Chapter 3.8, Problem 66PFA

a.

To determine

To graph the function $g(x)=\frac{1}{5}\left|x-20\right|+2.$

Explanation of Solution

Given:

Function: $g(x)=\frac{1}{5}\left|x-20\right|+2$

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:

Consider $g(x)=\frac{1}{5}\left|x-20\right|+2,$

Comparing it with the general form,$g(x)=a\left|x-h\right|+k$

Here, the value of $a=\frac{1}{5}$, so the graph opens upwards with a slope of $\frac{1}{5}$.

The value of $h=20$ and the value of $k=2$, so the vertex of the graph is $\left(20,2\right)$.

Calculation for graph:

Consider $g(x)=\frac{1}{5}\left|x-20\right|+2$

  

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

Graph:

  

b.

To determine

To determine the vertex of the function $g(x)=\frac{1}{5}\left|x-20\right|+2$.

Answer to Problem 66PFA

  $\left(20,2\right)$

Explanation of Solution

Given:

Function: $g(x)=\frac{1}{5}\left|x-20\right|+2$

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.

Calculation:

Consider $g(x)=\frac{1}{5}\left|x-20\right|+2$

Compare with the general form:$g(x)=a\left|x-h\right|+k$

Here, the value of $a=\frac{1}{5}$, so the graph opens upwards with a slope of $\frac{1}{5}$.

The value of $h=20$ and the value of $k=2$, so the vertex of the graph is $\left(20,2\right)$.

The value of $h=20$ represents the graph is translated horizontally to the right side of the vertex by 20 units and the value of $k=2$ represents that the graph is translated up by 2 units.

Conclusion:

Therefore, the vertex of the graph is $\left(20,2\right)$.

c.

To determine

To determine the distance of ship from the lighthouse when it is spotted first time by the captain.

2

Given:

Function: $g(x)=\frac{1}{5}\left|x-20\right|+2$

Explanation:

Since the ship was spotted initially in 20 minutes by the captain, the distance of the ship from the lighthouse is $g(20)=2$miles.

Conclusion:

Therefore, the distance of the ship from the lighthouse is 2 miles, when the captain first spots the ship.

Answer to Problem 66PFA

The graph of $f(x)=\left|x\right|$ is compressed by $\frac{1}{5}$, translated $20$units horizontally to the right and $2$units vertically up to get the graph of $g(x)=\frac{1}{5}\left|x-20\right|+2$.

Explanation of Solution

Function: $g(x)=\frac{1}{5}\left|x-20\right|+2$

Since the ship was spotted initially in 20 minutes by the captain, the distance of the ship from the lighthouse is $g(20)=2$miles.

Conclusion:

Therefore, the distance of the ship from the lighthouse is 2 miles, when the captain first spots the ship.

d.

To determine

To determine the relation of $g(x)=\frac{1}{5}\left|x-20\right|+2$ with the graph of the function $f(x)=\left|x\right|$.

Answer to Problem 66PFA

The graph of $f(x)=\left|x\right|$ is compressed by $\frac{1}{5}$, translated $20$units horizontally to the right and $2$units vertically up to get the graph of $g(x)=\frac{1}{5}\left|x-20\right|+2$.

Explanation of Solution

Function: $g(x)=\frac{1}{5}\left|x-20\right|+2$

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

The graph of $f(x)=\left|x\right|$ is compressed by $\frac{1}{5}$, translated $20$units horizontally to the right and $2$units vertically up to get the graph of $g(x)=\frac{1}{5}\left|x-20\right|+2$.

Conclusion:

Therefore, the transformation of $g(x)$ as the graph of$f(x)=\left|x\right|$ is that the graph of $f(x)=\left|x\right|$ is compressed by $\frac{1}{5}$, translated $20$units horizontally to the right and $2$units vertically up to get the graph of $g(x)=\frac{1}{5}\left|x-20\right|+2$.