Answer
Vertex: $(0,3)$
Axis of Symmetry: $x=0$
Domain: all real numbers
Range: $\{y|y\ge3\}$
Work Step by Step
In the form $f(x)=a(x-h)^2+k,$ the given equation, $
f(x)=x^2+3
,$ is equivalent to
\begin{align*}\require{cancel}
f(x)=(x-0)^2+3
.\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 $
(0,3)
$.
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+3
$. By substitution,
\begin{array}{l|r}
\text{If }x=1: & \text{If }x=2:
\\\\
y=x^2+3 & y=x^2+3
\\
y=(1)^2+3 & y=(2)^2+3
\\
y=1+3 & y=4+3
\\
y=4 & y=7
.\end{array}
Hence, the points $
(1,4)
$ and $
(2,7)
$ are on the given parabola.
Reflecting these points about the axis of symmetry, then the points $
(-1,4)
$ and $
(-2,7)
$ are also on the given parabola.
Using the points $\{(x,y)=
(1,4),(2,7),(0,3),(-1,4),(-2,7)
\}$ 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\ge3\}
$.
