Notice that the equation is of the form $h = a(t-h_0)^2 + k$. In this form, what are the coordinates $(t, h)$ of the vertex in terms of $h_0$ and $k$?

The number multiplying $(t-2.1)^2$ is $-4.9$. Does a negative coefficient make the parabola open up or down, and does that make the vertex a highest point or a lowest point?

Once you know the coordinates of the vertex, remember that the horizontal coordinate is a time in seconds and the vertical coordinate is a height in meters. Which answer choice describes that time and height, and notes that it is a maximum?

In Desmos, enter the equation using $x$ for time and $y$ for height: y = -4.9(x - 2.1)^2 + 9. This will display the parabola modeling the droplet’s height over time.

Click on the highest point of the graph (or use the maximum/vertex feature). Desmos will show its coordinates $(x, y)$. Interpret $x$ as the time in seconds and $y$ as the height in meters, then choose the answer choice that describes that time and height as the maximum.

The equation $h = -4.9(t-2.1)^2 + 9$ gives height $h$ (in meters) as a function of time $t$ (in seconds). On the $t$-$h$ graph, the vertex of this parabola is its highest or lowest point; since the graph models height over time, that point corresponds to a specific time and height of the droplet.

The equation is already in vertex form:

Here, $a = -4.9$, $h_0 = 2.1$, and $k = 9$. So the vertex of the parabola in the $t$-$h$ plane is at the point $(2.1, 9)$, where $t = 2.1$ and $h = 9$.

Because the coefficient of the squared term is negative ($-4.9$), the parabola opens downward. That means the vertex is the highest point on the graph, so it represents the maximum height the droplet reaches.

The vertex $(2.1, 9)$ tells us that at $t = 2.1$ seconds, the height is $h = 9$ meters, and this is the maximum height. In words, this is: "The droplet reaches a maximum height of 9 meters 2.1 seconds after leaving the nozzle." This corresponds to choice D.