Answer
Standard form: $f(x)=(x-3)^2-9$
Vertex: $(3,-9)$
Axis of symmetry: $x=3$
The x-intercepts: $(0,0)$ and $(6,0)$
Work Step by Step
$f(x)=x^2-6x~~$ ($a=1,~b=-6,~c=0$)
$f(x)=x^2-6x+9-9$
$f(x)=x^2-2(3)x-3^2-9$
$f(x)=(x-3)^2-9$
$-\frac{b}{2a}=-\frac{-6}{2(1)}=3$
$f(3)=3^2-6(3)=9-18=-9$
Vertex: $(-\frac{b}{2a},f(-\frac{b}{2a}))=(3,-9)$
So, the axis of symmetry is $x=-\frac{b}{2a}=3$.
The x-intercepts:
$f(x)=0$
$x^2-6x=0$
$x(x-6)=0$
$x=0$ or $x-6=0~→~x=6$
So, the x-intercept points are: $(0,0)$ and $(6,0)$
