Chapter 2.6, Problem 36PPSE

Solution

To write the equations for $f\left(x\right)$ and $g\left(x\right)$ represented in the graph and to identify the reflection that transforms the graph of $f\left(x\right)$ to the graph of $g\left(x\right)$ .

The equation of the line $f\left(x\right)$ is $f\left(x\right)=-2x+1$ .

The equation of the line $g\left(x\right)$ is $g\left(x\right)=2x-1$ .

The reflection that transforms $f\left(x\right)$ to $g\left(x\right)$ is the one about the $x$ -axis.

Given:

The graph representing $f\left(x\right)$ and $g\left(x\right)$:

  

Concept Used:

Equation of a line:

The equation of the line $y=f\left(x\right)$ passing through the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ :

  $\left(f\left(x\right)-y_1\right)=\frac{y_2-y_1}{x_2-x_1}\cdot \left(x-x_1\right)$ .

Reflection about the $x$-axis:

Given a function $y=f\left(x\right)$ .

Then, the transformation $y=-f\left(x\right)$ reflects $y=f\left(x\right)$ about the $x$ -axis.

Calculation:

Find the equation of the line $f\left(x\right)$:

Observe that $f\left(x\right)$ passes through the points $\left(x_1,y_1\right)=\left(0,1\right)$ and $\left(x_2,y_2\right)=\left(1,-1\right)$ .

Then, the equation of $f\left(x\right)$ :

  $\begin{array}{c}\left(f\left(x\right)-y_1\right)=\frac{y_2-y_1}{x_2-x_1}\cdot \left(x-x_1\right)\\ \left(f\left(x\right)-1\right)=\frac{\left(-1\right)-1}{1-0}\cdot \left(x-0\right)\\ \left(f\left(x\right)-1\right)=\frac{-2}{1}\cdot \left(x\right)\\ f\left(x\right)=-2x+1\end{array}$ .

Thus, the equation of the line $f\left(x\right)$ :

  $f\left(x\right)=-2x+1$ .

Find the equation of the line $g\left(x\right)$:

Observe that $g\left(x\right)$ passes through the points $\left(x_1,y_1\right)=\left(0,-1\right)$ and $\left(x_2,y_2\right)=\left(1,1\right)$ .

Then, the equation of $g\left(x\right)$ :

  $\begin{array}{c}\left(g\left(x\right)-y_1\right)=\frac{y_2-y_1}{x_2-x_1}\cdot \left(x-x_1\right)\\ \left(g\left(x\right)-\left(-1\right)\right)=\frac{1-\left(-1\right)}{1-0}\cdot \left(x-0\right)\\ g\left(x\right)+1=\frac{2}{1}\cdot \left(x\right)\\ g\left(x\right)=2x-1\end{array}$ .

Thus, the equation of the line $g\left(x\right)$ :

  $g\left(x\right)=2x-1$ .

Identify the reflection that transforms $f\left(x\right)$to $g\left(x\right)$:

Observe that:

  $\begin{array}{l}g\left(x\right)=2x-1\\ g\left(x\right)=-\left(-2x+1\right)\\ g\left(x\right)=-f\left(x\right)\end{array}$ .

Thus, the reflection that transforms $f\left(x\right)$ to $g\left(x\right)$ is $y=-f\left(x\right)$ .

That is, the reflection that transforms $f\left(x\right)$ to $g\left(x\right)$ is the one about the $x$ -axis.

Conclusion:

The equation of the line $f\left(x\right)$ is $f\left(x\right)=-2x+1$ .

The equation of the line $g\left(x\right)$ is $g\left(x\right)=2x-1$ .

The reflection that transforms $f\left(x\right)$ to $g\left(x\right)$ is the one about the $x$ -axis.