Answer
Vertex: $(4,0)$
Axis of Symmetry: $x=4$
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-4)^2
,$ is equivalent to
\begin{align*}\require{cancel}
f(x)=(x-4)^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 $
(4,0)
$.
The axis of symmetry is given by $x=h$. With $h$ given above as $h=0,$ then the axis of symmetry is $
x=4
$.
Let $y=f(x).$ Then, $y=
-x^2+2
$. By substitution,
\begin{array}{l|r}
\text{If }x=3: & \text{If }x=2:
\\\\
y=(x-4)^2 & y=(x-4)^2
\\
y=(3-4)^2 & y=(2-4)^2
\\
y=(-1)^2 & y=(-2)^2
\\
y=1 & y=4
.\end{array}
Hence, the points $
(3,1)
$ and $
(2,4)
$ are on the given parabola.
Reflecting these points about the axis of symmetry, then the points $
(5,1)
$ and $
(6,4)
$ are also on the given parabola.
Using the points $\{(x,y)=
(3,1),(2,4),(4,0),(5,1),(6,4)
\}$ 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\}
$.