一名土木工程师正在设计一座长 131 米、高 40 米的桥梁，桥身将包含四个完全相同的抛物线拱。每个拱高 34 米，相邻两拱的底端之间相距 1 米，如图所示。建立坐标系，以第一个拱底端的左端点为原点。

1. Find the width of a single arch

Let $w$ be the width of one arch.
The total length of the bridge consists of 4 arches and 3 gaps of 1 m each.
$4w + 3(1) = 131$
$4w = 128$
$w = 32$

2. Determine the $x$-intercepts of Arch 1

Since Arch 1 starts at the origin ($x = 0$) and has a width of $32$ m, its $x$-intercepts are at $x = 0$ and $x = 32$.

1. Substitute the $x$-intercepts of Arch 1 into the factored form of the quadratic equation.

From part (a), the $x$-intercepts of Arch 1 are $0$ and $32$.
Thus, $p = 0$ and $q = 32$.
The equation is $y = a(x - 0)(x - 32) = ax(x - 32)$.

2. Use the maximum height of the arch to find the value of $a$.

The maximum height occurs at the midpoint of the $x$-intercepts, which is $x = \frac{0 + 32}{2} = 16$.
The height of the arch is $34$ m, so the vertex is at $(16, 34)$.
Substitute $x = 16$ and $y = 34$ into the equation:
$34 = a(16)(16 - 32)$
$34 = a(16)(-16)$
$34 = -256a$
$a = -\frac{34}{256} = -\frac{17}{128}$ (shown)

1. Determine the $x$-intercepts of Arch 1 from the context

Let the width of each arch be $w$. There are 4 arches and 3 gaps of 1 m each.
$4w + 3(1) = 131$
$4w = 128 \implies w = 32$
The $x$-intercepts of Arch 1 are $x = 0$ and $x = 32$.

2. Determine the $x$-intercepts of Arch 2

Arch 2 starts after Arch 1 and a 1 m gap.
Start of Arch 2: $x = 32 + 1 = 33$
End of Arch 2: $x = 33 + 32 = 65$
The $x$-intercepts of Arch 2 are $x = 33$ and $x = 65$.

3. Form the equation of Arch 2

Since the arches are identical, the value of $a$ is the same as Arch 1, $a = -\frac{17}{128}$.
The equation of Arch 2 is $y = -\frac{17}{128}(x - 33)(x - 65)$.

4. Check if the point $(50, 30)$ satisfies the equation of Arch 2

Substitute $x = 50$ into the equation of Arch 2:
$y = -\frac{17}{128}(50 - 33)(50 - 65)$
$y = -\frac{17}{128}(17)(-15)$
$y = \frac{4335}{128} = 33.867...$
Since $y \neq 30$, the point $(50, 30)$ does not lie on Arch 2.