91Ó°ÊÓ

We have:

$y=-x^2-8x+16$

The equation of a vertical parabola with vertex $\left(h,k\right)$and focus $\left(h,k+\frac{1}{4a}\right)$is given by:

$y=a{\left(x+h\right)}^2+k$

Convert the given equation into standard form:

role="math" localid="1645425227927" $$\begin{gathered} x^2+8x=16-y \\ {x}^2+8x=-y+16 \\ {x}^2+8x+16=-y+16+16 \\ {\left(x+4\right)}^2=-y+32 \\ {\left(x+4\right)}^2=-\left(y-32\right) \end{gathered}$$

Compare the equation ${\left(x+4\right)}^2=-\left(y-32\right)$with standard form, we get:

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

Here, $a=-1$

A vertical parabola in standard form with a negative $a$opens downward.

Therefore, the parabola opens downward.

The vertex of a vertical parabola in standard form:

$y=-1{\left(x-\left(-4\right)\right)}^2+32$is:

$\left(-4,32\right)$

The axis of symmetry of a vertical parabola is given by:

$x=h$

$x=-4$

We have:

$y=-1{\left(x-\left(-4\right)\right)}^2+32$

By comparing the equation with standard parabola $y=a{\left(x-h\right)}^2+k$, we have:

$\left(h,k\right)=\left(-4,32\right)$

We have:

$y=-x^2-8x+16$

To find x-intercept: Substitute $y=0:$

$-x^2-8x+16=0$

For a quadratic equation of the form $a{x}^2+bx+c=0$, the solutions are:

$x_{1, 2}=\frac{-bÂ\pm \sqrt{b^2-4ac}}{2a}$

$$\begin{gathered} \mathrm{For} a=-1, b=-8, c=16 \\ {x}_{1, 2}=\frac{-\left(-8\right)Â\pm \sqrt{{\left(-8\right)}^2-4\left(-1\right)· 16}}{2\left(-1\right)} \\ {x}_{1, 2}=\frac{-\left(-8\right)Â\pm  8\sqrt{2}}{2\left(-1\right)} \\ x=-4\left(1+\sqrt{2}\right), x=4\left(\sqrt{2}-1\right) \end{gathered}$$

$\mathrm{X} \mathrm{Intercepts}: \left(-4\left(1+\sqrt{2}\right), 0\right), \left(4\left(\sqrt{2}-1\right), 0\right)$

Now, find the $y-$intercepts.

Substitute $x=0:$

$y=-{\left(0\right)}^2-8\left(0\right)+16$

$\mathrm{Y} \mathrm{Intercepts}: \left(0, 16\right)$

We have:

$y=-x^2-8x+16$

By comparing the equation with quadratic equation $y=a{x}^2+bx+c$, we get:

$a=-1$

If $a<0$then the vertex is a minimum value.

If $a>0$then the vertex is a maximum value.

Here, $a=-1$

Therefore,$\mathrm{Maximum}\left(-4, 32\right)$