AP Precalculus Quiz: Function Model Construction And Application
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Function Model Construction And ApplicationQuestion 1 of 20

A company's profit PP (in thousands of dollars) is modeled by P(x)=−2x3+15x2−24x+10P(x) = -2x^3 + 15x^2 - 24x + 10, where xx is the number of years since 2020. What is the company's profit in 2023?

77 thousand dollars
1313 thousand dollars
1919 thousand dollars
2525 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.

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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.

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Question 1

A company's profit PP (in thousands of dollars) is modeled by P(x)=−2x3+15x2−24x+10P(x) = -2x^3 + 15x^2 - 24x + 10, where xx is the number of years since 2020. What is the company's profit in 2023?

  1. 77 thousand dollars
  2. 1313 thousand dollars
  3. 1919 thousand dollars (correct answer)
  4. 2525 thousand dollars

Explanation: In 2023, x=3x = 3. Substituting: P(3)=−2(27)+15(9)−24(3)+10=−54+135−72+10=19P(3) = -2(27) + 15(9) - 24(3) + 10 = -54 + 135 - 72 + 10 = 19 thousand dollars. Choice A uses x=2x = 2. Choice B uses x=1x = 1. Choice D uses x=4x = 4.

Question 2

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

  1. The population approaches 500 bacteria
  2. The population approaches 5000 bacteria (correct answer)
  3. The population grows without bound
  4. The population approaches 50 bacteria

Explanation: As t→∞t \to \infty, P(t)=5000tt+10→5000tt=5000P(t) = \frac{5000t}{t + 10} \to \frac{5000t}{t} = 5000. The horizontal asymptote is y=5000y = 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 TT (in °F) in a building tt hours after midnight is modeled by T(t)=t2+38t+400t+10T(t) = \frac{t^2 + 38t + 400}{t + 10} for 0≤t≤240 \leq t \leq 24. What is the temperature at 2:00 AM?

  1. 3737 °F
  2. 3939 °F
  3. 4040 °F (correct answer)
  4. 4242 °F

Explanation: At 2:00 AM, t=2t = 2. Substituting: T(2)=22+38(2)+4002+10=4+76+40012=48012=40T(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=1t = 1. Choice D uses wrong denominator calculation.

Question 4

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

  1. 2525 ohms (correct answer)
  2. 5050 ohms
  3. 100100 ohms
  4. 200200 ohms

Explanation: Substituting L=100L = 100 and A=2A = 2: R(100)=0.5(100)2=502=25R(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 ll feet, what function represents the area AA of the pool?

  1. A(l)=l(40−l)A(l) = l(40 - l) for 0<l<400 < l < 40 (correct answer)
  2. A(l)=l(80−l)A(l) = l(80 - l) for 0<l<800 < l < 80
  3. A(l)=l(20−l)A(l) = l(20 - l) for 0<l<200 < l < 20
  4. A(l)=2l(40−l)A(l) = 2l(40 - l) for 0<l<400 < l < 40

Explanation: Perimeter: 2l+2w=802l + 2w = 80, so w=40−lw = 40 - l. Area: A(l)=l(40−l)A(l) = l(40 - l). Domain: 0<l<400 < 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 EE (as a percentage) of a solar panel depends on temperature TT (in °C) according to E(T)=−T2+20T+1200T+30E(T) = \frac{-T^2 + 20T + 1200}{T + 30}. What is the efficiency when the temperature is 10°C?

  1. 27.5%27.5\%
  2. 30.0%30.0\%
  3. 32.5%32.5\% (correct answer)
  4. 35.0%35.0\%

Explanation: Substituting T=10T = 10: E(10)=−(10)2+20(10)+120010+30=−100+200+120040=130040=32.5%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=20T = 20. Choice D uses wrong denominator calculation.

Question 7

The cost CC (in dollars) to produce xx items is given by C(x)=x3−12x2+45x+100C(x) = x^3 - 12x^2 + 45x + 100. The revenue RR (in dollars) from selling xx items is R(x)=60xR(x) = 60x. What function represents the profit P(x)P(x)?

  1. P(x)=−x3+12x2+15x−100P(x) = -x^3 + 12x^2 + 15x - 100 (correct answer)
  2. P(x)=x3−12x2+105x+100P(x) = x^3 - 12x^2 + 105x + 100
  3. P(x)=−x3+12x2−15x−100P(x) = -x^3 + 12x^2 - 15x - 100
  4. P(x)=x3+12x2−45x+160P(x) = x^3 + 12x^2 - 45x + 160

Explanation: Profit is revenue minus cost: P(x)=R(x)−C(x)=60x−(x3−12x2+45x+100)=−x3+12x2+15x−100P(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 ww feet, what function gives the area AA of the pen?

