1. **State the problem:** Given the height function $$h = -5t^2 + 290t + 600$$, find the height at $$t=10$$, the times when height is zero, and rewrite the function in vertex form. 2. **Find height at $$t=10$$:** Substitute $$t=10$$ into the equation: $$h = -5(10)^2 + 290(10) + 600$$ $$h = -5(100) + 2900 + 600$$ $$h = -500 + 2900 + 600$$ $$h = 3000$$ 3. **Find times when height is zero:** Set $$h=0$$: $$0 = -5t^2 + 290t + 600$$ Divide both sides by $$-5$$: $$0 = \cancel{-5}t^2 \cancel{/ -5} - \frac{290}{-5}t - \frac{600}{-5}$$ $$0 = t^2 - 58t - 120$$ Factor the quadratic: $$0 = (t - 60)(t + 2)$$ So, $$t = 60 \text{ or } t = -2$$ 4. **Rewrite in vertex form:** Complete the square: Start with $$h = -5t^2 + 290t + 600$$ Factor out $$-5$$ from the first two terms: $$h = -5(t^2 - 58t) + 600$$ Complete the square inside the parentheses: $$t^2 - 58t + \left(\frac{58}{2}\right)^2 - \left(\frac{58}{2}\right)^2 = (t - 29)^2 - 841$$ Substitute back: $$h = -5\left((t - 29)^2 - 841\right) + 600$$ Distribute $$-5$$: $$h = -5(t - 29)^2 + 4205 + 600$$ $$h = -5(t - 29)^2 + 4805$$ **Final answers:** - Height at $$t=10$$ is $$3000$$. - Height is zero at $$t=60$$ and $$t=-2$$. - Vertex form of the height function is $$h = -5(t - 29)^2 + 4805$$.