Answer
Vertex: $\left(-2,-3\right)$
Axis of Symmetry: $y=-3$
Domain: $\{x|x\ge-2\}$
Range: set of all real numbers
Graph of $x=(y+3)^2-2$
Work Step by Step
Since the vertex of a parabola defined by $x=(y-k)^2+h$ is given by $(h,k),$ then the vertex of $
x=(y+3)^2-2
,$ is $
\left(-2,-3\right)
$.
The axis of symmetry is given by $y=k$. Hence, the axis of symmetry of the parabola with the given equation is $
y=-3
$.
To graph the parabola, find points on the parabola by substituting values of of $y$ and then solving for $x$. That is,
\begin{array}{l|r}
\text{If }y=-1: & \text{If }y=-2:
\\\\
x=(y+3)^2-2 & x=(y+3)^2-2
\\
x=(-1+3)^2-2 & x=(-2+3)^2-2
\\
x=(2)^2-2 & x=(1)^2-2
\\
x=4-2 & x=1-2
\\
x=2 & x=-1
.\end{array}
Thus the points $
(2,-1)
$ and $
(-1,-2)
$ are points on the parabola. Reflecting these points about the axis of symmetry, then $
(-1,-4)
$ and $
(2,-5)
$ are also points on the parabola.
Using the points $\{
(2,-1),(-1,-2),
\left(-2,-3\right),
(-1,-4),(2,-5)
\}$ the graph of the given equation is derived (see graph above).
Using the graph above, the domain (all $x$-values used in the graph) is $
\{x|x\ge-2\}
$. The range (all $y$-values used in the graph) is the set of all real numbers.
