Answer
Vertex: $\left(-1,3\right)$
Axis of Symmetry: $y=3$
Domain: $\{x|x\le-1\}$
Range: set of all real numbers
Graph of $x=-(y-3)^2-1$
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)^2-1
,$ is $
\left(-1,3\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)^2-1 & x=-(y-3)^2-1
\\
x=-(1-3)^2-1 & x=-(2-3)^2-1
\\
x=-(-2)^2-1 & x=-(-1)^2-1
\\
x=-4-1 & x=-1-1
\\
x=-5 & x=-2
.\end{array}
Thus the points $
(-5,1)
$ and $
(-2,2)
$ are points on the parabola. Reflecting these points about the axis of symmetry, then $
(-2,4)
$ and $
(-5,5)
$ are also points on the parabola.
Using the points $\{
(-5,1),(-2,2),
\left(-1,3\right),
(-2,4),(-5,5)
\}$ 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\le-1\}
$. The range (all $y$-values used in the graph) is the set of all real numbers.
