A bridge shaped like a parabolic arch has a horizontal distance of 26 feet the center and the highest point of the arch is 5 feet above the ground. What is the height of the arch at a point of 5 feet from the center 

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.

A bridge shaped like a parabolic arch has a horizontal distance of 26 feet. The center and the highest point of the arch is 5 feet above the ground. What is the height of the arch at a point of 5 feet from the center?

Learn how to calculate the height of a parabolic arch at a specific point using quadratic equations. This example models a bridge with a span of 26 feet and a height of 5 feet at its highest point. Suitable for students in Grades 9-11, this step-by-step solution involves solving for a in quadratic equations and using the vertex form.

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