Answer
Vertex: $\left(2,-4\right)$
Axis of Symmetry: $y=-4$
Domain: $\{x|x\le2\}$
Range: set of all real numbers
Graph of $x=-\dfrac{1}{2}y^2-4y-6$
Work Step by Step
To find the properties of the given equation, $
x=-\dfrac{1}{2}y^2-4y-6
,$ convert the equation in the form $x=a(y-k)^2+h$.
Grouping the $y$-variables together and making the coefficient of $y^2$ equal to $1$, the given equation is equivalent to
\begin{align*}
x&=\left(-\dfrac{1}{2}y^2-4y\right)-6
\\\\
x&=-\dfrac{1}{2}\left(y^2+8y\right)-6
.\end{align*}
Completing the square of the right-side expression by adding $\left(\dfrac{b}{2}\right)^2,$ the equation above is equivalent to
\begin{align*}
x&=-\dfrac{1}{2}\left(y^2+8y+\left(\dfrac{8}{2}\right)^2\right)+\left[-6-\left(-\dfrac{1}{2}\right)\left(\dfrac{8}{2}\right)^2\right]
\\\\
x&=-\dfrac{1}{2}\left(y^2+8y+16\right)+\left[-6+8\right]
\\\\
x&=-\dfrac{1}{2}\left(y+4\right)^2+2
.\end{align*}(Note that $a\left(\dfrac{b}{2}\right)^2\Rightarrow
\left(-\dfrac{1}{2}\right)\left(\dfrac{8}{2}\right)^2
$ should be subtracted as well to cancel out the term that was added to complete the square.)
Since the vertex of a parabola defined by $x=(y-k)^2+h$ is given by $(h,k),$ then the vertex of of the equation above is $
\left(2,-4\right)
$.
The axis of symmetry is given by $y=k$. Hence, the axis of symmetry of the parabola with the given equation is $
y=-4
$.
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=-8: & \text{If }y=-6:
\\\\
x=-\dfrac{1}{2}y^2-4y-6 & x=-\dfrac{1}{2}y^2-4y-6
\\\\
x=-\dfrac{1}{2}(-8)^2-4(-8)-6 & x=-\dfrac{1}{2}(-6)^2-4(-6)-6
\\\\
x=-\dfrac{1}{2}(64)+32-6 & x=-\dfrac{1}{2}(36)+24-6
\\\\
x=-32+32-6 & x=-18+24-6
\\
x=-6 & x=0
.\end{array}
Thus the points $
(-6,-8)
$ and $
(0,-6)
$ are points on the parabola. Reflecting these points about the axis of symmetry, then $
(0,-2)
$ and $
(-6,0)
$ are also points on the parabola.
Using the points $\{
(18,-8),(0,-6),
\left(2,-4\right)
(0,-2),(18,0)
\}$ 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\le2\}
$. The range (all $y$-values used in the graph) is the set of all real numbers.
