Chapter 3.8, Problem 32PPS

To determine

To determine the function from the given graph.

Answer to Problem 32PPS

  $f(x)=\left|x-5\right|$

Explanation of Solution

Given:

  

Concept used:

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

The vertex of the function is $\left(h,k\right)$ .

The variables $h$ and $k$ tell how far the graph shifts horizontally and vertically.

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 $f(x)=a\left|x-h\right|+k$ .

Calculation:

Let the function be $f(x)=a\left|x-h\right|+k$ .

From the graph, it can be observed that the vertex of the function is $\left(5,0\right)$ , so the graph is translated $5$ units horizontally right and $0$ units vertically.

So, $h=5$ and $k=0$

To determine the stretch factor, input the known pair of value. From the graph, $\left(2,3\right)$ satisfies the given function.

So, substituting the values in the function $f(x)=a\left|x-h\right|+k$

  $\begin{array}{l}\Rightarrow 3=a\left|2-5\right|+0\\ \Rightarrow 3=a\left|-3\right|\\ \Rightarrow 3=3a\\ \Rightarrow a=1\end{array}$

Hence, the stretch factor $a=1$ .

The equation of the given function is $f(x)=\left|x-5\right|$

Conclusion:

Therefore, the given function is $f(x)=\left|x-5\right|$ .