  1. A(w)=w(200−2w)A(w) = w(200 - 2w) for 0<w<1000 < w < 100 (correct answer)
  2. A(w)=w(200−w)A(w) = w(200 - w) for 0<w<2000 < w < 200
  3. A(w)=2w(200−2w)A(w) = 2w(200 - 2w) for 0<w<1000 < w < 100
  4. A(w)=w(100−w)A(w) = w(100 - w) for 0<w<1000 < w < 100

Explanation: Three sides use fencing: two widths and one length. So 2w+l=2002w + l = 200, giving l=200−2wl = 200 - 2w. Area is A(w)=w(200−2w)A(w) = w(200 - 2w). Domain: w<100w < 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 rr feet and height 10 feet. If the tank is filled to a depth of hh feet, what function gives the volume VV of water in the tank when r=3r = 3 feet?

  1. V(h)=9πhV(h) = 9\pi h for 0≤h≤100 \leq h \leq 10 (correct answer)
  2. V(h)=3πhV(h) = 3\pi h for 0≤h≤100 \leq h \leq 10
  3. V(h)=6πhV(h) = 6\pi h for 0≤h≤100 \leq h \leq 10
  4. V(h)=18πhV(h) = 18\pi h for 0≤h≤100 \leq h \leq 10

Explanation: Volume of water is V=πr2hV = \pi r^2 h. With r=3r = 3: V(h)=π(32)h=9πhV(h) = \pi(3^2)h = 9\pi h. Domain is 0≤h≤100 \leq h \leq 10 since the tank is 10 feet tall. Choice B uses rr instead of r2r^2. Choice C uses 2r2r instead of r2r^2. Choice D uses 2r22r^2 instead of r2r^2.

Question 10

A box manufacturer wants to create an open-top box by cutting squares of side length xx from each corner of a 2020 by 1616 inch rectangular piece of cardboard and folding up the sides. What function represents the volume VV of the box?

  1. V(x)=x(20−2x)(16−2x)V(x) = x(20-2x)(16-2x) for 0<x<80 < x < 8 (correct answer)
  2. V(x)=x(20−x)(16−x)V(x) = x(20-x)(16-x) for 0<x<160 < x < 16
  3. V(x)=2x(20−2x)(16−2x)V(x) = 2x(20-2x)(16-2x) for 0<x<80 < x < 8
  4. V(x)=x(20−2x)(16−2x)V(x) = x(20-2x)(16-2x) for 0<x<100 < x < 10

Explanation: After cutting squares of side xx, the dimensions are height xx, length (20−2x)(20-2x), and width (16−2x)(16-2x). Volume is V(x)=x(20−2x)(16−2x)V(x) = x(20-2x)(16-2x). Domain: x<8x < 8 since 16−2x>016-2x > 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 ww meters, which function represents the area AA of the garden in square meters?

  1. A(w)=2w2+4wA(w) = 2w^2 + 4w (correct answer)
  2. A(w)=w2+4wA(w) = w^2 + 4w
  3. A(w)=2w2+8wA(w) = 2w^2 + 8w
  4. A(w)=w2+8wA(w) = w^2 + 8w

Explanation: The length is 2w+42w + 4 meters. The area is length times width: A(w)=w(2w+4)=2w2+4wA(w) = w(2w + 4) = 2w^2 + 4w. Choice B incorrectly uses w+4w + 4 for length. Choice C incorrectly uses 2w+82w + 8 for length. Choice D incorrectly uses w+8w + 8 for length.

Question 12

The velocity vv (in m/s) of a particle is given by v(t)=t3−6t2+9tv(t) = t^3 - 6t^2 + 9t, where tt is time in seconds. At what time is the velocity zero?

  1. t=0t = 0 and t=3t = 3 seconds only (correct answer)
  2. t=0t = 0, t=3t = 3, and t=6t = 6 seconds
  3. t=1t = 1 and t=3t = 3 seconds only
  4. t=0t = 0 and t=6t = 6 seconds only

Explanation: Setting v(t)=0v(t) = 0: t3−6t2+9t=t(t2−6t+9)=t(t−3)2=0t^3 - 6t^2 + 9t = t(t^2 - 6t + 9) = t(t-3)^2 = 0. Solutions: t=0t = 0 and t=3t = 3 (double root). Choice B incorrectly includes t=6t = 6. Choice C misses t=0t = 0 solution. Choice D has wrong second solution.

Question 13

A company's daily production cost is C(x)=0.01x3−0.6x2+15x+500C(x) = 0.01x^3 - 0.6x^2 + 15x + 500 dollars for producing xx units. If they want to minimize cost per unit, what function represents cost per unit?

