Answer
Standard form:
$f(x)=(x-4)^2-4$
Vertex: $(4,-4)$
Axis of symmetry: $x=4$.
The x-intercepts: $(6,0)$ and $(2,0)$
Work Step by Step
$f(x)=x^2-8x+12~~$ ($a=1,~b=-8,~c=12$)
$f(x)=x^2-8x+16-16+12$
$f(x)=(x^2-2(4)x+4^2)-4$
$f(x)=(x-4)^2-4$
$-\frac{b}{2a}=-\frac{-8}{2(1)}=4$
$f(-1)=(4-4)^2-4=-4$
Vertex: $(-\frac{b}{2a},f(-\frac{b}{2a}))=(4,-4)$
So, the axis of symmetry is $x=-\frac{b}{2a}=4$.
The x-intercepts:
$f(x)=0$
$(x-4)^2-4=0$
$(x-4)^2=4$
$x-4=±2$
$x=6$ or $x=2$
So, the x-intercept points are: $(6,0)$ and $(2,0)$
