Chapter 5.3, Problem 46AYU

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

$f\left(x\right)=4\cdot {\left(\frac{1}{3}\right)}^x$

To determine

The graph of the function $f(x)=4\cdot {\left(\frac{1}{3}\right)}^x$ using transformation, domain, range, horizontal asymptote $y$- intercept of function.

Answer to Problem 46AYU

Solution:

The graph of $f(x)=4\cdot {\left(\frac{1}{3}\right)}^x$ is

Domain of the function $f(x)=4\cdot {\left(\frac{1}{3}\right)}^x$ is $\left(-\infty ,\infty \right)$

Range of function $f(x)=4\cdot {\left(\frac{1}{3}\right)}^x$ is $\left(0,\infty \right)$

Horizontal asymptote of $f(x)=4\cdot {\left(\frac{1}{3}\right)}^x$ is $y=0$

$y$- intercept of $f(x)=4\cdot {\left(\frac{1}{3}\right)}^x$ is $4$

Explanation of Solution

Given information:

The function $f(x)=4\cdot {\left(\frac{1}{3}\right)}^x$

Explanation:

To graph a function, begin with a graph of $f(x)=3^x$

Find some points satisfying $f(x)=3^x$

$\begin{array}{cccccc}x& -1& 0& 1& 2& -2\\ f\left(x\right)& \begin{array}{l}f\left(-1\right)=3^{-1}\\ =\frac{1}{3}\end{array}& \begin{array}{l}f\left(0\right)=3^0\\ =1\end{array}& \begin{array}{l}f\left(1\right)=3^1\\ =3\end{array}& \begin{array}{l}f\left(2\right)=3^2\\ =9\end{array}& \begin{array}{l}f\left(-2\right)=3^{-2}\\ =\frac{1}{9}\end{array}\\ \left(x,f\left(x\right)\right)& \left(-1,\frac{1}{3}\right)& \left(0,1\right)& \left(1,3\right)& \left(2,9\right)& \left(-2,\frac{1}{9}\right)\end{array}$

Plot these points on the graph.

The graph is

Using the graph of $f(x)=3^x$, reflect the graph about $y$ axis to get the graph of $f(x)={\left(\frac{1}{3}\right)}^x$

Therefore, the graph of $f(x)={\left(\frac{1}{3}\right)}^x$ is

To graph the function $f(x)=4\cdot {\left(\frac{1}{3}\right)}^x$, stretch the graph of $f(x)={\left(\frac{1}{3}\right)}^x$ vertically by $4$ units

The graph of the function $f(x)=4\cdot {\left(\frac{1}{3}\right)}^x$ is

From the graph, domain of the function $f(x)=4\cdot {\left(\frac{1}{3}\right)}^x$ 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 $f(x)=4\cdot {\left(\frac{1}{3}\right)}^x$ is $\left(0,\infty \right)$

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

Therefore, $f(x)=4\cdot {\left(\frac{1}{3}\right)}^x$ have horizontal asymptote $y=0$

To find $y$- intercept, substitute $x=0$ in $f(x)=4\cdot {\left(\frac{1}{3}\right)}^x$

$\Rightarrow f(0)=4\cdot {\left(\frac{1}{3}\right)}^0=4$.

Therefore, $y$- intercept is $4$