Super-Hard SAT Math Question 154 | Free SAT Prep | Aniko

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Question 154·200 Super-Hard SAT Math Questions·Advanced Math

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A quadratic function models the height, in feet, of an object above the ground in terms of time, in seconds, after the object is launched off an elevated surface. The model indicates that at a time of 4 seconds, the object is 255 feet above the ground. At a time of 11 seconds, the object is 444 feet above the ground. If the object was at a height of 15 feet when it was launched, what is the height, in feet, of the object above the ground 18 seconds after being launched?

For quadratic modeling problems, immediately write $h(t)=at^2+bt+c$ and use any “at launch” information to get $c$ by substituting $t=0$ . Then plug the other two points into the equation to create a two-equation system in $a$ and $b$ . Look for quick elimination by simplifying the equations (dividing by a common factor) before subtracting, and only after you have $a$ and $b$ should you substitute the target time.

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Start with a general quadratic

Write the model as $h(t)=at^2+bt+c$ . Use the launch height to find $c$ right away.

Make two equations in two unknowns

Plug in $t=4$ and $t=11$ to get two equations involving only $a$ and $b$ .

Eliminate a variable

Try simplifying the two equations (for example, by dividing) so that subtracting them quickly eliminates the same variable.

Don’t forget the constant term when evaluating

When you compute $h(18)$ , make sure you include the $+15$ from the launch height.

Desmos Guide

Create equations for the coefficients

In Desmos, use variables $a$ and $b$ and enter the two equations:

16a+4b=240
121a+11b=429

Find the intersection point

Desmos will graph these as lines in the $(a,b)$ -plane. Click the intersection point to read the values of $a$ and $b$ .

Compute the height at 18 seconds

Define the function using the intersection values:

h(t)=a t^2 + b t + 15

Then evaluate h(18) (by typing h(18)), using your found values of $a$ and $b$ .

Step-by-step Explanation

Write the quadratic model and use the launch height

Let the height be $h(t)=at^2+bt+c$ .

At launch, $t=0$ and the height is 15, so:

h(0)=c=15

Thus $h(t)=at^2+bt+15$ .

Use the two given heights to create equations

At $t=4$ , the height is 255:

16a+4b+15=255 \Rightarrow 16a+4b=240

At $t=11$ , the height is 444:

121a+11b+15=444 \Rightarrow 121a+11b=429

Solve for $a$ and $b$ efficiently

Divide the first equation by 4:

4a+b=60

Divide the second equation by 11:

11a+b=39

Subtract to eliminate $b$ :

(11a+b)-(4a+b)=39-60

So $7a=-21$ , which gives $a=-3$ .

Then from $4a+b=60$ :

4(-3)+b=60 \Rightarrow b=72

Evaluate the height at 18 seconds

Now compute $h(18)$ :

\begin{aligned} h(18) &= -3(18)^2+72(18)+15 \\ &= -3(324)+1296+15 \\ &= -972+1296+15 \\ &= 339 \end{aligned}

So the height after 18 seconds is 339 feet.

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Desmos Guide

Step-by-step Explanation

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Desmos Guide

Step-by-step Explanation