1. **State the problem:** We are given the quadratic function $g(t) = -16t^2 + 30t$ which models height in feet as a function of time $t$ in seconds. 2. **Identify the coefficients:** The function is in the form $f(t) = at^2 + bt + c$ with $a = -16$, $b = 30$, and $c = 0$. 3. **Graph characteristics:** Since $a = -16 < 0$, the parabola opens downward. 4. **Find the vertex:** The vertex formula for $t$ is $t = -\frac{b}{2a}$. $$t = -\frac{30}{2 \times -16} = -\frac{30}{-32} = \frac{15}{16} = 0.9375$$ 5. **Calculate the vertex height:** Substitute $t = 0.9375$ into $g(t)$: $$g(0.9375) = -16(0.9375)^2 + 30(0.9375)$$ $$= -16(0.8789) + 28.125 = -14.0625 + 28.125 = 14.0625$$ So the vertex is approximately at $(0.94, 14.06)$, close to the given $(0.9, 14.1)$. 6. **Find the zeros (roots):** Solve $-16t^2 + 30t = 0$. Factor out $t$: $$t(-16t + 30) = 0$$ Set each factor to zero: $$t = 0$$ $$-16t + 30 = 0 \Rightarrow t = \frac{30}{16} = 1.875$$ So zeros are at $t=0$ and $t=1.875$, close to the given $(0,0)$ and $(1.9,0)$. 7. **Y-intercept:** At $t=0$, $g(0) = 0$, so the y-intercept is at $(0,0)$. **Summary:** The parabola opens downward with vertex near $(0.94,14.06)$, zeros at $0$ and $1.875$, and y-intercept at $(0,0)$. This matches the description given.