Answer
Domain: $\text{ the interval } [-2,\infty)$
Work Step by Step
The radicand of the given function, $
g(x)=\sqrt{x+2}
,$ should be nonnegative. Hence,
\begin{array}{l}\require{cancel}
x+2\ge0
\\
x\ge-2
.\end{array}
Hence, the domain of the given function is $
\text{ the interval } [-2,\infty)
.$
Substituting $x=-2$ in the given function results to
\begin{array}{l}\require{cancel}
g(x)=\sqrt{-2+2}
\\
g(x)=\sqrt{0}
\\
g(x)=0
.\end{array}
Substituting $x=-1$ in the given function results to
\begin{array}{l}\require{cancel}
g(x)=\sqrt{-1+2}
\\
g(x)=\sqrt{1}
\\
g(x)=1
.\end{array}
Substituting $x=2$ in the given function results to
\begin{array}{l}\require{cancel}
g(x)=\sqrt{2+2}
\\
g(x)=\sqrt{4}
\\
g(x)=2
.\end{array}
Substituting $x=7$ in the given function results to
\begin{array}{l}\require{cancel}
g(x)=\sqrt{7+2}
\\
g(x)=\sqrt{9}
\\
g(x)=3
.\end{array}
The completed table is shown below.
Using the table of values, the graph is shown above.
