AP Precalculus Quiz: Function Model Construction And Application

20 questions · exam conditions

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Function Model Construction And ApplicationQuestion 1 of 20

A company's profit $P$ (in thousands of dollars) is modeled by $P(x) = -2x^3 + 15x^2 - 24x + 10$ , where $x$ is the number of years since 2020. What is the company's profit in 2023?

7

thousand dollars

13

thousand dollars

19

thousand dollars

25

thousand dollars

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AP Precalculus Quiz

AP Precalculus Quiz: Function Model Construction And Application

Practice Function Model Construction And Application in AP Precalculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Function Model Construction And Application, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Precalculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A company's profit $P$ (in thousands of dollars) is modeled by $P(x) = -2x^3 + 15x^2 - 24x + 10$ , where $x$ is the number of years since 2020. What is the company's profit in 2023?

$7$ thousand dollars
$13$ thousand dollars
$19$ thousand dollars (correct answer)
$25$ thousand dollars

Explanation: In 2023,

x = 3

. Substituting:

P(3) = -2(27) + 15(9) - 24(3) + 10 = -54 + 135 - 72 + 10 = 19

thousand dollars. Choice A uses

x = 2

. Choice B uses

x = 1

. Choice D uses

x = 4

Question 2

A population of bacteria grows according to $P(t) = \frac{5000t}{t + 10}$ , where $t$ is time in hours. What happens to the population as $t$ approaches infinity?

The population approaches 500 bacteria
The population approaches 5000 bacteria (correct answer)
The population grows without bound
The population approaches 50 bacteria

Explanation: As

t \to \infty

P(t) = \frac{5000t}{t + 10} \to \frac{5000t}{t} = 5000

. The horizontal asymptote is

y = 5000

. Choice A divides 5000 by 10 incorrectly. Choice C ignores the rational function behavior. Choice D uses wrong calculation entirely.

Question 3

The temperature $T$ (in °F) in a building $t$ hours after midnight is modeled by $T(t) = \frac{t^2 + 38t + 400}{t + 10}$ for $0 \leq t \leq 24$ . What is the temperature at 2:00 AM?

$37$ °F
$39$ °F
$40$ °F (correct answer)
$42$ °F

Explanation: At 2:00 AM,

t = 2

. Substituting:

T(2) = \frac{2^2 + 38(2) + 400}{2 + 10} = \frac{4 + 76 + 400}{12} = \frac{480}{12} = 40

°F. Choice A uses wrong arithmetic in numerator. Choice B uses

t = 1

. Choice D uses wrong denominator calculation.

Question 4

The resistance $R$ (in ohms) of a wire is given by $R(L) = \frac{0.5L}{A}$ , where $L$ is length in meters and $A$ is cross-sectional area in square millimeters. For a wire with area $A = 2$ square millimeters, what is the resistance when $L = 100$ meters?

$25$ ohms (correct answer)
$50$ ohms
$100$ ohms
$200$ ohms

Explanation: Substituting

L = 100

and

A = 2

R(100) = \frac{0.5(100)}{2} = \frac{50}{2} = 25

ohms. Choice B forgets to divide by area. Choice C uses wrong coefficient calculation. Choice D doubles the correct answer.

Question 5

A rectangular swimming pool is being designed with a perimeter of 80 feet. If the length is $l$ feet, what function represents the area $A$ of the pool?

$A(l) = l(40 - l)$ for $0 < l < 40$ (correct answer)
$A(l) = l(80 - l)$ for $0 < l < 80$
$A(l) = l(20 - l)$ for $0 < l < 20$
$A(l) = 2l(40 - l)$ for $0 < l < 40$

Explanation: Perimeter:

2l + 2w = 80

, so

w = 40 - l

. Area:

A(l) = l(40 - l)

. Domain:

0 < l < 40

since width must be positive. Choice B doesn't divide perimeter correctly. Choice C uses wrong perimeter division. Choice D incorrectly doubles the area formula.

