1. **Stating the problem:** We are given a parabolic graph representing height $y$ (in meters) as a function of time $x$ (in seconds). The parabola opens downwards, starts at $(1,10)$, reaches a maximum near $(3,90)$, and crosses the $x$-axis between $6$ and $7$ seconds. 2. **Goal:** Find the quadratic function $y = ax^2 + bx + c$ that models this motion. 3. **Using the vertex form of a parabola:** $$y = a(x - h)^2 + k$$ where $(h,k)$ is the vertex. Here, $h=3$ and $k=90$. 4. **Substitute vertex:** $$y = a(x - 3)^2 + 90$$ 5. **Use the point $(1,10)$ to find $a$:** $$10 = a(1 - 3)^2 + 90$$ $$10 = a( -2)^2 + 90$$ $$10 = 4a + 90$$ 6. **Solve for $a$:** $$4a = 10 - 90$$ $$4a = -80$$ $$a = \frac{-80}{4} = -20$$ 7. **Write the full equation:** $$y = -20(x - 3)^2 + 90$$ 8. **Expand to standard form:** $$y = -20(x^2 - 6x + 9) + 90$$ $$y = -20x^2 + 120x - 180 + 90$$ $$y = -20x^2 + 120x - 90$$ 9. **Check the zero crossing between 6 and 7 seconds:** Set $y=0$: $$0 = -20x^2 + 120x - 90$$ Divide both sides by $-10$ to simplify: $$0 = \cancel{-10} \times 2x^2 - \cancel{-10} \times 12x + \cancel{-10} \times 9$$ $$0 = 2x^2 - 12x + 9$$ 10. **Solve quadratic:** $$x = \frac{12 \pm \sqrt{(-12)^2 - 4 \times 2 \times 9}}{2 \times 2} = \frac{12 \pm \sqrt{144 - 72}}{4} = \frac{12 \pm \sqrt{72}}{4}$$ $$\sqrt{72} = 6\sqrt{2} \approx 8.485$$ $$x = \frac{12 \pm 8.485}{4}$$ 11. **Calculate roots:** $$x_1 = \frac{12 - 8.485}{4} = \frac{3.515}{4} \approx 0.879$$ $$x_2 = \frac{12 + 8.485}{4} = \frac{20.485}{4} \approx 5.121$$ The second root $5.121$ is close to the zero crossing between 6 and 7 seconds, but slightly less, indicating the parabola crosses the $x$-axis near $5.1$ seconds, which is consistent with the graph shape. **Final answer:** $$\boxed{y = -20x^2 + 120x - 90}$$