MAT3791-ABED_1: L7.2 B3 Practice Solutions

Question 4, page 102

C

A horizontal stretch about the

y

- axis by a factor of

b

is shown by

\frac{1}{b}

in the transformed function.

If

b = 6

, then

y = \sqrt{\frac{1}{6} x}

.

Question 5, page 102

D

Additional Notes

The graph of the function

y = \sqrt{x}

has been reflected horizontally in the

y

-axis; therefore,

x

-values become negative. Then, the function was shifted vertically

2

units down.

y = \sqrt{- x} - 2

Question 6, page 102

B

Additional Notes

The stretch factor on

x

does not change the domain. It is a multiple of the domain, which was

x \geq 0

. By multiplying each component of

x \geq 0

, the domain remains as

x \geq 0

. The range is different because each component of the original range has

8

added to it.

Question 2, page 286

D

Additional Notes

Compare to

y = a \sin b (x - c) + d

.

The phase shift is

c = \frac{π}{3}

.

The period is

\frac{2 π}{b}

, where

b = 2

.

\begin{array}{rcl} period & = & \frac{2 π}{2} \\ = & π \end{array}

The amplitude is

a = 3

.

Question 3, page 286

C

Additional Notes

The graph of the function

y = \sin x

is a horizontal shift of

y = \cos x

by

\frac{π}{2}

units right, or

\frac{2 π}{4}

.

A value of

c = \frac{π}{4}

in the sine function means another shift of

\frac{π}{4}

units right on the cosine function. In total,

\frac{2 π}{4} + \frac{π}{4} = \frac{3 π}{4}

.

Question 4, page 286

D

Additional Notes

Compare to

y = a \cos b (x - c) + d

Determine the amplitude.

\begin{array}{rcl} a & = & \frac{\max - \min}{2} \\ = & \frac{14 - 2}{2} \\ = & 6 \end{array}

Determine the vertical displacement.

\begin{array}{rcl} d & = & \frac{\max + \min}{2} \\ = & \frac{14 + 2}{2} \\ = & 8 \end{array}

Determine the period, and use it to determine the value of

b

.

The period is double the distance between a subsequent maximum and minimum.

\begin{array}{l} 8 - 3 = 5 \\ 5 \times 2 = 10 \end{array}

period = 10

\begin{array}{rcl} |b| & = & \frac{2 π}{period} \\ = & \frac{2 π}{10} \\ = & \frac{π}{5} \end{array}

The first maximum occurred at

x = 3

instead of

x = 0

as in

y = \cos x

.

\begin{array}{l} c = 3 \\ y = 6 \cos \frac{π}{5} (x - 3) + 8 \end{array}

Question 8, page 287

\frac{π}{2}

Additional Notes

On the graph of

g (x) = \sin x

there are zeros at

n π, n \in I

. The graph of

f (x) = \sin 2 x

is a horizontal stretch of

g (x)

by a factor of

\frac{1}{2}

, so zeros will occur at

\frac{n π}{2}, n \in I

. Each of these zeros is

\frac{π}{2}

apart.

Question 9, page 287

Examine the function using technology.

asymptotes:

θ = \frac{π}{2} + n π, n \in I

domain:

\{θ ∣ θ \neq \frac{π}{2} + n π, θ \in R, n \in I\}

range:

\{y ∣ y \in R\}

period:

π

Question 12, page 287

Based on the equation of the function, the amplitude is

3

, the period is

12

, and the vertical displacement is

5

.

\begin{array}{rcl} \max y & = & vertical diplacement + amplitude \\ = & 5 + 3 \\ = & 8 \end{array}

\begin{array}{rcl} \min y & = & vertical diplacement - amplitude \\ = & 5 - 3 \\ = & 2 \end{array}

There is no phase shift and

a

is negative, so a minimum occurs when

t = 0

. Other minimums will occur every

12

hours, so

0, 12

, and

24

hours represent minimums.

A maximum will occur half a period past each minimum, so maximums will occur at

6

and

18

hours.

Question 2, page 368

C

Compare to

y = 3^{b (x - h)} + k

.