Question 6

The efficiency $E$ (as a percentage) of a solar panel depends on temperature $T$ (in °C) according to $E(T) = \frac{-T^2 + 20T + 1200}{T + 30}$ . What is the efficiency when the temperature is 10°C?

$27.5\%$
$30.0\%$
$32.5\%$ (correct answer)
$35.0\%$

Explanation: Substituting

T = 10

E(10) = \frac{-(10)^2 + 20(10) + 1200}{10 + 30} = \frac{-100 + 200 + 1200}{40} = \frac{1300}{40} = 32.5\%

. Choice A uses wrong numerator calculation. Choice B uses

T = 20

. Choice D uses wrong denominator calculation.

Question 7

The cost $C$ (in dollars) to produce $x$ items is given by $C(x) = x^3 - 12x^2 + 45x + 100$ . The revenue $R$ (in dollars) from selling $x$ items is $R(x) = 60x$ . What function represents the profit $P(x)$ ?

$P(x) = -x^3 + 12x^2 + 15x - 100$ (correct answer)
$P(x) = x^3 - 12x^2 + 105x + 100$
$P(x) = -x^3 + 12x^2 - 15x - 100$
$P(x) = x^3 + 12x^2 - 45x + 160$

Explanation: Profit is revenue minus cost:

P(x) = R(x) - C(x) = 60x - (x^3 - 12x^2 + 45x + 100) = -x^3 + 12x^2 + 15x - 100

. Choice B adds instead of subtracts cost. Choice C has incorrect sign on the linear term. Choice D has multiple sign and coefficient errors.

Question 8

A farmer has 200 feet of fencing to enclose three sides of a rectangular pen against a barn (the barn serves as the fourth side). If the width perpendicular to the barn is $w$ feet, what function gives the area $A$ of the pen?

$A(w) = w(200 - 2w)$ for $0 < w < 100$ (correct answer)
$A(w) = w(200 - w)$ for $0 < w < 200$
$A(w) = 2w(200 - 2w)$ for $0 < w < 100$
$A(w) = w(100 - w)$ for $0 < w < 100$

Explanation: Three sides use fencing: two widths and one length. So

2w + l = 200

, giving

l = 200 - 2w

. Area is

A(w) = w(200 - 2w)

. Domain:

w < 100

. Choice B doesn't account for two width sides. Choice C doubles the area incorrectly. Choice D uses wrong constraint on total fencing.

Question 9

A cylindrical tank has a circular base with radius $r$ feet and height 10 feet. If the tank is filled to a depth of $h$ feet, what function gives the volume $V$ of water in the tank when $r = 3$ feet?

$V(h) = 9\pi h$ for $0 \leq h \leq 10$ (correct answer)
$V(h) = 3\pi h$ for $0 \leq h \leq 10$
$V(h) = 6\pi h$ for $0 \leq h \leq 10$
$V(h) = 18\pi h$ for $0 \leq h \leq 10$

Explanation: Volume of water is

V = \pi r^2 h

. With

r = 3

V(h) = \pi(3^2)h = 9\pi h

. Domain is

0 \leq h \leq 10

since the tank is 10 feet tall. Choice B uses

r

instead of

r^2

. Choice C uses

2r

instead of

r^2

. Choice D uses

2r^2

instead of

r^2

Question 10

A box manufacturer wants to create an open-top box by cutting squares of side length $x$ from each corner of a $20$ by $16$ inch rectangular piece of cardboard and folding up the sides. What function represents the volume $V$ of the box?

$V(x) = x(20-2x)(16-2x)$ for $0 < x < 8$ (correct answer)
$V(x) = x(20-x)(16-x)$ for $0 < x < 16$
$V(x) = 2x(20-2x)(16-2x)$ for $0 < x < 8$
$V(x) = x(20-2x)(16-2x)$ for $0 < x < 10$

Explanation: After cutting squares of side

x

, the dimensions are height

x

, length

(20-2x)

, and width

(16-2x)

. Volume is

V(x) = x(20-2x)(16-2x)

. Domain:

x < 8

since

16-2x > 0

. Choice B doesn't account for cutting from both sides. Choice C incorrectly doubles the height. Choice D has wrong domain constraint.

