AP Precalculus Quiz: Function Model Construction And Application

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.

Question 1

1. $7$ thousand dollars
2. $13$ thousand dollars
3. $19$ thousand dollars (correct answer)
4. $25$ 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

1. $15$ mg/L
2. $20$ mg/L
3. $25$ mg/L
4. $30$ 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 3

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 $t \to \infty$, $P(t) = \frac{5000t}{t + 10} \to \frac{5000t}{t} = 5000$. 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 4

1. $37$ °F
2. $39$ °F
3. $40$ °F (correct answer)
4. $42$ °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 5

1. $t = 3.12$ seconds (correct answer)
2. $t = 2.87$ seconds
3. $t = 3.25$ seconds
4. $t = 3.00$ 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 6

1. $25$ ohms (correct answer)
2. $50$ ohms
3. $100$ ohms
4. $200$ 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 7

1. $A(l) = l(40 - l)$ for $0 < l < 40$ (correct answer)
2. $A(l) = l(80 - l)$ for $0 < l < 80$
3. $A(l) = l(20 - l)$ for $0 < l < 20$
4. $A(l) = 2l(40 - l)$ 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 8

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

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 9

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

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 10

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

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 11

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

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 12

1. $A(w) = w(200 - 2w)$ for $0 < w < 100$ (correct answer)
2. $A(w) = w(200 - w)$ for $0 < w < 200$
3. $A(w) = 2w(200 - 2w)$ for $0 < w < 100$
4. $A(w) = w(100 - w)$ 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 13

1. $t = 0$ and $t = 3$ seconds only (correct answer)
2. $t = 0$, $t = 3$, and $t = 6$ seconds
3. $t = 1$ and $t = 3$ seconds only
4. $t = 0$ 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 14

1. $V(h) = 9\pi h$ for $0 \leq h \leq 10$ (correct answer)
2. $V(h) = 3\pi h$ for $0 \leq h \leq 10$
3. $V(h) = 6\pi h$ for $0 \leq h \leq 10$
4. $V(h) = 18\pi h$ 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 15

1. $1600$ dollars
2. $1800$ dollars (correct answer)
3. $2000$ dollars
4. $2200$ 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 16

1. $V(x) = x(20-2x)(16-2x)$ for $0 < x < 8$ (correct answer)
2. $V(x) = x(20-x)(16-x)$ for $0 < x < 16$
3. $V(x) = 2x(20-2x)(16-2x)$ for $0 < x < 8$
4. $V(x) = x(20-2x)(16-2x)$ 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 17

1. $t = 1$ second
2. $t = 2$ seconds (correct answer)
3. $t = 3$ seconds
4. $t = 4$ 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 18

1. $15$ parts per million
2. $20$ parts per million (correct answer)
3. $25$ parts per million
4. $30$ 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 19

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

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.

Question 20

1. It decreases the maximum height because the parabola narrows
2. It increases the maximum height because the vertex rises (correct answer)
3. It leaves the maximum height unchanged because gravity is constant
4. It makes height linear because acceleration becomes negligible

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, the quadratic height model h(t) = -½gt² + v₀t + h₀ has its maximum at the vertex, which occurs at t = -v₀/(-g) = v₀/g with maximum height h_max = h₀ + v₀²/(2g). Choice B is correct because increasing initial velocity v₀ increases the maximum height quadratically - doubling v₀ quadruples the height gain v₀²/(2g). Choice A is incorrect because it confuses the effect on the parabola's shape with the effect on maximum height - higher initial velocity raises the vertex. To help students: Emphasize the vertex formula for quadratics and how coefficients affect vertex position. Practice analyzing how parameter changes affect key features of polynomial models.