Exponential Function Context and Data Modeling - AP Precalculus

Which of the following functions $P(t)$ best models the bird population, where $t$ is the number of years since 2015?
Answer: $P(t) = 80(1.5)^{t/5}$ Let the model be $P(t)=ab^t$ . The year 2015 corresponds to $t=0$ , so the initial population is $a=80$ . The year 2020 corresponds to $t=5$ . We have the point (5, 120). Plugging this into the model: $120 = 80(b)^5$ . Dividing by 80 gives $1.5 = b^5$ . Solving for $b$ gives $b=(1.5)^{1/5}$ . Therefore, the model is $P(t) = 80((1.5)^{1/5})^t = 80(1.5)^{t/5}$ .
A student invests money in a certificate of deposit (CD). The initial deposit is $P_0=\$1{,}800$ , and the CD earns $3.6\%$ annual interest compounded quarterly. The balance after $t$ years is modeled by $A(t)=P_0\left(1+\frac{0.036}{4}\right)^{4t}.$ A table of modeled balances (rounded to the nearest dollar) is shown below. $t$ (years): 0, 2, 4, 6 $A(t)$ (USD): 1800, 1934, 2078, 2233 In this context, $P_0$ is the starting amount deposited, and the factor $\left(1+\frac{0.036}{4}\right)$ represents the growth each quarter. The exponent $4t$ counts the number of compounding periods, so increasing $t$ increases the number of times the balance is multiplied by the quarterly growth factor. Students compare the table to the formula to confirm that the dollar increase is not constant; instead, the balance grows by a constant percentage per year. They also discuss how changing the interest rate or compounding frequency would change the long-term balance. Based on the scenario above, what does the initial value $P_0$ represent in this context?
Answer: The starting deposit before interest is added This question tests AP Precalculus understanding of exponential functions and data modeling, specifically interpreting parameters in compound interest contexts. Exponential functions model situations where a quantity grows or decays at a constant percentage rate. Key parameters include the initial value (starting point) and the growth/decay rate, which determines how quickly the function changes. In this scenario, the function models a certificate of deposit with quarterly compounding, with an initial deposit of $1,800 and 3.6% annual interest. Choice C is correct because P₀ = $1,800 explicitly represents the initial deposit amount before any interest is added, as stated directly in the problem and shown at t = 0 in the table. Choice B is incorrect because it confuses P₀ with the interest rate parameter 0.036, which appears separately in the growth factor. To help students: Emphasize understanding the meaning of each parameter in context before solving. Encourage students to verify parameter meanings by checking values at t = 0. Watch for: confusing different parameters or misinterpreting financial terminology.
Which of the following statements accurately compares the values of the two investments over time?
Answer: The value of Plan B is initially less than Plan A, but it will eventually surpass the value of Plan A. Plan A is modeled by $A(t) = 1000(1.05)^t$ and Plan B by $B(t) = 800(1.06)^t$ . Initially, at $t=0$ , $A(0)=1000$ and $B(0)=800$ , so Plan A is greater. However, because Plan B has a higher growth factor (1.06 > 1.05), it grows at a faster percentage rate. Therefore, the value of Plan B will eventually overtake and surpass the value of Plan A.
Which of the following functions $P(t)$ best models the population of bacteria in the petri dish after $t$ hours?
Answer: $P(t) = 500(1.30)^t$ The situation describes exponential growth. The initial population is $a=500$ . A 30% increase per hour corresponds to a growth factor of $b = 1 + 0.30 = 1.30$ . The exponential model is of the form $P(t) = ab^t$ , which gives $P(t) = 500(1.30)^t$ .
According to the model, approximately how many subscribers, in millions, did the service have at the beginning of 2018?
Answer: 0.77 The beginning of 2018 is two years before the beginning of 2020, so we need to evaluate the model at $t=-2$ . $S(-2) = 1.5(1.4)^{-2} = \frac{1.5}{1.4^2} = \frac{1.5}{1.96} \approx 0.765$ . So, there were approximately 0.77 million subscribers.
Which of the following functions $N(t)$ models the number of infected computers after $t$ days?
Answer: $N(t) = 5^t$ The condition that the rate of increase is proportional to the current number implies an exponential model of the form $N(t) = ab^t$ . We are given two points: (0, 1) and (1, 5). From (0, 1), we have $N(0) = a(b^0) = a = 1$ . From (1, 5), we have $N(1) = (1)b^1 = b = 5$ . Therefore, the model is $N(t) = 1 \cdot 5^t = 5^t$ .
Which of the following statements is the best interpretation of the number 0.88 in the context of the model?
