product functions Composite functions and functional equations 2J Modelling

function (including families of transformed functions for a single transformation parameter) Graphs of sum, difference, product and composite functions of f and g, where f and g are functions of the types above Recognition of the general form of possible models for data presented in graphical or tabular form, using polynomial and power functions Applications of simple combinations of the above functions, and interpretation of features of the graphs of these functions in modelling practical situations The relationship of f (x ± y), f (xy) and f to values of f (x) and f (y) for different functions f Composition of functions, where f composition g is defined by f (g(x)), given rang ⊆ domf ; for 4 example x 2 + 5 , x 3 + 2 x −1 eBook plus

2A

Transformations and the parabola

Digital doc

10 Quick Questions

Transformations In this chapter we consider the basic graphs of the quadratic and cubic functions, the hyperbola and truncus, square root and absolute value functions. The following transformations of the above graphs are discussed: dilation, reflection and translation.

Dilation A dilation is the stretching or compressing of a graph. Let the basic graph be y = f (x).

Chapter 2

Functions and transformations

57

Dilation away from the x-axis: y = af (x (x) 1. Stretches or compresses the graph f (x) by a factor of a from the x-axis. 2. Each y-value of the basic graph is multiplied by a factor of a, that is (x, y) → (x, ay). 3. When | a | > 1, the graph of f (x) is stretched and becomes narrower. 4. When 0 < | a | < 1, the graph of f (x) is compressed and becomes wider. Dilation away from the y-axis: y = f (nx) 1. Stretches or compresses the graph f (x) by a factor of 1 from the y-axis. n x 2. Each x-value of the basic graph is multiplied by a factor of 1 , that is (x, y) → ( , y). n n 3. When |n| > 1, the graph of f (x) is compressed from the y-axis and becomes narrower. 4. When 0 < |n| < 1, the graph of f (x) is stretched from the y-axis and becomes wider. Note: For the graphs we will be looking at in this chapter a horizontal dilation can be expressed as a vertical dilation. For example, (2 (2x + 1)3 can be written as 23 ( x + 12 )3 = 8( 8( x + 12 )3. So in this 1 case a horizontal dilation from the y-axis by a factor of 2 is the same as a vertical dilation from the x-axis by a factor of 8. This can simplify the process of describing transformations for these particular graphs. The concept of dilation is illustrated in the following diagram: y

y

x

x

Original graph dilated from the x-axis

Original graph dilated from the y-axis

x

Original graph

y

Ref lection Reflection provides a ‘mirror image’ of a graph. Reflection can take place in one or both axes. Let the basic graph again be y = f (x). Ref lection in the x-axis: y = −f (x (x) 1. The mirror image of the original graph appears across the x-axis (the mirror line). 2. Each y-value is the negative of the original, the x-value is unchanged, that is (x, y) → (x, −y). Reflection in the y-axis: y = f (−x) 1. The mirror image of the original graph appears across the y-axis (the mirror line). 2. Each x-value is the negative of the original, the y-value is unchanged, that is (x, y) → (−x, y). Reflection in both axes: y = −f (−x) 1. The basic graph is reflected in the x-axis and then the y-axis (or vice versa). 2. Both the x- and y-values are the negatives of the original, that is (x, y) → (−x, −y). The concept of reflection is shown in the diagram below. The red star is the original graph. y

y

x

y

x

x

Reflection in the y-axis

Reflection in the x-axis

58

Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad

Reflection in both axes

Translation: y = f (x − b) + c A translation slides the graph. Translation can be horizontal (to the right or left along the x-axis), or vertical (up or down along the y-axis). Consider our basic graph y = f (x). 1. If y = f (x – b) the basic graph is translated b units parallel to the x-axis: (a) in the positive direction (i.e. to the right) when b > 0 (b) in the negative direction (i.e. to the left) when b < 0. Each x-value has b added to it, that is (x, y) → (x + b, y). 2. If y = f (x) + c, the basic graph is translated c units parallel to the y-axis: (a) in the positive direction (i.e. up) when c > 0, (b) in the negative direction (i.e. down) when c < 0. Each y-value has c added to it, that is (x, y) → (x, y + c). 3. If y = f (x – b) + c the basic graph is translated both horizontally and vertically. y

y

y

x x Original graph

x Vertical translation up y

Vertical translation down

y

x

x

Horizontal translation to the left

Horizontal translation to the right

Naturally, the graph can be subject to a combination of two or more transformations. eBoo k plus eBook Digital doc

Spreadsheet 132 Transformations

Combination of transformations When describing transformations that have been applied to a basic graph f (x), it is best to put the graph into the format y = af (x − b) + c. The order of transformations is important as dilations and reflections are applied before translations, so ensure that you describe the transformations in this order (remember D-R-T). In this chapter we shall consider graphs, derived from basic curves, using single transformations — dilations, reflections or translations as well as combinations of those. Modelling of data will also be considered.

The quadratic function in power form The graph of y = x2 is a parabola with the turning point at the origin. The domain of the function is R and the range is R+ ∪ {0}. Throughout this section we refer to the graph of y = x2 as the basic parabola. Let us now consider the effect of various transformations on the graph of this basic parabola.

y

0

Chapter 2

x

Functions and transformations

59

Quadratic functions are also power functions. Power functions are functions of the form f (x) = xn, n ∈ R. The value of the power, n, determines the type of function. When n = 1, f (x) = x, and the function is linear. When n = 2, f (x) = x2 and the a=2 function is quadratic. Other power functions will be discussed later. y a=1 1 a = –2 Under a sequence of transformations of f (x) = xn, n ∈ R, the general form of a power function, is f (x) = a(x − b)n + c (where a, b, c and n ∈ R). All linear and quadratic polynomials are also linear and x quadratic power functions, because all linear and quadratic functions 0 are transformations of f (x) = x and f (x) = x2, respectively. When a quadratic function is written in turning point form it is written in power form. For example, the quadratic function y = ax2 y = x2 + 4x + 6 can also be represented as the power function y = (x + 2)2 + 2.

Dilation In power form, a is the dilation factor. It dilates the graph in the y direction. The larger | a | is, the thinner the graph of the parabola. If | a | is a proper fraction, that is, 0 < | a | < 1, the graph is wider than the basic parabola.

Reflection If a is negative, the graph of the basic parabola is reflected in the x-axis, that is, the graph is ‘flipped’ upside down. If x is replaced with −x, the graph of the basic parabola is reflected in the y-axis, that is, the graph is ‘flipped’ sideways. Due to its symmetry, this effect cannot be seen on the basic parabola, but it is more obvious with a parabola that has already been translated. For example, the graphs of y = (x − 3)2 and y = (−x − 3)2 are reflections of each other across the y-axis.

y = (−x − 3)2 y

y = (x − 3)2 (0, 9)

x

(−3, 0)0 (3, 0)

Translation Horizontal translation If b > 0, the graph of the basic parabola is translated horizontally to the right, and if b < 0, the graph of the basic parabola is translated horizontally to the left. For example, a graph with the equation y = (x − 2)2 is a basic parabola that has been translated 2 units to the right, and a graph with the equation y = (x + 3)2 is a basic parabola that has been translated 3 units to the left. If the coefficient of x is not 1, the equation must be rewritten in the form y = a(x − b)2 + c in order to be able to work out the value of b. For example, y = (4x + 3)2 is translated 43 of a unit to the left, since

y b=

−3

0 −3 2 y = (x − b)

b=2

2

x

y = (4 x + 3)2 3 = [4( x + 4 )]2 3

= 16( x + 4 )2 Vertical translation If c > 0, the graph is translated vertically upward, and if c < 0, the graph is translated vertically downward. For example, the graph with equation y = x2 + 2 is a basic parabola that has been translated 2 units up, and the graph with equation y = x2 − 1 is a basic parabola that has been translated 1 unit down.

60

Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad

y

c=2 c = −1

2 x 0 −1 y = x2 + c

eBoo k plus eBook Digital doc

Spreadsheet 108 The quadratic function in power form

Combination of transformations

y y = a(x − b)2 + c

The graph of y = a(x − + c shows the combination of the transformations shown above. The turning point of the graph is (b, c). The domain of the parabola is R and the range is [c, ∞) if a > 0 or [−∞, c) if a < 0. b)2

(b, c) 0

x

WORKED EXAMPLE 1

State the changes required to transform the graph of y = x2 into the graph of y = 2(x 2( + 3)2 − 4. THINK

WRITE

1

Write the general formula for the parabola.

y = a(x − b)2 + c

2

Identify the value of a.

a=2

3

State the effect of a on the graph.

The graph of y = x2 is dilated by the factor of 2 from the x-axis.

4

Identify the value of b.

b = −3

5

State the effect of b on the graph.

The graph is translated 3 units to the left.

6

Identify the value of c.

c = −4

7

State the effect of c on the graph.

The graph is translated 4 units down.

We can use transformations to find the equation of the function from its graph by first examining the new position of the turning point. WORKED EXAMPLE 2

Use transformations to find the equation of this function.

y

(4, 2)

x

0 THINK

WRITE

1

Write the general formula of the parabola.

y = a(x − b)2 + c

2

From the graph state the horizontal translation and hence the value of b.

Translated 4 units to the right, so b = 4.

3

From the graph, state the vertical translation and hence the value of c.

Translated 2 units up, so c = 2.

4

Substitute the values of b and c into the general formula. y = a(x − 4)2 + 2

5

The graph of the parabola passes through the origin. Substitute x = 0 and y = 0 into the formula.

Using (0, 0): 0 = a(0 − 4)2 + 2

6

Solve for a, which is the dilation factor.

0 = 16a + 2 16a = −2 −2 a = 16 =

−1 8

Chapter 2

Functions and transformations

61

7

Substitute the value of a into y = a(x − 4)2 + 2 and write your answer.

The equation of the parabola shown is: y = − 81 ( x − 4)2 + 2

WORKED EXAMPLE 3

Given the equation y = kx2, determine the effect on the graph y = x2, when k = {2, 3, 4}. Sketch the graphs. THINK

WRITE/DISPLAY

1

On the Graph & Tab page complete the function entry line as: y1 = x2 Then tick the y1 box and tap !.

2

Complete the function entry line as: y2 = 2x2 y3 = 3x2 y4 = 4x2 Tick the boxes of each of the equations and tap !.

3

Answer the question by describing the changes in words.

As the value of k increases the graph becomes thinner and stretches away from the x-axis.

REMEMBER

1. The graph of y = x2 is called a basic parabola. 2. The graph of y = a(x − b)2 + c is the basic parabola, dilated by the factor of a from the x-axis, translated b units horizontally (to the right if b > 0 or to the left if b < 0) and c units vertically (up if c > 0 or down if c < 0). 3. If a is negative, the graph is reflected in the x-axis. 4. If x is replaced with −x, the graph is reflected in the y-axis. 5. The turning point of the parabola is (b, c). 6. The domain of the parabola is R. 7. The range is y ≥ c if a > 0 or y ≤ c if a < 0.

62

Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad

y y = a(x − b)2 + c

(b, c) 0

x

EXERCISE

2A

Transformations and the parabola 1 WE 1 State the changes required to transform the graph of y = x2 into the graph of each of the following. a y = 2x b y = 13 x 2 2 c e g i k

y = −3x2 1 y = 1 − 2 x2 y = −(x + 3)2 y = (x + 2)2 − 1 y = 1 − 2(3 + x)2

y = x2 − 6

d f h j l

y = (x − 2)2 y = 2(3 − x)2 y = (x − 0.5)2 + 2 1

y = 3(2 x − 3)2 − 4

−1 2 MC The equation of a parabola is given by (2 − x )2 + 3 . Increasing m will result in the m graph being: A translated further to the left B translated further up C thinner D wider E reflected in the y-axis

3 Match the graphs of the parabolas with the following equations. a y = x2 + 2 b y = −2(xx − 2)2 c

y = 2 − (x + 2)2

d

e y = (2 + x)2 + 2

iii

ii y = x2

y

1

y = 2 (2 − x ) 2

iv

2

i

x

0 2 −2 −2 v

4 WE2 Use transformations to find the equation of each function. a

y

b

(2, 2)

y 0

x

0

x

(−1, −2) 2)

c

y

(1, 3)

0

2 x

d

y −4

0

x

(−2, −4)

5 MC The equation of the graph shown opposite is best given by: A y = (x − c)2 + d B y = c − (x − b)2 − 2 C y = (x + c) + b D y = (c − x)2 + d 2 E y = d − (x − c)

y d b a0

c

e

x

6 Find the equation of the image of y = x2 under each of the following transformations: a dilation by the factor of 12 from the x-axis b a reflection in the x-axis c a translation by 2 units to the right and 1 unit down

Chapter 2

Functions and transformations

63

d a dilation by the factor of 3 from the x-axis, followed by translation of 2 units down e a reflection in the x-axis, followed by translation of 3 units to the left. 7 Find the equations of these graphs. a

b

y

y 1

0

3

5

x

x

0 −1

−1

−4

c

d

y −3

y 2

x −1 0

x

0 2

−4

e

f

y

y 8

9 6 0

1

x

− 2 −2

−2

0 x

8 WE3 Find the equation of y = x2 under the following sequential transformations (in order): a dilation by a factor of 2 from the x-axis b reflection in the x-axis c translation of −1 parallel to the x-axis d translation of 3 parallel to the y-axis. 9 Find the image of the point ((x, y) under each of the following transformations: a reflection in the y-axis b reflection in the x-axis c dilation by a factor of 3 from the x-axis d dilation by a factor of 2 from the y-axis e dilation by a factor of 13 from the y-axis f translation of 2 units horizontally in the positive direction g translation of −1 unit parallel to the y-axis. 10

The parabola has a turning point at (z, ( −8); it intersects the y-axis at y = 10 and one of the x-intercepts is x = 5. Find: a the value of z b the equation of the parabola.

11 For the parabola whose range is y ≤ 3, whose x-coordinate of the turning point is −4 and whose 1 y-intercept is y = − 2 3 , find: a the y-coordinate of the turning point b the equation of the parabola c the coordinates of the x-intercepts. 64

Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad

12 The design shown in the diagram at right can be obtained by taking the red portion of the parabola and transforming it to form each of the other 9 fragments. (One or more transformations may be used to form each fragment.) If the highlighted fragment is given by f (x (x), −2 ≤ x ≤ 2, define the other 9 fragments in terms of f (x (x) and specify their domains.

2B

y

3 −2 0

2

x

The cubic function in power form The graph of the function y = x3 is shown at right: Both the domain and y range of the function are R. The function is constantly increasing and has a stationary point of inflection (where the gradient is 0) at the origin (0, 0). x 0 Throughout this section we shall refer to the shape of the graph of 3 y = x as a positive cubic, or a basic cubic curve. Cubic functions are also power functions. Power functions are functions of the form f (x) = xn, n ∈ R. The value of the power, n, determines the type of function. When n = 1, f (x) = x and the function is linear. When n = 2, f (x) = x2 and the function is quadratic. When n = 3, f (x) = x3 and the function is cubic. When n = 4, f (x) = x4 and the function is quartic. Other power functions will be discussed later. Under a sequence of transformations of f (x) = xn, n ∈ R, the general form of a power function is f (x) = a(x − b)n + c (where a, b, c and n ∈ R). All linear and quadratic polynomials are also linear and quadratic power functions, but this is not the case for cubic functions (or quartic functions). For example, a cubic power function in the form f (x) = a(x − b)3 + c has exactly one x-intercept and one stationary point of inflection. A cubic polynomial in the form f (x) = ax3 + bx2 + cx + d can have one, two or three x-intercepts and is therefore not a power function. All cubic power functions are also cubic polynomials, but not all cubic polynomials are cubic power functions. For example, the cubic function y = 2(x − 3)3 + 1 is a polynomial and a power function. It is the graph of y = x3 under a sequence of transformations.

Dilation

a=2 a=1 a = 1–2

y

The value a is the dilation factor; it dilates the graph from the x-axis. The larger a is, the thinner the graph. 0

x

y = ax3 eBoo k plus eBook Digital doc

Spreadsheet 015 Cubic function — y = a( x − b ) 3 + c

y = (−x − 1)3 y

Reflection If a is negative, the graph of the basic cubic is reflected in the x-axis, that is, the graph is ‘flipped’ upside down. If x is replaced with −x, the graph of the basic cubic is reflected in the y-axis, that is, the graph is ‘flipped’ sideways. For example, the graphs y = (x − 1)3 and y = (−x − 1)3 are reflections of each other across the y-axis.

