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