Chapter 1.2, Problem 42E

a.

To determine

To describe: The effect of reflection of the graph of $f$ in the $y-axis$ on the slope of the line and the intercepts of the graph.

Answer to Problem 42E

Slope and $x-\mathrm{int}ercept$ change their signs whereas $y-\mathrm{int}ercept$ remains the same.

Explanation of Solution

Given Information: Graph of $f(x)=mx+b$ is given.

Calculation:

After reflection of the graph of $f$ in the $y-axis$ , the transformed function is:

  $\begin{array}{l}g(x)=f(-x)\\ =m(-x)+b\\ -mx+b\end{array}$

Slope of $g(x)$ is $-m$ .

To find $x-\mathrm{int}ercept$ , putting $g(x)=0$ ,

  $\begin{array}{l}-mx+b=0\\ \Rightarrow x=\frac{b}{m}\end{array}$

To find $y-\mathrm{int}ercept$ , putting $x=0$ in $g(x)$ ,

  $\begin{array}{l}g(0)=-m(0)+b\\ =b\end{array}$

Thus, slope and $x-\mathrm{int}ercept$ change their signs whereas $y-\mathrm{int}ercept$ remains the same.

b.

To determine

To describe: The effect when the graph of $f$ is shrunk vertically by a factor of $\frac{1}{3}$ on the slope of the line and the intercepts of the graph.

Answer to Problem 42E

The slope and the

  $y-\mathrm{int}ercept$ become $\frac{1}{3}rd$ of the slope and $y-\mathrm{int}ercept$ of $f$ respectively whereas the $x-\mathrm{int}ercept$ remains the same.

Explanation of Solution

Given Information: Graph of $f(x)=mx+b$ is given.

Calculation:

When the graph of $f$ is shrunk vertically by a factor of $\frac{1}{3}$ , then the transformed function is:

  $\begin{array}{l}h(x)=\frac{1}{3}f(x)\\ =\frac{1}{3}(mx+b)\\ =\frac{m}{3}x+\frac{b}{3}\end{array}$

Slope of $h(x)$ is $\frac{m}{3}$ .

To find $x-\mathrm{int}ercept$ , putting $h(x)=0$ ,

  $\begin{array}{l}\frac{m}{3}x+\frac{b}{3}=0\\ \Rightarrow \frac{m}{3}x=-\frac{b}{3}\\ \Rightarrow mx=-b\\ \Rightarrow x=-\frac{b}{m}\end{array}$

To find $y-\mathrm{int}ercept$ , putting $x=0$ in $h(x)$ ,

  $h(0)=\frac{b}{3}$

Thus, the slope and the $y-\mathrm{int}ercept$ become $\frac{1}{3}rd$ of the slope and $y-\mathrm{int}ercept$ of $f$ respectively whereas the $x-\mathrm{int}ercept$ remains the same.

c.

To determine

To describe: The effect when the graph of $f$ is stretched horizontally by a factor of $2$ on the slope of the line and the intercepts of the graph.

Answer to Problem 42E

The slope becomes one-half of the slope of $f$ , $x-\mathrm{int}ercept$ becomes twice of the $x-\mathrm{int}ercept$ of $f$ whereas $y-\mathrm{int}ercept$ remains the same.

Explanation of Solution

Given Information: Graph of $f(x)=mx+b$ is given.

Calculation:

When the graph of $f$ is stretched horizontally by a factor of $2$ , then the transformed function is:

  $\begin{array}{l}k(x)=f(\frac{1}{2}x)\\ =m(\frac{1}{2}x)+b\\ =\frac{m}{2}x+b\end{array}$

Slope of $k(x)$ is $\frac{m}{2}$ .

To find $x-\mathrm{int}ercept$ , putting $k(x)=0$ ,

  $\begin{array}{l}\frac{m}{2}x+b=0\\ \Rightarrow \frac{m}{2}x=-b\\ \Rightarrow x=-\frac{2b}{m}\end{array}$

To find $y-\mathrm{int}ercept$ , putting $x=0$ in $k(x)$ ,

  $\begin{array}{l}k(0)=\frac{m}{2}0+b\\ =b\end{array}$

Thus, the slope becomes one-half of the slope of $f$ , $x-\mathrm{int}ercept$ becomes twice of the $x-\mathrm{int}ercept$ of $f$ whereas $y-\mathrm{int}ercept$ remains the same.