Chapter 2

(−1, 0) 0

y = (x − 1)3

(1, 0)

x

(0, −1)

Functions and transformations

65

Translation Horizontal translation If b > 0, the graph of the basic cubic is translated horizontally to the right, and if b < 0, the graph of the basic cubic is translated horizontally to the left. For example, the graph with equation y = (x − 2)3 is a basic cubic translated 2 units to the right, and the graph of y = (x + 3)3 is a basic cubic, translated 3 units to the left, that is, parallel to the x-axis in the negative direction. If the coefficient of x is not 1, the equation must be rewritten in the form y = a(x − b)3 + c in order to be able to work out the value of b. For example, the graph of y = (2x (2 − 5)3 is translated 25 units to the 3 right, since y = (2x (2 − 5)

y

b = −3

b=2

0

−3

2

x

y = (x − b)3

= [2( x − 25 )]3 = 8( x − 25 )3 Vertical translation The value of c translates the graph vertically or along the y-axis. If c > 0, the graph is translated vertically up, and if c < 0, the graph is translated vertically down. The coordinates of the stationary point of inflection are (b, c). For example, if y = x3 is translated 1 unit up, the equation of the resulting graph is y = x3 + 1 and the point of inflection is (0, 1); if it is translated 2 units down, the equation of the resulting graph is y = x3 − 2 and the point of inflection is (0, −2).

Combination of transformations The graph of y = a(x − b)3 + c shows the combination of the transformations described above. Finally, the domain and range of y = a(x − b)3 + c are R (all real numbers).

y 1

c=1 c = −2 x

0 −22

y = x3 + c

y

y = a(x − b)3 + c

(b, c) 0

WORKED EXAMPLE 4

State the changes necessary to transform the graph of y = x3 into the graph of y = 2(x 2( + 1)3 − 4. THINK

66

WRITE

1

Write the general equation of the cubic function.

y = a(x − b)3 + c

2

Identify the value of a.

a=2

3

State the effect of a on the graph.

The graph is dilated by the factor of 2 in the y direction.

4

Identify the value of b.

b = −1

5

State the effect of b on the graph.

The graph is translated 1 unit to the left.

6

Identify the value of c.

c = −4

7

State the effect of c on the graph.

The graph is translated 4 units down.

Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad

x

WORKED EXAMPLE 5

For each of the following graphs: i state the coordinates of the stationary point of inflection ii find the x- and y-intercepts iii sketch the graph iv state the transformations that the graph of y = x3 has undergone to form each new equation. a y = −(x ( + 3)3 − 1 b y = (4 − x)3 + 6 THINK a Write the equation.

WRITE/DRAW a y = −(x + 3)3 − 1 i b = −3, c = −1

i Since the rule is of the form y = a(x − b)3 + c,

Stationary point of inflection: (−3, −1)

identify the values of b and c and hence write the coordinates of the stationary point of inflection (b, c).

ii Find the y-intercept by letting x = 0.

i i y-intercept: x = 0,

y = −(0 + 3)3 − 1 = −27 − 1 = −28 x-intercept: y = 0 (x + 3)3 − 1 = 0 (x + 3)3 = −1 x + 3 = −1 x = −4

Find the x-intercept by letting y = 0.

iii To sketch the graph on a set of labelled axes,

iii

mark the stationary point of inflection and the x- and y-intercepts, then sketch the positive cubic passing through the points marked.

(−4, 0) 0 (−3, −1)

iv State the kind of reflection and the vertical and

Write the equation. i Since the rule is of the form y = a(x − b)3 + c,

identify the values of b and c and hence write the coordinates of the stationary point of inflection (b, c).

ii Find the y-intercept by letting x = 0.

Find the x-intercept by letting y = 0. Note: Do not round off until the very last step; for graphing purposes, round off your final answer to 1 decimal place.

x

i v The graph is reflected in the x-axis.

horizontal translations.

b

y

There is a horizontal translation of 3 units to the left and a vertical translation of 1 unit down. b

y = (4 − x)3 + 6 i b = 4, c = 6

Stationary point of inflection: (4, 6)

i i y-intercept: x = 0,

y = (4 − 0)3 + 6 = 64 + 6 = 70 x-intercept: y = 0 (4 − x)3 + 6 = 0 (4 − x)3 = −6 4−x=



3

6 3

x=4+ 6 ≈ 5.8

Chapter 2

Functions and transformations

67

iii To sketch the graph on a set of labelled

iii

y

axes, mark the stationary point of inflection and the x- and y-intercepts, then sketch the positive cubic passing through the points marked.

(4, 6)

0 iv State the kind of reflection and the vertical

(5.8, 0)

x

i v The graph is reflected in the y-axis.

and horizontal translations.

There is a horizontal translation of 4 units to the right and a vertical translation of 6 units up.

To find the equation of the curve from a given graph, we need to establish exactly what transformations were applied to the basic cubic curve. This is best done by observing the shape of the graph and the position of the stationary point of inflection. WORKED EXAMPLE 6

Find the equation of the curve, if it is of the form y = a(x ( − b)3 + c. (x

y 5 3

THINK

0

1

1

Write the general equation of the cubic function.

y = a(x − b)3 + c

2

Write the coordinates of the stationary point of inflection (b, c) and hence state the values of b and c. Substitute the values of b and c into the general formula.

The stationary point of inflection is (1, 3). So b = 1, c = 3.

4

The graph passes through the point (0, 5) (y-intercept). Substitute the coordinates of this point into the equation.

Using (0, 5): 5 = a(0 – 1)3 + 3

5

On the Main page, complete entry line by either typing the line as shown, or type the equation: 5 = a(0 − 1)3 + 3. Highlight it and tap: • Interactive • Advanced • solve Change the variable to a and tap: • OK

6

Write the solution for the equation.

a = −2

7

Substitute the value of a into y = a(x − 1)3 + 3.

y = −2(x − 1)3 + 3

3

68

WRITE/DISPLAY

y = a(x − 1)3 + 3

Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad

x

REMEMBER

1. The graph of y = x3 is a basic cubic curve. 2. The graph of y = a(x − b)3 + c where a > 0 is the basic cubic, dilated by the factor of a from the x-axis, translated b units along the x-axis (to the right if b > 0, or to the left if b < 0) and c units along the y-axis (up if c > 0, or down if c < 0). y = a(x − b)3 + c

y

(b, c)

x

0

3. If a < 0, the graph is reflected in the x-axis. 4. The stationary point of inflection is at (b, c). 5. Both the domain and range are R. EXERCISE

2B

The cubic function in power form 1 WE4 State the changes necessary to transform the graph of y = x3 into the graph of each of the following. b y=

a y = 7x3 c e g i

y = x3 + 4 y = (x − 1)3 y = 4(2 − x)3 y = 3(x + 3)3 − 2

k y = 1 (2 x + 5)3 4 2

−2 3

x3

d f h j

y = 6 − x3 y = −(x + 3)3 y = −6(7 − x)3 1 y = 6 − 2 ( x − 1)3

l

y = 3 − 2(4 + 2 x )3

1

Which of these transformations were applied to the graph of y = x3 to obtain each of the graphs below? i reflection in the x-axis i i translation to the left i i i translation to the right i v translation up v translation down a

b

y

c

y x

0 x

0

d

y

0

e

y

0

x

f

y

0

y

0

x

Chapter 2

x

x

Functions and transformations

69

3 WE5 For each of the following graphs: i find the stationary point of inflection ii find the x- and y-intercepts iii state the transformation(s) that the graph y = x3 has undergone to produce the given graph iv sketch the graph.

EXAM TIP Be careful when sketching graphs — use appropriate scales on the axes, clearly draw the graph and use the correct domain. Show at least one scale point on each axis and two coordinate points on each curve (including intercepts). [Authors’ advice]

eBoo k plus eBook

a y = 3 x3 4

b y = 1 − 2x 2 3

c y = 2 x3 − 6 3

d y = 2(x − 4)3

−1

f y = 4(1 − x)3

e y=

2

( x − 2)3

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

Digital doc

Spreadsheet 236 Function grapher

h y = 3 − (x + 2)3

Questions 4 to 6 refer to the function y = 2(mx − 4)3 − 3. 4 MC The coordinates of the stationary point of inflection are: − A ( 4 , 3) m

4 −3 C ( , ) m m

B (4m, −3)

4 E ( , − 3) m

4 D ( , − 3m3 ) m

5 MC The graph of y = x3 is dilated in the y direction by the factor of: A 2

C 2m3

B 2m

2 m

D

2 m3

E

6 MC If m > 1, increasing m will cause the graph to become: A wider B thinner and translated not as far to the right C shifted further to the left D shifted further to the right E shifted further down 7 Find the equation of the graph resulting from each of the following transformations of the graph of y = x3: a a dilation by the factor of 12 from the x-axis b a reflection in the x-axis and a translation by 5 units to the left c a translation by 3 units to the right and 1 unit down d a dilation in the y direction by the factor of 2, followed by the vertical translation of 3 units e a reflection in the x-axis, then a translation of 1 unit to the left and 1 unit down. 8 Find the equation of the graph resulting from the following sequential transformations of the graph of y = x3: a dilation by a factor of 2 from the x-axis b reflection in the y-axis c translation of 2 in the positive direction parallel to the x-axis d translation of 1 in the negative direction parallel to the y-axis. 9 WE6 Find the equations of these curves, if they are of the form y = a(x ( − b)3 + c. (x a

y

b

y

4

y

c

1

(1, 2) −1

0

70

2

x

0

x

d 0

−2

Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad

y

e

9

x 0

3

x

y − 1–2 −1 1 0 − –2

x

10 MC The graph of y = 2(x 2( + 3)3 + 1 has been reflected in the x-axis, shifted 3 units to the right and 1 unit up. The equation of the resulting graph is: A y = 2(3 − x)3 + 2 B y = −2(x + 2)3 + 2 C y = 2(2 − x)3 D y = −2(x − 3)3 + 1 E y = −2x 2 3 11 The graph of a cubic function of the form y = a(x ( − b)3 + c has a stationary point of inflection (x at (−1, −4) and cuts the y-axis at y = −2. Find the equation of the function. 12

The graph of y = a(b − x)3 + c has a stationary point of inflection at (2, 1) and passes through the point (1, 12 ). a Find the equation of the curve. b State the shape of the curve (that is, whether it is positive or negative cubic).

13 The graph of y = a(x ( − b)3 + c cuts the x-axis at x = −4 and the y-axis at y = 28. If it is known (x that the dilation factor is equal to 1: a find the position of the stationary point of inflection b sketch the graph.

2C

The power function (the hyperbola) The graph shown at right is called a hyperbola and is given by the y equation y = 1 . x Power functions are functions of the form f (x) = xn, n ∈ R. The y=0 value of the power, n, determines the type of function. We saw earlier x 0 that when n = 1, f (x) = x and the function is linear. When n = 2, f (x) = x2 and the function is quadratic. When n = 3, f (x) = x3 and the function is cubic. When n = 4, f (x) = x4 and the function is quartic. x=0 The power function that produces the graph of a hyperbola has a 1 value of n = −1. Thus, the function f ( x ) = can also be expressed as x the power function f (x) = x−1. The graph exhibits asymptotic behaviour. That is, as x becomes very large, the graph approaches the x-axis, but never touches it, and as x becomes very small (approaches 0), the graph approaches the y-axis, but never touches it. So the line x = 0 (the y-axis) is a vertical asymptote and the line y = 0 (the x-axis) is the horizontal asymptote. Both the domain and the range of the function are all real numbers, except 0; that is, R\{0}. The graph of y = 1 can be subject to a number of transformations. x a − Consider y = + c or y = a(x − b) 1 + c x−b

Dilation

y

The value a is a dilation factor. It dilates the graph from the x-axis.

y=0 0

Reflection If a is negative, the graph of the basic hyperbola is reflected in the x-axis. If x is replaced with −x, the graph of the basic hyperbola is reflected in the y-axis.

Chapter 2

x=0

a=2 a=1 a = 1–2 x y = a–x

Functions and transformations

71

y

1 1 and y = − x−3 x−3 are reflections of each other across the y-axis. For example, the graphs of y =

x = −3 y=

Translation

x=3 y = x −1 3

1

−x − 3

0

Horizontal translation x y=0 The value b translates the graph b units horizontally, that is, parallel to the x-axis. If b > 0, the graph is translated to the right, and if b < 0, the graph is (0, − 13 ) translated to the left. For example, the graph with 1 equation y = is a basic hyperbola translated x−3 3 units to the right. This graph has a vertical asymptote of x = 3 and domain R\{3} (and a horizontal asymptote y = 0). If a basic hyperbola is translated 3 units to the left, it becomes 1 y= , with a vertical asymptote of x = −3 and domain R\{−3}. Hence, the equation of the x+3 vertical asymptote is x = b and the domain is R\{b}. The horizontal asymptote and the range remain the same, x = 0 and R \ {0}, respectively. Vertical translation The value c translates the graph c units vertically, that is, parallel to the y-axis. If c > 0, the graph is translated upward, and if c < 0, the graph is translated c units downward. The graph 1 with equation y = + 3 is a basic hyperbola translated 3 units up. This graph has a horizontal x asymptote of y = 3 and a range of R\{3} (and a vertical asymptote x = 0). If a basic hyperbola is 1 translated 3 units down, it becomes y = − 3, with a horizontal asymptote of y = −3 and a range x of R\{−3} (and a vertical asymptote x = 0). Hence the equation of the horizontal asymptote is y = c and the range is R\{c}. Always draw the asymptote as a dotted line and label it with its equation (for example, y = 3) at the end of the asymptote. Ensure that the graph continues to approach the asymptote getting closer but not touching or crossing the asymptote or bouncing away from the asymptote.

Combination of transformations The graph of y =

a + c shows the combination of these transformations. x−b y

eBoo k plus eBook Digital doc

Spreadsheet 051 The hyperbola

y=c

a

y = x— c −b+

c 0

b

x

x=b

Finally, if the coefficient of x is a number other than 1, to obtain the value of h the equation should be rearranged first. For example, 4 4 y= = 3 x + 6 3( x + 2) Therefore, b = −2 (not −6 as it may seem at first); that is, the graph is translated 2 units to the left.

72

Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad

WORKED EXAMPLE 7

1 State the changes that should be made to the graph of y = in order to obtain the graph of x − y = 4 − 1. x+2 THINK

WRITE

a +c x−b

1

Write the general equation of the hyperbola.

y=

2

Identify the value of a.

a = −4

3

1 State the changes to y = , caused by a. x

1 The graph of y = x is dilated by the factor of 4 from the x-axis and reflected in the x-axis.

4

Identify the value of b.

b = −2

5

State the effect of b on the graph.

The graph is translated 2 units to the left.

6

Identify the value of c.

c = −1

7

State the changes to the graph, caused by c.

The graph is translated 1 unit down.

WORKED EXAMPLE 8

For the graph of y =

2 + 2, state: x−3

a the equations of the asymptotes c the range.

b the domain

THINK a

WRITE

1

Write the general equation of the hyperbola.

2

Identify the values of b and c and hence write the equations of the asymptotes: Horizontal asymptote: y = c Vertical asymptote: x = b

a y=

a +c x−b

b = 3, c = 2 Horizontal asymptote: y = 2 Vertical asymptote: x = 3

b

State the domain of the hyperbola: R\{b}.

b Domain: R\{3}

c

State the range of the hyperbola: R\{c}.

c Range: R\{2}

Sketching the graph of the hyperbola by hand can be easily done by following these steps: 1. Find the position of the asymptotes. 2. Find the values of the intercepts with the axes. 3. Decide whether the hyperbola is positive or negative. 4. On the set of axes draw the asymptotes (using dotted lines) and mark the intercepts with the axes. 5. Treating the asymptotes as the new set of axes, sketch either the positive or negative hyperbola, making sure it passes through the intercepts that have been previously marked.

Chapter 2

Functions and transformations

73

WORKED EXAMPLE 9

eBoo k plus eBook

2 −4 , clearly showing the intercepts x+2 with the axes and the position of the asymptotes. Sketch the graph of y = THINK 1

a +c Compare the given equation with y = x − b and state the values of a, b and c. Write a short statement about the effects 1 of a, b and c on the graph of y = . x

3

Write the equations of the asymptotes. The horizontal asymptote is at y = c. The vertical asymptote is at x = b. Find the value of the y-intercept by letting x = 0.

5

int-0522 Worked example 9

WRITE/DRAW

2

4

Tutorial

Find the value of the x-intercept by making y = 0.

b = −2,

a = 2,

c = −4

1 is dilated by the factor of x 2 from the x-axis, translated 2 units to the left and 4 units down. Asymptotes: x = −2; y = −4 The graph of y =

y-intercept: x = 0 2 y= −4 0+2 = 1− 4 = −3 Point (0, −3) x-intercept: y = 0 2 0= −4 x+2 2 =4 x+2 2 = 4( x + 2) = 4x + 8 4x = 2 − 8 = −6 x= = Point

6

(

−6 4 −3 2

−3 2

,0

)

To sketch the graph:

y

a Draw the set of axes and label them. b Use dotted lines to draw the asymptotes. The

(− 3–2 , 0) 0

asymptotes are x = −2 and y = −4.

(0, −3)

c Mark the intercepts with the axes. The

intercepts are y = −3 and x =

−3 . 2

d Treating the asymptotes as your new set of

x y = −4

x = −2

axes, sketch the graph of the hyperbola (as a is positive, the graph is not reflected); make sure the upper branch passes through the x- and y-intercepts previously marked. The next example shows how to find the equation of the hyperbola from its graph.

