Exercise 37 Page 624

Let's first graph the given square root function and find its domain and range. Then, we will compare the graph to the graph of its parent function.

Graph, Domain, and Range

Let's start by finding the domain of $y=\frac{1}{2}\sqrt{x+2}.$ To do so, recall that the radicand of a square root is always greater than or equal to $0.$ Therefore, the domain of the given function is all real numbers greater than or equal to $\text{-}2.$ With this in mind, we will make a table of values to graph the function.

$x$	$\frac{1}{2}\sqrt{x+2}$	$y=\frac{1}{2}\sqrt{x+2}$
$\text{-}2$	$\frac{1}{2}\sqrt{\text{-}2+2}$	$0$
$\text{-}1$	$\frac{1}{2}\sqrt{\text{-}1+2}$	$0.5$
$2$	$\frac{1}{2}\sqrt{2+2}$	$1$
$7$	$\frac{1}{2}\sqrt{7+2}$	$1.5$

Let's plot and connect the obtained points. Remember, the domain is all real numbers greater than or equal to $\text{-}2,$ so we do not want to extend the function any farther to the left.

We can see that the function takes values of $y$ that are greater than or equal to $0.$ This tells us the range.

Comparison with the Parent Function

To compare the graph of our function with the graph of the parent function $f(x)=\sqrt{x},$ we will consider some possible transformations.

Transformations of $f(x)$
Horizontal Translations	$$\begin{array}{c}\text{Translation right }h\text{ units, }h>0\\ y=f(x-h)\end{array}$$
	$$\begin{array}{c}\text{Translation left }h\text{ units, }h>0\\ y=f(x+h)\end{array}$$
Vertical Stretch or Compression	$$\begin{array}{c}\text{Vertical stretch, }a>1\\ y=af(x)\end{array}$$
	$$\begin{array}{c}\text{Vertical compression, }0<a<1\\ y=af(x)\end{array}$$

Let's now identify the transformations in our function. The graph of the given function is a vertical compression by a factor of $\frac{1}{2}$ followed by a translation $2$ units left of the graph of $f(x)=\sqrt{x}.$