Unit 3

Transformations

Practice 3.2A Solutions

Use these solutions to correct your work. When finished, give yourself a grade using the Practice Assessment rubric.

Pages 72 to 74, questions 2a to 2d, 4a to 4d, 5b, 5d, 5f, 6a to 6d, 10a, 10d, and 12a to 12d.

Questions 2a to 2d, page 72

Vertical stretch about the

x

-axis by a factor of

7

Horizontal translation

9

units to the right
Domain:

\{x ∣ x \geq 9, x \in R\}

Range:

\{y ∣ y \geq 0, y \in R\}

Reflection in the

y

-axis
Vertical translation

8

units up
Domain:

\{x ∣ x \leq 0, x \in R\}

Range:

\{y ∣ y \geq 8, y \in R\}

Reflection in the

x

-axis
Horizontal stretch about the

y

-axis by a factor of

5

Domain:

\{x ∣ x \geq 0, x \in R\}

Range:

\{y ∣ y \leq 0, y \in R\}

Vertical stretch about the

x

-axis by a factor of

\frac{1}{3}

Horizontal translation

6

units to the left
Vertical translation

4

units down
Domain:

\{x ∣ x \geq - 6, x \in R\}

Range:

\{y ∣ y \geq - 4, y \in R\}

Additional Notes

Refer to

y = a f (b (x - h)) + k

to help determine the transformations.

\begin{matrix} a = 7 \\ b = 1 \\ h = 9 \\ k = 0 \end{matrix}

Because

a = 7

, there is a vertical stretch about the

x

-axis by a factor of

7

. All the

y

-values will be multiplied by

7

.

Because

h = 9

, there is a horizontal shift to the right

9

units.

The domain of

y = \sqrt{x}

is affected by the horizontal translation. The vertical stretch does not affect the range.

Refer to

y = a f (b (x - h)) + k

to help determine the transformations.

\begin{array}{l} a = 1 \\ b = - 1 \\ h = 0 \\ k = 8 \end{array}

Because the

b

-value is negative, there is a reflection in the

y

-axis. All the

x

-values will change to the opposite sign. There is no horizontal stretch because

|b| = 1

.

Because

k = 8

, there is a vertical shift

8

units up.

The range of

y = \sqrt{x}

is affected by the vertical translation.

The domain of

y = \sqrt{x}

is affected by the reflection in the

y

-axis.

Refer to

y = a f (b (x - h)) + k

to help determine the transformations.

\begin{array}{l} a = - 1 \\ b = 0.2 \\ h = 0 \\ k = 0 \end{array}

Because

a = - 1

, there is no vertical stretch, but the graph is reflected in the

x

-axis. All

y

-values will change to the opposite sign.

Because the value of

b

0.2

, or

\frac{1}{5}

, there is a horizontal stretch about the

y

-axis by a factor of

5

.

Only the range of

y = \sqrt{x}

is affected by the vertical reflection in the

x

-axis.

Refer to

y = a f (b (x - h)) + k

to help determine the transformations.

\begin{array}{l} a = \frac{1}{3} \\ b = 1 \\ h = - 6 \\ k = - 4 \end{array}

Because

a = \frac{1}{3}

, there is a vertical stretch about the

x

-axis by a factor of

\frac{1}{3}

. All

y

-values will be multiplied by

\frac{1}{3}

.

Because

h = - 6

, there is a horizontal shift

6

units to the left.

Because

h = - 4

, there is a vertical shift

4

units down.

Both the domain and the range of

y = \sqrt{x}

are affected by the horizontal and vertical translations.

Questions 4a to 4d, page 73

y = 4 \sqrt{x + 6}

y = \sqrt{8 x} - 5

y = \sqrt{- (x - 4)} + 11

y = \sqrt{- x + 4} + 11

y = - 0.25 \sqrt{\frac{1}{10} x}

Additional Notes

Vertical stretch about the

x

-axis by a factor of

4 \to a = 4

Horizontal translation

6

units left

\to h = - 6

\begin{array}{l} y = a \sqrt{x - h} \\ y = 4 \sqrt{x - (- 6)} \\ y = 4 \sqrt{x + 6} \end{array}

Horizontal stretch about the

y

-axis by a factor of

\frac{1}{8} \to b = 8

Vertical translation

5

units down

\to k = - 5

\begin{array}{l} y = \sqrt{b x} + k \\ y = \sqrt{8 x} + (- 5) \\ y = \sqrt{8 x} - 5 \end{array}

Horizontal reflection in the

y

-axis

\to b

is negative
Horizontal translation

4

units right

\to h = 4

Vertical translation

11

units up

\to k = 11

y = \sqrt{- b (x - h)} + k

y = \sqrt{- (x - 4)} + 11

y = \sqrt{- x + 4} + 11

Vertical stretch about the

x

-axis by a factor of

0.25 \to |a| = 0.25

Vertical reflection in the

x

-axis

\to a

is negative
Horizontal stretch about the

y

-axis by a factor of

10 \to b = \frac{1}{10}

\begin{array}{l} y = - a \sqrt{b x} \\ y = - 0.25 \sqrt{\frac{1}{10} x} \end{array}

Questions 5b, 5d, and 5f, page 73

The domain is

\{x ∣ x \geq - 1, x \in R\}

.
The range is

\{y ∣ y \geq 0, y \in R\}

The domain is

\{x ∣ x \leq 2, x \in R\}

.
The range is

\{y ∣ y \leq 1, y \in R\}

The domain is

\{x ∣ x \leq - 2, x \in R\}

.
The range is

\{y ∣ y \geq - 1, y \in R\}

Additional Notes

r (x) = 3 \sqrt{x + 1}

corresponds to

r (x) = a \sqrt{x - h}

, where

a = 3

and

h = - 1

. There was a vertical stretch about the

x

-axis by a factor of

3

and a horizontal translation

1

unit left.

y - 1 = - \sqrt{- 4 (x - 2)}

can be written as

y = - \sqrt{- 4 (x - 2)} + 1

, which corresponds to

y = a \sqrt{b (x - h)} + k

, where

a

is negative. Therefore, there is a vertical reflection in the

x

-axis.

