Chapter 3.6, Problem 48E

To determine

To describe: the transformation from the graph of $P\left(x\right)=8x-150$ to the graph of $Q\left(x\right)=16x-200$

Answer to Problem 48E

Vertical stretch by a factor of 2 and vertical shift upwards by 100 units.

Explanation of Solution

Concept used:

Vertical translation:

The graph of $g\left(x\right)=f\left(x\right)+k$ is a vertical translation of the graph of f .

The vertical shift is k units up if $k>0$ and k units down if $k<0$ .

Vertical transformation:

Relative to the graph of $f\left(x\right)$ , the transformation of graph of $g\left(x\right)=cf\left(x\right)$ is a vertical shrink by a factor of $c$ when $0<c<1$ and is a vertical stretch by a factor of $c$ when $c>1$ .

Consider the function:

  $\begin{array}{c}Q\left(x\right)=16x-200\\ =16x-300+100\\ =2\left(8x-150\right)+100\\ =2P\left(x\right)+100\end{array}$

Since, $Q\left(x\right)=2P\left(x\right)+100$

So, there is no horizontal translation from the graph of P to the graph of Q.

There is a vertical stretch of the graph of P by a factor of 2.

Finally, there is a vertical shift 100 units up.