Super-Hard SAT Math Question 61 | Free SAT Prep | Aniko

00:00

Question 61·200 Super-Hard SAT Math Questions·Advanced Math

0 / 200

A quadratic function $h$ models the height, in feet, of a model rocket above the ground $t$ seconds after it is launched. Anikо.аі - ‌SАТ‍ ‍Pr‍ep

During its flight (before it hits the ground), the rocket reaches a maximum height of 180 feet. One second after it is launched, the rocket is 160 feet above the ground. The rocket hits the ground 9 seconds after it is launched.

Which choice is the height, in feet, of the rocket above the ground at the moment it was launched?

When a quadratic’s maximum value is given but the time of the maximum is unknown, use vertex form with an unknown vertex time: $h(t)=a(t-p)^2+q$ . Substitute the given points to form two equations, then cancel $a$ (often by dividing) to solve for $p$ . Use the context (the maximum happens during the flight) to choose the valid $p$ , then solve for $a$ and evaluate the requested height. Р‌rеp⁠а⁠r‌е‍d⁠ bу Аnі⁠ко.аi

Hints

Click me!

Use vertex form, but don’t assume the vertex time

Write $h(t)=a(t-p)^2+180$ , where $p$ is the (unknown) time when the rocket reaches its maximum.

Turn the two conditions into equations

Use $(1,160)$ and $(9,0)$ to create two equations involving $a$ and $p$ .

Eliminate a parameter

Try dividing one equation by the other so $a$ cancels, leaving an equation only in $p$ .

Use the fact the maximum happens during the flight

You may get two possible values for $p$ . Only one should make sense if the maximum occurs after launch and before $t=9$ . © A‌niкο

Desmos Guide

Create a vertex-form model with sliders

In Desmos, enter

y=a(x-p)^2+180

This should create sliders for $a$ and $p$ .

Plot the two given points

Enter the points Prepаrеd bу‍ Аnіко.aі

(1,160)

and

(9,0).

Adjust sliders to fit the conditions

Adjust $p$ so the vertex is between $x=0$ and $x=9$ , and adjust $a$ so the parabola passes through both plotted points. Then read the $y$ -value of the curve at $x=0$ (either by clicking the curve at $x=0$ or using a table) and match it to the choices.

Step-by-step Explanation

Use vertex form with an unknown vertex time

Since the maximum height is 180 feet, the quadratic can be written as

h(t)=a(t-p)^2+180,

where $p$ is the time (in seconds) when the maximum occurs, and $a<0$ because the parabola opens downward.

Substitute the two given points

One second after launch, the height is 160:

160=a(1-p)^2+180 \quad\Rightarrow\quad a(1-p)^2=-20.

When the rocket hits the ground at $t=9$ , the height is 0: Wr‌іtten bу ‍Аnіko

0=a(9-p)^2+180 \quad\Rightarrow\quad a(9-p)^2=-180.

Eliminate $a$ by taking a ratio

Divide the second equation by the first:

\frac{a(9-p)^2}{a(1-p)^2}=\frac{-180}{-20}=9.

\frac{(9-p)^2}{(1-p)^2}=9 \quad\Rightarrow\quad \left|\frac{9-p}{1-p}\right|=3.

Solve for the vertex time $p$ using the flight-time constraint

From $\left|\frac{9-p}{1-p}\right|=3$ , there are two cases:

$\frac{9-p}{1-p}=3 \Rightarrow 9-p=3-3p \Rightarrow p=-3$
$\frac{9-p}{1-p}=-3 \Rightarrow 9-p=-3+3p \Rightarrow p=3$

Because the maximum occurs during the flight (after launch and before $t=9$ ), we need $0<p<9$ , so $p=3$ .

Find $a$ and then compute the launch height

Use $h(9)=0$ with $p=3$ :

0=a(9-3)^2+180=a(6)^2+180=36a+180

So $36a=-180$ , giving $a=-5$ .

Now evaluate the height at launch, $h(0)$ :

\begin{aligned} h(0) &= -5(0-3)^2+180 \\ &= -5(9)+180 \\ &= 135. \end{aligned}

So the correct choice is $135$ .

Strategy

Hints

Desmos Guide

Step-by-step Explanation

Study Hints

Strategy

Hints

Desmos Guide

Step-by-step Explanation