we have that
![y=-x^2+36](https://img.qammunity.org/2023/formulas/mathematics/high-school/bwlfuhbn7k61thffq6b599jtpan7n9h862.png)
we know that
Is the equation of a vertical parabola open down
so
the vertex is a maximum
step 1
convert the equation of the parabola in the vertex form
![y=-x^2+36](https://img.qammunity.org/2023/formulas/mathematics/high-school/bwlfuhbn7k61thffq6b599jtpan7n9h862.png)
![y-36=-x^2](https://img.qammunity.org/2023/formulas/mathematics/high-school/kfynfe95vze8ykmt4gr745fj3uogt9d298.png)
the vertex (h,k) is the point (0,36)
Part a) The point on the graph where the height of the tunnel is a maximum is (0,36)
Part b) The points on the graph where the height of the tunnel is zero feet is when y=0
so
for y=0
![-x^2+36=0](https://img.qammunity.org/2023/formulas/mathematics/high-school/r1pgjk5fmjaja70tqfydttlim5hb9hkqwo.png)
![x^2=36](https://img.qammunity.org/2023/formulas/mathematics/college/meu6utumwh6khchfadaxh1yg2cgi3qkryd.png)
![x=(+/-)6](https://img.qammunity.org/2023/formulas/mathematics/high-school/uozm11y0u3ul3r7s7igngunn1bc4n3q0sq.png)
the points are (-6,0) and (6,0)
see the attached figure