Question 11

A rectangular garden has a length that is 4 meters more than twice its width. If the width is $w$ meters, which function represents the area $A$ of the garden in square meters?

$A(w) = 2w^2 + 4w$ (correct answer)
$A(w) = w^2 + 4w$
$A(w) = 2w^2 + 8w$
$A(w) = w^2 + 8w$

Explanation: The length is

2w + 4

meters. The area is length times width:

A(w) = w(2w + 4) = 2w^2 + 4w

. Choice B incorrectly uses

w + 4

for length. Choice C incorrectly uses

2w + 8

for length. Choice D incorrectly uses

w + 8

for length.

Question 12

The velocity $v$ (in m/s) of a particle is given by $v(t) = t^3 - 6t^2 + 9t$ , where $t$ is time in seconds. At what time is the velocity zero?

$t = 0$ and $t = 3$ seconds only (correct answer)
$t = 0$ , $t = 3$ , and $t = 6$ seconds
$t = 1$ and $t = 3$ seconds only
$t = 0$ and $t = 6$ seconds only

Explanation: Setting

v(t) = 0

t^3 - 6t^2 + 9t = t(t^2 - 6t + 9) = t(t-3)^2 = 0

. Solutions:

t = 0

and

t = 3

(double root). Choice B incorrectly includes

t = 6

. Choice C misses

t = 0

solution. Choice D has wrong second solution.

Question 13

A company's daily production cost is $C(x) = 0.01x^3 - 0.6x^2 + 15x + 500$ dollars for producing $x$ units. If they want to minimize cost per unit, what function represents cost per unit?

$\frac{C(x)}{x} = 0.01x^2 - 0.6x + 15 + \frac{500}{x}$ (correct answer)
$\frac{C(x)}{x} = 0.01x^3 - 0.6x^2 + 15x + 500$
$\frac{C(x)}{x} = 0.01x^2 - 0.6x + 15 + 500$
$\frac{C(x)}{x} = 0.01x - 0.6 + \frac{15}{x} + \frac{500}{x}$

Explanation: Cost per unit is

\frac{C(x)}{x} = \frac{0.01x^3 - 0.6x^2 + 15x + 500}{x} = 0.01x^2 - 0.6x + 15 + \frac{500}{x}

. Choice B doesn't divide by

x

. Choice C doesn't properly handle the constant term division. Choice D incorrectly divides each coefficient by

x

Question 14

Based on the situation described, determine the function that models the ball's height $h(t)$ over time $t$ .

$h(t)=-4.9t^2+18t+2$ (correct answer)
$h(t)=4.9t^2+18t+2$
$h(t)=-4.9t^2+2t+18$
$h(t)=-9.8t^2+18t+2$

Explanation: This question tests AP Precalculus skills: constructing and applying polynomial and rational function models. Polynomial and rational functions model various real-world phenomena, capturing relationships between variables under specific conditions. In this scenario, a ball's height over time under constant gravity follows a quadratic model h(t) = -½gt² + v₀t + h₀, where g ≈ 9.8 m/s², v₀ is initial velocity, and h₀ is initial height. Choice A is correct because it accurately captures the relationship with -4.9t² (half of -9.8), initial velocity 18 m/s, and initial height 2 m. Choice D is incorrect because it uses -9.8t² instead of -4.9t², which would double the effect of gravity. To help students: Emphasize that projectile motion uses -½gt² not -gt². Practice identifying initial conditions from word problems and matching them to function parameters.

Question 15

A rational function models the concentration $C$ (in mg/L) of a medication in the bloodstream $t$ hours after injection: $C(t) = \frac{120t}{t^2 + 4}$ . What is the concentration after 2 hours?

