Unit 3

Transformations

Practice 3.2C Solutions

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

Pages 354 to 356, questions 1a to 1d, 3e, 4a to 4d, and 9a to 9d.

Questions 1a to 1d, page 354

Additional Notes

Compare each function to the function

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

y = 2 {(3)}^{x}

a = 2

There is a vertical stretch about the

x

-axis by a factor of

2

. The function matches description C.

y = 3^{x - 2}

h = 2

There is a horizontal translation

2

units right. The function matches description D.

y = 3^{x} + 4

k = 4

There is a vertical translation up

4

units. The function matches description A.

y = 3^{\frac{x}{5}}

b = \frac{1}{5}

There is a horizontal stretch about the

y

-axis by a factor of

5

. The function matches description B.

Question 3e, page 354

Compare the function to

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

$a = - \frac{1}{2}$	There is a vertical stretch about the $x$ -axis by a factor of $\frac{1}{2}$ . Because $a$ is negative, there is a reflection in the $x$ -axis.
$b = 2$	There is a horizontal stretch about the $y$ -axis by a factor of $\frac{1}{2}$ .
$h = 4$	There is a horizontal translation $4$ units right.
$k = 3$	There is a vertical translation $3$ units up.

Questions 4a to 4d, page 355

Matches C: reflection in the

x

-axis, so

a

is negative, the graph was decreasing before it was reflected in the

x

-axis, so

0 < c < 1

, vertical translation

2

units up, so

k = 2

Matches A: horizontal translation

1

unit right, so

h = 1

, vertical translation

2

units down, so

k = - 2

Matches D: reflection in the

x

-axis, so

a

is negative, vertical translation

2

units up, so

k = 2

Matches B: horizontal translation

2

units right, so

h = 2

, vertical translation

1

unit up, so

k = 1

Additional Notes

Compare to the graph of

y = 2^{x}

The graph has been reflected in the

x

-axis. The value of

a

will be negative. The choices are equation C or D. There has been a vertical translation of

2

units upwards. Again, C or D are possible choices.

Either the base must be a fraction between

0

and

1

, or the

b

-value must be negative for the curve to be decreasing. (Think of how it looked before the reflection.)

The graph matches C.

The graph was shifted vertically

2

units down.

The graph matches A.

The graph has been reflected in the

x

-axis. The value of

a

will be negative.

The remaining choice with a negative

a

value is function D.

The graph shows a vertical translation of

1

unit up.

The graph matches B.

Questions 9a to 9d, page 356

The value of

0.79

is the base in the exponential function. After each

3

hour period, there will be

0.79

of the previous amount of the drug. So, after

3

hours, there will be

0.79 (79 %)

of the original amount, after

6

hours there will be

0 . 79^{2}

of the original amount, and so on.

The reciprocal of

\frac{1}{3}

can be used to determine how long it will take to decrease to

0.79

. After each

3

hour period, there will be

0.79

of the previous amount of the drug.

The

M

-intercept represents the drug dose taken at time

= 0

.

In the graph, it is

100 mg

The domain represents the values possible for time in the function.

Domain:

\{h ∣ h \geq 0, h \in R\}

The range represents the amount of the drug in the bloodstream.

Range:

\{M ∣ 0 < M \leq 100, M \in R\}

L3.2 C3 Practice Solutions

Unit 3

Transformations

Practice Solutions

Practice 3.2C Solutions

Questions 1a to 1d, page 354

Question 3e, page 354

Questions 4a to 4d, page 355

Questions 9a to 9d, page 356