Answer: The value of the car decreases by 12% each year. In an exponential model $V(t) = ab^t$ , the base $b$ is the growth or decay factor. Since $b=0.88$ is less than 1, it's a decay factor. The rate of decay $r$ is found by $b=1-r$ , so $0.88 = 1-r$ , which gives $r=0.12$ . This means the value decreases by 12% each year.
Which of the following functions $B(t)$ models the person's credit card balance after $t$ years?
Answer: $B(t) = 2000(1.02)^{12t}$ The formula for compound interest is $A = P(1 + r/n)^{nt}$ . The principal $P$ is 2000. The annual rate $r$ is 0.24. Since it's compounded monthly, $n=12$ . The monthly interest rate is $0.24/12 = 0.02$ . The number of compounding periods in $t$ years is $12t$ . So the model is $B(t) = 2000(1+0.02)^{12t} = 2000(1.02)^{12t}$ .
Which of the following functions models the amount of the substance remaining after $m$ months?
Answer: $A(m) = 100(0.95^{1/12})^m$ Since there are 12 months in a year, the relationship between years $t$ and months $m$ is $t = m/12$ . Substitute this into the original function: $A(m) = 100(0.95)^{m/12}$ . Using the properties of exponents, this can be rewritten as $A(m) = 100((0.95)^{1/12})^m$ .
Which of the following functions $W(t)$ models the city's daily water consumption, in millions of gallons, $t$ years from now?
Answer: $W(t) = 50(0.96)^t$ This describes exponential decay. The initial value is $a=50$ . A decrease of 4% per year means the decay factor is $b = 1 - 0.04 = 0.96$ . The model is $W(t) = ab^t$ , which becomes $W(t) = 50(0.96)^t$ .
A country's population is modeled by $P(t)=9.50\times10^6(1.012)^t$ , where $t$ is years after 2015. Estimated values (rounded) are: 2015: $9.50\times10^6$ , 2020: $1.01\times10^7$ , 2025: $1.07\times10^7$ . Using the provided data, using the exponential model, predict the population in 2035.
Answer: Approximately $1.20\times10^7$ This question tests AP Precalculus understanding of exponential functions and data modeling, specifically interpreting parameters and predicting outcomes. Exponential functions model situations where a quantity grows or decays at a constant percentage rate. Key parameters include the initial value (starting point) and the growth/decay rate, which determines how quickly the function changes. In this scenario, the function models a country's population growth, with an initial value of 9.50× $10^6$ and a growth rate of 1.2% per year. Choice A is correct because it accurately identifies the predicted population in 2035 as approximately 1.20× $10^7$ , consistent with the passage details. Choice B is incorrect because it overestimates the growth, a common error when students miscalculate the exponent. To help students: Emphasize understanding of exponential growth vs. linear growth, and practice interpreting data through graph analysis. Encourage identifying key function parameters and their real-world implications. Watch for: confusing initial values with growth rates or misreading graphical data.
If an ancient artifact originally contained $A_0$ grams of Carbon-14, which of the following functions $M(c)$ models the mass remaining after $c$ centuries? (Note: 1 century = 100 years)
Answer: $M(c) = A_0(0.5)^{10c/573}$ The model for mass remaining after $t$ years is $A(t) = A_0(0.5)^{t/5730}$ . To create a model in terms of centuries, $c$ , we use the conversion $t = 100c$ . Substituting this into the year-based model gives $M(c) = A_0(0.5)^{100c/5730}$ . This simplifies to $M(c) = A_0(0.5)^{10c/573}$ .
Which of the following functions $M(t)$ represents the amount of medication, in mg, remaining in the bloodstream after $t$ hours?
Answer: $M(t) = 200e^{-0.15t}$ Continuous decay is modeled by the function $A(t) = A_0e^{kt}$ . The initial amount $A_0$ is 200 mg. A continuous decay rate of 15% means $k = -0.15$ . Therefore, the function is $M(t) = 200e^{-0.15t}$ . The function in choice B represents discrete decay, not continuous.
What is the physical meaning of the value -10 in the model?
Answer: It is the long-term temperature of the bar as it cools in the freezer. The model is a transformation of an exponential decay function. As time $t$ increases without bound, the term $110(0.4)^t$ approaches 0 because $0 < 0.4 < 1$ . Therefore, $T(t)$ approaches $-10$ . This represents the limiting temperature, which is the temperature of the surrounding environment (the freezer).
If $t$ is the number of months after the launch, which of the following functions $U(t)$ models the number of users on the platform?
