Answer
Vertex: $\left(0,-2\right)$
Axis of Symmetry: $x=0$
Domain: set of all real numbers
Range: $\{y|y\ge-2\}$
Graph of $f(x)=\dfrac{1}{2}x^2-2$
Work Step by Step
In the form $f(x)=a(x-h)^2+k$, the given function, $
f(x)=\dfrac{1}{2}x^2-2
,$ is equivalent to
\begin{align*}
f(x)=\dfrac{1}{2}(x-0)^2-2
.\end{align*}
Since the vertex of the function $f(x)=a(x-h)^2+k$ is given by $(h,k)$, then the vertex of the function above is $
\left(0,-2\right)
$.
The axis of symmetry of the function $f(x)=a(x-h)^2+k$ is given by $x=h$. With $h=
0
$ then the axis of symmetry is $
x=0
$.
To graph the parabola, find points that are on the parabla. This can be done by substituting values of $x$ and solving the corresponding value of $y$. Let $y=f(x).$ Then $
y=\dfrac{1}{2}x^2-2
$. Substituting values of $x$ and solving $y$ results to
\begin{array}{l|r}
\text{If }x=-4: & \text{If }x=-2:
\\\\
y=\dfrac{1}{2}(-4)^2-2 & y=\dfrac{1}{2}(-2)^2-2
\\\\
y=\dfrac{1}{2}(16)-2 & y=\dfrac{1}{2}(4)-2
\\\\
y=8-2 & y=2-2
\\
y=6 & y=0
.\end{array}
Hence, the points $
(-4,6)
$ and $
(-2,0)
$ are on the parabola. Reflecting these points about the axis of symmetry, the points $
(2,0)
$ and $
(4,6)
$ are also on the parabola.
Using the points $\{
(-4,6), (-2,0),
\left(0,-2\right),
(2,0), (4,6)
\}$ 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-2\}
$.
