Chapter 3.7, Problem 30E

To determine

To graph the functions $w(x)=\left|x-3\right|$ and $y(x)=\left|5x-3\right|+4$ and compare them.

Explanation of Solution

Given:

Functions: $w(x)=\left|x-3\right|$ and $y(x)=\left|5x-3\right|+4$

Concept used:

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

The graph opens up if $a>0$ and opens down if $a<0$ . If $0<a<1$ , the graph is compressed, and if $a>1$ , it is stretched.

The graph of $f(x)=\left|x\right|$ is translated $h$ units horizontally and $k$ units vertically to get the graph of $f(x)=a\left|x-h\right|+k$ .

For $w(x)=\left|x-3\right|$ ,

Compare with the general form:

  $f(x)=a\left|x-h\right|+k$

Here, the value of $a=1$ , so the graph opens upwards with a slope of $1$ .

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

Calculation for graph:

Consider $w(x)=\left|x-3\right|$

Values of x	Values of w ( x )
0	$3$
1	$2$
-1	$4$
2	$1$
-2	$5$

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

Graph:

  

For $y(x)=\left|5x-3\right|+4$ ,

Compare with the general form:

  $f(x)=a\left|x-h\right|+k$

Here, the value of $a=5$ , so the graph opens upwards with a slope of $5$ .

The value of $h=\frac{3}{5}$ and the value of $k=4$ , so the vertex of the graph is $\left(\frac{3}{5},4\right)$ .

Calculation for graph:

Consider $y(x)=\left|5x-3\right|+4$

Values of x	Values of y ( x )
0	$7$
$1$	$6$
-1	$12$
$2$	$11$
-2	$17$

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

Graph:

  

Interpretation:

From the two graphs, it can be observed that the graph of $y(x)$ is compressed and translated up by $4$ units as compared to the graph of $w(x)$ .