The value of

b

is negative, so there is a horizontal reflection in the

y

-axis.

|b| = 4

, so there is a horizontal stretch about the

y

-axis by a factor of

\frac{1}{4}

h = 2

, so there is a horizontal translation

2

units right.

k = 1

, so there is a vertical translation up

1

unit.

y + 1 = \frac{1}{3} \sqrt{- (x + 2)}

can be written as

y = \frac{1}{3} \sqrt{- (x + 2)} - 1

, which corresponds to

y = a \sqrt{b (x - h)} + k

, where

a = \frac{1}{3}

.

Therefore, there is a vertical stretch about the

x

-axis by a factor of

\frac{1}{3}

b

is negative, so there is a reflection in the

y

-axis.

h = - 2

, so there is a horizontal translation

2

units to the left.

k = - 1

, so there is a vertical translation down

1

unit.

Questions 6a to 6d, page 73

a = \frac{1}{4}

indicates a vertical stretch about the

x

-axis by a factor of

\frac{1}{4}

b = 5

indicates a horizontal stretch about the

y

-axis by a factor of

\frac{1}{5}

y = \frac{\sqrt{5}}{4} \sqrt{x}

and

y = \sqrt{\frac{5}{16} x}

a = \frac{\sqrt{5}}{4}

indicates a vertical stretch about the

x

-axis by a factor of

\frac{\sqrt{5}}{4}

b = \frac{5}{16}

indicates a horizontal stretch about the

y

-axis by a factor of

\frac{16}{5}

The three transformed graphs look the same, hence why it appears as though there is only one.

Questions 10a and 10d, page 74

y = - \sqrt{x + 3} + 4

Use the graph of

y = \sqrt{x}

and mapping to determine the equation of the transformed graph.

Because the transformed graph shows a vertical reflection in the

x

-axis, the value of

a

is negative. The point

(0, 0) \to (- 3, 4)

, which indicates a horizontal shift

3

units left. Therefore,

h = - 3

. It also indicates a vertical shift

4

units up. Therefore,

k = 4

. Apply those transformations to the point

(4, 2)

, and the result is

(1, 2)

. There are no vertical or horizontal stretches.

y = - 4 \sqrt{- (x - 4)} + 5

y = - 4 \sqrt{- x + 4} + 5

Use the graph of

y = \sqrt{x}

and mapping to determine the equation of the transformed graph.

There is a reflection in the

x

-axis, so all the

y

-values become the opposite sign. The value of

a

will be negative.

There is a reflection in the

y

-axis, so all the

x

-values will become the opposite sign. The value of

b

is negative.

(0, 0)

is shifted

4

units to the right and

5

units up.

(0, 0) \to (4, 5)

Therefore,

h = 4, k = 5

\begin{array}{l} y = - a \sqrt{- (x - h)} + k \\ y = - a \sqrt{- (x - 4)} + 5 \end{array}

If the stretch is vertical, use the image point

(0, - 3)

in the equation

y = - a \sqrt{- (x - 4)} + 5

to determine the value of

a

\begin{array}{rcl} y & = & - a \sqrt{- (x - 4)} + 5 \\ - 3 & = & - a \sqrt{- (0 - 4)} + 5 \\ - 3 & = & - a \sqrt{4} + 5 \\ - 8 & = & - a \sqrt{4} \\ - 8 & = & - 2 a \\ 4 & = & a \end{array}

The vertical stretch factor would be

4

. The equation is

y = - 4 \sqrt{- (x - 4)} + 5

y = - 4 \sqrt{- x + 4} + 5

.

If the stretch is horizontal, use image point

(0, - 3)

in the equation

y = - \sqrt{- b (x - 4)} + 5

to determine the value of

b

\begin{array}{rcl} - 3 & = & - \sqrt{- b (0 - 4)} + 5 \\ - 8 & = & - \sqrt{- b (0 - 4)} \\ 8 & = & \sqrt{- b (0 - 4)} \\ 8 & = & \sqrt{4 b} \\ 8 & = & 2 \sqrt{b} \\ 4^{2} & = & {(\sqrt{b})}^{2} \\ 16 & = & b \end{array}

The horizontal stretch factor would be

\frac{1}{16}

. The equation would be

y = - \sqrt{- 16 (x - 4)} + 5

, which simplifies to

y = - 4 \sqrt{- (x - 4)} + 5

y = - 4 \sqrt{- x + 4} + 5

Questions 12a to 12d, page 74

The vertical stretch factor is

760

.
The value of

k

2 000

, so there is a vertical shift

2 000

units up.

Domain:

\{n ∣ n \geq 0, n \in R\}

Range:

\{Y (n) ∣ Y (n) \geq 2 000, Y (n) \in R\}

The

Y

values are the yield per hectare. In this case the minimum yield per hectare is

2 000 kg / ha

. The domain and range are not restricted with an upper limit. It indicates that the yield would increase without bound, which is not realistic. It is also not realistic for an unlimited amount of nitrogen to be applied to the crop.

L3.2 A4 Practice Solutions

Unit 3

Transformations

Practice Solutions

Practice 3.2A Solutions

Questions 2a to 2d, page 72

Questions 4a to 4d, page 73

Questions 5b, 5d, and 5f, page 73

Questions 6a to 6d, page 73

Questions 10a and 10d, page 74

Questions 12a to 12d, page 74