Answer: $U(t) = 250(2)^t$ Let the model be $U(t)=ab^t$ . We have two points: (2, 1000) and (5, 8000). This gives two equations: $1000 = ab^2$ and $8000 = ab^5$ . Dividing the second equation by the first gives $\frac{8000}{1000} = \frac{ab^5}{ab^2}$ , which simplifies to $8 = b^3$ , so $b=2$ . Substituting $b=2$ into the first equation gives $1000 = a(2^2) = 4a$ , so $a=250$ . The model is $U(t) = 250(2)^t$ .
Assuming the number of trees follows an exponential decay model, which function $T(t)$ represents the number of trees in the forest $t$ years after 2010?
Answer: $T(t) = 50000(0.8)^{t/8}$ Let the model be $T(t)=ab^t$ . The year 2010 is $t=0$ , so the initial value is $a=50000$ . The year 2018 is $t=8$ . We have the point (8, 40000). So, $40000 = 50000(b)^8$ . Dividing by 50000 gives $0.8 = b^8$ . The annual decay factor is $b=(0.8)^{1/8}$ . The function is $T(t) = 50000((0.8)^{1/8})^t = 50000(0.8)^{t/8}$ .
A savings account earns 4.2% annual interest compounded monthly. An initial deposit of $P_0=2500$ is made, and the balance is modeled by $A(t)=2500\left(1+\frac{0.042}{12}\right)^{12t}$ , where $t$ is in years. A bank statement shows the following balances (rounded): Year 0: $2500$ , Year 1: $2607$ , Year 2: $2719$ , Year 3: $2836$ , Year 4: $2958$ . Using the provided data, what does the initial value in the exponential function represent in this context?
Answer: The starting account balance at $t=0$ years This question tests AP Precalculus understanding of exponential functions and data modeling, specifically interpreting parameters and predicting outcomes. Exponential functions model situations where a quantity grows or decays at a constant percentage rate. Key parameters include the initial value (starting point) and the growth/decay rate, which determines how quickly the function changes. In this scenario, the function models a savings account balance with monthly compounded interest, with an initial value of 2500 and a growth rate based on 4.2% annual interest. Choice C is correct because it accurately identifies the initial value as representing the starting account balance at t=0 years, consistent with the passage details. Choice A is incorrect because it misinterprets the initial value as interest earned, a common error when students confuse initial conditions with rate parameters. To help students: Emphasize understanding of exponential growth vs. linear growth, and practice interpreting data through graph analysis. Encourage identifying key function parameters and their real-world implications. Watch for: confusing initial values with growth rates or misreading graphical data.
A lab models radioactive decay with $M(t)=M_0\left(\tfrac12\right)^{t/8}$ , where $t$ is in days and the half-life is 8 days. A sample starts with $M_0=120$ mg. Measurements (rounded) are: Day 0: 120 mg, Day 8: 60 mg, Day 16: 30 mg, Day 24: 15 mg. Based on the scenario above, using the exponential model, predict the mass after 32 days.
Answer: About 7.5 mg This question tests AP Precalculus understanding of exponential functions and data modeling, specifically interpreting parameters and predicting outcomes. Exponential functions model situations where a quantity grows or decays at a constant percentage rate. Key parameters include the initial value (starting point) and the growth/decay rate, which determines how quickly the function changes. In this scenario, the function models radioactive decay with a half-life of 8 days, with an initial value of 120 mg and a decay factor of 1/2 every 8 days. Choice A is correct because it accurately identifies the predicted mass after 32 days as about 7.5 mg, consistent with the passage details. Choice B is incorrect because it misinterprets the decay progression, a common error when students confuse half-lives with linear subtraction. To help students: Emphasize understanding of exponential growth vs. linear growth, and practice interpreting data through graph analysis. Encourage identifying key function parameters and their real-world implications. Watch for: confusing initial values with growth rates or misreading graphical data.
What is the physical meaning of the value 150 in the model?
Answer: The initial concentration of the contaminant before filtration is 150 ppm. In the exponential model $f(x) = ab^x$ , the parameter $a$ represents the initial value, or the value of the function when $x=0$ . In this context, $x=0$ means the water has not yet passed through any length of the filter. Therefore, 150 is the initial concentration of the contaminant.
Which of the following represents the annual decay factor for this isotope?
Answer: $(0.25)^{1/10}$ Let the model be $A(t) = A_0 b^t$ , where $b$ is the annual decay factor. We are given that $A(10) = 0.25 A_0$ . So, $0.25 A_0 = A_0 b^{10}$ . Dividing by $A_0$ gives $0.25 = b^{10}$ . To find the annual decay factor $b$ , we take the 10th root of both sides: $b = (0.25)^{1/10}$ .

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