  1. C(x)x=0.01x2−0.6x+15+500x\frac{C(x)}{x} = 0.01x^2 - 0.6x + 15 + \frac{500}{x} (correct answer)
  2. C(x)x=0.01x3−0.6x2+15x+500\frac{C(x)}{x} = 0.01x^3 - 0.6x^2 + 15x + 500
  3. C(x)x=0.01x2−0.6x+15+500\frac{C(x)}{x} = 0.01x^2 - 0.6x + 15 + 500
  4. C(x)x=0.01x−0.6+15x+500x\frac{C(x)}{x} = 0.01x - 0.6 + \frac{15}{x} + \frac{500}{x}

Explanation: Cost per unit is C(x)x=0.01x3−0.6x2+15x+500x=0.01x2−0.6x+15+500x\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 xx. Choice C doesn't properly handle the constant term division. Choice D incorrectly divides each coefficient by xx.

Question 14

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

  1. h(t)=−4.9t2+18t+2h(t)=-4.9t^2+18t+2 (correct answer)
  2. h(t)=4.9t2+18t+2h(t)=4.9t^2+18t+2
  3. h(t)=−4.9t2+2t+18h(t)=-4.9t^2+2t+18
  4. h(t)=−9.8t2+18t+2h(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 CC (in mg/L) of a medication in the bloodstream tt hours after injection: C(t)=120tt2+4C(t) = \frac{120t}{t^2 + 4}. What is the concentration after 2 hours?

  1. 1515 mg/L
  2. 2020 mg/L
  3. 2525 mg/L
  4. 3030 mg/L (correct answer)

Explanation: Substituting t=2t = 2: C(2)=120(2)22+4=2404+4=2408=30C(2) = \frac{120(2)}{2^2 + 4} = \frac{240}{4 + 4} = \frac{240}{8} = 30 mg/L. Choice A incorrectly computes 1208\frac{120}{8}. Choice B uses wrong denominator calculation (22+8)(2^2 + 8). Choice C uses wrong numerator calculation 2008\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)=−16t2+48t+6h(t) = -16t^2 + 48t + 6. When does the ball hit the ground?

  1. t=3.12t = 3.12 seconds (correct answer)
  2. t=2.87t = 2.87 seconds
  3. t=3.25t = 3.25 seconds
  4. t=3.00t = 3.00 seconds

Explanation: The ball hits ground when h(t)=0h(t) = 0: −16t2+48t+6=0-16t^2 + 48t + 6 = 0. Using the quadratic formula: t=−48±482+4(16)(6)−32=−48±2688−32t = \frac{-48 \pm \sqrt{48^2 + 4(16)(6)}}{-32} = \frac{-48 \pm \sqrt{2688}}{-32}. Taking the positive root: t≈3.12t \approx 3.12 seconds. Choices B, C, and D result from computational errors in the quadratic formula.

Question 17

A store's weekly revenue RR (in dollars) from selling xx items is R(x)=−2x2+120xR(x) = -2x^2 + 120x. What is the maximum weekly revenue the store can achieve?

  1. 16001600 dollars
  2. 18001800 dollars (correct answer)
  3. 20002000 dollars
  4. 22002200 dollars

Explanation: Maximum occurs at vertex: x=−1202(−2)=30x = -\frac{120}{2(-2)} = 30. Maximum revenue: R(30)=−2(900)+120(30)=−1800+3600=1800R(30) = -2(900) + 120(30) = -1800 + 3600 = 1800 dollars. Choice A uses x=20x = 20 instead. Choice C uses x=25x = 25 instead. Choice D uses incorrect vertex formula calculation.

Question 18

The height hh (in feet) of a projectile is given by h(t)=−16t2+64t+80h(t) = -16t^2 + 64t + 80, where tt is time in seconds. At what time does the projectile reach its maximum height?

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

Explanation: For a quadratic f(t)=at2+bt+cf(t) = at^2 + bt + c, the vertex occurs at t=−b2at = -\frac{b}{2a}. Here, t=−642(−16)=6432=2t = -\frac{64}{2(-16)} = \frac{64}{32} = 2 seconds. Choice A uses incorrect calculation 6464\frac{64}{64}. Choice C uses incorrect calculation 6416−1\frac{64}{16} - 1. Choice D uses incorrect calculation 6416\frac{64}{16}.

Question 19

The concentration of a pollutant in a lake tt days after cleanup begins is C(t)=500t2+25C(t) = \frac{500}{t^2 + 25} parts per million. What is the initial concentration of the pollutant?

  1. 1515 parts per million
  2. 2020 parts per million (correct answer)
  3. 2525 parts per million
  4. 3030 parts per million

Explanation: Initial concentration occurs at t=0t = 0: C(0)=50002+25=50025=20C(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 12,00012{,}000 and P(0)=3,000P(0)=3{,}000.​

  1. P(t)=120001+3e−0.4tP(t)=\dfrac{12000}{1+3e^{-0.4t}} (correct answer)
  2. P(t)=120001+e−0.4tP(t)=\dfrac{12000}{1+e^{-0.4t}}
  3. P(t)=12000(1+3e−0.4t)P(t)=12000(1+3e^{-0.4t})
  4. P(t)=30001+3e−0.4tP(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.