To solve this problem, let's model the parabolic arch using a quadratic equation in the form:

$y = ax^2 + bx + c$

Given the conditions of the problem:

• The vertex of the parabola is at the highest point of the arch. This is at the center of the bridge, so at $x = 0$, the height is 5 feet. Hence, the vertex is $(0, 5)$.
• The horizontal distance of 26 feet means that the span of the bridge is 26 feet. Therefore, the parabola intersects the x-axis at $x = 13$ and $x = -13$.

Since the parabola is symmetric and has its vertex at the center, the equation simplifies to:

$y = a(x - h)^2 + k$

Where $h = 0$ and $k = 5$. Therefore, the equation becomes:

$y = ax^2 + 5$

We now use the information that at $x = 13$, $y = 0$ (since the height is zero at the edges of the bridge):

$0 = a(13)^2 + 5$

Now, solve for $a$:

$0 = 169a + 5$ $169a = -5$ $a = -\frac{5}{169}$

Now that we have the equation of the parabola:

$y = -\frac{5}{169}x^2 + 5$

Next, we need to find the height of the arch at a point 5 feet from the center. This corresponds to $x = 5$. Substituting $x = 5$ into the equation:

$y = -\frac{5}{169}(5)^2 + 5$ $y = -\frac{5}{169}(25) + 5$ $y = -\frac{125}{169} + 5$ $y = 5 - \frac{125}{169}$ $y = \frac{845}{169} - \frac{125}{169}$ $y = \frac{720}{169}$

Finally, divide:

$y \approx 4.26 \text{ feet}$

So, the height of the arch at a point 5 feet from the center is approximately 4.26 feet.

---

Would you like more details on any specific step? Here are some related questions you might find interesting:

1. How would the equation change if the highest point of the arch was 10 feet instead of 5 feet?
2. What would the height of the arch be at 10 feet from the center?
3. How can we calculate the horizontal distance from the center where the arch is 3 feet high?
4. What is the significance of the vertex in the parabola equation in real-world applications like this one?
5. How would the equation change if the arch were inverted, forming a downward-facing parabola?

Tip: When dealing with parabolas, always start by identifying the vertex, as it gives you a strong foundation for forming the equation.