Answer
Standard form: $f(x)=(x-4)^2$
Vertex: $(4,0)$
Axis of symmetry: $x=4$.
The x-intercept: $(4,0)$
Work Step by Step
$f(x)=x^2-8x+16~~$ ($a=1,~b=-8,~c=16$)
$f(x)=x^2-2(4)x+4^2$
$f(x)=(x-4)^2$
$-\frac{b}{2a}=-\frac{-8}{2(1)}=4$
$f(4)=(4-4)^2=0$
Vertex: $(-\frac{b}{2a},f(-\frac{b}{2a}))=(4,0)$
So, the axis of symmetry is $x=-\frac{b}{2a}=4$.
The x-intercepts:
$f(x)=0$
$(x-4)^2=0$
$x-4=0$
$x=4$
So, the x-intercept point is: $(4,0)$
