Unit 3

Transformations

Part 3.2C corresponds to section 7.2, starting on page 346 of your Pre-Calculus 12 textbook.

Exponential functions can be transformed in much the same way as the previous functions.

Recall the exponential function has a variable in the exponent.

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

is the equation of a transformed exponential function.

In this equation,

c

is the base of the exponential function,

|a|

is the vertical stretch factor,

\frac{1}{|b|}

is the horizontal stretch factor,

h

is the horizontal translation, and

k

is the vertical translation.

y = 2^{x}

is a basic exponential function.

The function

y = 3 {(2)}^{4 (x - 1)} - 6

can be thought of as a transformation of

y = 2^{x}

because both are of base

2

$a = 3$	There is a vertical stretch about the $x$ -axis by a factor of $3$ . All the $y$ -values will be multiplied by $3$ .
$b = 4$	There is a horizontal stretch about the $y$ -axis by a factor of $\frac{1}{4}$ . All the $x$ -values will be multiplied by $\frac{1}{4}$ .
$h = 1$	There is a horizontal translation $1$ unit to the right. All the $x$ -values will increase by $1$ .
$k = - 6$	There is a vertical translation $6$ units down. All the $y$ -values will decrease by $6$ .

Example 1

Use the function

y = 4^{x}

and transformations to sketch the graph of

y = 4^{- 2 (x + 5)} - 3

Compare the domain and range of

y = 4^{x}

to that of

y = 4^{- 2 (x + 5)} - 3

Solution

First, graph

y = 4^{x}

and identify key points.

There is one key point at

(0, 1)

and another at

(1, 4)

Parameter and Transformation	New point mapped on the graph of the transformed function
$a = 1$ . There is no vertical stretch and no reflection in the $x$ -axis.	$(0, 1)$ and $(1, 4)$ remain the same.
$b = - 2$ . There is a horizontal stretch about the $y$ -axis by a factor of $\frac{1}{2}$ . All the $x$ -values will be multiplied by $\frac{1}{2}$ . Because $b$ is negative, there is a reflection in the $y$ -axis. All the $x$ -values will change to the opposite sign.	$(0, 1) \to (0, 1)$ and $(1, 4) \to (- \frac{1}{2}, 4)$
$h = - 5$ . There is a horizontal shift $5$ units to the left. All the $x$ -values will decrease by $5$ .	$(0, 1) \to (- 5, 1)$ and $(- \frac{1}{2}, 4) \to (- 5 \frac{1}{2}, 4)$
$k = - 3$ . There is a vertical shift down of $3$ units. All the $y$ -values will decrease by $3$ .	$(- 5, 1) \to (- 5, - 2)$ and $(- 5 \frac{1}{2}, 4) \to (- 5 \frac{1}{2}, 1)$

The two new points on the graph of the transformed function are

(- 5, - 2)

and

(- 5 \frac{1}{2}, 1)

.

Note: Only two key points have been shown, but the more points used, the more accurate your graph will be.

The domain of

y = 4^{x}

\{x ∣ x \in R\}

.

The domain of

y = 4^{- 2 (x + 5)} - 3

is also

\{x ∣ x \in R\}

.

The domain did not change.

The range of

y = 4^{x}

\{y ∣ y > 0, y \in R\}

.

The range of

y = 4^{- 2 (x + 5)} - 3

\{y ∣ y > - 3, y \in R\}

.

Note: Both graphs have a horizontal asymptote that can be used to help determine the range.

There is a horizontal asymptote for the graph of

y = 4^{x}

y = 0

.

There is a horizontal asymptote for the graph of

y = 4^{- 2 (x + 5)} - 3

y = - 3

L3.2 C1 Transforming Exponential Functions

Unit 3

Transformations

Read

Part 3.2C corresponds to section 7.2, starting on page 346 of your Pre-Calculus 12 textbook.

Example 1