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 + 1}-3 )
graph ( h(x)=e^{x + 1}-3 ) and its asymptote. graph the asymptote as a dashed line. use the graphing tool to graph the function.

Explanation:

Step1: Analyze the transformation of the function

The parent function is \(y = e^{x}\). For the function \(h(x)=e^{x + 1}-3\), we use the transformation rules. The transformation \(y = f(x + c)\) shifts the graph of \(y = f(x)\) to the left by \(c\) units (here \(c = 1\)), and the transformation \(y=f(x)-d\) shifts the graph of \(y = f(x)\) down by \(d\) units (here \(d = 3\)).

Step2: Find the asymptote

The horizontal asymptote of the parent function \(y = e^{x}\) is \(y = 0\). For the function \(y=e^{x+1}-3\), using the vertical - shift rule, if \(y = f(x)\) has a horizontal asymptote \(y = k\), then \(y=f(x)-d\) has a horizontal asymptote \(y=k - d\). Since \(k = 0\) and \(d = 3\), the horizontal asymptote of \(h(x)=e^{x + 1}-3\) is \(y=-3\).

Step3: Determine the domain

The domain of the exponential function \(y = e^{x}\) is \((-\infty,\infty)\). Since the transformations \(y = e^{x+1}-3\) (horizontal and vertical shifts) do not restrict the input values of \(x\), the domain of \(h(x)=e^{x + 1}-3\) is \((-\infty,\infty)\)

Step4: Determine the range

The range of the parent function \(y = e^{x}\) is \((0,\infty)\). After a vertical shift down by 3 units (\(y=e^{x+1}-3\)), we use the rule: if \(y = f(x)\) has a range \((a,\infty)\), then \(y=f(x)-d\) has a range \((a - d,\infty)\). Here \(a = 0\) and \(d = 3\), so the range of \(h(x)=e^{x + 1}-3\) is \((-3,\infty)\)

Answer:

• Asymptote: \(y=-3\)
• Domain: \((-\infty,\infty)\)
• Range: \((-3,\infty)\)