74

Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad

WORKED EXAMPLE 10

Find the equation of the graph shown.

y 6 3 0 2

THINK

4

x

WRITE

a +c x−b

1

Write the general equation of the hyperbola.

y=

2

From the graph, identify the values of b and c Remember that the equation of the horizontal asymptote is y = c and of the vertical asymptote is x = b.

b = 2,

3

Substitute the values of b and c into the formula.

y=

4

Substitute the coordinates of any of the 2 known points of intersection with the axes into the formula (say, x-intercept).

5

Solve for a.

Substitute (4, 0): a 0= +3 4−2 a 0= +3 2 a − = 3 2 a = −6

6

Substitute the value of a into y =

7

Transpose (optional).

a + 3. x−2

y=

c=3

a +3 x−2



6 +3 x−2

y = 3−

6 x−2

REMEMBER

1 is called a hyperbola. x a 2. The graph of y = + c is the graph of the basic x−b hyperbola, dilated by the factor of a from the x-axis, translated b units horizontally (to the right if b > 0, or to the left if b < 0) and c units vertically (up if c > 0, or down if c < 0). If a < 0, the graph is reflected in the x-axis. The equations of the asymptotes are: x = b and y = c. The domain of the function is R\{b} and its range is R\{c}. 1. The graph of y =

Chapter 2

y

y= a

x−b

+ c y=c x

0 x=b

Functions and transformations

75

EXERCISE

2C

The power function (the hyperbola) 1

2

1 State the changes that should be made to the graph of y = in order to obtain the x graph of each of the following. −3 1 2 a y= b y= c y= x x−6 x 2 1 2 d y= e y= +7 f y= −5 x+4 x x −4 1 2 g y= −3 +6 h y= i y= −4 4+x x−3 x −1 WE7

1 Which of the following transformations were applied to the graph of y = to obtain each x of the graphs shown below? i translation to the right i i translation to the left iii translation up i v translation down v reflection in the x-axis a

0

d

g

0

0

h

x

0

x

f

y

x

y

y

x

0

y

x

0

x

WE8 For each of the following, state:

i the equations of the asymptotes

a y=

76

0

e

y

c

y

x

y

0

3

b

y

−2

x

i i the domain

i i i the range.

b y=

1 x+6

c y=

d y=

2 3− x

e y=

3 +4 x

f

g y=

4 −2 x+6

h y=

5 +1 2− x

i

Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad



3 x−2 −

1 −5 x 1 y= −m n+ x y=

x

4 For each of the following graphs, state: i the equations of the asymptotes i i the domain i i i the range. a

b

y

c

y

y 2

0

d

4

e

y −1 0

2 0

x

−1

x

f

y

y

n 0

a m

x

0

b

eBoo k plus eBook

−4 1 2 3 5 On the same set of axes sketch the graphs of y = , y = , y = and y = . x x 3x 3x

Digital doc

6

Spreadsheet 236 Function grapher

7

x

WE9 Sketch each of the following, clearly showing the position of the asymptotes and the intercepts with the axes. Check your answers, using a CAS calculator. a y= 1 b y = 1 −1 c y= 3 −3 x+3 x+2 x −1 4 −2 − 6 d y= e y= f y = 3 +6 −3 1− x x+5 x−2 1 2 4 g y = 1− h y= + i y= 1 +4 2− x 5 1+ x 2x + 3 2 x + 3 j y= k y= l y = 4x + 3 −1 x−2 x −1 3 − 4x MC The equation of the graph shown is likely to be:

1 x−4 1 C y = 3− 4−x A y = 3+

E 8

x

3

0

x

y = 3−

1 +4 x−3 1 D y= −3 4−x

y

B y=

3 0

1 x−4

MC Which of the following is a true statement for the graph of y =

A B C D E

The domain is R\{1}. The range is R\{3}. The equation of the horizontal asymptote is y = −3. The equation of the vertical asymptote is x = 2. None of the above.

Chapter 2

4

x

2 − 3? x +1

Functions and transformations

77

9 WE10 Find the equation for each of the following hyperbolas, if they are of the form a y= + c. x−b a

b

y

−1

d

0

1

eBoo k plus eBook Digital docs

WorkSHEET 2.1 History of mathematics The history of some major curves

2D

1

x

0

e

y

−4

10

2

c

y

0

−4

x

f

y 2 11–2 0

x

3

y

x

− 3–4

y 5

3 4

x

−11 −1

5

x

1 If a function is given by f ( x ) = , sketch each of the following, labelling the asymptotes x and the intercepts with the axes. a f (x + 2) b f (x) − 1 c −f (x) − 2 − d f (1 − x) + 2 e f (x − 1) − 1 f 1 − f (x − 2)

11 Sketch the graph of yx − 3x + 1 = 0, and state its domain and range. (Hint: First transpose the equation to make y the subject.)

EXAM TIP When you are asked to give the domain and range in a question, be sure to demonstrate clearly which answer is which. For example, for y = x2, dom f : R, ran f : [0, ∞]. [Authors’ advice]

The power function (the truncus) The graph shown at right is known as a truncus. The equation of the graph is given by:

y

y = 12 x Power functions are functions of the form f (x) = xn, n ∈ R. y=0 x 0 The value of the power, n, determines the type of function. We saw earlier that when n = 1, f (x) = x and the function is linear. x=0 When n = 2, f (x) = x2 and the function is quadratic. When n = 3, f (x) = x3 and the function is cubic. When n = 4, f (x) = x4 and the function is quartic. When n = −1, f (x) = x−1 and the power function produces the graph of a hyperbola. The power function that produces the graph of a truncus has a value of n = −2. Thus, the function f ( x ) = 12 can also be expressed as the power function f (x) = x−2. x The function is undefined for x = 0. Hence, the equation of the vertical asymptote is x = 0 and the domain of the function is R\{0}.

78

Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad

We can also observe that the graph approaches the x-axis very closely, but never touches it. So y = 0 is the horizontal asymptote. Since the whole graph of the truncus is above the x-axis, its range is R+ (that is, all positive real numbers). 1 Similar to the graphs of the functions, discussed in the previous sections, the graph of y = 2 x can undergo various transformations. a + c , or y = a (x − b)−2 + c. Consider the general formula y = ( x − b) 2

Dilation

y

The value a is the dilation factor. It dilates the graph from the x-axis. The dilation factor does not affect the domain, range or asymptotes.

y=0

a=3 a=2 x

0

x=0 y

Reflection If a is negative, the graph of a basic truncus is reflected in the x-axis. The range becomes R− (that is, all negative real numbers).

a y = –2 x

y = 12 x

y=0

x

0

− y = 12

If x is replaced with −x, the graph of the basic truncus is reflected in the y-axis. The effect of this reflection cannot be seen in the basic graph, but it becomes more obvious if the graph has been translated horizontally first. For example, 1 1 the graphs of y = and y = (− x − 3)2 ( x − 3)2 are reflections across the y-axis. The vertical asymptote changes from x = 3 to x = −3 and the domain changes from R \{3} to R \{−3}.

x

x=0 y 1 ( x − 3)2

1 y= − ( x − 3)2

y=

y=0

x

0

x = −3

x=3

Translation Horizontal translation The value b translates the graph b units horizontally. If b > 0, the graph is translated to the right, and if b < 0, the graph is translated left. For example, the graph of the equation 1 results from translating a basic truncus 3 units y= ( x − 3)2 to the right. The vertical asymptote is x = 3

y b = −2 y=0

−2

b=3 0

3

x

1 y = ——— 2 (x − b)

Chapter 2

Functions and transformations

79

and the domain is R \{3}. If a basic truncus is translated 2 units to the left, it becomes 1 y= , where the vertical asymptote is x = −2 and the domain is R \{−2}. Hence, the ( x + 2)2 equation of the vertical asymptote is x = b and the domain is R \{b}. The range is still R+ and the equation of the horizontal asymptote is y = 0. Vertical translation y 1 The value c translates the graph c units vertically. If c > 0 the graph y = –– +c x2 is translated upward, and if c < 0, the graph is translated c units 1 c=1 downward. For example, the graph with equation y = 2 + 1 results x y=1 1 when a basic truncus is translated 1 unit upward. The horizontal 0 c = −1 x asymptote is y = 1 and the range is (1, ∞). If a basic truncus is 1 −1 y = −1 translated 1 unit down, it becomes y = 2 − 1, with y = −1 as x=0 x the horizontal asymptote and (−1, ∞) as the range. Hence the equation of the horizontal asymptote is y = c and the range is (c, ∞). Note: If a is positive (see graph below), the whole graph of the truncus is above the line y = c (the horizontal asymptote) and hence its range is y > c, (c, ∞). If a is negative, the whole graph is below its horizontal asymptote and therefore the range is y < 0, or (−∞, c). y

y y=3 y=3

3 0

The graph of y =

3

x x

a + c shows the combination of these ( x − b) 2

y

y=

a + c (x − b)2

transformations. y=c

c 0

b

x

x=b WORKED EXAMPLE 11

State the transformations required to change the graph of y = THINK

80

−1 1 − 1. into the graph of y = 2 ( x − 2 )2 x

WRITE

a +c ( x − b) 2

1

Write the general formula for the truncus.

y=

2

Identify the value of a.

a = −1

Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad

1

3

State the effect of a on the graph.

The graph of y =

4

Identify the value of b.

b=2

5

State the effect of b on the graph.

The graph is translated 2 units to the right.

6

Identify the value of c.

c = −1

7

State the effect of c on the graph.

The graph is translated 1 unit down.

x2

is reflected in the x-axis.

WORKED EXAMPLE 12

2 − 4 , state: ( 3 + x) 2 a the equations of the asymptotes For the function y =

b the domain

THINK a

b c

c the range. WRITE

a y=

a +c ( x − b) 2

1

Write the general formula for the truncus.

2

Write the general equations of the asymptotes.

Vertical asymptote: x = b Horizontal asymptote: y = c

3

Identify the values of b and c.

b = −3, c = −4

4

State the equations of the asymptotes by substituting the values of b and c into corresponding formulas.

Asymptotes: x = −3 and y = −4

Write the domain of the truncus, which is R\{b}. 1

Check whether a is positive or negative.

2

Write the range (which for a > 0 is y > c).

b Domain: R\{−3} c a>0

Range: y > −4

a + c, then compare the given ( x − b) 2 equation with the general formula to see what changes should be made to the basic curve (the 1 graph of y = ) to transform it to the one you want. This should give you an idea of how the x2 graph will look. To sketch the graph of a truncus, first put it in the form y =

The following algorithm can then be used: 1. Find the position of the asymptotes. 2. Find the intercepts with the axes. 3. On the set of axes, draw the asymptotes (using dotted lines), label with the equation and mark the x- and y-intercepts. a>0 4. Treating the asymptotes as the new set of axes, sketch the basic truncus curve. 5. Make sure the curve passes through the points marked on the axes.

Chapter 2

a 0 or to the left if b < 0) and c units along the y-axis (up if c > 0 or down if c < 0). If a is negative, the graph is reflected in the x-axis. The vertical asymptote is x = b. The horizontal asymptote is y = c. The domain is R\{b}. The range is y > c if a > 0, or y < c if a < 0.

Chapter 2

y

y=

a (x − b)2 + c

c 0

b

x

Functions and transformations

83

EXERCISE

2D

The power function (the truncus) 1 1 WE11 State the transformations required to change the graph of y = 2 into the graph of x each of the following: − 1 2 a y= b y= 3 c y= 2 2 ( x + 2)2 x x d y=

2 ( x − 3)2

e y=

−5

(4 +

x )2

4 +1 ( x − 3)2 1 2 MC To obtain the graph shown, the graph of y = 2 was: x A reflected in the x-axis and translated 2 units down B translated 2 units to the left C reflected in the x-axis and translated 2 units to the left D reflected in the x-axis and translated 2 units to the right E reflected in the x-axis and translated 2 units up g y = 3− 1 x2

h y=

2 +6 x2

f

y=

i

y = 5−

1 ( x + 2)2 y

0

x

x=2

3 MC Which of the following translations took place, so that the 1 graph of y = 2 was changed into the one shown at right? x A m units to the left and p units up B m units to the right and p units up C m units to the left and n units up D m units to the right and n units up E m units to the left and n units down

y p y=n x x=m

4 WE12 For each of the following state: i the equations of the asymptotes ii the domain iii the range. − a y= 2 b y= 4 x2 3x 2 −5 2 d y= e y = ( x + 1)2 (4 + x ) 2 4 1 1 3 g y= + 2 h y= − 2 5 x 2 x

c y= f i

1 ( x − 2)2

2 −3 x2 2 y= +4 ( x − 1)2 y=

Questions 5 tο 7 refer to the following diagrams. i

ii

y

iii

y

y

−3 0

84

3

x

−3

Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad

0

x

0

−3

x

iv

v

y

Digital doc Function grapher

x

3

0 −3

−3

0

Spreadsheet 236

y

x

0

3

eBoo k plus eBook

vi

y

x

3

5 MC Which of the above functions has the domain R\{−3}? A i i only B i i i only D i i , i i i and v i E v and vi

C

i i and i i i

6 MC Which of the above functions has the range y < −3? A i , i i and i v B i i i , v and vi E i i i and vi D v only

C

i v only

7 MC Which of the graphs has asymptotes y = 0 and x = −3? A i B ii D iv E v

C

iii

8 WE13 Sketch each of the following, clearly showing the position of the asymptotes and the intercepts with the axes (correct to 1 decimal place where appropriate). − 1 2 a y= 2 b y= c y= 2 2 ( x − 3) (4 + x ) 2 5x d y=

−1

( x − 1)2

2 −2 ( x − 1)2 3 1 y= − 2 4 4( x + 1)

g y= j

e y = 1− 4 x2

f

2 (3 + x )2 1 +3 k y= (2 − x ) 2

i

h y = 4−

9 Find the equations for each of the following curves, if it is known that all of them are a of the form y = + c and a is ( x − b) 2 + − either 1 or 1. a

b

y

l

1 −3 2x2 2 1 y= + 3 ( x − 2)2 4 y= −1 (2 x − 4) 2 y=

EXAM TIP Always check you have met the requirements of the answer format, for example, ensure that you have rounded correctly and show the correct number of decimal places. [Assessment report 1 2006 © VCAA]

c

y

y

p −q

n m

0

d

x

0

r

0

x

x

e

y

f

y

−cc −d −a 0

t 0

−s

x

x

−b

Chapter 2

y

−f −f

g

h

x

−e

Functions and transformations

85

g

h

y

y x

0 −l −i

j

0

x

−k [© VCAA 2006]

10 WE14 Find the equation for each of the following. a b y y

−1

1

0

x

−2

y=

− 3–4

−2

0

x=0

d

c

1

2

x y=0

e

f

y

y=4

11

x

0

−2.5

0

y 0 −2

4 x=

−5

y=1 x

x = −2

7

−1

−2

x=2

y

−1 0

y

x

4

y = −3

−3 x=4

1

x

y = −2

−7 x=1

The domain of a truncus is R\{−2}; its range is y > −3 and its graph cuts the x-axis at 1 and x = −3. Find the equation of the function.

12 The domain of a truncus is R\{1}; its range is (2, ∞) and its graph cuts the y-axis at y = 5. Find the equation of the function.

2E

The square root function in power form The square root function is given by y = x (or y = x ). y Power functions are functions of the form f (x) = xn, n ∈ R. The value of the power, n, determines the type of function. We saw earlier that when n = 1, f (x) = x and the function is linear. When n = 2, f (x) = x2 and the function is quadratic. When n = 3, f (x) = x3 and the x 0 function is cubic. When n = 4, f (x) = x4 and the function is quartic. When n = −1, f (x) = x−1 and the power function produces the graph of a hyperbola. When n = −2, f (x) = x−2 and the power function produces the graph of a truncus. 1 The power function that produces the graph of the square root function has a value of n = . 2 1 2

Thus, the function f ( x ) = x can also be expressed as the power function f ( x ) = f ( x ) = x . The function is defined for x ≥ 0; that is, the domain is R+ ∪ {0}, or [0, ∞). 1 2

86

Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad

As can be seen from the graph, the range of the square root function is also R+ ∪ {0}, or [0, ∞). Throughout this section we will refer to the graph of y = x as ‘the basic square root curve’. Let us now investigate the effects of various transformations on the basic square root curve. a( x − b) + c. Consider the function y = a x − b + c, or y = a( 1 2

Dilation

a=3

y

a=2

The value a is a dilation factor; it dilates the graph from the x-axis. The domain is still [0, ∞).

a=1 a = 1–2 0

y=a x

x

Reflection If a is negative, the graph of a basic square root curve is reflected in the x-axis. The range becomes (−∞, 0]. The domain is still [0, ∞).

y y=√ √x (1, 1)

(0, 0)

x

(1, −1) y = −√ √x

If x is replaced with −x, the graph is reflected in the y-axis. For example, the graphs with equations y = x and y = − x are reflected across the y-axis. The domain becomes (−∞, 0] and the range is [0, ∞).

y y = √−x

y=√ √x

(−1, 1)

(1, 1) x

(0, 0)

Translation Horizontal translation The value b translates the graph horizontally. If b > 0, the graph is translated to the right, and if b < 0, the graph is translated to the left. The graph with the equation y = x − 3 results when the basic curve is translated 3 units to the right. This translated graph has domain [3, ∞) and range [0, ∞). If the basic curve is translated 2 units to the left, it becomes y = x + 2 and has domain [−2, ∞) and range [0, ∞). The domain of a square root function after a translation is given by [b, ∞). Vertical translation The value c translates the graph vertically. If c > 0, the graph is translated vertically up, and if c < 0, the graph is translated vertically down. If y = x is translated 2 units vertically up, the graph obtained is y = x + 2, with domain [0, ∞) and range [2, ∞). If the basic curve is translated 4 units down, it becomes y = x − 4, with domain [0, ∞) and range [−4, ∞). The range of the square root function is [c, ∞) for a > 0.

