1. **Stating the problem:** We have data for the height $h$ of a ball at different times $t$ in seconds, and we want to find the quadratic function that models this height, find the maximum height, and estimate the height at $t=2.5$ seconds. 2. **Given data:** $$\begin{array}{c|ccccc} t (s) & 2 & 2.2 & 2.4 & 2.6 & 3 \\ h (\text{dyuym}) & 2 & 16 & 26 & 33 & 42 \end{array}$$ 3. **Modeling with a quadratic function:** We assume height $h$ depends on time $t$ as $$h = at^2 + bt + c$$ where $a$, $b$, and $c$ are constants to be found. 4. **Using three points to find $a$, $b$, and $c$:** Choose points $(2,2)$, $(2.4,26)$, and $(3,42)$. Set up equations: $$\begin{cases} 4a + 2b + c = 2 \\ 5.76a + 2.4b + c = 26 \\ 9a + 3b + c = 42 \end{cases}$$ 5. **Subtract first from second:** $$\cancel{c} + 5.76a + 2.4b - (\cancel{c} + 4a + 2b) = 26 - 2$$ $$1.76a + 0.4b = 24$$ 6. **Subtract first from third:** $$\cancel{c} + 9a + 3b - (\cancel{c} + 4a + 2b) = 42 - 2$$ $$5a + b = 40$$ 7. **Solve system:** From step 6, $$b = 40 - 5a$$ Substitute into step 5: $$1.76a + 0.4(40 - 5a) = 24$$ $$1.76a + 16 - 2a = 24$$ $$-0.24a = 8$$ $$a = -\frac{8}{0.24} = -33.33$$ 8. **Find $b$:** $$b = 40 - 5(-33.33) = 40 + 166.65 = 206.65$$ 9. **Find $c$ using first equation:** $$4(-33.33) + 2(206.65) + c = 2$$ $$-133.32 + 413.3 + c = 2$$ $$c = 2 - 279.98 = -277.98$$ 10. **Quadratic function:** $$h = -33.33t^2 + 206.65t - 277.98$$ 11. **Find maximum height:** Vertex time $$t_{max} = -\frac{b}{2a} = -\frac{206.65}{2(-33.33)} = 3.1 \text{ seconds}$$ 12. **Maximum height:** $$h_{max} = -33.33(3.1)^2 + 206.65(3.1) - 277.98 = 44.9 \text{ dyuym}$$ 13. **Estimate height at $t=2.5$ seconds:** $$h(2.5) = -33.33(2.5)^2 + 206.65(2.5) - 277.98 = 34.3 \text{ dyuym}$$ **Final answers:** - Quadratic function: $$h = -33.33t^2 + 206.65t - 277.98$$ - Maximum height: $$44.9$$ dyuym at $$t=3.1$$ seconds - Height at $$t=2.5$$ seconds: $$34.3$$ dyuym