1. **State the problem:** We need to write a quadratic function representing a bridge with the following conditions: - Width (distance between x-intercepts) is 5 - Maximum height is 10 - The parabola opens downwards 2. **Choose the form:** The vertex form of a quadratic is useful here: $$y = a(x-h)^2 + k$$ where $(h,k)$ is the vertex. 3. **Identify vertex:** The maximum height is 10, so the vertex is at $(h,k) = (\frac{5}{2}, 10)$ because the parabola is symmetric and the width between x-intercepts is 5, so the midpoint is at $x=\frac{5}{2}$. 4. **Use x-intercepts:** The x-intercepts are 0 and 5, so the parabola passes through points $(0,0)$ and $(5,0)$. 5. **Plug vertex form and point to find $a$:** $$0 = a(0 - \frac{5}{2})^2 + 10$$ $$0 = a\left(\frac{-5}{2}\right)^2 + 10$$ $$0 = a \cdot \frac{25}{4} + 10$$ 6. **Solve for $a$:** $$a \cdot \frac{25}{4} = -10$$ $$a = \frac{-10 \times 4}{25} = \frac{-40}{25} = -\frac{8}{5}$$ 7. **Write the function:** $$y = -\frac{8}{5} \left(x - \frac{5}{2}\right)^2 + 10$$ 8. **Summary:** The quadratic function representing the bridge is $$y = -\frac{8}{5} \left(x - \frac{5}{2}\right)^2 + 10$$ which has x-intercepts at 0 and 5, maximum height 10, and opens downwards.