Chapter 2

y b = −2 (−2, 0) 0

(3, 0)

b=3 x

y= x−b

y (0, 2) 0 (0, −4)

c=2

c = −4

x

y= x+c

Functions and transformations

87

Combination of transformations

y

The graph of y = a x − b + c shows the combination of these transformations. The point (b, c) is the end point of the square root curve. For example, the end point of y = x − 2 + 1 is (2, 1). It is always good practice to label the end point with its coordinates. Make sure it is an open circle if the x-value is not in the required domain and a closed circle if its x-value is within the function’s domain. Consider the function y = a b − x + c. The graph of y = a x + b + c has (−b, c) as its end point. If this function is reflected in the y-axis, it becomes y = a − x + b + c with end point (b, c). The equation y=a

−x

y=a x−b+c

(b, c) x

0

y

y=a b−x +c y=a x+b +c

+ b + c can then be rewritten as (−b, c)

y = a b − x + c.

(b, c) x

For example, the graph of y = 2 − x + 1 can be −x

rewritten as y =

+ 2 + 1, which has an

end point of (2, 1) and bends to the left. The domain is (−∞, 2] and the range is [1, ∞). The equation y = x + 2 + 1 results in y = changes from

[−2,

∞) to

(−∞,

be rewritten as y = 2( x +

3 ) 2

−x

+ 2 + 1 when it is reflected in the y-axis. The domain

2] and the range remains [1, ∞). The equation y = 2 x + 3 − 1 can −

3

− 1; 1 the domain is [ 2 , ∞) and the range is [−1, ∞).

WORKED EXAMPLE 15

State the transformations required to change y = x to y = − 3 x + 5 + 3. THINK

WRITE

1

Write the general formula for the square root curve.

y=a x−b +c

2

Identify the value of a.

a = −3

3

State the effect of a on the graph.

The graph is dilated by a factor of 3 from the x-axis and reflected in the x-axis.

4

Identify the value of b.

b = −5

5

State the effect of b on the graph.

The graph is translated 5 units to the left.

6

Identify the value of c.

c=3

7

State the effect of c on the graph.

The graph is translated 3 units up.

WORKED EXAMPLE 16

eBoo k plus eBook

For each of the following functions find the domain and range.

88

a

y = 2 x − 3 +1

c

y= 4− x +2

b

y = −4 3x + 2 − 4

Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad

Tutorial

int-0523 Worked example 16

THINK

WRITE

Write the general formula.

y=a x−b +c

a

a y = 2 x − 3 +1

b

c

1

Write the question.

2

Identify the values of b and c.

b = 3, c = 1

3

State the domain x ≥ b.

The domain is [3, ∞).

4

State the range (y ≥ c for a > 0).

The range is [1, ∞). b y = − 4 3x + 2 − 4

1

Write the question.

2

Factorise the expression under the square root sign.

y = − 4 3( x + 23 ) − 4

3

State the domain.

∞). The domain is [ 3 , ∞)

4

Identify the value of c and check whether a is positive or negative.

c = −4, a < 0

5

State the range.

The range is (−∞, − 4].

1

Write the question.

2 3



c y=

2

4− x +2

Identify the values of b and c.

b = 4, c = 2

Since the function is of the form

The domain is (−∞, 4].

y = a b − x + c, the domain is x ≤ b. 4

State the range (y ≥ c).

The range is [2, ∞).

To sketch the graph of the square root function, we a>0 a>0 need to compare the given formula with y=a b−x+c y=a x−b+c y = a x − b + c. This will give us an idea of the (b, c) a 0

B y = a x − b + c, a < 0

C y = a b − x + c, a > 0

D y = a b − x + c, a < 0

(−2, 2) 0

x

E could be either B or C 7 MC The domain and range (in that order) of the function are: A (−∞, −2] and (−∞, −2] B (−∞, −2) and (−∞, 2) C (−∞, 2} and {−2, +∞) − − − − − − D ( ∞, 2] and ( ∞, 2] E ( ∞, 2] and ( ∞, 2] 8 WE17 Sketch the graph of each of the following, clearly marking intercepts and end points. a y= x+2 b y = 13 x + 3 c y = 2− x

d y = x − 6 +1

e y= 3+ x +2

f y = 12 − 4 + x

g y = 2x − 3

h y = 6 + 3x + 2

i y = 2 − x −1

EXAM TIP Graphs should be drawn showing correct features, such as smoothness and end points. [Assessment report 2 2007]

[© VCAA 2007]

Chapter 2

Functions and transformations

91

eBoo k plus eBook

9 WE18 A

y = 2− x −1

B

y = 2− 2 1− x

Spreadsheet 236

C

y = x −1−2

D

y = x − 2 +1

Function grapher

E

y = 2−2 x −1

Digital doc

y

MC The equation of the graph shown at right is:

0

(1, 2) x

2

The graph of y = x was dilated by the factor of 2 from the x-axis and translated m units to the right and 4 units down. It intersects with the x-axis at x = 5. Find: a the value of m b the equation of the curve. 11 The end point of the square root curve is at (4, 3) and its y-intercept is 9. Sketch the graph of the curve and hence establish its equation. 10

12 The graph of y = x was dilated by the factor of 4 from the x-axis, reflected in the x-axis, translated 1 unit to the left and p units up. Find: a the value of p, if the graph cuts the y-axis at y = 4 b the equation of the curve c the x-intercept d the domain e the range. f Hence, sketch the graph, showing the coordinates of the end points and the intercepts with the axes.

2F

The absolute value function The function f (x) = | x | is called an absolute value function or modulus function. The domain of this function is R and its range is R+ ∪ {0}. Its graph is symmetrical in the y-axis and has a cusp (a sharp point) at the origin. The symbol | x | represents the magnitude of x, (that is, the size of x), regardless of its sign.  x, if x ≥ 0 Therefore, |x| =  −  x, if x < 0 Compare the graphs of y = x and y = | x |. For x ≥ 0, the graphs of the two functions are identical, while for x < 0 the graph of y = | x | is the reflection of y = x in the x-axis. In general, any graph of the form y = |f|f (x)| is called an absolute value function. To sketch the graph of y = |f|f (x)|, we need to sketch the graph of y = f (x) first and then reflect in the x-axis the portion of the graph which is below the x-axis.

y

y

x

0 y = |x| y

x

0 y=x

x

0 y = |x|

WORKED EXAMPLE 19

Sketch the graph of y = | x2 − 1|. THINK 1

92

We first need to sketch the graph of y = x2 − 1. State the shape of this graph.

WRITE/DRAW

Let y = x2 − 1 Shape: positive parabola, translated 1 unit down

Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad

2

State the coordinates of the turning point (note that it is also the y-intercept).

Turning point: (0, −1)

3

Find the x-intercept by letting y = 0.

x-intercept: y = 0 x2 − 1 = 0 x2 = 1 x=± 1 x = +1 or −1

4 5

Sketch the graph of the parabola (Figure A).

y

y

−1

Reflect the portion of the parabola for < x < 1 in the x-axis; mark the new y-intercept (Figure B).

(0, 1) (−1, 1, 0)

0

(1, 0) x

(−1, 0) 0 (1, 0)

(0, −1)

x

y = |x2 − 1|

Figure A

Figure B

Similar to the graphs discussed in the previous sections, the graph of the absolute value function can be transformed through dilations, translations and reflections. If y = a|| f (x)| + c, a is the dilation factor. It dilates the graph from the x-axis. The larger a is, the thinner the graph. If a < 0, the graph is reflected in the x-axis.

y

a=2 a=1

0

The value c translates the graph along the y-axis. If c > 0, the graph is moved c units up and if c < 0, it is moved c units down.

y

a = −1

x

c=1 c = −2

1 0 −2

WORKED EXAMPLE 20

y = |x| + c

eBoo k plus eBook

Sketch the graph of y = |x − 2| + 1. THINK

x

Tutorial

WRITE/DRAW

int-0524 Worked example 20

1

Compare the given function with y = a|f|f (x)| + c, and write a short comment.

The graph of y = |x − 2| is translated 1 unit up.

2

To sketch the required shape we first need to sketch y = x − 2 (a straight line). Find the y-intercept by letting x = 0.

Let y = x − 2. y-intercept: x = 0 y=0−2 = −2

3

Find the x-intercept by letting y = 0.

x-intercept: y = 0 x−2=0 x=2

Chapter 2

Functions and transformations

93

4

y

Sketch the line.

(2, 0) x

0 (0, −2)

5

Reflect the portion below the x-axis in the x-axis; mark the new y-intercept.

y (0, 2) 0

6

Move the graph 1 unit up; mark the new y-intercept and the coordinates of the cusp.

(2, 0)

x

y (0, 3) (2, 1) 0

2

x

Absolute value functions as hybrid functions An absolute value expression can be thought of as two separate expressions, depending on whether it is negative or positive. |2x 2 + 3| can be written as (2 2x (2x + 3) or −(2x (2 + 3) depending upon the value that x takes. To determine these particular values of x, we need to solve the two inequalities −3 −3 2 + 3 > 0 and 22x + 3 < 0, giving x > 2x and x < , respectively. 2 2 This gives us a domain for the two expressions above, so we can write a representation for the absolute value expression as: −3   2 x + 3, wh ere x ≥ 2 |2x 2 + 3| =  2x −  − (2 x + 3), wh ere x < 3  2 This is a useful process when used to rewrite an absolute value function that is to be graphed, because it gives us a rule for each part of the graph in the form of a hybrid function. It is also important when needing to differentiate a function of this type in a later chapter. WORKED EXAMPLE 21

k plus eBook eBoo

Express f (x) = |5x 5 − 4| as a hybrid function, defining the domain of each part 5x and graphing the function. THINK 1

Break the function into two parts: a negative and positive part.

2

Simplify the domain and function for each.

Tutorial

int-0525 Worked example 21

WRITE/DRAW

 5 x − 4, wh ere 5 x − 4 ≥ 0 f ( x) = | 5x − 4 | =  −  (5 x − 4), wh ere 5 x − 4 < 0 First function: 5x − 4 First domain: 5x − 4 ≥ 0 4

x≥ 5 − Second function: (5x − 4) = −5x + 4 Second domain: 5 x − 4 < 0 ∴x<

94

Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad

4 5

3

Rewrite the function in hybrid form with the two rules with their respective domains.

4

Graph the two functions for the specific domains.

4   5 x − 4, wh ere x ≥ 5 f ( x) =   − 5 x + 4, wh ere x < 4  5 y

5 4 (0, 4) 3 2 1 −2

5

−1

0

f x) = 5x − 4 f(

1 ( –45 , 0)

x

2

The absolute value function can easily be drawn using a CAS calculator. On the Graph & Tab page, complete the function entry line as: y1 = abs(5x − 4) Tick the y1 box and tap !.

REMEMBER y

1. The symbol | x | denotes the magnitude of x. 2. | x | = x, if x ≥ 0 | x | = −x, if x < 0 3. To sketch the graph of y = |f|f (x)|: (a) sketch the graph of y = f (x) (b) reflect the portion of the graph which is below the x-axis in the x-axis. EXERCISE

2F

x

0 y = |x|

The absolute value function 1

Sketch the graph of each of the following, showing exact values of intercepts

WE19

of axes. a y = |2x 2 | 2x d y=

|x2

b y = |x − 1|

− 6|

e y = |4 −

g y = |3x3| h i

y = |(x + y=

2)3

2 x −1

c

x2|

y = |3 − 6x|

f y = |(x − 3)2 − 4| EXAM TIP Always give answers in exact form unless the question asks for a numerical approximation (for example, ‘give your answer to 2 decimal places’).

− 1|

[Assessment report 2 2007]

[© VCAA 2007]

Chapter 2

Functions and transformations

95

2

y

MC Which of the following functions best describes this graph?

A y = |(x − 1)3|

B y = |(x + 1)3|

C y = |x3 + 1|

D y = |x3 − 1|

E y = |(x +

1)3

1

+ 1|

–1 0

3 For each of the following functions state the domain and range.

eBoo k plus eBook

a y = 2|x|

b y = |x| + 1

d y = |x2 − 3| − 2

e

4

Spreadsheet 002

d y=

Absolute value function

1 +1 x +1

y = 4 − 3|x|

f

y= 2−

1 x2

WE20 Sketch the graphs of each of the following.

a y = −2|x|

Digital doc

y=

c

x

|x2

b y = |x + 5| − 6

− 1| + 1

1 3 − x 4

g

y=

j

y=

l

y=2−



1 −1 x2

c

y = 2|3 − x| + 1

− 2|

f

y = |(x + 1)2 − 1| − 2

2 +3 6− x

i

y=

e y=2− h

y=

k

y=

|x2

1 1 −4 4 x2

2− x −2 + 3

x +1 − 8

EXAM TIP

Use a ruler for linear and absolute

value graphs. [Assessment report 2 2007 © VCAA]

5

WE21 Given the function f (x (x) = | 3x − 1|:

a rewrite the function as a hybrid function with appropriate domains b find f (0) and f (2) c sketch the graph, labelling any significant points.

6 Given the function f (x (x) = | x2 − 3x |+ 2: a rewrite the function as a hybrid function with appropriate domains b find f (−1) and f (2) c sketch the graph, labelling any significant points.

eBoo k plus eBook Digital doc

WorkSHEET 2.2

2G eBoo eBook k plus Interactivity

int-0247 Transformations with matrices

96

7 The design shown at right is to be embroidered on the outer side of a pair of children’s socks. The total length of the design is 12 cm and its width is 8 cm. y 6 If we draw the set of axes through the centre of the design, the red section can be thought of as the absolute value function on a restricted domain. a Find the rule for the red section and specify the domain. b Using your knowledge of the transformations, and the rule for 4 4x the red section, find the rules for the blue, green and yellow sections of the design. c Using a graphics calculator, sketch the 4 functions that were obtained in a and b . Have you obtained the right design? −6

Transformations with matrices Transformations of graphs can be described using matrices as an alternative to function notation. The transformations that have been considered so far (dilations, reflections and translations) can be represented in matrix form. This describes how a particular point on a graph will be moved (or mapped) to a resultant location by the application of a dilation, reflection or translation, or a combination of the three. Remember the definition of a transformation is a rule that ‘links’ each point in the Cartesian plane to another point. So the matrix can be used for any

Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad

point on a curve, or in fact used to determine the new rule for a function that has undergone one or a series of transformations. The new points or rules are termed ‘images’ of the original.

Reflections and dilations

  x  We can summarise the use of matrices to map these transformations, T    , of points on a   y  curve as follows: (Let x' be the transformed value of x, and y' be the transformed value of y.)   x   x'   − 1 0   x   − x  T    =   =     =   represents a reflection in the y-axis.   y   y'   0 1   y   y    x   x'   1 T    =   =    y   y'   0

0  x

 x =  −  represents a reflection in the x-axis.     y

−1   y 

  x   x'   a 0   x   ax  T    =   =     =   represents a dilation of a factor of a from the y-axis.   y   y'   0 1   y   y    x   x'   1 0   x   x  T    =   =     =   represents a dilation of a factor of a from the x-axis.   y   y'   0 a   y   ay  These operations can be combined to represent more than one transformation, for example   x   x'   − 2 0   x   − 2 x  T    =   =    =     y   y'   0 3   y   3 y  represents a reflection in the y-axis, a dilation of a factor of 2 from the y-axis, and a dilation of a factor of 3 from the x-axis. WORKED EXAMPLE 22

Using matrices, find the location of the point ((x′, y′) under the following transformations of the point (1, 3): • dilation by a factor of 2 from the y-axis • reflection in the x-axis. THINK 1

Construct the correct matrix that represents the transformations described.   x   x'   2 T    =   =    y   y'   0

2

WRITE/DISPLAY

0  x × −    1  y

2  0

0

−1 



On the Main page, tap: • ) • 8 Enter the value into the 2 × 2 matrix and multiply the 2 × 1 matrix by tapping: • 7 then entering the matrix values and pressing: • E

Chapter 2

Functions and transformations

97

3

Write the matrix equation and interpret this to answer the question.

2  0

0  1 

 2 =−     3

−1   3 

The image of the point (1, 3) is at (2, −3). ∗

Matrix operations can be done using a CAS calculator, but as the matrix multiplication required here is simple, it is recommended it be done by hand.

