Answer
Vertex: $(0,0)$
Axis of Symmetry: $x=0$
Domain: all real numbers
Range: $\{y|y\ge0\}$
Work Step by Step
In the form $f(x)=a(x-h)^2+k,$ the given equation, $
f(x)=-2x^2
,$ is equivalent to
\begin{align*}\require{cancel}
f(x)=-2(x-0)^2+0
.\end{align*}
Since the vertex of $f(x)=a(x-h)^2+k$ is given by $(h,k)$ then the vertex of the equation above is $
(0,0)
$.
The axis of symmetry is given by $x=h$. With $h$ given above as $h=0,$ then the axis of symmetry is $
x=0
$.
Let $y=f(x).$ Then, $y=
-2x^2
$. By substitution,
\begin{array}{l|r}
\text{If }x=1: & \text{If }x=2:
\\
y=-2x^2 & y=-2x^2
\\
y=-2(1)^2 & y=-2(2)^2
\\
y=-2(1) & y=-2(4)
\\
y=-2 & y=-8
.\end{array}
Hence, the points $
(1,-2)
$ and $
(2,-8)
$ are on the given parabola.
Reflecting the points above about the axis of symmetry, then the points $
(-1,-2)
$ and $
(-2, -8)
$ are also on the given parabola.
Using the points $\{(x,y)=
(-1,-2),(-2,-8),(0,0),(1,-2),(2,-8)
\}$ the graph of the given parabola is determined (see graph above).
Using the graph above the domain of the given function is the set of all real numbers. The range is $
\{y|y\ge0\}
$.
