Chapter 3.8, Problem 35PPS

a.

To determine

To make the table of values for the function $f(x)=\left|2x+3\right|.$

Explanation of Solution

Given:

Function: $f(x)=\left|2x+3\right|$

Calculation:

Consider $f(x)=\left|2x+3\right|$

Values of x	Values of f (x)
$-5$	$7$
$-4$	$5$
$-3$	$3$
$-2$	$1$
$-1$	$1$
0	$3$
1	$5$
$2$	$7$
$3$	$9$
$4$	$11$
$5$	$13$

b.

To determine

To plot the points obtained in (a) on a coordinate grid.

Explanation of Solution

Given:

Table in subpart (a) of the function $f(x)=\left|2x+3\right|$ .

Graph:

Consider the table in (a).

  

c.

To determine

To graph the given function $f(x)=\left|2x+3\right|$

Explanation of Solution

Given:

Function: $f(x)=\left|2x+3\right|$

Concept used:

The general form of an absolute value function is $f(x)=a\left|x-h\right|+k$ .

The graph of $f(x)=a\left|x-h\right|+k$ is a “V” with vertex $\left(h,k\right)$ , slope $m=a$ on the right side of the vertex $\left(x>h\right)$ and slope $m=-a$ on the left side of the vertex $\left(x<h\right)$ .

The graph of $f(x)=-a\left|x-h\right|+k$ is an upside-down “V” with vertex $\left(h,k\right)$ , slope $m=-a$ for $\left(x>h\right)$ and slope $m=a$ for $\left(x<h\right)$ .

Calculation for graph:

Consider $f(x)=\left|2x+3\right|=2\left|x+\frac{3}{2}\right|$

Compare with the general form:

  $g(x)=a\left|x-h\right|+k$

Here, the value of $a=2$ , so the graph opens upwards with a slope of $2$ .

The value of $h=-\frac{3}{2}$ and the value of $k=0$ , so the vertex of the graph is $\left(-\frac{3}{2},0\right).$

Values of x	Values of f (x)
$-5$	$7$
$-4$	$5$
$-3$	$3$
$-2$	$1$
$-1$	$1$
0	$3$
1	$5$
$2$	$7$
$3$	$9$
$4$	$11$
$5$	$13$

By taking different values of x , the graph can be plotted.

Graph:

The graph of $f(x)=\left|2x+3\right|$ :

  

Interpretation:

From the graph, it can be observed that the horizontal extent of the graph is from $-\infty$ to $\infty$ , so the domain of $g(x)$ is $\left(-\infty ,\infty \right)$ .

The vertical extent of the graph is from $0$ to $\infty$ so the range of $g(x)$ is $\left[0,\infty \right)$ .