Answer
Vertex: $\left(4,2\right)$
Axis of Symmetry: $y=2$
Domain: $\{x|x\le4\}$
Range: set of all real numbers
Graph of $x=-(y-2)^2+4$
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-2)^2+4
,$ is $
\left(4,2\right)
$.
The axis of symmetry is given by $y=k$. Hence, the axis of symmetry of the parabola with the given equation is $
y=2
$.
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=0: & \text{If }y=1:
\\\\
x=-(y-2)^2+4 & x=-(y-2)^2+4
\\
x=-(0-2)^2+4 & x=-(1-2)^2+4
\\
x=-(-2)^2+4 & x=-(-1)^2+4
\\
x=-4+4 & x=-1+4
\\
x=0 & x=3
.\end{array}
Thus the points $
(0,0)
$ and $
(3,1)
$ are points on the parabola. Reflecting these points about the axis of symmetry, then $
(3,3)
$ and $
(0,4)
$ are also points on the parabola.
Using the points $\{
(0,0),(3,1),
\left(4,2\right),
(3,3),(0,4)
\}$ 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.
