For each parabola, find (a) which ways it opens, (b) the axis of symmetry, (c) the vertex, (d) the x-and y-intercepts, and (e) the maximum or minimum value.

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

The given function is$y=-x^2-8x+16$

Compare the given equation with the standard form $y=a{x}^2+bx+c$we get $a=-1,b=-8,andc=16$

Since the value of a is negative, so the parabola opens down.

The axis of symmetry is the vertical line given by:

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

The axis of symmetry is$x=-4$.

Since the vertex is a point on the line of symmetry so the x-coordinate of the vertex is $-4$.

Now determine the respective y-coordinate.

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

The vertex is $(-4,32)$.

For y-intercept put $x=0$into the given equation.

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

The y-intercept is point $(0,16)$.

For x-intercept put $y=0$into the given equation.

$$\begin{gathered} -x^2-8x+16=0 \\ x=\frac{-(-8)\pm \sqrt{{(-8)}^2-4(-1)\left(16\right)}}{2(-1)} \\ x=-4\pm 4\sqrt{2} \end{gathered}$$

The x-intercepts are$(-4-4\sqrt{2},0)and(-4+4\sqrt{2},0)$

Since the parabola opens down, the graph will have the maximum point at the vertex.

The maximum value is 32 at$x=-4$.