91Ó°ÊÓ

We have $f\left(x\right)=-x^2+8x-16$.

The parabola will open downwards here, because the leading coefficient is negative. Then by finding the axis of symmetry, vertex, the intercepts and a point with help of axis of symmetry, will help us to plot the points of the graph and get the required graph.

Axis of symmetry is a vertical line given by $x=\frac{-b}{2a}$where $a,b,c$are the coefficients of the equation when compared with the standard form of the quadratic equation, i.e., $a{x}^2+bx+c$.

So the axis of symmetry comes out to be,
$$\begin{gathered} x=\frac{-b}{2a} \\ x=-\frac{8}{2(-1)} \\ x=\frac{8}{2} \\ x=4 \end{gathered}$$
Thus the equation of axis of symmetry is$x=4$.

Vertex is a point on the line of axis of symmetry. Here axis of symmetry is $x=4$.

We will get the vertex by putting $x=4$ in the equation. This gives us,
$$\begin{gathered} =-x^2+8x-16 \\ =-{\left(4\right)}^2+8\left(4\right)-16 \\ =-16+32-16 \\ =-32+32 \\ =0 \end{gathered}$$
We get the vertex as$(4,0)$.

To find $y-$intercept, we need to put$x=0$ in the equation. This gives us,
role="math" localid="1645173778717" $$\begin{gathered} =-x^2+8x-16 \\ =-{\left(0\right)}^2+8\left(0\right)-16 \\ =0+0-16 \\ =-16 \end{gathered}$$
The intercept comes out to be$(0,-16)$.

To find $x-$intercept, we need to put $f\left(x\right)=0$. So,

localid="1664203339918" $$\begin{gathered} -x^2+8x-16=0 \\ -(x^2-8x+16)=0 \\ -{(x-4)}^2=0 \\ {(x-4)}^2=0 \\ x-4=0 \\ x=4 \end{gathered}$$
The intercept comes out to be$(4,0)$.

We have $y-$intercept as $(0,-16)$.

The axis of symmetry is $x=4$.

It means that the intercept is at a distance of$4$units to the left of axis of symmetry.
We can get a point on the graph using this information.
A point at the same distance, i.e.,$\mathit{4}$ units, but to the right will also lie on the graph. This point will be$(8,-16)$.

Plotting all the points we got and the axis of symmetry give us a graph. Joining these points will give us the graph as-