The difficult part is to correctly identify the transformation matrix. Once you have done that it is a matter of performing a matrix multiplication.

Translations Translations require a slightly different process. The transformation matrix is a 2 × 1 matrix, and finding the new image requires addition of the matrices rather than multiplication. b  Matrices describing translations are of the form   . c This represents: • a translation of b units in the positive direction of the x-axis and • a translation of c units in the positive direction of the y-axis. Note b and c > 0: • If either of the terms is negative, the translation is in the negative direction. • A zero entry indicates there is no translation in a particular direction. So a translation of a point (x, y) can be described as follows:   x   x'   x   b  T    =   =   +     y   y'   y   c  x + b =  y + c WORKED EXAMPLE 23

Find the location of the point ((x', y') under the following transformations of the point (−2, 4): • translation of 3 units in the x direction • translation of −5 units in the y direction. THINK 1

WRITE

Construct the correct matrix that represents the transformations described.   x   x'   x   3  T    =   =   +  −    y   y'   y   5 

2

Construct and solve the appropriate matrix equation.∗   x   x'   3 T    =   =   +  −  y y '       4   5   − 2

3

Interpret this to answer the question.

 3 −   5

 − 2   3   1  +− = −   4  5   1  − 2   3   1   + −  =  −   4   5  1  The image of the point (−2, 4) is at (1, −1).



Matrix operations can be done using a CAS calculator, but as the matrix addition required here is simple, it is recommended it be done by hand.

98

Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad

Note it is very important to use the correct mathematical language, both for the size and the direction of the transformation. For example, dilations are ‘by a factor of’, in other words a multiple of the original value, as distinct from translations, which are described as ‘of n units’, which is a set distance. In terms of direction, the expression ‘from the y-axis’ can also be expressed as ‘parallel to the x-axis’, ‘in the x direction’ or even ‘horizontally’. The same applies for ‘from the x-axis’.

Putting it all together The formal notation often used to describe a transformation begins as T : R2 → R2, which is saying ‘the transformation that maps a point (x, y) to another point (x, y) is ….’ and then the transformation is described. For example, a transformation involving a dilation by a factor of 3 from the x-axis followed by a translation of −2 in the x direction and 1 in the y direction could be defined as T : R2 → R2, T( T x, y) = (x − 2, 3y + 1), or simply (x, y) → (x − 2, 3y + 1). When more than one transformation is described, it is known as a composition of the transformations. When a series of transformations are described, they need to be done in the correct order as stated in the question. WORKED EXAMPLE 24

Find the location of the point ((x', y') under the following transformation of the point (3,−2): 1 from the y-axis 2 • dilation by a factor of 3 from the x-axis • reflection in the x-axis • translation of 3 units in the x direction. • dilation by a factor of

THINK

WRITE

1

Perform the first three transformations together. Construct the  1 correct matrix that represents these transformations described.  2  0   x   x'   1 0 x T    =   =  2 ×    y   y'   0 − 3   y 

0  −3  

2

Construct and solve the appropriate matrix equation.   x   x'   1 0  3 T    =   =  2 ×    y   y'   0 − 3   − 2 

1 2  0

0   3  3    = 2 −3  − 2    6 

3

Now perform the translation. Construct and solve the appropriate matrix equation. Note we are using the product matrix from step 2.

 3  3  9  2 +   = 2 6   0  6 

  x   x'   1 T    =   =  2   y   y'   0 4

0   3  3  × − +   − 3   2  0 

Interpret this to answer the question.

The image of the point (3, −2) is at

( , 6) 9 2

Remember that a transformation maps any point on a curve to another by the same rule. Rather than mapping a series of individual points on the same curve, we can simply find a new rule under a transformation (or a series of them) and use this new rule to determine the location of any points from the curve described by the original rule.

Chapter 2

Functions and transformations

99

WORKED EXAMPLE 25

k plus eBook eBoo

Write the resultant equation from the following transformations of the curve described by y = x3: • dilation by a factor of 2 from the y-axis • reflection in the x-axis • translation of 2 units in the negative y direction. THINK 1

3

2  0

−1 

Construct and solve the appropriate matrix equation.

2  0

−1   y 

Now perform the translation. Construct the correct matrix that represents this transformation described.

 0 −   2

Perform the first two transformations together. Construct the correct matrix that represents these transformations described.

0 

0  x ×   y

−1 

5

  x   x'   2 0   x   0  T    =   =  −  ×   +  −    y   y'   0 1   y   2  Construct and solve the appropriate matrix equation. Note we are using the product matrix from our first equation in step 2. The transformed values of x and y are 22xx and −y − 2.

6

Express x' and y' in terms of x and y.

7

State the resultant equation.

4

int-0526 Worked example 25

WRITE

  x   x'   2 T    =   =    y   y'   0 2

Tutorial

0  x

2x  = −      y

 x'   2 x   0   2 x    =  −  + −  =  −   y'   y   2   y − 2  Therefore x' = 2x 2x and y' = −y − 2. x' and 2 −y = y' + 2 y = −y' − 2 x=

3

− x y=   −2  2

So for any point on the graph of the original function, y = x3, we can map the corresponding point under the transformations above by substituting the values into this transformed equation. Let’s have a look at another example. This time we will complete reflections/dilations and translations in the one step. WORKED EXAMPLE 26

Find the image of the curve with equation y = x after a reflection in the x-axis, followed by a dilation of a factor of 2 from the x-axis, and then a translation by +3 in the x direction. THINK 1

100

Construct the correct matrices that represent the transformations described.   x   x'   1 0   x   3  T    =   =  −  ×   +     y   y'   0 2   y   0 

WRITE

1  0

Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad

0

3 and  0    

−2 

2

Construct and solve the appropriate matrix equation.

 x'   1    y'   0

0   x   3   x + 3  +   =  −    y  0   2y 

−2

3

The transformed values of x and y are x + 3 and −2y.

So x' = x + 3 and y' = −2y.

4

Express x' and y' in terms of x and y.

− x = x' − 3 and y = y' 2

5

Substitute new y- and x-terms for y = x, x and drop the primes.

−y

2 −y

= x−3 = 2 x−3

y = −2 x − 3 6

State the resultant equation showing all transformations.

y = −2 x − 3

REMEMBER

a 0  1.   represents a dilation of a from the y-axis and a dilation of b from the x-axis. 0 b   −1 0 1 2.   represents a reflection in the y-axis, and  0  0 1 the x-axis.

0

−1 



represents a reflection in

b  3.   represents a horizontal translation of b and a vertical translation of c. c EXERCISE

2G

Transformations with matrices 1

Identify the dilations and/or reflections described by the following matrices. i

 −1 0     0 2

ii

1 2  0

0  −4  

iii

 −2 0    0 3

iv

1   0

−1 

0  2

2 WE22 WE24 Find the image of the point (−3, 5) under the above transformations in question 1 . 3 Find the image of the graphs of the following equations under the transformations in 1 i and 1 ii. a y=

1 x2

b y = x3 − 5

c

y= x

4 Identify the translations described by the following matrices. i

3   2

ii

 2 −   2

iii

Chapter 2

 −1   5  0 

Functions and transformations

101

5 WE23 Find the image of the point (1, −2) under the transformations given in question 4 . 6 Find the image of the following equations under each of the transformations defined in questions 4 i and 4 ii. a y = |x| b y = x2 − 3x 7 The transformation T : R2 → R2 which maps the curve with the equation y = x3 to the curve with the equation y = (3x − 6)3 + 1, could have: A

  x   1 0   x   − 6  T    =   +     y   0 3   y   1 

B

  x   1 0   x   6  T    =  3  +     y   0 1   y  1 

C

  x   1 0   x   2  T    =  3  +     y   0 1   y   1 

D

  x   1 0   x  − 3  T    =   +     y   0 3   y   1 

E

  x   3 0   x   6  T    =   +     y   0 1   y   1 

For the following transformations, where T : R2 → R2, state what the transformation T 1 represents and determine the image of the equation f ( x ) = . x   x   − 1 0    x   6    x   2 0   x   3  a T    =  b T    =  − 1    +  −      +     y   0 2    y   1    y   0 2   y   1    x   1 c T    =  3   y   0

0    x   1     + −  −1      y   2 

9 A function g(x ( ) is mapped to the curve h(x (x ( ) = −g(4(x (x (4( + 1)) + 3. Create a matrix equation that (4(x will map g(x ( ) to h(x (x ( ). (x 10 WE25 The following transformations are applied, in order, to the graph of y = x3 − 4x 4 : • dilation by a factor of 2 from the x-axis • reflection in the y-axis • translation of −1 unit in the y direction. a Use matrices to determine the image equation under these transformations. b Find the image of the point (2, 0) and check whether this point lies on the curve of the equation from a. −3 1 + 1 , describe, in order, the transformations and g( x ) = 2 ( x − 2)2 x performed to the graph of f (x) to give g(x) and create a matrix equation which would map f (x) to g(x).

11 WE26

If f ( x ) =

12 If f (x (x) = −g(2(x (2( + 1)) + 1 and g( x ) = x , find f (x (2(x (x) in terms of x only, using: a an algebraic method without the use of a CAS calculator b matrices and a CAS calculator. 13 If f (x (x) = 2g(x ( − 1) − 2 and g(x (x ( ) = x2 − 3x, find f (x (x (x) in terms of x only, using: a an algebraic method without the use of a CAS calculator b matrices and a CAS calculator. 14 If

102



− x3 1 1 ( ) using matrix methods. (x h( x + 2) + 1 = − 3 x 2 − 6 x − , find h(x 2 2 2

Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad

2H

Sum, difference and product functions Sum functions

y

A sum function is of the form y = f (x) + g(x), or alternatively y = x2 4 y = (f ( + g)(x). 3 Many functions include two or more terms added (or 2 y = 1x 1 1 2 subtracted) together. For example the function y = x + can be x −4 −3 −2 −1−10 1 2 3 4 x 1 2 thought of as the sum of the functions y = x and y = . −2 x These graphs can be drawn by sketching the two individual functions on the same set of axes then adding the y-values (ordinates) for each x-value and plotting the resulting points. This is a useful method when we know the basic shape of the individual functions but do not recognise the whole function. We would not use this method for a familiar function such as y = x2 + 3x, as we have learnt ways of sketching this without breaking it up into parts. 1 y Using the example in the first paragraph, y = x 2 + , we 4 y = x2 x do not recognise the shape of this function, but we know the 3 2 two individual functions are the basic positive parabola and y = x2 + 1x 1 y = 1x the hyperbola. We could therefore sketch the graph of the parabola and the hyperbola and add the y-values together for −44 −3 −22 −11−10 1 2 3 4 x corresponding x-values to obtain points on the curve of the sum −2 function — which can be joined together to obtain the graph of the sum function. Note that the domain of the hyperbola is restricted to R \ {0} so the y-value at x = 0 is undefined. As you cannot add an undefined number, this x-value is also undefined for the sum function. A general rule is that the sum function is only defined for the domain over which both of the individual functions are defined. The domain of the sum function is, therefore, the intersection of the domains of the individual functions. If h(x) = f (x) + g(x), then the domain h(x) = domain f (x) ∩ domain g(x).

Difference functions A difference function is of the form y = f (x) − g(x), or alternatively y = (f ( − g)(x). It is essentially the same as a sum function except that one of the individual functions is subtracted from the 1 other. So y = x 2 − could be sketched by the same method as described above but instead of x adding the y-ordinates, we would subtract one from the other. The domain of a difference function is determined in the same way as a sum function. We could extend our rule above to include difference functions. y If h(x) = f (x) ± g(x), then the domain h(x) = 4 y = x2 domain f (x) ∩ domain g(x). 3 y = x2 − 1x 2 We can also think of a difference function as ‘adding a y = 1x 1  −1  negative’ and it could be written y = x 2 +   . With this  x −4 −3 −2 −1−10 1 2 3 4 x in mind, an alternative method of sketching the graph of a −2 difference function is to reflect the graph of the second function 1 (in this example, ) in the x-axis and then add the ordinates as for a sum function. x

Chapter 2

Functions and transformations

103

When sketching graphs of sum/difference functions, there are key points that can be found on either individual function to easily identify the value of the ordinate of the sum or difference function. These are the x-intercepts and any point of intersection of the individual functions. The x-intercept is where the ordinate of that particular function is zero, so the graph of the sum or difference function is actually the ordinate of y the other function for that value of x. 2.5 At the intersection, the ordinate of the sum function 2 will be double that of the two individual functions. 1.5 For a difference function, an intersection of the two individual functions corresponds to on x-intercept 1 (y = 0) of the difference function. 0.5 Another useful y-value to look for is where the graphs of individual functions have y-values that are −1.5 −1 −0.5 0 0.5 1 1.5 x of the same magnitude but one is positive and one −0.5 is negative. This point is an x-intercept of the sum function.

WORKED EXAMPLE 27

eBoo k plus eBook

Using addition of ordinates, sketch the graph of

Tutorial

f ( x ) = log log e ( x + 2 ) + x , x ∈[ ∈[ − 1, 2 ].

int-0527 Worked example 27

THINK 1

WRITE/DRAW

Sketch the graphs of y = loge (x + 2) and y = |x| on the same set of axes over the required domain x ∈ [−1, 2].

y 3 2 y = loge (x + 2) (−1, 1)

1

−1.5 −1 −.5 0 2

104

Moving from left to right, add the y-coordinates of the two graphs for the key points and plot the resultant points. The key points are the: • end points • y-intercepts • points of intersection. The new points on the graph are marked by an asterisk.

y =x (2, loge (4)) 0.5 1 1.5 2 2.5 x

y



3 (−1, 1)

∗ ∗

2 y = log (x∗ + 2) e 1

−1.5 −1 −0.5 0

(2, 2)

(2, 2)

y =x (2, loge (4)) 0.5 1 1.5 2 2.5 x

Left end points (−1, 0) and (−1, 1), so the new point will be at (−1, 1). Right end points (2, loge (4)) and (2, 2), so the new point will be at (2, 2 + loge (4)). y-intercepts (0, 0) and (0, loge (2)). Points of intersection (−0.44, 0.44) and (1.15, 1.15), so the new points will be (−0.44, 0.88) and (1.15, 2.30).

Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad

3

Join these points with a smooth curve to create f (x).

y f( f x) = loge (x + 2) +x 3 (2, 2) y = x (−1, 1) 1 y = loge (x + 2) (2, loge (4)) −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 x 2

4

Remove the two individual graphs to leave the sum function.

(2, 2 + loge (4))

y 3 2 (−1, 1)

f x) = loge (x + 2) +x f(

1

−1.5 −1 −0.5 0

0.5

1 1.5

2 2.5 x

Product functions A product function is of the form y = f (x) g(x), or alternatively y = (fg ( )(x). When graphing product functions, it is useful to graph the individual functions, f and g, and for any relevant values of x, to identify the y-values, or ordinates and multiply these together to obtain the y-value of the product function. If the y-value is undefined at a particular value of x for either of the individual functions, then the product function is undefined for that value. We cannot multiply by an undefined number. If h(x) = f (x)g(x), then the domain h(x) = domain f (x) ∩ domain g(x). When examining the graph of the two individual functions, it is useful to look at x-intercepts and points where the value of either function is ±1. The product function will also have an x-intercept at a point where either individual function has an intercept (as multiplying by zero gives zero). At a point where a function = ±1, the product function will have a value equal to the value of the other function, or its negative. It is also useful to observe that where the individual functions are both above the x-axis, or both below the x-axis, the value of the product function will be positive, that is, above the x-axis. This is because the product of two positive numbers or two negative numbers is positive. Alternatively, where one function is above and one below the x-axis, the value of the product function will be negative, that is, below the x-axis. WORKED EXAMPLE 28

If f (x (x) = 2x 2 and g( x ) = x + 1 , sketch the graph of f ( x ) g( x ) = 2 x x + 1 . THINK 1

Sketch the graphs of f (x) and g(x).

WRITE/DRAW y

y = 2x 2 y= x+1

(0, 1) (−1, 0) 0

Chapter 2

x

Functions and transformations

105

2

Find the domain of f (x) and the domain of g(x).

Dom f = R and dom g = [−1, ∞)

3

Find the domain of f (x) g(x).

Dom fg = [−1, ∞)

4

Find the x-intercepts of both f and g and hence find the x-intercepts of the product fg.

x-intercept for f (x) is when x = 0 and f (x) = 0 x-intercept for g(x) is when x = −1 and g(x) = 0 Hence, the x-intercepts for the product are when x = 0 and x = −1.

5

Find the values of x for which the product is negative.

f (x) is negative and g(x) is positive for x ∈ (−1, 0), so fg is negative for x ∈ (−1, 0).

6

Find the values of x for which the product is positive.

7

Find the turning point using a CAS calculator. Round the answer to 2 decimal places as appropriate.

f (x) and g(x) are both positive for x ∈ (0, ∞), so fg is positive for x ∈ (0, ∞). The turning point is ( 32 , −0.77).

