1. **State the problem:** We have a quadratic function $f(x) = ax^2 + bx + c$ representing the height of a ball at distance $x$. Given points $(3, 2.36)$, $(10, 5)$, and $(17, 7.16)$ lie on this curve. 2. **Write equations from points:** Substitute each point into $f(x) = ax^2 + bx + c$. From $(3, 2.36)$: $$2.36 = a(3)^2 + b(3) + c = 9a + 3b + c$$ From $(10, 5)$: $$5 = a(10)^2 + b(10) + c = 100a + 10b + c$$ From $(17, 7.16)$: $$7.16 = a(17)^2 + b(17) + c = 289a + 17b + c$$ 3. **Set up system of equations:** $$\begin{cases} 9a + 3b + c = 2.36 \\ 100a + 10b + c = 5 \\ 289a + 17b + c = 7.16 \end{cases}$$ 4. **Subtract first equation from second and third to eliminate $c$:** Second minus first: $$100a + 10b + c - (9a + 3b + c) = 5 - 2.36$$ $$\cancel{c} + 100a + 10b - 9a - 3b - \cancel{c} = 2.64$$ $$91a + 7b = 2.64$$ Third minus first: $$289a + 17b + c - (9a + 3b + c) = 7.16 - 2.36$$ $$\cancel{c} + 289a + 17b - 9a - 3b - \cancel{c} = 4.8$$ $$280a + 14b = 4.8$$ 5. **Simplify second equation by dividing by 7:** $$\frac{91a + 7b}{7} = \frac{2.64}{7}$$ $$13a + b = 0.3771$$ Simplify third equation by dividing by 14: $$\frac{280a + 14b}{14} = \frac{4.8}{14}$$ $$20a + b = 0.3429$$ 6. **Subtract the two simplified equations to solve for $a$:** $$(13a + b) - (20a + b) = 0.3771 - 0.3429$$ $$13a + b - 20a - b = 0.0342$$ $$-7a = 0.0342$$ $$a = -\frac{0.0342}{7} = -0.004886$$ 7. **Find $b$ using $13a + b = 0.3771$:** $$13(-0.004886) + b = 0.3771$$ $$-0.06352 + b = 0.3771$$ $$b = 0.3771 + 0.06352 = 0.4406$$ 8. **Find $c$ using $9a + 3b + c = 2.36$:** $$9(-0.004886) + 3(0.4406) + c = 2.36$$ $$-0.04397 + 1.3218 + c = 2.36$$ $$c = 2.36 - 1.2778 = 1.0822$$ 9. **Quadratic model:** $$f(x) = -0.004886x^2 + 0.4406x + 1.0822$$ 10. **Check if ball goes over 4-m fence at 80 m:** $$f(80) = -0.004886(80)^2 + 0.4406(80) + 1.0822$$ $$= -0.004886(6400) + 35.248 + 1.0822$$ $$= -31.27 + 35.248 + 1.0822 = 5.06$$ Since $5.06 > 4$, the ball goes over the fence. 11. **Find horizontal distance when ball hits ground ($f(x) = 0$):** Solve: $$-0.004886x^2 + 0.4406x + 1.0822 = 0$$ Use quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ $$a = -0.004886, b = 0.4406, c = 1.0822$$ Calculate discriminant: $$\Delta = (0.4406)^2 - 4(-0.004886)(1.0822) = 0.1941 + 0.0211 = 0.2152$$ $$x = \frac{-0.4406 \pm \sqrt{0.2152}}{2(-0.004886)}$$ $$= \frac{-0.4406 \pm 0.464}{-0.009772}$$ Two solutions: $$x_1 = \frac{-0.4406 + 0.464}{-0.009772} = \frac{0.0234}{-0.009772} = -2.39$$ (discard negative distance) $$x_2 = \frac{-0.4406 - 0.464}{-0.009772} = \frac{-0.9046}{-0.009772} = 92.56$$ So, the ball hits the ground at approximately 92.56 meters. **Final answers:** - Quadratic model: $$f(x) = -0.004886x^2 + 0.4406x + 1.0822$$ - Ball height at 80 m: 5.06 m (over 4 m fence) - Horizontal distance when ball hits ground: 92.56 m