Answer
Standard form: $f(x)=(x-3)^2-8$
Vertex: $(3,-8)$
Axis of symmetry: $x=3$.
x-intercepts: $(3-2\sqrt 2,0)$ and $(3+2\sqrt 2,0)$
Work Step by Step
$f(x)=x^2-6x+1~~$ ($a=1,~b=-6,~c=1$)
$f(x)=x^2-6x+9-9+1$
$f(x)=[x^2-2(3)x+3^2]-8$
$f(x)=(x-3)^2-8$
$-\frac{b}{2a}=-\frac{-6}{2(1)}=3$
$f(3)=(3-3)^2-8=-8$
Vertex: $(-\frac{b}{2a},f(-\frac{b}{2a}))=(3,-8)$
So, the axis of symmetry is $x=-\frac{b}{2a}=3$.
The x-intercepts:
$f(x)=0$
$(x-3)^2-8=0$
$(x-3)^2=8$
$x-3=±2\sqrt 2$
$x=3±2\sqrt 2$
So, the x-intercept points are: $(3-2\sqrt 2,0)$ and $(3+2\sqrt 2,0)$
