Chapter 9.6, Problem 17PPS

To determine

To plot the graph $y=\frac{1}{2}\left(2^x-8\right)$ , find the y-intercept and state the domain and range.

Answer to Problem 17PPS

  $\begin{array}{l}y-\text{intercept:}\left(0,-3.5\right)\\ \text{Domain:}\left(-\infty ,\infty \right)\\ \text{Range}:\left(-4,\infty \right)\end{array}$

Explanation of Solution

Given information:

  $y=\frac{1}{2}\left(2^x-8\right)$

Concept used:

• Domain of a function $y=f\left(x\right)$ is the set of x values for which the function is defined.
• Range of a function $y=f\left(x\right)$ is the set of y -values that the function can take.
• x -intercept is the point where the graph of the function crosses/touches the x -axis.
• y -intercept is the point where the graph of the function crosses/touches the y -axis.

Calculation:

In order to get the graph of $y=\frac{1}{2}\left(2^x-8\right)$ , first make a table by taking some random values of x and find the corresponding values of y . And then plot the ordered pairs and connect the points with a smooth curve, so the table is:

x	$y=\frac{1}{2}\left(2^x-8\right)$	y
- 2	$y=\frac{1}{2}\left(2^{-2}-8\right)$	-3.9
- 1	$y=\frac{1}{2}\left(2^{-1}-8\right)$	-3.75
0	$y=\frac{1}{2}\left(2^0-8\right)$	-3.5
1	$y=\frac{1}{2}\left(2^1-8\right)$	-3
2	$y=\frac{1}{2}\left(2^2-8\right)$	-2

Graph of the function is shown below:

  

From the above graph it is clear that the curve intersecting the y -axis at the point (0, -3.5). So, the y -intercept of the function is at $y=-3.5$ .

Since the graph is moving form left to right without any discontinuity, so the function is defined for all values of x . Thus, the domain of the function is all real numbers, that is, $\text{Domain}=\left(-\infty ,\infty \right)$ .

Also, the range of the function will be all real number greater than -4, that is, $\text{Range}=\left(-4,\infty \right)$ .