The parabola with equation $y=-x^{2}+16$ intersects the $x$-axis at points $A$ and $B$ and the horizontal line with equation $y=7$ at points $M$ and $N$, as shown.

(a) Determine the coordinates of $A$ and $B$.

(b) Determine the area of the trapezoid $M N B A$.

(c) Suppose that $O$ is the origin and $V$ is the vertex of the parabola. The line $y=-33$ intersects the parabola at points $P$ and $Q$. Determine the area of quadrilateral $V P O Q$, which is shaded in the diagram below.
![](https://cdn.mathpix.com/cropped/2024_04_17_5629b903ac0448ea13d7g-3.jpg?height=616&width=1134&top_left_y=1611&top_left_x=468)
(a) Points $A$ and $B$ are the $x$-intercepts of the parabola.

To find their coordinates, we set $y=0$ to obtain the equation $0=-x^{2}+16$, which gives $x^{2}=16$ and so $x= \pm 4$.

Therefore, $A$ has coordinates $(-4,0)$ and $B$ has coordinates $(4,0)$.

(b) We start by finding the coordinates of $M$ and $N$.

Since these points lie on the horizontal line with equation $y=7$ and on the parabola with equation $y=-x^{2}+16$, we equate expressions for $y$ to obtain $7=-x^{2}+16$ which gives $x^{2}=9$ and so $x= \pm 3$.

Thus, $M$ has coordinates $(-3,7)$ and $N$ has coordinates $(3,7)$.

Trapezoid $M N B A$ has parallel, horizontal sides $M N$ and $A B$.

Here, $M N=3-(-3)=6$ and $A B=4-(-4)=8$.

Also, the height of $M N B A$ is 7 since $M N$ lies along $y=7$ and $A B$ lies along $y=0$.

Therefore, the area of $M N B A$ is $\frac{1}{2}(6+8) \cdot 7=49$.

(c) The origin, $O$, has coordinates $(0,0)$.

The vertex, $V$, of the parabola has coordinates $(0,16)$, since it lies on the $y$-axis and substituting $x=0$ into the equation $y=-x^{2}+16$ gives $y=16$.

When we set $y=-33$ in the equation of the parabola $y=-x^{2}+16$, we obtain $x^{2}=49$ and so $x= \pm 7$.

The area of quadrilateral $V P O Q$ can be found by adding the areas of $\triangle V O P$ and $\triangle V O Q$. Each of these triangles can be thought of as having a vertical base $V O$ of length 16, and a horizontal height.

The length of each horizontal height is 7 , since the distance from each of $P$ and $Q$ to the $y$-axis is 7 .

Therefore, the area of $V P O Q$ is $2 \cdot \frac{1}{2} \cdot 16 \cdot 7=112$.