Answer
Vertex: $\left(2,-3\right)$
Axis of Symmetry: $x=2$
Domain: set of all real numbers
Range: $\{y|y\ge-3\}$
Graph of $f(x)=2(x-2)^2-3$
Work Step by Step
Since the vertex of the function $f(x)=a(x-h)^2+k$ is given by $(h,k)$, then the vertex of the given quadratic function, $
f(x)=2(x-2)^2-3
$, is $
\left(2,-3\right)
$.
The axis of symmetry of the function $f(x)=a(x-h)^2+k$ is given by $x=h$. With $h=
2
$ then the axis of symmetry is $
x=2
$.
To graph the parabola, find points that are on the parabola. This can be done by substituting values of $x$ and solving the corresponding value of $y$. Let $y=f(x).$ Then $
y=2(x-2)^2-3
$. Substituting values of $x$ and solving $y$ results to
\begin{array}{l|r}
\text{If }x=0: & \text{If }x=1:
\\\\
y=2(0-2)^2-3 & y=2(1-2)^2-3
\\
y=2(-2)^2-3 & y=2(-1)^2-3
\\
y=2(4)-3 & y=2(1)-3
\\
y=8-3 & y=2-3
\\
y=5 & y=-1
.\end{array}
Hence, the points $
(0,5)
$ and $
(1,-1)
$ are on the parabola. Reflecting these points about the axis of symmetry, the points $
(3,-1)
$ and $
(4,5)
$ are also on the parabola.
Using the points $\{
(0,5), (1,-1),
\left(2,-3\right),
(3,-1), (4,5)
\}$ the graph of the parabola is determined (see graph above).
Using the graph, the domain (values of $x$ used in the graph) is the set of all real numbers. The range (values of $y$ used in the graph) is $
\{y|y\ge-3\}
$.
