1. **State the problem:** We are given the quadratic function $h(t) = -5t^2 + 20t + 2$ and want to analyze it. 2. **Formula and rules:** This is a quadratic function of the form $h(t) = at^2 + bt + c$ where $a = -5$, $b = 20$, and $c = 2$. 3. **Find the vertex:** The vertex of a parabola $y = at^2 + bt + c$ is at $t = -\frac{b}{2a}$. Calculate: $$t = -\frac{20}{2 \times (-5)} = -\frac{20}{-10} = 2$$ 4. **Find the maximum height (value of $h(t)$ at vertex):** $$h(2) = -5(2)^2 + 20(2) + 2 = -5(4) + 40 + 2 = -20 + 40 + 2 = 22$$ 5. **Find the roots (when $h(t) = 0$):** Solve $-5t^2 + 20t + 2 = 0$. Divide entire equation by $-1$ for simplicity: $$5t^2 - 20t - 2 = 0$$ Use quadratic formula: $$t = \frac{20 \pm \sqrt{(-20)^2 - 4 \times 5 \times (-2)}}{2 \times 5} = \frac{20 \pm \sqrt{400 + 40}}{10} = \frac{20 \pm \sqrt{440}}{10}$$ Simplify $\sqrt{440} = \sqrt{4 \times 110} = 2\sqrt{110}$: $$t = \frac{20 \pm 2\sqrt{110}}{10} = 2 \pm \frac{\sqrt{110}}{5}$$ 6. **Summary:** - The parabola opens downward (since $a = -5 < 0$). - The vertex is at $t=2$ with maximum height $h(2) = 22$. - The roots are at $t = 2 \pm \frac{\sqrt{110}}{5}$. This completes the analysis of the function $h(t)$.