Answer
Vertex: $\left(4,6\right)$
Axis of Symmetry: $y=6$
Domain: $\{x|x\le4\}$
Range: set of all real numbers
Graph of $x=-\dfrac{1}{2}y^2+6y-14$
Work Step by Step
To find the properties of the given equation, $
x=-\dfrac{1}{2}y^2+6y-14
,$ convert the equation to 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+6y\right)-14
\\\\
x&=-\dfrac{1}{2}\left(y^2-12y\right)-14
.\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^2-12y+\left(\dfrac{-12}{2}\right)^2\right)+\left[-14-\left(-\dfrac{1}{2}\right)\left(\dfrac{-12}{2}\right)^2\right]
\\\\
x&=-\dfrac{1}{2}\left(y^2-12y+36\right)+\left[-14+18\right]
\\\\
x&=-\dfrac{1}{2}\left(y-6\right)^2+4
.\end{align*}(Note that $a\left(\dfrac{b}{2}\right)^2\Rightarrow
\left(-\dfrac{1}{2}\right)\left(\dfrac{-12}{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 $x=-\dfrac{1}{2}\left(y-6\right)^2+4$ is $
\left(4,6\right)
$.
The axis of symmetry is given by $y=k$. Hence, the axis of symmetry of the parabola with the given equation is $
y=6
$.
To graph the parabola, find points on the parabola by substituting values of $y$ and then solving for $x$. That is,
\begin{array}{l|r}
\text{If }y=2: & \text{If }y=4:
\\\\
x=-\dfrac{1}{2}(2)^2+6(2)-14 & x=-\dfrac{1}{2}(4)^2+6(4)-14
\\\\
x=-\dfrac{1}{2}(4)+6(2)-14 & x=-\dfrac{1}{2}(16)+6(4)-14
\\\\
x=-2+12-14 & x=-8+24-14
\\
x=-4 & x=2
.\end{array}
Thus the points $
(-4,2)
$ and $
(2,4)
$ are points on the parabola. Reflecting these points about the axis of symmetry, then $
(2,8)
$ and $
(-4,10)
$ are also points on the parabola.
Using the points $\{
(-4,2),(2,4),
\left(4,6\right)
(2,8),(-4,10)
\}$ 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\le4\}
$. The range (all $y$-values used in the graph) is the set of all real numbers.
