Graph$y=-x^2+8x-12$by using properties.

$y=-x^2+8x-12$.

Compare the given equation with the $y=a{x}^2+bx+c$.

we get, $a=-1,b=8,c=-12$.

Since $a$is $-1$, the parabola opens downward.

To find the axis of symmetry, $x=\frac{-b}{2a}$.

Substitute the known values in the formula.

$$\begin{gathered} x=\frac{-8}{2\left(-1\right)} \\ =\frac{8}{2} \\ =4 \end{gathered}$$

Substitute $x=4$in the given equation.

$$\begin{gathered} y=-{\left(4\right)}^2+8\left(4\right)-12 \\ =-16+32-12 \\ y=4 \end{gathered}$$

The vertex is,$\left(4,4\right)$.

The $y$intercept occurs when $x=0$.

So, substitute $x=0$in the given equation.

$$\begin{gathered} y=0+0-12 \\ y=-12 \end{gathered}$$

The$y$intercept is,$\left(0,-12\right)$.

The point $\left(0,-12\right)$is four times to the left of the line of symmetry. The point four units to the right of the line of symmetry is, $\left(4,-12\right)$.

The point symmetric to the $y$intercept is,$\left(4,-12\right)$.

The $x$- intercept occurs when $y=0$.

Substitute $y=0$in the given equation.

$0=-x^2+8x-12$.

Now factorize it.

$$\begin{gathered} \left(x-2\right)\left(x-6\right)=0 \\ x=2,6 \end{gathered}$$

The$x$intercepts are,$\left(2,0\right),\left(6,0\right)$.