8

Sketch the graph of the product.



y

y = 22xx x + 1

(−1, 0) (0, 0)

x

2 (−— , −0.77) 3

REMEMBER

1. For the sum/difference function, dom( dom(f (x) ± g(x)) = dom f (x) ∩ dom g (x). The graph of the sum/difference function can be obtained by using the addition of ordinates method. 2. For the product function, dom ((ff (x) g(x)) = dom f (x) ∩ dom g(x). Some features of the graph of the product function are as follows: (a) the x-intercepts of f (x) g(x) occur where either f (x) or g(x) have their x-intercepts (b) f (x) g(x) is above the x-axis where f (x) and g(x) are either both positive or both negative (c) f (x) g(x) is below the x-axis where one of the functions f (x) or g(x) is positive and the other is negative. EXERCISE

2H

Sum, difference and product functions 1

106

Sketch the graphs of f (x (x) = g(x ( ) + h(x (x ( ) using addition of ordinates, given the following (x functions g(x) and h(x). State the domain of f (x) in each case. a g(x) = x2, h(x) = x x3 b g( x ) = , h( x ) = x 2 c g(x) = 3x2, h(x) = | x | 1 x) = x + 2 d g( x ) = , h( x) x

Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad

Determine the equation of g(x ( ) − h(x (x ( ) in each of the following cases then, using addition (x of ordinates, sketch the graph of g(x) − h(x). − x + 1, h( x ) = x a g( x ) = b g(x) = | x |, h(x) = | x + 1| − 2 3 For each of the following, find the domain of f (x (x)g(x ( ). (x 2 a f ( x ) = x , g( x ) = 3 − x b f (x) = −x + 2, g(x) = |2x 2 + 1| 2x c f ( x ) = x , g( x ) = 1 − x d f (x) = | x |, g(x) = x2 − 1 e f ( x ) = x 3 , g( x ) = − x + 2 4

WE27 Sketch the graph of f ( x ) = x +

the general shape and any asymptotes.

2 for −4 ≤ x ≤ 4, by the addition of ordinates, showing x

5 Two functions are defined as f (x ( ) = −x2 and g( x ) = x . a Sketch the graph of each on the same set of axes for −2 ≤ x ≤ 2. b Find the smallest possible value of a given that the domain of the function h, where h(x) = (f ( + g)(x), is a ≤ x ≤ 2. c Find f (0) and g(0), and hence find h(0). d Find f (1) and g (1), and hence find h (1). e Find f (2) and g (2), and hence find h (2). f Using this information, sketch the graph of h(x ( ) (on the same set of axes as in a). (x Given the functions f (x ( ) = −x3, g(x ( ) = | x | and h(x (x ( ) = f (x (x ( ) + g(x ( ): (x − a Sketch the graph of each on the same set of axes for 2 ≤ x ≤ 2. b Find f (−2)and g (−2), and hence find h (−2). c Find f (0) and g (0), and hence find h (0). d Find f (1) and g (1), and hence find h (1). e Find f (2) and g (2), and hence find h (2). f What is the range of the function h (in exact form). g Using this information, sketch the graph of h(x) (on the same set of axes as in a). 7 WE28 Two functions are defined as f (x ( ) = x − 3 and g( x ) = x . Let h( x ) = f ( x ) g( x ). a Find the domain of h. b Sketch the graph of each on the same set of axes. c Find f (0) and g (0), and hence find h (0). d Find f (1) and g (1), and hence find h (1). e Find f (2) and g (2), and hence find h (2). f What is the range of the function h (in exact form)? g Using this information, sketch the graph of h(x) (on the same set of axes as in a). 6

8 Sketch the graphs of the functions f ( x ) = x + 5 and g( x ) = 8 − x and use these to find the domain of the function h(x ( ) = f (x (x ( ) + g (x ( ). On the same axes, sketch the graph of h(x ( ), including (x the coordinates of any end points. x2 − 4, for 9 Use a CAS calculator to view and sketch the graphs of f (x ( ) = | x + 2 | and g( x ) = 2 −2.5 ≤ x ≤ 2.5. Then, without using the calculator, use these graphs to sketch the graph of h, if h(x) = (f ( − g)(x) on the same set of axes. Using the calculator, check the shape of the graph you have drawn and use it to identify any significant points such as intercepts and cusp points to 2 decimal places. (You may need to adjust your window settings in order to clearly identify these points.)

Chapter 2

Functions and transformations

107

2I

Composite functions and functional equations Composite functions A composite function is formed from two functions in the following way. If f (x) = x + 5 and g(x) = 2x 2x are two functions, then we combine the two functions to form the composite function g(f ( (x)) = 2f (f 2 (x) = 2(x + 5). That is, f (x) replaces x in the function g(x). The composite function reads g of f and can be written g ° f. f Another composite function is f (g(x)) = g(x) + 5 = 2x 2 + 5. In this case, g(x) replaces x in f (x). This composite function reads f of g and can be written f ° g. For the composition function f (g(x)) to be defined, the range of g must be a subset of (or equal to) the domain of ff, that is ran g ⊆ dom ff. It is easiest to list the domain and function of both f (x) and g(x) first when dealing with composite function problems. For example: f (x) = x2 and g( x ) = x :

Domain Range

f (x ( )

g(x ( ) (x

R

[0, ∞)

[0, ∞)

[0, ∞)

Composite functions can be rather complex to graph by hand, so a CAS calculator can be used for assistance when sketching.

WORKED EXAMPLE 29

For the pair of functions f ( x ) = a show that f (g ( (x ( )) is defined c state its domain.

1 and g( x ) = x : x+2 b find f (g ( (x ( ))

THINK a

1

2

b

WRITE

Create a table showing the domain and range of both functions.

For f (g(x)) to exist the range of g must be a subset of f. f

Form the composition function f (g(x)) by substituting g(x) into f (x).

f (x (x)

g(x ( ) (x

Domain

R \{−2}

R+

Range

R \{0}

R+

ran g(x) ⊆ dom f (x) ∴ f (g(x)) is defined. f ( g( x )) = f ( x ) f ( g( x )) =

c

108

1 x +2

The domain of f (g(x)) must be the same as Domain of f (g(x)) = R+ the domain of g(x). Since the domain of g(x) is R+, it is the domain of f (g(x)).

Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad

Functional equations Sometimes we are required to solve or analyse equations that are in terms of unknown functions, for example, f (x) or f (y), rather than being in terms of unknown variables, for example x or y. An example of the type of problem you might encounter is to find a function that satisfies f (x + y) = f (x) + f (y). Alternatively, you might be required to determine if a particular function satisfies the rule f (2x (2x) = 2f 2f (x). Equations such as f (x + y) = f (x) + f (y) are called functional equations. There are generally two ways to solve these types of problems: algebraically or using a CAS calculator. WORKED EXAMPLE 30

Determine if f (x ( ) = 3x satisfies the equation f (x ( + y) = f (x ( ) × f (y ( ). THINK 1

WRITE

Substitute the function into the LHS and RHS of the equation separately. Simplify the LHS of the equation to determine if it equals the RHS of the equation. Answer the question.

2 3

LHS = f (x + y) = 3x + y RHS = f (x) × f (y) = 3x × 3y LHS = f (x + y) = 3x + y LHS = f (x + y) = 3x × 3y ∴LHS = RHS x ∴ff (x) = 3 satisfies the equation f (x + y) = f (x) × f (y)

WORKED EXAMPLE 31

Determine if g(x ( ) = 10x (x 10 satisfies the equation g(2x (2 ) = 2g (2x 2 (x ( ). THINK 1

On the Main page, define the equation by entering the equation as: 10x Highlight it and tap: • Interactive • Define Set: Func name: g Variable/s: x • OK

2

To calculate g(2x) and 2g(x), complete the entry lines as: g(2x) 2g(x) Press E after each entry. Examine the solutions to determine if the expressions are equal.

WRITE/DISPLAY

LHS = g(2x (2 ) = 20x (2x 20

Chapter 2

RHS = 2g(x) = 20x 20

Functions and transformations

109

3

Check that the equation holds true for all values of x by completing the entry line as: g(2x) = 2g(x) Then press E. Since the LHS = RHS, the statement is true and the equation holds for all values of x.

4

Write the solutions for g(2x (2 ) and 2g(x) (2x and answer the question.

g(2x (2 ) = 20x (2x 20 2g(x) = 20x 20 ∴g(2x (2 ) = 2g(x) (2x When g(x) = 10x 10 it satisfies the equation g(2x (2 ) = 2g(x). (2x

If we consider the same equation f (2x (2x) = 2f 2f (x (x) for a different function, for example, f ( x ) = x , we obtain two different equations, f (2 x ) = 2 x and 2 f ( x ) = 2 x , which are not equal. However, if we define this function on a CAS calculator and enter the statement f (2x (2x) = 2f 2f (x), the result is x = 0. This means this equation holds true when x = 0 but not for any other values of x. REMEMBER

1. For the composite function f (g(x)) to be defined, the range of g must be a subset of the domain of ff. Furthermore, if f (g(x)) is defined, the domain of f (g(x)) equals the domain of g(x). 2. For equations involving algebra of functions, determine if the equation is true for the particular function by considering the LHS and RHS of the equation separately to test if the equation holds true for all values of x. EXERCISE

2I

Composite functions and functional equations 1

WE29 For each of the following pairs of functions:

i show that f (g(x)) is defined

i i find f (g(x)) and state its domain.

a f (x) = 2x 2 − 1 and g( x ) = x + 3 c f (x) = 3(x − 2)3 and g(x) = x2 e f (x) = (x + 1)(x + 3) and g(x) = x2 2 Show that the function f (x ( ) = 3xx satisfies  x + y  f ( x) + f ( y) = the equation f  .  3  3 3

WE30 Show that f ( x ) =

equation

110

[© VCAA 2007]

2 satisfies the x

1 and g(x) = | x | + 1 x+2 d f (x) = | x | and g(x) = x3 b f ( x) =

EXAM TIP ‘Show that’ questions require detailed working out to be given, so provide as many steps as possible to attain full marks. Write ‘as required’ at the end of your solution once you have shown the problem is correct.

f ( x) + f ( y) = x + y. f ( xxyy )

Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad

[Assessment report 2 2007]

4 WE31 Determine which of i–v –v hold for the following functions: –v f ( x) ii f ( x − y ) = i f (x − y) = f (x) − f (y) f ( y)  x  f ( x) f = iii f (x) + f (y) = (x2 + y2) × f (xy) iv  y  f ( y ) v f (xy) = f (x) × f (y)

5 6 7 8

9

10

b f (x) = | x | a f ( x) = x 1 1 c f ( x) = d f ( x) = 2 x x e f (x) = x2 f f (x) = 2x 1 f ( x) = and g( x ) = x , determine the values of a such that f (g(x ( )) exists. (x ( x + a) 2 1 + 2, determine if f ° g If f: f R→R, where f ( x ) = x − 2 and g:R {−1} → R, where g( x ) = x +1 and g ° f exist and, if so, find the composition functions. If f: f R→R, where f ( x ) = 3 − x and g:R→R, where g(x ( ) = x2 − 1, show that f ° g is not defined. (x By restricting the domain of g, find a function h such that f ° h is defined. Given w(x ( ) = x + 3, x > −3 and v(x (x ( ) = | x | − 2, x ∈ R+, state the domain and range of each (x function. Hence, find if w ° v and v ° w exist and, if so, state their rules including their domains. Show that the equation g(x ( ) = x3 satisfies the equation g(−x) = −g(x (x ( ). Show that (x n this statement is true for all functions of the form g(x) = x , where n is an odd natural number. Show that g(x ( ) = x4 satisfies the equation g(xy (x ( ) = g(x (xy ( )g(y (x ( ). Show that this equation is true (y for all functions of the form g(x) = xn, where n is a natural number.

11 Given the two functions: 1 f : [0, 6] → R R,, f ( x) x ) = 4 x ( x − 6)2 g: [0, 6] → R, g (x) = −x2 + 6x find the maximum value of | f (x) − g(x)| for 0 ≤ x ≤ 6, correct to 2 decimal places, and determine the values of x in this interval for which the maximum occurs. [© VCAA 2007]

EXAM TIP Be careful to answer all aspects of the question (this question requires the x-value and the maximum). Always re-read the question to check you have met all requirements before moving on to the next question. [Assessment report 2 2007]

12 Consider f : [4, ∞] → R ( ) = 1 − x. What transformations are (x R,, f ( x) x ) = x − 4 and g: R → R, g(x required to obtain f (g(x ( )) from f (x (x ( )?

2J

Modelling People such as scientists, financial advisers, business analysts, economists, statisticians and others often have to deal with large and small sets of data. Once the data are collected, we are often interested in finding the rules that link features of the data. The process of finding such a rule is called modelling and the rule itself is known as the mathematical model. When finding the model, the best way to start is to plot the data, as the shape of the graph might suggest the type of relationship between the variables.

Types of graphs By recognising the shape of a graph, it is possible to find the rule or mathematical model that describes it. Throughout this chapter, several types of graphs have been investigated.

Chapter 2

Functions and transformations

111

The parabola: y = x2

The graph of a cubic function: y = x3

y

The hyperbola: y =

y

y

y=0

x

0

1 x

x

0

x

0

x=0

The truncus: y = y

y=0

The graph of a square root function:

1 x2

y= x y

x

0

x

0 x=0

Reflections and translations can be applied to each of these graphs, but the basic shape of each graph remains the same.

WORKED EXAMPLE 32

Match each of the following graphs with the appropriate model. i y = ax2 a

ii y = ax3 b

y x

iii y = c

y

a x

iv y = d

y

x

x

THINK

a x2

v e

y

y= a x y

x

x

WRITE

Match the graphs using the information in the summary above.

i is a parabola; it matches graph b . i i is a cubic; it matches graph e . i i i is a hyperbola (the graph is in opposite quadrants); it matches graph c . iv is a truncus (the graph is in adjacent quadrants); it matches graph a . v

112

is a square root function; it matches graph d .

Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad

WORKED EXAMPLE 33

x

0

1

2

3

4

5

y

0

2.5

3.54

4.33

5

5.59

The data in the above table exactly fit one of these rules: y = ax 2 , y = ax 3 , y = a Plot the values of y against x.

b Select the appropriate rule and state the value of a.

THINK

WRITE/DRAW

a Plot the values of y against x.

a

y 5 4 3 2 1 0

b

a a , y= or y = a x . x x2

1 2 3 4 5

x

b Assume that y = a x .

1

Study the graph. It appears to be a square root curve. Write the appropriate rule.

2

To find the value of a: select any pair of corresponding values of x and y. (Since we need to take a square root, the best to choose is the one where x is a perfect square.)

Using (1, 2.5):

3

Substitute selected values into the rule and solve for a.

2.5 = a 1 =a×1 a = 2.5

4

We need to make sure that the selected rule is the right one. Replace a with 2.5 in the rule. Substitute the values of x from the table into the formula and check if you will obtain the correct values of y.

Verifying: y = 2.5 x

As the values of y obtained by using the rule match those in the table, the choice of model is correct.

The rule that fits the data is y = a x , where a = 2.5.

5

6

y = 2.5 0 =0 (2, 3.54): y = 2.5 2 = 3.54 (3, 4.33): y = 2.5 3 = 4.33 y = 2.5 4 (4, 5): =5 (5, 5.59): y = 2.5 5 = 5.59 (0, 0):

Chapter 2

Functions and transformations

113

The process of fitting a straight line to a set of points is often referred to as regression. Statistical data is easiest to deal with in linear form. If the data is not linear, then a linear relationship can still be found by transforming the x scale. A regression line can then be fitted. m For example, y = x + c is a hyperbola. However, if we substitute X for 1 , the rule becomes x linear: y = mX + c. The graph of y versus X will be a straight line with a gradient of m and a y-intercept of c. These values (m and c) can then be established from the graph and thus the hyperbolic model can be determined. Note: In a quadratic relationship, X is substituted for x2; in a cubic relationship, X is substituted for x3. WORKED EXAMPLE 34

It is believed that, for the data in the table below, the relationship between x and y can be modelled by y = aaxx 2 + bbxx + c. x

0

1

2

y

4

5.3

8.6

3

4

5

14.8

23

34.4

a Plot the values of y against x.

THINK a

114

b Calculate the values of a, b and c (correct to 3 decimal places) and write the equations. WRITE/DISPLAY

1

On the Statistics page, enter x-values into column A and y-values into column B. Label each column x and y respectively.

2

To graph the data, tap: • SetGraph • Setting Set: Type: Scatter XList: main\x YList: main\y Freq: 1 Mark: square • Set • y Use 6 if the window needs adjusting.

Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad

b

1

Tap: • Calc • Quadratic Reg Set: XList: main\x YList: main\y • OK

2

Interpret the variables given on the screen.

a = 1.252 b = −0.222 c = 4.096 ∴ y = 1.252x 1.252 2 − 0.222x 0.222 + 4.096 Correct to 3 decimal places.

