AP PreCalculus Practice Quiz: Exponential Function Context and Data Modeling

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 16 questions to check your progress.

Question 1 of 16

An exponential function is the most appropriate model for a quantity that, over equal-length input-value intervals, changes by:

A)adding the same constant amount.B)multiplying by the same constant factor.C)alternating between increasing and decreasing by a fixed amount.D)approaching a fixed horizontal asymptote from both directions.

All Questions (16)

An exponential function is the most appropriate model for a quantity that, over equal-length input-value intervals, changes by:

A) adding the same constant amount.

B) multiplying by the same constant factor.

C) alternating between increasing and decreasing by a fixed amount.

D) approaching a fixed horizontal asymptote from both directions.

Correct Answer: B

Based on the provided content, exponential functions model growth patterns where successive output values over equal-length input-value intervals are proportional, which means they are multiplied by a constant factor.

The population of a city is modeled by the function P(t) = 120,000(1.03)^t, where t is the number of years since 2010. In this model, what does the value 1.03 represent?

A) The initial population of the city in 2010.

B) The annual population growth rate of 103%.

C) The population of the city after 1.03 years.

D) The annual growth factor, corresponding to a 3% increase.

Correct Answer: D

In an exponential model f(x) = ab^x, the base 'b' is the growth factor. A value of b = 1.03 can be interpreted as 1 + 0.03, which corresponds to a 3% increase for each unit change in the input (in this case, each year).

A scientist starts an experiment with 500 bacteria. The number of bacteria is found to double every 3 hours. Which function models N(t), the number of bacteria after t hours?

A) N(t) = 500(2)^t

B) N(t) = 500(3)^t

C) N(t) = 500(2)^(t/3)

D) N(t) = 500(3)^(t/2)

Correct Answer: C

An exponential model can be constructed from an initial value and a ratio. The initial value 'a' is 500. The growth factor 'b' is 2 (doubling). Since this doubling occurs every 3 hours, the exponent must be scaled by t/3 to correctly model the growth over time.

The value of a rare collectible grows exponentially. In 2015 (t=0), its value was $800. In 2020 (t=5), its value was $1600. Which of the following functions V(t) models the value of the collectible t years after 2015?

A) V(t) = 800(2)^(t/5)

B) V(t) = 800(2)^t

C) V(t) = 800(1.5)^t

D) V(t) = 1600(0.5)^t

Correct Answer: A

An exponential model can be constructed from two input-output pairs. We have the points (0, 800) and (5, 1600). The initial value is a=800. We can set up the equation 1600 = 800 * b^5. Solving for b gives b^5 = 2, so b = 2^(1/5). The model is V(t) = 800 * (2^(1/5))^t = 800(2)^(t/5).

A researcher is analyzing a data set that shows the concentration of a medication in a patient's bloodstream over several hours. The researcher suspects the decay is exponential. What is the most appropriate technological method to construct a model for this data?

A) Linear regression

B) Quadratic regression

C) Exponential regression

D) Calculating the simple average of the concentrations.

Correct Answer: C

The provided content states that exponential function models can be constructed for a data set with technology using exponential regressions. Since the decay is suspected to be exponential, this is the correct tool.

The value of a machine depreciates according to the model V(t) = 40,000(0.90)^t, where t is the number of years. What is the annual percent change in the machine's value?

A) A 90% decrease

B) A 10% increase

C) A 90% increase

D) A 10% decrease

Correct Answer: D

The base of the exponent, b=0.90, is the decay factor. The percent change 'r' is related by b = 1 + r. So, 0.90 = 1 + r, which gives r = -0.10. This represents a 10% decrease in value each year.

An investment grows according to the function A(t) = 5000(1.2)^t, where t is in years. Which of the following equivalent functions reveals the approximate growth factor for each half-year period?

A) A(h) = 5000(1.1)^h, where h is half-years

B) A(h) = 5000(1.095)^h, where h is half-years

C) A(h) = 5000(1.44)^h, where h is half-years

D) A(h) = 5000(0.6)^h, where h is half-years

Correct Answer: B

Equivalent forms of an exponential function can reveal different properties. To find the half-year growth factor from the annual factor of 1.2, we calculate (1.2)^(1/2), which is approximately 1.095. The new model would be A(h) = 5000(1.095)^h, where h is the number of half-year periods.

In many models for continuous growth or decay, such as in finance or physics, a particular irrational number is used as the base of the exponential function. This number is known as the natural base and is represented by the letter:

A) p

B) i

C) π

D) e

Correct Answer: D

The provided content explicitly states that the natural base e, which is approximately 2.718, is often used as the base in exponential functions that model contextual scenarios.

