1. **Problem 7: Write the quadratic equation of the parabolic arch in vertex form.** The arch is 6 m high at the vertex and 11 m wide at the base. The vertex is at the highest point, so the vertex coordinates are $\left(\frac{11}{2}, 6\right)$ or $\left(5.5, 6\right)$ since the arch is centered over the doorway. 2. **Vertex form of a parabola:** $$y = a(x - h)^2 + k$$ where $(h, k)$ is the vertex. 3. **Use the vertex:** $$y = a(x - 5.5)^2 + 6$$ 4. **Use the point $(0,0)$ which lies on the parabola to find $a$:** $$0 = a(0 - 5.5)^2 + 6$$ $$0 = a(30.25) + 6$$ $$a(30.25) = -6$$ $$a = \frac{-6}{30.25} = -\frac{24}{121}$$ 5. **Final quadratic equation in vertex form:** $$y = -\frac{24}{121}(x - 5.5)^2 + 6$$ --- 6. **Problem 8a: Determine the season ticket price that will maximize revenue.** Let $x$ be the number of $10$ decreases in price. - Original price: 1400 - New price: $1400 - 10x$ - Tickets sold: $2600 + 50x$ 7. **Revenue function:** $$R(x) = (1400 - 10x)(2600 + 50x)$$ 8. **Expand:** $$R(x) = 1400 \times 2600 + 1400 \times 50x - 10x \times 2600 - 10x \times 50x$$ $$R(x) = 3,640,000 + 70,000x - 26,000x - 500x^2$$ $$R(x) = 3,640,000 + 44,000x - 500x^2$$ 9. **Rewrite in standard form:** $$R(x) = -500x^2 + 44,000x + 3,640,000$$ 10. **Find vertex $x$ to maximize revenue:** $$x = -\frac{b}{2a} = -\frac{44,000}{2 \times (-500)} = \frac{44,000}{1,000} = 44$$ 11. **Price that maximizes revenue:** $$1400 - 10 \times 44 = 1400 - 440 = 960$$ --- 12. **Problem 8b: Find the maximum revenue.** 13. **Calculate $R(44)$:** $$R(44) = -500(44)^2 + 44,000(44) + 3,640,000$$ $$= -500 \times 1936 + 1,936,000 + 3,640,000$$ $$= -968,000 + 1,936,000 + 3,640,000 = 4,608,000$$ **Final answers:** - Problem 7: $$y = -\frac{24}{121}(x - 5.5)^2 + 6$$ - Problem 8a: Price to maximize revenue is 960 - Problem 8b: Maximum revenue is 4,608,000