If the relationship between the variables is not given, we have to make an assumption of a model from the graph of the data. We then have to transform the data according to our assumption. If the assumption was correct, the transformed data, when plotted, will produce a perfectly straight, or nearly straight, line. Note: In this section we will consider only the rules of the type y = ax2 + b, y = ax3 + b, a y = + b and so on (we will not allow for a horizontal translation), so that the appropriate x substitution can be made. WORKED EXAMPLE 35

x

1

2

3

4

5

6

y

35

21

16

12

11

10

Establish the rule connecting x and y that fits these data. THINK 1

Using either graph paper or a CAS calculator, plot y against x.

WRITE/DRAW y 35 30 25 20 15 10 5 0

2

The scatterplot appears to be a hyperbola. Write the appropriate formula (remember that we do not consider horizontal translations in this section).

1 2 3 4 5 6

Assumption: y =

x

a +b x

Chapter 2

Functions and transformations

115

3

4

Check your assumption: prepare a new 1 table by replacing values of x with x (leave the values of y unchanged). 1 Plot y against . x

1 x

1

y

35

116

Comment on the shape of the graph.

6

If we replace

1 with X, the rule x becomes y = aX + b, which is the equation of the straight line, where a is the gradient and b is the y-intercept. These (a and b) can be found from the graph as follows: draw in the line of best fit.

21

0.33

0.25

16

12

0.2 11

0.17 10

y 35 30 25 20 15 10 5 0

5

0.5

0.2 0.4 0.6 0.8

1

1 x–

The graph is very close to a straight line, therefore the assumption of a hyperbolic model is correct. y 35 30 25 20 15 10 5 0

0.2 0.4 0.6 0.8

1

X

y2 − y1 x2 − x1

7

Write the formula for the gradient.

m=

8

Select any 2 points on the line.

Using (0.17, 10) and (1, 35):

9

Substitute the coordinates of the points into the formula and evaluate.

m=

10

Write the value of a.

Since a is the gradient, a = m = 30.12.

11

Write the general equation of the straight line.

y = mx + c

12

Substitute the value of m and the coordinates of any of the 2 points, say (1, 35) into the equation.

35 = 30.12 × 1 + c

13

Solve for c. (Alternatively, read the y-intercept directly from the graph.)

14

State the value of b.

35 = 30.12 + c c = 35 − 30.12 = 4.88 Since b is the y-intercept, b = c = 4.88 88.

15

Substitute the values of a and b into a y = + b to obtain the rule that fits x the given data.

35 − 10 1 − 0.17 25 = 0.883 = 30.12

The rule for the given data is: y=

30.12 + 4.888 x

Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad

REMEMBER

1. 2. 3. 4.

Modelling is the process of finding the rule that fits the given data. The rule itself is called a mathematical model. The best way to start modelling is to produce a scatterplot of the original data. Use the scatterplot of the data to make an assumption of the model of the relationship. a It should be of the type y = ax2 + b, y = ax3 + b, y = + b and so on. To test the x assumption, transform the data accordingly. If the assumption is correct, the transformed data when plotted will produce a straight, or nearly straight, line. 5. To find the values of a and b in the model, draw a line of best fit; a is the gradient of the line and b is the y-intercept. EXERCISE

2J

Modelling 1

WE32 Match each of the graphs with the appropriate model:

i y = ax2 + b a iii y = + b x

ii y = ax3 + b a iv y = 2 + b x

v y=a x +b a

b

y

c

y

x

x

d

y

e

y

x

y

x x

2 eBoo k plus eBook Digital doc

Spreadsheet 076 Modelling

WE33 The data in each of the tables below exactly fit one of these rules: y = ax2, y = ax3,

a a ,y = or y = a x . For each set of data, plot the values of y against x and draw the x x2 graph. Select the most appropriate rule, and find the value of a. y=

a

b

x

−3

−2

−1

0

1

2

3

y

−8.1

−2.4

−0.3

0

0.3

2.4

8.1

x

−2

−1

0

1

2

3

y

−24

−6

0

−6

−24

−54

Chapter 2

Functions and transformations

117

c

−5

x

d

e

f

−2

−1

1

2

5

2

2

0.5

0.08

y

0.08

0.5

x

0

0.5

1

1.5

2

y

0

1.13

1.6

1.96

2.26

x

1

2

4

5

y

5

2.5

1.25

1

x

−3

−2

−1

0

1

0

−1.5

y

40.5

12

1.5

10 0.5 2 −12

a + b? x2 iii y

3 MC Which of the graphs below could be modelled by y = i

ii

y

y

x

x

iv

v

y

y x

x

A i only D i , i i and i v 4

x

B i , i i and i i i E i , i v , and v

C i v and v

WE34 It is believed that for the data in the table below, the relationship between x and y can be modelled by y = ax2 + b.

x

0

1

y

−3.2

−1

2

3

4

5

4.9

14.5

29

46.8

a Plot the values of y against x. b Plot the values of y against x2 and draw the line of best fit. c Find the values of a and b and hence the equation describing the original data. 5 The table below shows the values of 2 variables, x and y. x

−4

y

−28

−2

0

−13.5

−12.5

2 −10

4

6

4.3

41

Establish the mathematical model of the relationship between the variables, if it is known that it is of the form y = ax3 + b. 6 The table below shows the results, obtained from an experiment, investigating the frequency of a sound, ff, and the length of the sound wave, λ.

λ f

118

0.3 1130

0.5 680

1

3

5

8

10

340

110

70

40

35

Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad

a Plot f against λ. b From the following relationships select the one which you think is suggested by the plot: a f = aλ2, f = , f = a λ . λ 1 c Based on your choice in part b, plot f against either λ2, or λ , draw in the line of best λ fit and use it to find the rule that connects the 2 variables. 7 For her science assignment, Rachel had to find the relationship between the intensity of the light, II, and the distance between the observer and the source of light, d. From the experiments she obtained the following results. d

1

I

270

1.5

2

120

2.5

68

3

43

30

3.5 22

4 17

a Use a graphics calculator to plot the values of I against d. What form of relationship does the graph suggest? b Nathan (Rachel’s older brother) is a physics student. He tells Rachel that from his studies a he is certain that the relationship is of the type I = 2 . Use this information to help d Rachel to find the model for the required relationship. 8 WE35 The table below gives the values of 2 variables, x and y. Establish the rule, connecting x and y, that fits these data. x

0

1

3

5

7

9

y

4

7

9

11

12

13

9 Joseph is a financial adviser. He is studying the prices of shares of a particular company over the last 10 months. Months

1

2

3

4

5

6

7

8

Price, $

6.00

6.80

7.45

8.00

8.50

8.90

9.30

9.65

9 10.00

10 10.30

a Represent the information graphically. b Establish a suitable mathematical model, which relates the share price, P, and the number of the month, m. c Use your model to help Joseph predict the share price for the next 2 months. eBoo k plus eBook Digital doc

Investigation Goal accuracy

Chapter 2

Functions and transformations

119

SUMMARY Graphs of the power functions

Name Parabola

Equation y = a(x − b)2 + c

Basic shape y

(b, c) 0

Cubic

y = a(x − b)3 + c

y=

a +c x−b

or y = a( x − b) − 1 + c Truncus

a +c ( x − b) 2 or y = a( x − b) − 2 + c

If a > 0 y≥c If a < 0 y≤c

Turning point at (b, c)

R

R

Stationary point of inflection at (b, c)

R \{b}

R \{c}

Horizontal asymptote y = c, vertical asymptote x=b

R \{b}

If a > 0 y>c If a < 0 y 0 y≥c If a < 0 y≤c

End point at (b, c)

c x

0 b y

y=

y=a x−b +c or y = a( x − b) + c

Special feature

(b, b, c) c x

y

c 0

Square root

Range

R

x

y 0

Hyperbola

Domain

b

x

y

1 2

0

(b, c)

x

• The equation for any graph y = f (x) above can be written in the general form: y = af (x − b) + c. This form can be used to describe transformations of all of the functions considered. • For all of the above functions: 1. a is the dilation factor: it dilates the graph from the x-axis. 2. When an equation for these types of graphs is put into its general form of y = af (x − b) + c, the horizontal dilation can be described in terms of a vertical dilation. 3. If a < 0, the basic graph is reflected in the x-axis. 4. f (b − x) or f (−x + b) is the reflection of f (x + b) in the y-axis. 5. b translates the graph b units along the x-axis (to the right if b > 0, or to the left if b < 0). 6. c translates the graph c units along the y-axis (up if c > 0, or down if c < 0).

120

Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad

• To put equations into general form: 7. If the coefficient of x is a number other than 1, to find the value of b and a, the equation should be transposed to make the coefficient of x equal to 1. For example, y = (3x + 5)2 + 4 = [3( x + 35 )]2 + 4 = 32 ( x + 35 )2 + 4 = 9( x + 35 )2 + 4 Hence, a = 9, b =



5 . 3

The absolute value function

• y = |x| means •

y = x, if x ≥ 0 and y = −x, if x < 0 y y = |x| x

Domain: R Range: R+ ∪ {0} • To sketch y = | f (x)|: 1. Sketch the graph of y = f (x). 2. Reflect the portion of the graph that is below the x-axis in the x-axis. Or: 1. Express the function in hybrid form with specific domains where the absolute value expression is positive and negative. 2. Sketch each rule for the specified domain. • For functions of the form y = a|| f (x)| + c, a and c have the same impact on the graph of the absolute value function, as on the graphs of all other functions discussed in this section. Transformations with matrices

• The use of matrices to map transformations of points and equations can be summarised as follows, where (x′, y′) is the image of the point (x, y) under the transformation.   x  T      y    x  T      y    x  T      y    x  T      y 

 x'   −1 = =  y'   0  x'   1 = =  y'   0

0 x  −x    =   represents a reflection in the y-axis. 1 y  y  0  x

 x =  −  represents a reflection in the x-axis.    y   x'   a 0   x   ax  = =    =   represents a dilation of a factor of a from the y-axis.  y'   0 1   y   y   x'   1 0   x   x  = =    =   represents a dilation of a factor of a from the x-axis.  y'   0 a   y   ay  −1   y 

• Transformations can be combined to represent more than one transformation. For example,  4x + 2   x '  4 0   x   2    = + =     0 1      −y  y' y 3 + 3      2       2  −

∴ x′ = 4x + 2 −y y′ = +3 2

Chapter 2

Functions and transformations

121

1

describes the following: dilation by a factor of 4 from the y-axis, a dilation by a factor of 2 from the x-axis, reflection in the x-axis, a horizontal translation of +2 and a vertical translation of +3. Sum and difference of functions

• For the graph of the sum/difference function, dom ((f (x) ± g(x)) = dom f (x) ∩ dom g(x). The graph of the sum/difference function can be obtained by using the addition of ordinates method. • For the product function, dom ((ff (x)g(x)) = dom f (x) ∩ dom g(x). Some features of the graph of the product function are as follows: ° the x-intercepts of f (x)g(x) occur where either f (x) or g(x) have their x-intercepts ° f (x)g(x) is above the x-axis where f (x) and g(x) are either both positive or both negative ° f (x)g(x) is below the x-axis where one of the functions f (x) or g(x) is positive and the other is negative. Composite functions and functional equations

• For the composite function f (g(x)) to be defined, the range of g must be a subset of the domain of f. f Furthermore, if f (g(x)) is defined, the domain of f (g(x)) equals the domain of g(x). • Equations involving algebra of functions, for example f (2x (2 ) = 2f 2 (x), are generally tested to determine if they are true for particular functions. ° To determine if an equation is true for a particular function, consider the LHS and RHS of the equation separately to determine if the equation holds true for all values of x. ° Alternatively, you may find a particular x-value for which the equation does not work; that is, a counterexample. ° These types of equations can be investigated by defining the functions on the CAS calculator and then testing the algebraic function equation. Modelling

• Modelling is the process of finding the rule (mathematical model) that fits the given data. • To model: 1. Plot the original data on graph paper or use a CAS calculator. 2. Make an assumption of the model. 3. Transform the data in accordance with your assumption. 4. Check the assumption by plotting the transformed data (if correct, the graph will be a straight or nearly straight line). 5. Draw in a line of best fit. 6. Find the equation of the line (y = mx + c). 1 7. Replace x in the equation with the transformed variable (for example, x2, ). x

122

Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad

CHAPTER REVIEW SHORT ANSWER

1 For the function x − 3)2 − 4: a state the coordinates of the turning point b state the domain and range c sketch the graph. a 2 The graph of y = + c has a vertical x−b asymptote at x = 2 and a horizontal asymptote at y = −1. a Find the values of b and c. This graph then undergoes the following transformations: • reflection in the x-axis • dilation by a factor of 3 from the x-axis • horizontal shift of 2 units right. b If the intersection of the two graphs is at (m, 2), find the value of m. c Hence find the equation of the transformed graph. 6x − 5 3 Express the function f ( x ) = in the form 3x + 1 b a+ . Hence describe the transformations x+c required to produce this curve from the graph of 1 y= . x 4 The graph of a cubic function has a stationary point of inflection at (1, 1). It cuts the y-axis at y = 4. Find the equation of the graph. 1 5 The graph of y = was dilated by the factor of x 4 from the x-axis, reflected in the x-axis and then translated 2 units to the left and 1 unit down. a State the equation of the asymptotes. b State the domain and range. c State the equation of the new graph. d Sketch the graph. 6 a State the changes necessary to transform the 1 y graph of y = 2 into x the one shown. b Find the equation of the graph.

−2

−1 − –23

x

7 The domain of a truncus is R \{1}, the range is (−∞, 2) and the graph cuts the y-axis at y = −3. Find the equation of the function. 8 The basic square root curve was reflected in both axes and then translated so that its intercepts at the axes were (0, 1) and (−5, 0). Find the size and the direction of the translations; hence, find the equation of the new graph. 2 − 2, 9 a Sketch the graph of y = 2 − ( x + 2)2 clearly showing the coordinates of the cusps, the intercepts with the axes and the position of the asymptotes. b State the domain and range of the graph in a. 10 The graph of ff: [−5, 1] → R where f (x ( ) = x3 + 6x 6 2 + 9x is shown. y

−3

0

x

a On the same set of axes sketch the graph of y = /f (x)) /. b State the range of the function with the rule y = /f (x)) / and domain [−5, 1]. [© VCAA 2006]

11 The point (−1, 3) undergoes a translation given by a  the matrix   to (2, 0). Find a and b and describe b  the transformations involved. 12 The point (1, 2) undergoes a series of a 0   transformations given by the matrices   and 0 b   1  then   to (−7, 4). 2 a Find the values of a and b. b Find the image under the transformations of: i y=2 x ii y = x3 + x 13 A point on a curve ((x, y) undergoes a a 0  transformation descibed by   to (x', y'), 0 2  where a is a real constant such that a > 0.

Chapter 2

Functions and transformations

123

a Find the values of x and y in terms of a, x' and y'. b If the point is on the curve y = 2x 2 2 − x, find the image of the curve in terms of a under this transformation. c If the point (3, 6) is on the transformed curve, find the value(s) of a and hence the rule of this image. 14 The graph of the function f : (−2, 1) → R, f (x ( ) = x3 + 2x 2 2 is shown below. y

f x) = x3 + 2x f( 2 2

2 1

(−2, 0)

−2 −1.5 −1 −0.5 0 −1

0.5

1

x

Let g(x) = f (x) + 1, and sketch this graph on the same set of axes. Hence, sketch ((f + g)(x). 15 The graph of the function f:( f −2, 2), f (x ( ) = /x / is shown below. (−2, 2)

f x) =x y f( 2 (2, 2) 1

0 −2 −1 −1

1 2 x

−2

Let g(x) = −f (x) + 1 and sketch the graph of this function on the same set of axes. Hence, sketch f (g(x)). 16 Let f (x ( ) = x2 determine which of the following relationships are true. a f (x) − f (−y) = f (x) − f (y) b −f (x) − f (y) = f (x) + f (y) c f (−x) + f (−y) = f (x) + f (y) d f (x) − f (−y) = f (x) + f (y) 17 The data in the table below exactly fit one of these a models: y = ax3, y = 2 or y = a x . x x

2

y

25

4

5

10

6.25

4

1

20 0.25

25 0.16

a Plot the values of y against x and use the scatterplot to choose a suitable model. b Plot the values of y against either x3, 1 or x x2 (depending on your choice in part a ). Did you choose the right model? Explain your answer. c Find the value of a.

