At what times is the height 0? A quadratic with zeros at those times can be written in a factored form like $h(t)=a(t-r_1)(t-r_2)$. What would $r_1$ and $r_2$ be here?

Once you write $h(t)$ in factored form using the zeros, plug in the time and height of the maximum point to solve for the constant $a$.

After you have the full equation for $h(t)$, substitute $t=4.5$ and simplify carefully. Watch your arithmetic with decimals or fractions.

In Desmos, type the function using the zeros at 0 and 6: h(t) = 8t(6 - t). This defines the height $h$ as a function of time $t$.

On a new line in Desmos, type h(4.5) and let Desmos compute the value. This number is the model’s predicted height (in meters) 4.5 seconds after takeoff.

The problem says the height is modeled by a quadratic function and that:

• The drone reaches its maximum height of 72 meters at $t=3$ seconds. This gives the vertex $(3,72)$.
• The drone touches down on the stage at $t=6$ seconds, so $h(6)=0$.

Because the height is measured above the stage, the drone is also at height 0 at takeoff, so $h(0)=0$.

So the quadratic goes through the points $(0,0)$, $(3,72)$, and $(6,0)$, and opens downward (since it has a maximum).

Since the height is 0 at $t=0$ and $t=6$, $t=0$ and $t=6$ are the zeros (roots) of the quadratic. A quadratic with these zeros can be written in factored form as

for some constant $a$.

Now use the vertex point $(3,72)$ to find $a$:

So

Thus the model is

Now substitute $t=4.5$ into the function:

First compute the parentheses:

So

Multiply $4.5$ and $1.5$:

Now multiply by 8:

So the height of the drone $4.5$ seconds after takeoff is 54 meters.