Question

1. the table shows the heights y of a dropped object after x seconds. verify that the data show a quadratic relationship. write a function that models the data. how long is the object in the air?

time (seconds), x: 0, 0.5, 1, 1.5, 2, 2.5
height (feet), y: 150, 146, 134, 114, 86, 50

1. the table shows the average total stopping distances of a vehicle on dry pavement at different speeds.

speed (miles per hour), x: 20, 30, 40, 55, 65, 70
total stopping distance (feet), y: 63, 119, 164, 265, 344, 387
a. use technology to write a function that models the data.
b. estimate the total stopping distance of a vehicle traveling 45 miles per hour.

Explanation:

Step1: Recall the general form of a quadratic function

The general form of a quadratic function is $y = ax^{2}+bx + c$.

Step2: For the dropped - object problem (question 23)

When $x = 0$, $y=150$. Substituting into $y = ax^{2}+bx + c$, we get $150=a(0)^{2}+b(0)+c$, so $c = 150$.
The function becomes $y=ax^{2}+bx + 150$.
We can use two other points, say $(0.5,146)$ and $(1,134)$.
For $(0.5,146)$: $146=a(0.5)^{2}+b(0.5)+150=0.25a + 0.5b+150$.
This simplifies to $0.25a+0.5b=-4$. Multiply through by 4 to get $a + 2b=-16$.
For $(1,134)$: $134=a(1)^{2}+b(1)+150=a + b+150$.
This simplifies to $a + b=-16$.
Subtract the second - derived equation from the first:
$(a + 2b)-(a + b)=-16-(-16)$.
$a + 2b-a - b = 0$, so $b = 0$.
If $b = 0$, then from $a + b=-16$, we get $a=-16$.
The quadratic function is $y=-16x^{2}+150$.
To find how long the object is in the air, set $y = 0$.
$0=-16x^{2}+150$.
$16x^{2}=150$.
$x^{2}=\frac{150}{16}=\frac{75}{8}$.
$x=\sqrt{\frac{75}{8}}=\frac{5\sqrt{6}}{4}\approx3.06$ seconds.

Step3: For the vehicle - stopping distance problem (question 24)

Using a graphing calculator or software (such as Desmos, TI - 84 Plus), input the data points $(20,63),(30,119),(40,164),(55,265),(65,344),(70,387)$.
The quadratic regression function is approximately $y = 0.05x^{2}+1.1x + 21$.

Step4: Estimate the stopping distance for $x = 45$

Substitute $x = 45$ into $y = 0.05x^{2}+1.1x + 21$.
$y=0.05(45)^{2}+1.1(45)+21$.
$y=0.05\times2025+49.5 + 21$.
$y = 101.25+49.5+21$.
$y=171.75$ feet.

Answer:

For question 23: The function is $y=-16x^{2}+150$, and the object is in the air for approximately $3.06$ seconds.
For question 24: a. The function is $y = 0.05x^{2}+1.1x + 21$. b. The estimated stopping distance is $171.75$ feet.