Function Model Construction and Application - AP Precalculus

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?
Answer: $t = 3.12$ seconds 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.
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 $P(x)$ ?
Answer: $P(x) = -x^3 + 12x^2 + 15x - 100$ 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.
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?
Answer: $32.5\%$ 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.
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?
Answer: $\frac{C(x)}{x} = 0.01x^2 - 0.6x + 15 + \frac{500}{x}$ 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 $x$ .
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?
Answer: $A(w) = w(200 - 2w)$ for $0 < w < 100$ 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.
Using the provided data, what type of function best models height under constant gravity for a ball thrown upward from $2$ m?
Answer: Quadratic function, because height depends on $t^2$ under constant acceleration 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 thrown upward from 2 m under constant gravity follows the kinematic equation h(t) = h₀ + v₀t - ½gt², which is a quadratic function in t. Choice A is correct because constant acceleration (gravity) leads to a quadratic position function - this is a fundamental result from physics where integrating constant acceleration twice gives a t² term. Choice D is incorrect because constant speed would mean zero acceleration, contradicting the presence of gravity. To help students: Emphasize the connection between physics and mathematics - constant acceleration always produces quadratic position functions. Practice deriving polynomial models from physical principles.
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?
Answer: $40$ °F 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.
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?
Answer: $V(x) = x(20-2x)(16-2x)$ for $0 < x < 8$ 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.
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?
Answer: $t = 2$ seconds 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 $\frac{64}{16}$ .
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?
Answer: $A(l) = l(40 - l)$ for $0 < l < 40$ 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.
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?
Answer: $30$ mg/L 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 $\frac{200}{8}$ .
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?
Answer: $25$ ohms 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.
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?
Answer: The population approaches 5000 bacteria 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.
Based on the situation described, how does increasing initial upward velocity affect the quadratic height model's vertex?
Answer: The vertex height increases and occurs later 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, for projectile motion h(t) = -16t² + v₀t + h₀, increasing initial velocity v₀ affects both the vertex height and time. Choice A is correct because increasing v₀ increases the maximum height (vertex y-coordinate = h₀ + v₀²/64) and delays when it occurs (vertex time = v₀/32). Choice B is incorrect because it claims the vertex height decreases with increased initial velocity, contradicting the physics of projectile motion. To help students: Emphasize how parameters affect quadratic behavior. Practice analyzing how changing coefficients impacts vertex location and parabola shape.
Using the provided data, determine the rational model $T(t)$ for cooling from $90^ $ toward $20^ $ with $T(10)=40$ .
Answer: $T(t)=20+\dfrac{70}{1+0.25t}$ 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, cooling from 90°F toward 20°F follows Newton's cooling law, modeled as T(t) = 20 + 70/(1 + kt), where the temperature difference decreases over time. Choice A is correct because when we substitute t = 10, we get T(10) = 20 + 70/(1 + 0.25(10)) = 20 + 70/3.5 = 20 + 20 = 40°F, matching the given condition. Choice B is incorrect because it would give T(10) = 20 + 70/6 ≈ 31.7°F, not matching the required 40°F at t = 10. To help students: Emphasize verifying models by substituting known data points. Practice setting up cooling/warming models where temperature approaches ambient temperature asymptotically.
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?
Answer: $t = 0$ and $t = 3$ seconds only 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.
Based on the situation described, determine the function that models the ball's height $h(t)$ over time $t$ .
Answer: $h(t)=-4.9t^2+18t+2$ 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.
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?
Answer: $V(h) = 9\pi h$ for $0 \leq h \leq 10$ 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 $r^2$ .
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?
Answer: $20$ parts per million 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.
Based on the situation described, what does the rational population model imply about population as $t\to\infty$ ?
Answer: It approaches a finite carrying capacity set by the model 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 rational population model typically has the form P(t) = L/(1 + $ae^{-kt}$ ) or P(t) = K + N/(1 + bt), where the population approaches a horizontal asymptote as time increases. Choice B is correct because rational functions modeling population growth have horizontal asymptotes representing carrying capacity - the maximum sustainable population. Choice A is incorrect because population models don't decrease without bound; they approach a positive carrying capacity from below. To help students: Emphasize that rational functions with positive numerators and denominators approach positive horizontal asymptotes. Practice analyzing long-term behavior of rational models in real-world contexts.

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