Chapter 3.5, Problem 64PFA

a.

To determine

To find:Transformation of given equation relating to graph of $f\left(x\right)=x$ .

Answer to Problem 64PFA

The function $f\left(x\right)$ is Horizontally compressed and vertically shifted 8 units downwards to get $g\left(x\right)$

Explanation of Solution

Given information:

Given function passes through the points $\left(4,2\right)$ and $\left(2,-2\right)$

Concept:

A Translation is a movement of the graph either horizontally parallel to x-axis or vertically parallel to y-axis.

The graph of $y=f\left(x\right)$ where $f\left(x\right)=x$ is same as the graph of $y=x$ . Writing the equations as functions in the form of $f\left(x\right)$ is useful when applying translations and reflections to graphs.

If $g\left(x\right)=f\left(x\right)+k$ it is known as vertical shift.

If $k>0$ it moves upwards.

If $k<0$ it moves downwards.

If $g\left(x\right)=f\left(x+h\right)$ it is known as horizontal shift.

If $h<0$ moves it right.

If $h>0$ moves it left.

If $g\left(x\right)=-f\left(x\right)$ it is known as Vertical reflection it means reflects about X-axis

If $g\left(x\right)=f\left(-x\right)$ it is known as Horizontal reflection it means reflects about Y-axis

If $g\left(x\right)=af\left(x\right)$

If $a>1$ it is known as Vertical stretch.

If $0<a<1$ it is known as Vertical compression.

If $g\left(x\right)=f\left(bx\right)$

If $b>1$ it is known as Horizontal compression.

If $0<b<1$ it is known as Horizontal Stretch.

From the given Graph we can say that the line passed through the points

  $\left(x_1,y_1\right)=\left(4,2\right)$

  $\left(x_2,y_2\right)=\left(2,-2\right)$

Using two-point formula

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

On substituting values

We get

  $\begin{array}{l}y-y_1=\left(\frac{y_2-y_1}{x_2-x_1}\right)\left(x-x_1\right)\\ \Rightarrow y-2=\left(\frac{-2-2}{2-4}\right)\left(x-4\right)\\ \Rightarrow y-2=\left(\frac{-4}{-2}\right)\left(x-4\right)\\ \Rightarrow y=\left(2\right)\left(x-4\right)\\ \Rightarrow y=2x-\left(2\times 4\right)\\ \Rightarrow y=2x-8\end{array}$

Here $y$ can be replaced with $g\left(x\right)$

Therefore

  $\begin{array}{l}g\left(x\right)=\left(2x\right)-8\\ \Rightarrow g\left(x\right)=f\left(2x\right)-8\end{array}$

Therefore, from above concept

If $g\left(x\right)=f\left(bx\right)$

If $b>1$ it is known as Horizontal compression.

If $0<b<1$ it is known as Horizontal Stretch.

Here $b=2$ as $2>1$ there will be Horizontal compression.

  $\Rightarrow g\left(x\right)=f\left(2x\right)-8$

If $g\left(x\right)=f\left(x\right)+k$ it is known as vertical shift.

If $k>0$ it moves upwards.

If $k<0$ it moves downwards.

Here $k=-8$ as $-8<0$ it is known as Vertical shift with 8 units downwards.

Therefore, the function $f\left(x\right)$ is Horizontal compression and vertically shifted 8 units downwards to get $g\left(x\right)$ .

b.

To determine

To find: Value of $k$ .

Answer to Problem 64PFA

  $k=-8$

Explanation of Solution

Given information:

  $g\left(x\right)=\left(ax\right)+k$

From the given Graph we can say that the line passed through the points

  $\left(x_1,y_1\right)=\left(4,2\right)$

  $\left(x_2,y_2\right)=\left(2,-2\right)$

Using two-point formula

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

On substituting values

We get

  $\begin{array}{l}y-y_1=\left(\frac{y_2-y_1}{x_2-x_1}\right)\left(x-x_1\right)\\ \Rightarrow y-2=\left(\frac{-2-2}{2-4}\right)\left(x-4\right)\\ \Rightarrow y-2=\left(\frac{-4}{-2}\right)\left(x-4\right)\\ \Rightarrow y=\left(2\right)\left(x-4\right)\\ \Rightarrow y=2x-\left(2\times 4\right)\\ \Rightarrow y=2x-8\end{array}$

Here $y$ can be replaced with $g\left(x\right)$

Therefore

  $g\left(x\right)=\left(2x\right)-8$

On comparing with $g\left(x\right)=\left(ax\right)+k$

We get

  $k=-8$

c.

To determine

To find:Value of $a$ .

Answer to Problem 64PFA

  $a=2$

Explanation of Solution

Given information:

  $g\left(x\right)=\left(ax\right)+k$

From the given Graph we can say that the line passed through the points

  $\left(x_1,y_1\right)=\left(4,2\right)$

  $\left(x_2,y_2\right)=\left(2,-2\right)$

Using two-point formula

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

On substituting values

We get

  $\begin{array}{l}y-y_1=\left(\frac{y_2-y_1}{x_2-x_1}\right)\left(x-x_1\right)\\ \Rightarrow y-2=\left(\frac{-2-2}{2-4}\right)\left(x-4\right)\\ \Rightarrow y-2=\left(\frac{-4}{-2}\right)\left(x-4\right)\\ \Rightarrow y=\left(2\right)\left(x-4\right)\\ \Rightarrow y=2x-\left(2\times 4\right)\\ \Rightarrow y=2x-8\end{array}$

Here $y$ can be replaced with $g\left(x\right)$

Therefore

  $g\left(x\right)=\left(2x\right)-8$

On comparing with $g\left(x\right)=\left(ax\right)+k$

We get

  $a=2$

d.

To determine

To find:Function $g\left(x\right)$ .

Answer to Problem 64PFA

  $g\left(x\right)=\left(2x\right)-8$

Explanation of Solution

Given information:

  $g\left(x\right)=\left(ax\right)+k$

From the given Graph we can say that the line passed through the points

  $\left(x_1,y_1\right)=\left(4,2\right)$

  $\left(x_2,y_2\right)=\left(2,-2\right)$

Using two-point formula

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

On substituting values

We get

  $\begin{array}{l}y-y_1=\left(\frac{y_2-y_1}{x_2-x_1}\right)\left(x-x_1\right)\\ \Rightarrow y-2=\left(\frac{-2-2}{2-4}\right)\left(x-4\right)\\ \Rightarrow y-2=\left(\frac{-4}{-2}\right)\left(x-4\right)\\ \Rightarrow y=\left(2\right)\left(x-4\right)\\ \Rightarrow y=2x-\left(2\times 4\right)\\ \Rightarrow y=2x-8\end{array}$

Here $y$ can be replaced with $g\left(x\right)$

Therefore

  $g\left(x\right)=\left(2x\right)-8$