Answer
Vertex: $\left(2,2\right)$
Axis of Symmetry: $y=2$
Domain: $\{x|x\le2\}$
Range: set of all real numbers
Graph of $x=-(y-2)^2+2$
Work Step by Step
Since the vertex of the function $x=a(y-k)^2+h$ is given by $(h,k)$, then the vertex of the given equation, $
x=-(y-2)^2+2
$, is $
\left(2,2\right)
$.
The axis of symmetry of the equation $x=a(y-k)^2+h$ is given by $y=k$. With $k=
2
$ then the axis of symmetry is $
y=2
$.
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=0: & \text{If }y=1:
\\\\
x=-(0-2)^2+2 & x=-(1-2)^2+2
\\
x=-(-2)^2+2 & x=-(-1)^2+2
\\
x=-4+2 & x=-1+2
\\
x=-2 & x=1
.\end{array}
Hence, the points $
(-2,0)
$ and $
(1,1)
$ are on the parabola. Reflecting these points about the axis of symmetry, the points $
(1,3)
$ and $
(-2,4)
$ are also on the parabola.
Using the points $\{
(-2,0), (1,1),
\left(2,2\right),
(1,3), (-2,4)
\}$ 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\le2\}
$. The range (values of $y$ used in the graph) is the set of all real numbers.
