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Q13.

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Found in: Page 341

### Precalculus Mathematics for Calculus

Book edition 7th Edition
Author(s) James Stewart, Lothar Redlin, Saleem Watson
Pages 948 pages
ISBN 9781337067508

# Graphing Exponential Functions Graph the function, not by plotting points, but by starting from the graph of $y=ex$ in Figure 1. State the domain, range, and asymptote. $f\left(x\right)={e}^{x-2}$

The graph is:

The domain is $\left(-\infty ,\infty \right)$.

The range is $\left(0,\infty \right)$.

The horizontal asymptote is $y=0$.

See the step by step solution

## Step 1. Given.

The given function is $f\left(x\right)={e}^{x-2}$.

## Step 2. To determine.

We have to graph the given function, not by plotting points, but by starting from the graph of $y={e}^{x}$ in Figure 1. Then we have to state the domain, range, and asymptote.

## Step 3. Calculation.

The parent function of $f\left(x\right)={e}^{x-2}$ is $y=g\left(x\right)={e}^{x}$.

So,

$f\left(x\right)={e}^{x-2}\phantom{\rule{0ex}{0ex}}=g\left(x-2\right)$

It means we will move $y={e}^{x}$ by 2 units to the right to graph $f\left(x\right)={e}^{x-2}$.

Therefore, the graph is:

The possible x-values are all real numbers. So, the domain is $\left(-\infty ,\infty \right)$.

The y-values are from 0 to $\infty$. So, the range is $\left(0,\infty \right)$.

From the graph the horizontal asymptote is $y=0$.