Unit 3

Transformations

Practice 3.2D Solutions

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

Pages 442 and 443, questions 2a to 2d, 3c, 3d, 4a, 5c, 7c, 8a and 8b.

Questions 2a to 2d, page 442

The base function is

y = \frac{1}{x}

.

The vertical asymptote is at

x = - 2

.

The horizontal asymptote is at

y = 0

The base function is

y = \frac{1}{x}

.

The vertical asymptote is at

x = 3

.

The horizontal asymptote is at

y = 0

The base function is

y = \frac{1}{x^{2}}

.

The vertical asymptote is at

x = - 1

.

The horizontal asymptote is at

y = 0

The base function is

y = \frac{1}{x^{2}}

.

The vertical asymptote is at

x = 4

.

The horizontal asymptote is at

y = 0

Questions 3c and 3d, page 442

Compare to the equation

y = \frac{a}{x - h} + k

The value of

a

2

. There is a vertical stretch about the

x

-axis by a factor of

2

h = 4

means there is a horizontal translation

4

units right.

k = - 5

means there is a vertical translation

5

units down.

There is a horizontal asymptote at

y = - 5

and a vertical asymptote at

x = 4

.

Domain:

\{x ∣ x \neq 4, x \in R\}

Range:

\{y ∣ y \neq - 5, y \in R\}

$x$ -intercept at $y = 0$	$y$ -intercept at $x = 0$
$\begin{array}{rcl} 0 & = & \frac{2}{x - 4} - 5 \\ 5 & = & \frac{2}{x - 4} \\ 5 (x - 4) & = & 2 \\ 5 x - 20 & = & 2 \\ 5 x & = & 22 \\ x & = & 4.4 \end{array}$	$\begin{array}{l} y = \frac{2}{0 - 4} - 5 \\ y = \frac{2}{- 4} - 5 \\ y = - 0.5 - 5 \\ y = - 5.5 \end{array}$

The value of

a

- 8

. There is a vertical stretch about the

x

-axis by a factor of

8

. There is a reflection in the

x

-axis.

h = 2

means there is a horizontal translation

2

units right.

k = 3

means there is a vertical translation

3

units up.

There is a horizontal asymptote at

y = 3

and a vertical asymptote at

x = 2

.

Domain:

\{x ∣ x \neq 2, x \in R\}

Range:

\{y ∣ y \neq 3, y \in R\}

$x$ -intercept at $y = 0$	$y$ -intercept at $x = 0$
$\begin{array}{rcl} y & = & - \frac{8}{x - 2} + 3 \\ 0 & = & - \frac{8}{x - 2} + 3 \\ - 3 & = & - \frac{8}{x - 2} \\ 3 & = & \frac{8}{x - 2} \\ 3 (x - 2) & = & 8 \\ 3 x - 6 & = & 8 \\ 3 x & = & 14 \\ x & = & \frac{14}{3} \end{array}$	$\begin{array}{l} y = - \frac{8}{0 - 2} + 3 \\ y = 4 + 3 \\ y = 7 \end{array}$

Question 4a, page 442

The asymptotes are

x = 4

and

y = 2

.

The

x

-intercept is

x = - 0.5

.

The

y

-intercept is

y = - 0.25

Question 5c, page 442

Write the equation of the function in the form

y = \frac{a}{x - h} + k

\begin{array}{l} y = \frac{- x - 2}{x + 6} \\ y = \frac{- 2}{x + 6} + \frac{- x}{x + 6} \\ y = \frac{- 2}{x + 6} + \frac{(- x + 6 - 6)}{x + 6} \\ y = \frac{- 2}{x + 6} + \frac{- (x + 6) + 6}{x + 6} \\ y = \frac{- 2}{x + 6} + \frac{- (x + 6)}{x + 6} + \frac{6}{x + 6} \\ y = \frac{- 2}{x + 6} - 1 + \frac{6}{x + 6} \\ y = \frac{- 2 + 6}{x + 6} - 1 \\ y = \frac{4}{x + 6} - 1 \end{array}

Asymptotes:	$x$ -intercept	$y$ -intercept
$\begin{array}{l} x = - 6 \\ y = - 1 \end{array}$	$\begin{array}{rcl} y & = & \frac{4}{x + 6} - 1 \\ 0 & = & \frac{4}{x + 6} - 1 \\ 1 & = & \frac{4}{x + 6} \\ x + 6 & = & 4 \\ x & = & - 2 \end{array}$	$\begin{array}{rcl} y & = & \frac{4}{x + 6} - 1 \\ = & \frac{4}{0 + 6} - 1 \\ = & \frac{2}{3} - 1 \\ = & - \frac{1}{3} \end{array}$

Question 7c, page 443

Read the asymptotes from the graph. The horizontal asymptote indicates the vertical shift, or value of

k

.

The horizontal asymptote is at

y = 4

. Therefore,

k = 4

.

The vertical asymptote is at

x = 2

. Therefore,

h = 2

.

Now, determine if there is a vertical stretch.

Use a point on the graph of the function to determine the vertical stretch.

y = \frac{a}{x - h} + k

Use point

(6, 6)

\begin{array}{l} 6 = \frac{a}{6 - 2} + 4 \\ 6 = \frac{a}{4} + 4 \\ 2 = \frac{a}{4} \\ 8 = a \end{array}

The equation of the transformed function is

y = \frac{8}{x - 2} + 4

Questions 8a and 8b, page 443

a = - 15, k = 6

Additional Notes

Use both points in

y = \frac{a}{x - 7} + k

to solve the system of two equations in two unknowns.

Use point

(10, 1)

\begin{array}{rcl} 1 & = & \frac{a}{10 - 7} + k \\ 1 - (\frac{a}{3}) & = & k \end{array}

Use point

(2, 9)

\begin{array}{rcl} 9 & = & \frac{a}{2 - 7} + k \\ 9 - (\frac{a}{- 5}) & = & k \\ 9 + (\frac{a}{5}) & = & k \end{array}

\begin{array}{rcl} 1 - (\frac{a}{3}) & = & 9 + (\frac{a}{5}) \\ - 8 & = & (\frac{a}{5}) + (\frac{a}{3}) \\ - 8 & = & (\frac{3 a}{15}) + (\frac{5 a}{15}) \\ - 8 & = & (\frac{8 a}{15}) \\ \frac{(- 8) (15)}{8} & = & a \\ - 15 & = & a \end{array}

Use

a = - 15

and the point

(10, 1)

to determine the value of

k

\begin{array}{rcl} 1 & = & \frac{- 15}{10 - 7} + k \\ 1 - (\frac{- 15}{3}) & = & k \\ 1 + 5 & = & k \\ 6 & = & k \end{array}

L3.2 D3 Practice Solutions

Unit 3

Transformations

Practice Solutions

Practice 3.2D Solutions

Questions 2a to 2d, page 442

Questions 3c and 3d, page 442

Question 4a, page 442

Question 5c, page 442

Question 7c, page 443

Questions 8a and 8b, page 443