Answer
Vertex: $(-1,0)$
Axis of Symmetry: $x=-1$
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)=(x+1)^2
,$ is equivalent to
\begin{align*}\require{cancel}
f(x)=(x-(-1))^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 $
(-1,0)
$.
The axis of symmetry is given by $x=h$. With $h$ given above as $
h=-1
,$ then the axis of symmetry is $
x=-1
$.
Let $y=f(x).$ Then, $y=
(x+1)^2
$. By substitution,
\begin{array}{l|r}
\text{If }x=-3: & \text{If }x=-2:
\\\\
y=(x+1)^2 & y=(x+1)^2
\\
y=(-3+1)^2 & y=(-2+1)^2
\\
y=(-2)^2 & y=(-1)^2
\\
y=4 & y=1
.\end{array}
Hence, the points $
(-3,4)
$ and $
(-2,1)
$ are on the given parabola.
Reflecting these points about the axis of symmetry, then the points $
(0,1)
$ and $
(1,4)
$ are also on the given parabola.
Using the points $\{(x,y)=
(-3,4),(-2,1),(-1,0),(0,1),(1,4)
\}$ 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\ge0\}
$.