$15$ mg/L
$20$ mg/L
$25$ mg/L
$30$ mg/L (correct answer)

Explanation: Substituting

t = 2

C(2) = \frac{120(2)}{2^2 + 4} = \frac{240}{4 + 4} = \frac{240}{8} = 30

mg/L. Choice A incorrectly computes

\frac{120}{8}

. Choice B uses wrong denominator calculation

(2^2 + 8)

. Choice C uses wrong numerator calculation

\frac{200}{8}

Question 16

A ball is thrown upward from a height of 6 feet with an initial velocity of 48 feet per second. The height function is $h(t) = -16t^2 + 48t + 6$ . When does the ball hit the ground?

$t = 3.12$ seconds (correct answer)
$t = 2.87$ seconds
$t = 3.25$ seconds
$t = 3.00$ seconds

Explanation: The ball hits ground when

h(t) = 0

-16t^2 + 48t + 6 = 0

. Using the quadratic formula:

t = \frac{-48 \pm \sqrt{48^2 + 4(16)(6)}}{-32} = \frac{-48 \pm \sqrt{2688}}{-32}

. Taking the positive root:

t \approx 3.12

seconds. Choices B, C, and D result from computational errors in the quadratic formula.

Question 17

A store's weekly revenue $R$ (in dollars) from selling $x$ items is $R(x) = -2x^2 + 120x$ . What is the maximum weekly revenue the store can achieve?

$1600$ dollars
$1800$ dollars (correct answer)
$2000$ dollars
$2200$ dollars

Explanation: Maximum occurs at vertex:

x = -\frac{120}{2(-2)} = 30

. Maximum revenue:

R(30) = -2(900) + 120(30) = -1800 + 3600 = 1800

dollars. Choice A uses

x = 20

instead. Choice C uses

x = 25

instead. Choice D uses incorrect vertex formula calculation.

Question 18

The height $h$ (in feet) of a projectile is given by $h(t) = -16t^2 + 64t + 80$ , where $t$ is time in seconds. At what time does the projectile reach its maximum height?

$t = 1$ second
$t = 2$ seconds (correct answer)
$t = 3$ seconds
$t = 4$ seconds

Explanation: For a quadratic

f(t) = at^2 + bt + c

, the vertex occurs at

t = -\frac{b}{2a}

. Here,

t = -\frac{64}{2(-16)} = \frac{64}{32} = 2

seconds. Choice A uses incorrect calculation

\frac{64}{64}

. Choice C uses incorrect calculation

\frac{64}{16} - 1

. Choice D uses incorrect calculation

\frac{64}{16}

Question 19

The concentration of a pollutant in a lake $t$ days after cleanup begins is $C(t) = \frac{500}{t^2 + 25}$ parts per million. What is the initial concentration of the pollutant?

$15$ parts per million
$20$ parts per million (correct answer)
$25$ parts per million
$30$ parts per million

Explanation: Initial concentration occurs at

t = 0

C(0) = \frac{500}{0^2 + 25} = \frac{500}{25} = 20

parts per million. Choice A uses wrong denominator value. Choice C confuses the constant 25 with the answer. Choice D uses incorrect numerator calculation.

Question 20

Using the provided data, determine the rational function modeling population with carrying capacity $12{,}000$ and $P(0)=3{,}000$ .

$P(t)=\dfrac{12000}{1+3e^{-0.4t}}$ (correct answer)
$P(t)=\dfrac{12000}{1+e^{-0.4t}}$
$P(t)=12000(1+3e^{-0.4t})$
$P(t)=\dfrac{3000}{1+3e^{-0.4t}}$

Explanation: This question tests AP Precalculus skills: constructing and applying polynomial and rational function models. Polynomial and rational functions model various real-world phenomena, capturing relationships between variables under specific conditions. In this scenario, we need a logistic growth model with carrying capacity 12,000 and initial population P(0) = 3,000. Choice A is correct because P(t) = 12000/(1 + 3e^(-0.4t)) satisfies both conditions: as t→∞, P(t)→12,000 (carrying capacity), and P(0) = 12000/(1 + 3) = 3,000. Choice D is incorrect because it has the wrong carrying capacity of 3,000 instead of 12,000, failing to match the given parameters. To help students: Emphasize checking initial conditions and limiting behavior. Practice verifying that proposed models satisfy all given constraints.