Question

use transformations of the graph of $f(x)=e^{x}$ to graph the given function. be sure to give the equation of the asymptote. use the graphs to determine each functions domain and range. if applicable, use a graphing utility to confirm the hand - drawn graphs.
$h(x)=e^{x + 2}-3$
shifted to the right by 2 units and shift $f(x)$ upward by 3 units.
the graph of $f(x)=e^{x}$ should be shifted to the left by 2 units and shift $f(x)$ upward by 3 units.
the graph of $f(x)=e^{x}$ should be shifted to the left by 2 units and shift $f(x)$ downward by 3 units.
the graph of $f(x)=e^{x}$ should be shifted to the right by 2 units and shift $f(x)$ downward by 3 units.
graph $h(x)=e^{x + 2}-3$ and its asymptote. graph the asymptote as a dashed line. use the graphing tool to graph the function.
find the equation of the asymptote for $h(x)=e^{x + 2}-3$ using the graph.
$y=-3$
(type an equation.)
observe the graph and find the domain of $h(x)=e^{x + 2}-3$.
$(-\infty,\infty)$
(type your answer in interval notation.)
observe the graph and find the range of $h(x)=e^{x + 2}-3$.
(type your answer in interval notation.)

Explanation:

Step1: Analyze the transformation of the exponential function

For the function \(y = e^{x + 2}-3\), compared with the parent function \(y = e^{x}\), according to the transformation rule of the function \(y = f(x + a)+b\). When \(a>0\), the graph of \(y = f(x)\) is shifted to the left by \(a\) units; when \(b<0\), the graph of \(y = f(x)\) is shifted downward by \(|b|\) units. Here \(a = 2\) and \(b=-3\), so the graph of \(y = e^{x}\) is shifted to the left by 2 units and shifted downward by 3 units.

Step2: Determine the range of the function

The range of the parent - function \(y = e^{x}\) is \((0,\infty)\). For the function \(y=e^{x + 2}-3\), let \(t=x + 2\), then \(y = e^{t}-3\). Since \(e^{t}>0\) for all real \(t\), then \(e^{t}-3>-3\).

Answer:

The range of \(h(x)=e^{x + 2}-3\) is \((-3,\infty)\)