Chapter 5.3, Problem 45AYU

To determine

The graph of function $f(x)=3\cdot {\left(\frac{1}{2}\right)}^x$ using transformation, domain, range, horizontal asymptote, $y$ - intercept of function.

Answer to Problem 45AYU

Solution:

The graph of $f(x)=3\cdot {\left(\frac{1}{2}\right)}^x$ is

  

Domain of the function $f(x)=3\cdot {\left(\frac{1}{2}\right)}^x$ is $\left(-\infty ,\infty \right)$

Range of function $f(x)=3\cdot {\left(\frac{1}{2}\right)}^x$ is $\left(0,\infty \right)$

Horizontal asymptote of $f(x)=3\cdot {\left(\frac{1}{2}\right)}^x$ is $y=0$

  $y$ - intercept of $f(x)=3\cdot {\left(\frac{1}{2}\right)}^x$ is $3$

Explanation of Solution

Given information:

The function $f(x)=3\cdot {\left(\frac{1}{2}\right)}^x$

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

  

Using the above graph, take the reflection about $y$ axis to get graph of $f(x)={\left(\frac{1}{2}\right)}^x$

The graph is

  

To graph a function $f(x)=3\cdot {\left(\frac{1}{2}\right)}^x$ , stretch the graph of $f(x)={\left(\frac{1}{2}\right)}^x$ vertically by $3$ units

The graph of the function $f(x)=3\cdot {\left(\frac{1}{2}\right)}^x$ is

  

From the graph, domain of the function $f(x)=3\cdot {\left(\frac{1}{2}\right)}^x$ 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 $f(x)=3\cdot {\left(\frac{1}{2}\right)}^x$ is $\left(0,\infty \right)$

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

Therefore, $f(x)=3\cdot {\left(\frac{1}{2}\right)}^x$ have horizontal asymptote $y=0$

To find $y$ - intercept, substitute $x=0$ in $f(x)=3\cdot {\left(\frac{1}{2}\right)}^x$

  $\Rightarrow f(0)=3\cdot {\left(\frac{1}{2}\right)}^0=3$ .

Therefore, $y$ - intercept is $3$