Chapter 6.1, Problem 3E

To determine

To graph: The functions $y=2{\left(5\right)}^x$ and $y=5^x$ then compare both the graphs.

Explanation of Solution

Given information:

The functions are $y=2{\left(5\right)}^x$ and $y=5^x$ .

Graph:

The exponential function is the continuous function that is defined as $f\left(x\right)=a{b}^x$ where $b$ is the base such that $a\ne 0,b>0,b\ne 1$ and $x$ is the independent variable.

Now find the coordinates to draw the graph of the function $f\left(x\right)=2{\left(5\right)}^x$ .

Substitute $x=0$ in the function $y=2{\left(5\right)}^x$ to obtain the $y$ -coordinate.

  $\begin{array}{c}y=2{\left(5\right)}^0\\ =2\end{array}$

Substitute $x=1$ in the function $y=2{\left(5\right)}^x$ to obtain the $y$ -coordinate.

  $\begin{array}{c}y=2{\left(5\right)}^1\\ =10\end{array}$

Substitute $x=2$ in the function $y=2{\left(5\right)}^x$ to obtain the $y$ -coordinate.

  $\begin{array}{c}y=2{\left(5\right)}^2\\ =27\end{array}$

Now find the coordinates to draw the graph of the function $g\left(x\right)={\left(5\right)}^x$ .

Substitute $x=0$ in the function $y={\left(5\right)}^x$ to obtain the $y$ -coordinate.

  $\begin{array}{c}y={\left(5\right)}^0\\ =1\end{array}$

Substitute $x=1$ in the function $y={\left(5\right)}^x$ to obtain the $y$ -coordinate.

  $\begin{array}{c}y={\left(5\right)}^1\\ =5\end{array}$

Substitute $x=2$ in the function $y={\left(5\right)}^x$ to obtain the $y$ -coordinate.

  $\begin{array}{c}y={\left(5\right)}^2\\ =25\end{array}$

Now draw the graph $f\left(x\right)=2{\left(5\right)}^x$ with coordinates $\left(0,2\right)\left(1,10\right)\text{ and }\left(2,27\right)$ and graph $g\left(x\right)=5^x$ with coordinates $\left(0,1\right)\left(1,1\right)\text{ and }\left(2,25\right)$ as shown in Figure 1.

  

Now compare both the graphs.

It can be observed from Figure 1 that the graph of $g\left(x\right)$ is the graph of $f\left(x\right)$ vertically stretched with a factor 2. Each value of the function of $g$ two times of the value of the function $f$ .

The $y$ -intercept of $f\left(x\right)$ is $\left(0,2\right)$ and the $y$ -intercept of $g\left(x\right)$ is $\left(0,1\right)$ .

Interpretation:

It can be observed from Figure 1 that both the functionsconverge to 0 when $x$ tends to $-\infty$ .