Answer
Vertex: $(0,-1)$
Axis of Symmetry: $x=0$
Domain: all real numbers
Range: $\{y|y\ge-1\}$
Work Step by Step
In the form $f(x)=a(x-h)^2+k,$ the given equation, $
f(x)=x^2-1
,$ is equivalent to
\begin{align*}\require{cancel}
f(x)=(x-0)^2-1
.\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,-1)
$.
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=
x^2-1
$. By substitution,
\begin{array}{l|r}
\text{If }x=1: & \text{If }x=2:
\\\\
y=x^2-1 & y=x^2-1
\\
y=(1)^2-1 & y=(2)^2-1
\\
y=1-1 & y=4-1
\\
y=0 & y=3
.\end{array}
Hence, the points $
(1,0)
$ and $
(2,3)
$ are on the given parabola.
Reflecting these points about the axis of symmetry, then the points $
(-1,0)
$ and $
(-2,3)
$ are also on the given parabola.
Using the points $\{(x,y)=
(1,0),(2,3),(0,-1),(-1,0),(-2,3)
\}$ the graph of the given parabola is determined (see graph above).
Using the graph above, the domain (set of all $x$'s used in the graph) of the given function is the set of all real numbers. The range (set of all $y$'s used in the graph) is $
\{y|y\ge-1\}
$.
