Answer
Vertex: $\left(1,-2\right)$
Axis of Symmetry: $y=-2$
Domain: $\{x|x\ge1\}$
Range: set of all real numbers
Graph of $x=(y+2)^2+1$
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+2)^2+1
,$ is $
\left(1,-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+1 & x=(y+2)^2+1
\\
x=(0+2)^2+1 & x=(-1+2)^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 $
(5,0)
$ and $
(2,-1)
$ are points on the parabola. Reflecting these points about the axis of symmetry, then $
(2,-3)
$ and $
(5,-4)
$ are also points on the parabola.
Using the points $\{
(5,0),(2,-1),
\left(1,-2\right),
(2,-3),(5,-4)
\}$ 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\ge1\}
$. The range (all $y$-values used in the graph) is the set of all real numbers.
