Chapter 5.3, Problem 44AYU

In problems $45-56,$ use the transformations to the graph each function. Determine the domain, range, horizontal asymptote and y-intercept of each function,

$f\left(x\right)=2^{x+2}$

To determine

The graph of function $f(x)=2^{x+2}$ using transformation, domain, range, horizontal asymptote, $y$- intercept of function.

Answer to Problem 44AYU

Solution:

The graph of $f(x)=2^{x+2}$ is

Domain of the function $f(x)=2^{x+2}$ is $\left(-\infty ,\infty \right)$

Range of function $f(x)=2^{x+2}$ is $\left(0,\infty \right)$

Horizontal asymptote of $f(x)=2^{x+2}$ is $y=0$.

$y$- intercept of $f(x)=2^{x+2}$ is $4$

Explanation of Solution

Given information:

The function $f(x)=2^{x+2}$

Explanation:

To graph a function begin with a graph of $f(x)=2^x$

Find some points satisfying $f(x)=2^x$

$\begin{array}{cccccc}x& -1& 0& 1& 2& 3\\ f\left(x\right)& \begin{array}{l}f\left(-1\right)=2^{-1}\\ =\frac{1}{2}\end{array}& \begin{array}{l}f\left(0\right)=2^0\\ =1\end{array}& \begin{array}{l}f\left(1\right)=2^1\\ =2\end{array}& \begin{array}{l}f\left(2\right)=2^2\\ =4\end{array}& \begin{array}{l}f\left(3\right)=2^3\\ =8\end{array}\\ \left(x,f\left(x\right)\right)& \left(-1,\frac{1}{2}\right)& \left(0,1\right)& \left(1,2\right)& \left(2,4\right)& \left(3,8\right)\end{array}$

Plot these points on the graph.

Therefore, the graph is

To graph a function $f(x)=2^{x+2}$, shift the graph of $f(x)=2^x$ by $2$ units to the left..

The graph of function $f(x)=2^{x+2}$ is

From the graph, domain of the function $f(x)=2^{x+2}$ is $\left(-\infty ,\infty \right)$

Range of the function is the set of values of the dependent variable for which a function is defined.

Therefore, the range of $f(x)=2^{x+2}$ is $\left(0,\infty \right)$

As $x\to -\infty$, $f(x)$ approaches to $0$

Therefore, $f(x)=2^{x+2}$ has horizontal asymptote $y=0$

To find $y$- intercept, substitute $x=0$ in $f(x)=2^{x+2}$

$\Rightarrow f(0)=2^{0+2}=4$

Therefore, $y$- intercept is $4$