Chapter 3.4, Problem 12CFU

To determine

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

Answer to Problem 12CFU

  

Explanation of Solution

Given information:

  $y=3\pm \sqrt{x+1}$

  $p\left(x\right)=x^2$

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

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

The graph of $p\left(x\right)\text{ and }{p}^{-1}\left(x\right)$ 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 left by 1unit and up by 3 units to get the resulting graph $y=3\pm \sqrt{x+1}$ as follows: