Chapter 5.3, Problem 49AYU

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

$f\left(x\right)=2+4^{x-1}$

To determine

The graph of function $f(x)=2+4^{x-1}$ using transformation. Also domain, range, horizontal asymptote, $y$- intercept of function.

Answer to Problem 49AYU

Solution:

The graph of $f(x)=2+4^{x-1}$ is

Domain of the function $f(x)=2+4^{x-1}$ is $\left(-\infty ,\infty \right)$

Range of function $f(x)=2+4^{x-1}$ is $\left(2,\infty \right)$

Horizontal asymptote of $f(x)=2+4^{x-1}$ is $y=2$

$y$- intercept of $f(x)=2+4^{x-1}$ is $\frac{9}{4}$

Explanation of Solution

Given information:

The function $f(x)=2+4^{x-1}$

Explanation:

To graph a function $f(x)=2+4^{x-1}$, begin with a graph of $f(x)=4^x$

Find some points satisfying $f(x)=4^x$

$\begin{array}{cccccc}x& -1& 0& 1& 2& -2\\ f\left(x\right)& \begin{array}{l}f\left(-1\right)=4^{-1}\\ =\frac{1}{4}\end{array}& \begin{array}{l}f\left(0\right)=4^0\\ =1\end{array}& \begin{array}{l}f\left(1\right)=4^1\\ =4\end{array}& \begin{array}{l}f\left(2\right)=4^2\\ =16\end{array}& \begin{array}{l}f\left(-2\right)=4^{-2}\\ =\frac{1}{16}\end{array}\\ \left(x,f\left(x\right)\right)& \left(-1,\frac{1}{4}\right)& \left(0,1\right)& \left(1,4\right)& \left(2,16\right)& \left(-2,\frac{1}{16}\right)\end{array}$

Plot these points on the graph.

Therefore, the graph is

To graph a function $f(x)=4^{x-1}$, shift the graph of $f(x)=4^x$ to the right by $1$ unit.

The graph of function $f(x)=4^{x-1}$ is

To graph the function $f(x)=2+4^{x-1}$, shift the graph of $f(x)=4^{x-1}$ by $2$ units upward.

The graph of $f(x)=2+4^{x-1}$ is

From the graph, domain of the function $f(x)=2+4^{x-1}$ 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, range of $f(x)=3^{x-1}$ is $\left(2,\infty \right)$

As $x\to -\infty$, $f(x)$ approaches to $2$.

Therefore, $f(x)=2+4^{x-1}$ has a horizontal asymptote $y=2$

To find $y$- intercept, substitute $x=0$ in $f(x)=2+4^{x-1}$

$\Rightarrow f(0)=2+4^{0-1}=2+\frac{1}{4}=\frac{9}{4}$.

Therefore, $y$- intercept is $\frac{9}{4}$.