Answer
Vertex: $\left(-4,-3\right)$
Axis of Symmetry: $y=-3$
Domain: $\{x|x\ge-4\}$
Range: set of all real numbers
Graph of $x=2(y+3)^2-4$
Work Step by Step
Since the vertex of the equation $
x=a(y-k)^2+h
$ is given by $(h,k)$, then the vertex of the quadratic equation, $
x=2(y+3)^2-4
$, is $
\left(-4,-3\right)
$.
The axis of symmetry of the equation $x=a(y-k)^2+h$ is given by $y=k$. With $k=
-3
$ then the axis of symmetry is $
y=-3
$.
To graph the parabola, find points that are on the parabola. This can be done by substituting values of $y$ and solving the corresponding value of $x$. That is
\begin{array}{l|r}
\text{If }y=-5: & \text{If }y=-4:
\\\\
x=2(-5+3)^2-4 & x=2(-4+3)^2-4
\\
x=2(-2)^2-4 & x=2(-1)^2-4
\\
x=2(4)-4 & x=2(1)-4
\\
x=8-4 & x=2-4
\\
x=4 & x=-2
.\end{array}
Hence, the points $
(4,-5)
$ and $
(-2,-4)
$ are on the parabola. Reflecting these points about the axis of symmetry, the points $
(-2,-2)
$ and $
(4,-1)
$ are also on the parabola.
Using the points $\{
(4,-5), (-2,-4),
\left(-4,-3\right),
(-2,-2), (4,-1)
\}$ the graph of the parabola is determined (see graph above).
Using the graph, the domain (values of $x$ used in the graph) is $
\{x|x\ge-4\}
$. The range (values of $y$ used in the graph) is the set of all real numbers.
