Chapter 3.4, Problem 36E

To determine

To graph: the equation using parent graph $p\left(x\right)$ .

Explanation of Solution

Given information:

  $y=1\pm \sqrt{x-2}$

  $p\left(x\right)=x^2$

Calculation:

The $p^{-1}\left(x\right)$ can be evaluated as follows:

  $\begin{array}{c}y=x^2\left[\text{Let }p\left(x\right)=y\right]\\ x=y^2\left[\text{Interchange }x\text{ and }y\right]\\ y=\pm \sqrt{x}\left[\text{Solve for }y\right]\\ p^{-1}\left(x\right)=\pm \sqrt{x}\left[\text{Replace }y\text{ with }{p}^{-1}\left(x\right)\right]\end{array}$

Now make a table of values for $p^{-1}\left(x\right)$ as shown below.

$p\left(x\right)$	$x$	$-3$	$-2$	$-1$	0	1	2	3	4
	$y$	9	4	1	0	1	4	9	16

$p^{-1}\left(x\right)$	$x$	$-1$	0	1	2	3	4
	$y$	$\pm \sqrt{-1}$	0	$\pm 1$	$\pm \sqrt{2}$	$\pm \sqrt{3}$	$\pm 2$

Since, there is only positive side of $f\left(x\right)$ is given. So, only graph of $p^{-1}\left(x\right)=\sqrt{x}$ and graph of $p\left(x\right)$ is shown below.

  

The red line represents $p\left(x\right)$ and blue line represents $p^{-1}\left(x\right)$ . Shift the $p^{-1}\left(x\right)$ graph to right by 2 units and up by 1 unit to get the resulting graph $y$ as follows: