Answer
Vertex: $(3,0)$
x-intercept: $(3,0)$
y-intercept: $(0,-9)$
Work Step by Step
$f(x)=-x^2+6x-9$
$a=-1$, $b=6$, $c=-9$
Vertex is at $x=-b/2a$
$x=-6/2*-1$
$x=-6/-2$
$x=3$
$f(x)=-x^2+6x-9$
$f(3)=-3^2+6*3-9$
$f(3)=-9+18-9$
$f(3)=0$
$(3,0)$ is the vertex and is the x-intercept.
$x=0$
$f(x)=-x^2+6x-9$
$f(0)=-0^2+6*0-9$
$f(0)=-0+0-9$
$f(0)=-9$
