Question

y = -x² + 36
that negative x - values indicate the distance to the left of the center line, and positive x - values indicate the distance to the right of the center line.
the points on the graph where the height of the tunnel is a maximum and where the height of the tunnel is zero feet

Explanation:

Step1: Identify the function

The given function is $y = -x^2 + 36$.

Step2: Find maximum height

This is a downward-opening parabola ($a=-1<0$), so vertex is the maximum. For $y=ax^2+bx+c$, vertex x-coordinate is $x=-\frac{b}{2a}$. Here $b=0$, so $x=0$. Substitute $x=0$:
$y = -(0)^2 + 36 = 36$
Maximum point is $(0, 36)$.

Step3: Find height = 0 points

Set $y=0$ and solve for $x$:
$0 = -x^2 + 36$
$x^2 = 36$
$x = \pm\sqrt{36} = \pm6$
Points are $(-6, 0)$ and $(6, 0)$.

Answer:

Maximum height point: $(0, 36)$
Zero height points: $(-6, 0)$ and $(6, 0)$