Because

b = \frac{1}{4}

, there is a horizontal stretch about the

y

-axis by a factor of

4

.

Because

h = 5

, there is a horizontal translation

5

units right.

Because

k = - 2

, there is a vertical translation

2

units down.

Questions 6a and 6b, page 368

a.

y = 5^{x + 3} + 2

b.

y = - 0.5 {(2)}^{x - 1} - 4

Additional Notes

a.

Use points and mapping to determine the translations.

There are points on the graph of

y = 5^{x}

at

(1, 5)

and

(2, 25)

.

(1, 5) \to (- 2, 7)

(2, 25) \to (- 1, 27)

The

x

-values decreased by

3

units and the

y

-values increased by two units.

Compare to

y = 5^{(x - h)} + k

h = - 3

and

k = 2

y = 5^{x + 3} + 2

b.

Again, use points and mapping.

There are points at

(0, 1)

,

(1, 2)

, and

(2, 4)

on the graph of

y = 2^{x}

.

\begin{array}{l} (0, 1) \to (1, - \frac{9}{2}) \\ (1, 2) \to (2, - 5) \\ (2, 4) \to (3, - 6) \end{array}

First, consider stretches and reflections.

There is a reflection in the

x

-axis, which means all the

y

-values changed sign, or

y = - f (x)

.

There is also a vertical stretch factor. The first two points on the graph of

y = 2^{x}

are separated by a vertical distance of

1

unit. In the graph of the transformed function, the first two points are separated by a vertical distance of

\frac{1}{2}

unit.

The second two points on the graph of

y = 2^{x}

are separated by a vertical distance of

2

units, and in the transformed graph, the corresponding points are separated by a vertical distance of

1

unit.

The vertical distance between the corresponding points is reduced by a factor of

\frac{1}{2}

. The vertical stretch factor is

|a| = \frac{1}{2}

.

There is a horizontal translation to the right

1

unit because the

x

-values increased by

1

unit for each point that was mapped to the transformed function. Thus,

h = 1

.

Finally, there is a vertical translation

4

units down because, after the

y

-values change sign and are stretched by a factor of

\frac{1}{2}

, the

y

-values are decreased by

4

units.

Compare to

y = a {(2)}^{x - h} + k

y = - 0.5 {(2)}^{x - 1} - 4

Question 3, page 470

C

Additional Notes

Use technology to graph the function.

The denominator cannot be equal to zero.

\begin{array}{rcl} x - 2 & \neq & 0 \\ x & \neq & 2 \end{array}

There is a vertical asymptote at

x = 2

. Select values of

x

that are close to

x = 2

, and on either side of

x = 2

, to determine the behaviour of the graph as it approaches the vertical asymptote.

Question 4, page 470

B

Question 12, page 471

a \to D

The denominator cannot be equal to zero. As such,

x

cannot equal zero. However, the factor

x

is common to numerator and denominator, and therefore is eliminated during simplification of the function. The result is a point of discontinuity at

x = 0

on the linear graph of the function.

b \to A

The denominator factors to

(x - 3) (x + 3)

and the factors are not eliminated during simplification. There are vertical asymptotes at

x = 3

and

x = - 3

.

c \to B

The denominator contains only

x^{2}

. As such,

x

cannot equal

0

. There is a vertical asymptote at

x = 0

because the function cannot be simplified.

d \to C

The denominator factors to

x (x - 9)

. As such,

x

cannot equal

0

or

9

. However, the factor

x

is common to numerator and denominator, and therefore is eliminated during simplification of the function. The result is a point of discontinuity at

x = 0

and a vertical asymptote at

x = 9

.

L7.2 B3 Practice Solutions

Unit 7

Units 1 to 6 Review

Practice Solutions

Practice 7.2B Solutions

Question 4, page 102

Question 5, page 102

Question 6, page 102

Question 2, page 286

Question 3, page 286

Question 4, page 286

Question 8, page 287

Question 9, page 287

Question 12, page 287

Question 2, page 368

Questions 6a and 6b, page 368

Question 3, page 470

Question 4, page 470

Question 12, page 471