Answer
Standard form: $f(x)=(x-4)^2-16$
Vertex: $(4,-16)$
Axis of symmetry: $x=4$.
The x-intercepts: $(0,0)$ and $(8,0)$
Work Step by Step
$f(x)=x^2-8x~~$ ($a=1,~b=-8,~c=0$)
$f(x)=x^2-8x+16-16$
$f(x)=x^2-2(4)x-4^2-16$
$f(x)=(x-4)^2-16$
$-\frac{b}{2a}=-\frac{-8}{2(1)}=4$
$f(3)=4^2-8(4)=16-32=-16$
Vertex: $(-\frac{b}{2a},f(-\frac{b}{2a}))=(4,-16)$
So, the axis of symmetry is $x=-\frac{b}{2a}=4$.
The x-intercepts:
$f(x)=0$
$x^2-8x=0$
$x(x-8)=0$
$x=0$ or $x-8=0~→~x=8$
So, the x-intercept points are: $(0,0)$ and $(8,0)$
