Question: The graph shows a downward-opening parabola on an h-versus-t coordinate grid, with labeled points $(0, 1.6)$ and $(4, 80)$; the vertex is at approximately $(4, 80)$, and the curve returns to the t-axis near $(8, 0)$. Ontario at a Canada Day fireworks display. It is a parabola, where $h$ is the height of the rocket above the water in metres, and $t$ is the time in seconds. Find an equation to describe the flight of the rocket. Show your work. Find the equation for this parabola.
1. **State the problem:** We need to find the equation of a parabola describing the height $h$ of a rocket above water as a function of time $t$. The parabola opens downward, has vertex at $(4, 80)$, passes through $(0, 1.6)$, and returns to the $t$-axis near $(8, 0)$.
2. **Formula for a parabola with vertex:** The vertex form of a parabola is $$h = a(t - h)^2 + k$$ where $(h, k)$ is the vertex. Here, vertex is $(4, 80)$, so
$$h = a(t - 4)^2 + 80$$
3. **Use the point $(0, 1.6)$ to find $a$:** Substitute $t=0$, $h=1.6$:
$$1.6 = a(0 - 4)^2 + 80$$
$$1.6 = a(16) + 80$$
4. **Solve for $a$:**
$$a(16) = 1.6 - 80$$
$$a(16) = -78.4$$
Divide both sides by 16:
$$\cancel{16}a = \frac{-78.4}{\cancel{16}}$$
$$a = -4.9$$
5. **Write the final equation:**
$$h = -4.9(t - 4)^2 + 80$$
6. **Check with the point near $(8, 0)$:**
$$h = -4.9(8 - 4)^2 + 80 = -4.9(4)^2 + 80 = -4.9(16) + 80 = -78.4 + 80 = 1.6$$
This is close to zero, indicating the rocket returns near the water surface at $t=8$ seconds.
**Answer:** The equation describing the flight of the rocket is
$$h = -4.9(t - 4)^2 + 80$$