124

1 The equation of a parabola is given by y = m − 2(x 2( + 3)2, where m > 0. The increase in m will result in: A the graph being thinner B the graph being wider C the increase of the domain D the increase of the range E the graph being shifted further to the right 2 The coordinates of the turning point of the parabola y = 2(3x + 6)2 − 3 are: A (6, −3) B (−6, −3) C (2, −3) − − − D ( 2, 3) E ( 2, 3)

(1, 3)

3

MULTIPLE CHOICE

bx − 3)3 + 1 is dilated in the 3 The graph of y = 23 (bx y direction by the factor of: 3 C 2b B 23 b A 23 3 2 3 D b E 3b3 4 The graph of y = 2 − (3 + 4x 4 )3 has a stationary point of inflection at: A ( 43 , 2)

B (



3 − , 2) 4 − 3 , 2) 4

C (−3, 2) D ( E (−3, −2) 2 5 If f ( x ) = + 1, then f (x ( ) + 2 will have: x A the horizontal asymptote y = 2 B the horizontal asymptote y = 1 C the horizontal asymptote y = 3 D the vertical asymptote x = 2 E the vertical asymptote x = 1 6 The equation of the graph shown is likely to be: −2 y A y= −1 x−2 −2 B y =1 x+2 −2 x −2 −1 C y= −2 x +1 2 −1 D y= x+2 −2 E y= −1 x+2 7 If the graph of y = 1 is reflected in the y-axis, x translated 43 units to the right and 2 units up, the resulting graph would have the equation: 1 3 A y= + x−2 4 3 +2 B y= 4x

Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad

1 3 − 4x −1 +2 D y= 4x − 3 1 +2 E y= 3 − 4x C y = 2−

8 Which of the following is not true for the graph at right? y A The vertical asymptote is x = 2. B The horizontal asymptote is y = −4. x 2 C The domain is R\{2}. D The range is R\{−4}. −4 E The value of the y-intercept is greater than −4. 9 To obtain the graph shown, we need to: A translate the graph of y 1 y = 2 one unit to the left x and reflect in the x-axis x −1 B translate the graph of 1 y = 2 one unit to the left x and reflect in the y-axis 1 C translate the graph of y = one unit to the x2 left and reflect in both axes 1 D translate the graph of y = 2 one unit to the x left and dilate it in the x direction E none of the above 10 The equation of this graph is of the form: A y = a x − m + n, a > 0 B y = a m − x + n, a > 0

y (m, n) x

C y = a x − m + n, a < 0 D y = a m − x + n, a < 0 E y = a x + m + n, a < 0 11 The equation of this graph could be: A y=d− a−x

y d

B y= x−a +d C y=d− x−a D y = c− a − x E y= a− x −c

b a c

x

12 The domain of the function f ( x ) = 2 x − 1 + 3 is: A [1, ∞) B [3, ∞) C [0.5, ∞) D [2, ∞) E [−1, ∞) 1 13 The range of the function y = − 3 − − 2 is: x2 A (1, ∞) B (−∞, −5] C [3, ∞) D (−∞, 2) − − E ( ∞, 2] 14 The equation of the graph shown in the diagram below is best described by: y A y = |x + 2| + 2 2 B y = 2 − |x + 2| C y = |2 − x| + 2 x −2 D y = 2 − |2 − x| E y = |x + 2| − 2 15 The value(s) of k for which |2k + 1| = k + 1 are: − −2 2 B A 0 only C 0 or 3 3 − 1 2 E or 0 D 0 or 2 3 [© VCAA 2006] 16 Under the transformation T : R2 → R2 of the plane   x   2 0    x   1  defined by T    =      +   , the   y   0 1    y   − 3  image of the curve y = | x | has the equation: A B C D E

y=

1 2

x −1 + 3

1 2

x −1 − 3

y = | 22x + 1| − 3 y = 2 | x + 1| − 3 y = 2 | x − 1| + 3 y=

17 If g:[2, 4] → R, where g(x ( ) = x2(x (x ( − 3), and h:(0, 3] → R, where h(x ( ) = 3 − x, then the (x function f (x ( ) such that f (x ( ) = g(x ( ) × h(x (x ( ) is defined (x by the rule: A f: f R → R, where f (x) = −x2(x − 3)2 B ff: (0, 3] → R, where f (x) = −x2(x − 3)2 C ff: [2, 3] → R, where f (x) = −x2(x − 3)2 D ff: (0, 4] → R, where f (x) = −x2(x − 3)2 E ff: [2, 3] → R, where f (x) = −x2(x − 3) (3 + x) 18 Given the function f ( x ) = x − 3 then f ° h could exist if h(x ( ) was defined as: (x A h(x) = −x + 2 B h(x) = x2 − 3 C h(x) = x3 D h(x) = x2 + 3 E h( x ) = x − 2

Chapter 2

Functions and transformations

125

19 The data in the following table exactly fit one of these models: y = ax2, y = ax3 or y = a x . x

1

2

3

y

0.3

2.4

8.1

The value of a is: A 2.4 C 2.7 E 0.3

4 19.2

B 1.2 D 0.9

20 For certain data the values of y were plotted against 1 and the line of best fit was drawn as seen on x the diagram below. The model that relates the variables x and y is: y A y = 20x 20 − 1 (1, 19) B y = 19x + 1 x −1 C y= 20 19 D y = −1 x (0.1, 1) 20 1 −1 E y= x– x

EXTENDED RESPONSE

1 The graph of y = f (x ( ) is shown at right. a Sketch the graph of each of the following functions on the same set of axes with the original graph and give the coordinates of the points A, B, C and D. i y = −f (x) ii y = f (−x) iii y = f (xx − 2) iv y = f (x) + 3 v y = 2f 2 (x) vi y = 1 − f (x + 1) b Maya, a fabric designer, wishes to use the curve of y = f (x) (red) to create a ‘wavy’ pattern as shown in the diagram at right. If she wants the waves to be 2 units apart vertically, suggest the best way she could alter the equation of y = f (x). (Remember a fabric has a fixed width!)

y

D(4, 6)

B C(2, 3) A 2 −3 −2 7 x

2 units apart

2 Consider the function f:R → R, f( f x) = (x − 1)2 (x − 2) + 1. a The coordinates of the turning points of the graph of y = f( f x) are (a, 1) 25 and (b, 27 ). Find the values of a and b. b Find the real values of p for which the equation f( f x) = p has exactly one solution. c For the following, k is a positive real number.  x i Describe a sequence of transformations that maps the graph of y = f (x) onto the graph of y = f   − 1.  k  x y = f − 1 ii Find the x-axis intercepts of the graph of in terms of k.  k  d Find the real values of h for which only one of EXAM TIP the solutions of the equation f (x + h) = 1 is positive. Students must ensure that they show [© VCAA 2004]

3 The graph of the function f: 1 f ( x ) = 2 is shown below. x (−0.5, 4)

y 4

(–2, –0.5)

∪ (0.5, 2) → R,

(0.5, 4)

3

[Assessment reports 1 and 2 2007 © VCAA]

f x) = 12 f(

2 (−2, 0.25)

126

x

1

−2.5 −2 −1.5 −1 −0.5 0

their working — if a question is worth more than one mark, students risk losing all available marks if only the answer is given and it is incorrect. The instructions at the beginning of the paper state that if more than one mark is available for a question, appropriate working must be shown. When students present working and develop their solutions, they should use conventional mathematical expressions, symbols, notation and terminology.

(2, 0.25) 0.5

1 1.5

2 2.5 x

Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad

The graph of y = f (x) is to be transformed to become the graph of y = f (2x (2 ) + 1. a Describe these two transformations. b Create matrices to represent these transformations. c Use these matrices to find the images of the points (1, 1) and (2, 0.25) under these transformations, and use these values to deduce the images of the points (−1, 1) and (−2, 0.25). d On the same axes, sketch this transformed function, showing the coordinates of the four points from c above. e Using any method, find a rule for this transformed function. 4 A proposed section of a ride at an amusement park is to be modelled on the curve y = 1 (600 x + 25 x 2 − x 3 ), 500 where y is the height (in metres) of the ride above ground level and x is the horizontal distance (in metres). The x-axis represents ground level. It will travel through a tunnel from A to C; B is the lowest point in the tunnel and D is the highest point on the ride. y

A

D

Ex

C B

a Find the horizontal distance from A to E. b Find the greatest depth below ground level and the maximum height above ground level that the rollercoaster will reach in this section (correct to 2 decimal places). c Describe the impact that a dilation by a factor greater than 1 from the x-axis would have on: i the maximum depth and maximum height from b ii the point at which the rollercoaster would emerge from the tunnel iii the gradient of the slope at this point. 5 Lena and Alex are planning to buy a new house. They’ve been watching the prices of 3-bedroom houses in a specific area, where they want to live, for the whole year. During each month they collected the data and then, at the end of the month, they calculated the average price for that month. The results of their calculations are shown in the table below. (The prices given are in thousands of dollars.) Month Price

1

2

3

4

5

6

7

8

9

10

11

12

240 248 255 261 266 271 273 274 275 274 272 270

a Plot the prices against the months. What model does the graph suggest? b If the model of the form y = a(x − b)2 + c is to be used for these data, what is (judging from the graph) the most suitable value for h? c Plot the values of y (the prices) against (x − b)2, where b is the value you’ve selected in part b . Comment on the shape of the graph. d Draw a line of best fit and find its equation. Hence, state the values of a and c in the model. e Write the equation of the model. f According to the Real Estate Institute, the property market is on a steady rise (that is, the prices are going up and are likely to rise further). Do the data collected by Lena and Alex support this theory? g Use the model to predict the average price for the next 2 months.

Chapter 2

Functions and transformations

127

h Lena and Alex were planning to spend no more than 250 000 for their new house. Several months ago the prices were in their range, but they could not find what they wanted. If the prices are going to behave according to our model, how long do they have to wait until the prices fall back into their range? 6 An eagle soars from the top of a cliff that is 48.4 metres above the ground and then descends towards unsuspecting prey below. The eagle’s height, h metres above the ground, at time t seconds can be modelled by a the equation h = 50 + , where 0 ≤ t < 25 and a is a constant. t − 25 a Find the value of a. b Find the eagle’s height above the ground after i 5 seconds ii 20 seconds. c After how many seconds will the eagle reach the ground? d Comment on the changes in speed during the eagle’s descent. e Sketch the graph of the equation. After 24 seconds, the eagle becomes distracted by another bird and reaches the ground exactly 2 seconds later. For this second part of the journey, the relationship between h and t can be modelled by the equation h = a(t − 24)2 + c. f Find the values of a and c. g Fully define the hybrid function that describes the descent of the eagle from the top of the cliff to the ground below. eBoo k plus eBook Digital doc

Test Yourself Chapter 2

128

Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad

eBook plus

ACTIVITIES

Chapter opener Digital doc

• 10 Quick Questions: Warm up with ten quick questions on functions and transformations. (page 57) 2A

Transformations of the parabola

Digital docs

• Spreadsheet 132: Investigate transformations. (page 59) • Spreadsheet 108: Investigate the quadratic function in power form. (page 61) 2B

The cubic function in power form

Digital docs

• Spreadsheet 015: Investigate the cubic function in power form. (page 65) • Spreadsheet 236: Investigate graphs of functions. (page 70) 2C

The power function (the hyperbola)

Tutorial

• WE 9 int-0522: Watch a worked example on sketching the graph of a hyperbola. (page 74) Digital docs

• Spreadsheet 051: Investigate the hyperbola. (page 72) • Spreadsheet 236: Investigate graphs of functions. (page 71) • WorkSHEET 2.1: Find the domain, range, coordinates of turning points and equations of asymptotes of various graphs. (page 78) • History of Mathematics: Investigate the history of major curves. (page 78) 2D

The power function (the truncus)

Digital doc

• Spreadsheet 236: Investigate graphs of functions. (page 85) 2E

The square root function in power form

Tutorial

• WE 16 int-0523: Watch a worked example on implied domain and range. (page 88) Digital doc

• Spreadsheet 236: Investigate graphs of functions. (page 92) 2F

The absolute value function

Tutorials

• WE 20 int-0524: Watch a worked example on sketching the graph of an absolute value function. (page 93) • WE 21 int-0525: Watch a worked example on expressing an absolute value function as a hybrid function. (page 94)

Digital docs

• WorkSHEET 2.2: Identify transformations, state domain and range, sketch graphs of power functions and absolute value functions. (page 96) • Spreadsheet 002: Investigate graphs of absolute value functions. (page 96) 2G

Transformations with matrices

Interactivity

• Transformations with matrices int-0247: Consolidate your understanding of using matrices to transform functions. (page 96) Tutorial

• WE 25 int-0526: Watch how to use matrices to determine the resultant equation after transformations. (page 100) 2H

Sum, difference and product functions

Tutorial

• WE 27 int-0257: Watch how to use addition of ordinates to sketch the sum of two functions. (page 104) 2J

Modelling

Digital docs

• Spreadsheet 076: Investigate modelling with functions. (page 117) • Investigation: Goal accuracy. (page 119) Chapter review Digital doc

• Test Yourself Chapter 2: Take the end-of-chapter test to test your progress. (page 128) To access eBookPLUS activities, log on to www.jacplus.com.au

Chapter 2

Functions and transformations

129

EXAM PRACTICE 1 SHORT ANSWER

15 minutes

1 The functions f g are graphed below. On the same axes sketch a graph of f + g. f

g

CHAPTERS 1 TO 2

1 −a x−b 1 C y= +b x+a 1 E y= +a x+b A y=

B y=

−1

+b a−x −1 D y= −a x−b

2 The graph sketched below is best represented by the rule:

0

1 mark

y

2 Write down the maximal domain of f : X → R where f ( x ) = x 2 − 3 .

a

1 mark

3 For y = (2x (2 − 3)2 − 1: a write down the y-coordinate of the turning point b determine the equation of the axis of symmetry. 2 marks

4 a Using the factor theorem, show that x + 2 is a factor of x3 − 7x − 6. b Given x3 − 7x − 6 = (x + 2)Q(x) where Q(x) is a quadratic factor, determine Q(x). 2 marks −2x 2

5 For what values of c does the graph of y = +c intersect the graph of y = −x2 + x − 2 at two distinct points? 2 marks

6 Sketch the graph of the function f :[−2,5) − R, f (x ( ) = | 3(x 3( − 3)2 − 10 |. 2 marks

MULTIPLE CHOICE

10 minutus

Each question is worth 1 mark. 1 Given a, b ∈ R, this graph could have the rule:

A B C D E

b

c

x

y = (xx – a)(xx – b)2(xx – c)2 y = (xx – a)(xx – c)2 y = (xx – a)3(xx – c)2 y = (xx – a)(xx – c) y = (xx – a)(xx – b)(xx – c)

3 The coefficient of the term in x4 in the expansion of 3 (2 x 2 − )5 is: x A −240 B −72 C −60 D 72 E 720 4 What is the equation of the horizontal asymptote of the graph of y = A y = −4 3 D y= 4

3x − 2 ? 4−x B y = −3

C y = −1

E y=3

5 The quadratic function f : D → R, f (x ( ) = 2(x 2( + 1)2 − 5 has a domain, D,, of [0, 3]. The range of f is: A [−1, 3] C [−5, 27] B R D [−3, 27] E [−5, ∞)

y

y=b x=a 0

130

x

6 The simultaneous linear equations 3x + ay = 12 and ax + 3y = 4a have infinitely many solutions for: A a ∈ [−3,3] B a=3 C a = 3 or a = −3 D a=9 E a=0

Maths Quest 12 Mathematical Methods CAS for the Casio ClassPad

EXTENDED RESPONSE

40 minutes

1 Severe tropical cyclone Vance devastated the town of Exmouth in northern WA in 1999. It produced a measured wind gust of 267 km/h, which is the highest wind speed ever recorded on mainland Australia. In order for it to be first categorised as a cyclone, its wind speed needed to exceed 119 km/h. Several years later, it is a peaceful day, with no wind detectable at 1 pm. By 3 pm, however, the wind speed is gusting to 200 km/h, and the residents know they are in trouble. a Let t be the time in hours after noon and v be the wind speed. Establish a linear model of the form v = at + b to represent the relationship between the wind speed and time. 2 marks b Using this model, determine to the nearest minute when the wind speed will be high enough for classification as a cyclone. 1 mark c Predict to the nearest minute when the cyclone will break the record for the highest wind speed ever recorded. 1 mark

d Explain why the linear model is unsatisfactory as a model for the cyclone’s behaviour. e

1 mark

i The wind speed actually peaks at 256 km/h at 5 pm. Use this data and the wind speed at 1 pm to help to create a quadratic model relating wind speed (v) to hours after noon (t). How well does this model match all the data provided?

2 marks

ii The wind speed actually returns to zero at 1 am the following morning, 12 hours later. Discuss how well this result matches the quadratic model. f

1 mark

i Establish a second quadratic model that exactly matches the following data: Time

Wind speed

1 pm

0 km/h

5 pm

256 km/h

1 am

0 km/h 1 marks

ii Evaluate how well this model represents the relationship between time and wind speed. g It has been suggested that a cubic function would be a better model for the data. Use all the data provided in order to establish a cubic model.

1 mark 2 marks

1020 1010 1000 990 980 970 960 950 0 km

Eye

Pressure Wind speed

175 150 125 100 75 50 25

Wind speed (km/h)

Pressure (hPa)

h The graph below shows how air pressure in hectopascals (P hPa) and wind speed (v km/h) relate to 1 distance (x) across a cyclone. It has been suggested that a truncus (general form v ( x ) 2 ) could represent x the shape of the relationship between speed and distance across the cyclone.

500 km

i What is the equation to the vertical asymptote for this relationship? ii Determine a rule for the relationship between speed (v) and distance (x). iii Determine a rule for the relationship between air pressure (P) and distance (x).

1 mark 2 marks 2 marks

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Solutions Exam practice1

Exam practice 1

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