Answer
Vertex: $\left(1,5\right)$
Axis of Symmetry: $y=5$
Domain: $\{x|x\le1\}$
Range: set of all real numbers
Graph of $x=-\dfrac{1}{5}y^2+2y-4$
Work Step by Step
To find the properties of the given equation, $
x=-\dfrac{1}{5}y^2+2y-4
,$ 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}{5}y^2+2y\right)-4
\\\\
x&=-\dfrac{1}{5}\left(y^2-10y\right)-4
.\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}{5}\left(y^2-10y+\left(\dfrac{-10}{2}\right)^2\right)+\left[-4-\left(-\dfrac{1}{5}\right)\left(\dfrac{-10}{2}\right)^2\right]
\\\\
x&=-\dfrac{1}{5}\left(y^2-10y+25\right)+\left[-4+5\right]
\\\\
x&=-\dfrac{1}{5}\left(y-5\right)^2+1
.\end{align*}(Note that $a\left(\dfrac{b}{2}\right)^2\Rightarrow
\left(-\dfrac{1}{5}\right)\left(\dfrac{-10}{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(1,5\right)
$.
The axis of symmetry is given by $y=k$. Hence, the axis of symmetry of the parabola with the given equation is $
y=5
$.
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=-5: & \text{If }y=0:
\\\\
x=-\dfrac{1}{5}y^2+2y-4 & x=-\dfrac{1}{5}y^2+2y-4
\\\\
x=-\dfrac{1}{5}(-5)^2+2(-5)-4 & x=-\dfrac{1}{5}(0)^2+2(0)-4
\\\\
x=-\dfrac{1}{5}(25)-10-4 & x=-\dfrac{1}{5}(0)+0-4
\\\\
x=-5-10-4 & x=0+0-4
\\
x=-19 & x=-4
.\end{array}
Thus the points $
(-19,-5)
$ and $
(-4,0)
$ are points on the parabola. Reflecting these points about the axis of symmetry, then $
(-4,10)
$ and $
(-19,15)
$ are also points on the parabola.
Using the points $\{
(-19,-5),(-4,0),
\left(1,5\right)
(-4,10),(-19,15)
\}$ 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\le1\}
$. The range (all $y$-values used in the graph) is the set of all real numbers.
