Question 901·Hard·Math

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The height $h$ , in meters, of a stunt drone above a stage $t$ seconds after takeoff is modeled by a quadratic function.

The model estimates that the drone reached its maximum height of $72$ meters $3$ seconds after takeoff and then touched down on the stage $6$ seconds after takeoff.

According to the model, what was the height, in meters, of the drone $4.5$ seconds after takeoff?

Your answer

A stated maximum identifies the vertex, the parabola's peak, while a landing gives a height of zero. Write the model in vertex form, then use a one-unknown regression in Desmos to find its missing coefficient and evaluate the requested time. Don't split the height evenly because a time is halfway to landing; a quadratic's height does not change at a steady rate.

Hints

Hint 1
The vertex is the peak of the parabola. In $h(t)=a(t-p)^2+q$ , $p$ is the time of the peak and $q$ is its height. Which numbers belong in those two places?
Hint 2
Touching the stage means the height is zero. Substitute the landing time into your vertex-form expression to make an equation for the missing coefficient $a$ .
Hint 3
A regression uses $\sim$ to find a missing number in Desmos. Once you've found $a$ , evaluate the model at the requested time, not at the landing time.

Step-by-step

Approach 1: Build the model from its peak

Step 1
Write the height in vertex form
The vertex is the highest point here, so its coordinates are $(3,72)$ : time comes first, then height. Write $h(t)=a(t-3)^2+72$ . The square is $0$ at $t=3$ , making the height $72$ . Only the coefficient $a$ , which controls the drop away from the peak, remains unknown.
Step 2
Turn the touchdown into an equation
Touching the stage at $t=6$ means height zero, or $h(6)=0$ . Substitute that time into the model: $0=a(6-3)^2+72$ . The peak can't find $a$ because its squared term is $0$ for every value of $a$ ; the touchdown gives the new information you need.
Step 3
Find the missing coefficient
Type $0\sim a(6-3)^2+72$ . The regression symbol $\sim$ tells Desmos to find $a$ from the touchdown equation. Under PARAMETERS, Desmos shows $a=-8$ . Its negative sign fits a height that falls on either side of the maximum.
Step 4
Evaluate the requested height
Keep the regression and type $a(4.5-3)^2+72$ on the next line. Desmos uses its stored value of $a$ and prints $54$ . According to the model, the drone was $54$ meters above the stage $4.5$ seconds after takeoff. Grid in 54.

Approach 2: Compare distances from the peak

Step 1
Compare the two times with the peak
Count from the vertex, the peak, to touchdown: $6-3=3$ . Count from the peak to the requested time: $4.5-3=1.5$ . The requested time is half as far from the peak as touchdown is.
Step 2
Scale the drop, not the height
In a quadratic model, the drop below the peak depends on the square of the time from the peak. So half the distance means one-quarter of the drop. Touchdown is $72$ meters below the peak, so the drop at the requested time is: $\frac{72}{4}=18$ .
Step 3
Find the height above the stage
Subtract that drop from the maximum height: $72-18=54$ . The drone's modeled height after $4.5$ seconds is $54$ meters above the stage. Grid in 54.

Question 901·Hard·Math

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