Answer
$$
f(x)=-x^{2}+6x-6
$$
The $x$-intercepts are
$$
x_{1}=3-\sqrt{3} \approx 1.268 ,\:x_{2}=3+\sqrt{3} \approx 4.732
$$
The $y$-intercept is -6.
Vertex parabola is $\left(3 , 3\right)$.
The axis is $x=3$ , the vertical line through the vertex.
Plotting the vertex, the $y$-intercept,the $x$-intercepts. and the point $\left(3 , 3\right)$ gives the graph in Figure
Work Step by Step
$$
f(x)=-x^{2}+6x-6
$$
The $x$-intercepts can be found by letting $f(x)=0$ to get
$$
f(x)=-x^{2}+6x-6=0
$$
This does not appear to factor, so we’ll try the quadratic formula.
$$
\:x_{1,\:2}=\frac{-6\pm \sqrt{6^2-4\left(-1\right)\left(-6\right)}}{2\left(-1\right)}
$$
from which x=3-\sqrt{3},\:x=3+\sqrt{3}
$$
x_{1}=3-\sqrt{3} \approx 1.268 ,\:x_{2}=3+\sqrt{3} \approx 4.732
$$
are the $x$-intercepts.
To find the $y$-intercept , set $x=0 $
$$
f(x)=-(0)^{2}+6.(0)-6=-6.
$$
So the $y$-intercept is -6.
The $x$-coordinate of the vertex is :
$$
x=\frac{-b}{2a}=-\frac{-6}{-2}=3
$$
Substituting this into the equation gives
$$
f(x)=-(3)^{2}+6(3)-6=-9+18-6=3
$$
The vertex is $\left(3 , 3\right)$.
The axis is $x=3 $ , the vertical line through the vertex.
Plotting the vertex, the $y$-intercept,the $x$-intercepts. and the point $\left(3 , 3\right)$ gives the graph in Figure