The number of fish in a lake is modeled by F(t) = 800(0.75)^t, where t is the number of years from today. According to the model, what is the expected number of fish in the lake 2 years from today?

A) 1200

B) 600

C) 450

D) 300

Correct Answer: C

To apply the exponential model, we substitute the input value t=2 into the function. F(2) = 800(0.75)^2 = 800(0.5625) = 450. So, the model predicts there will be 450 fish in 2 years.

A population of a certain species is modeled by P(t) = Ae^(kt). The population was 2,000 in year 0 and 2,443 in year 5. Which function correctly models this situation?

A) P(t) = 2000e^(0.04t)

B) P(t) = 2000e^(0.2t)

C) P(t) = 2443e^(-0.04t)

D) P(t) = 2000e^(ln(1.2215)t)

Correct Answer: A

This model uses the natural base e. The initial value A is 2000. We use the second point (5, 2443) to find k: 2443 = 2000e^(k*5). This simplifies to 1.2215 = e^(5k). Taking the natural log of both sides gives ln(1.2215) = 5k. Since ln(1.2215) is approximately 0.2, we have 0.2 ≈ 5k, which means k ≈ 0.04.

Which of the following situations is best modeled by an exponential function?

A) The cost of buying x gallons of gas at $3.50 per gallon.

B) A savings account that earns 2% interest compounded annually.

C) The height of a rock dropped from a cliff over time.

D) The perimeter of a square as a function of its side length.

Correct Answer: B

A savings account with annually compounded interest involves a proportional increase in the output (the balance) over equal-length input intervals (years). The balance is multiplied by a constant factor (1.02) each year, which is the definition of an exponential model.

An exponential function is given by g(x) = 9^(x/2). Which of the following is an equivalent form of the function that most clearly reveals the growth factor for each unit increase in x?

A) g(x) = 3^x

B) g(x) = 4.5^x

C) g(x) = (9/2)^x

D) g(x) = 9 * (1/2)^x

Correct Answer: A

Equivalent forms can reveal different properties. To find the unit growth factor, we want the form ab^x. We can rewrite g(x) = 9^(x/2) as g(x) = (9^(1/2))^x. Since 9^(1/2) is the square root of 9, which is 3, the equivalent function is g(x) = 3^x. This form clearly shows the growth factor is 3.

A computer's processing power has grown exponentially. In 2010, a chip had 1 billion transistors. In 2014, it had 4 billion. Assuming the model is P(t) = ab^t where t is years since 2010, what is the value of the growth factor b?

A) 4

B) 2

C) √2

D) ⁴√4

Correct Answer: D

An exponential model can be constructed from two input-output pairs. Let t=0 be 2010. The initial value a=1 (billion). The second point is (4, 4). So, 4 = 1 * b^4. To solve for b, we take the fourth root of both sides, so b = ⁴√4. This is also equal to √2.

The model f(x) = 50(1.08)^x represents the value of an asset in x years. An equivalent model using the natural base e is f(x) = 50e^(kx). What is the approximate value of k?

A) 1.08

B) 0.08

C) 0.077

D) 2.718

Correct Answer: C

To find an equivalent form using base e, we set the growth factors equal: e^k = 1.08. To solve for k, we take the natural logarithm of both sides: ln(e^k) = ln(1.08). This gives k = ln(1.08), which is approximately 0.077.

A model is constructed in the general form f(x) = ab^x. What property must the base, b, have for the function to model exponential decay?

A) b > 1

B) 0 < b < 1

C) b < 0

D) b = 1

Correct Answer: B

In the form f(x) = ab^x, the base 'b' is the growth factor. If b > 1, the function models growth. If 0 < b < 1, the function models decay, as each successive output is a fraction of the previous one. If b=1, the function is constant, and b cannot be negative for real-valued exponential functions.

The amount of a radioactive isotope is modeled by A(t) = 12(e)^(-0.05t), where A(t) is the amount in grams and t is the time in years. According to the model, what is the approximate amount of the isotope remaining after 10 years?

A) 19.8 grams

B) 7.28 grams

C) 6.00 grams

D) 0.6 grams

Correct Answer: B

To apply this exponential model, substitute t=10 into the function. A(10) = 12e^(-0.05*10) = 12e^(-0.5). Using the approximation e ≈ 2.718, this is approximately 12 * (1/√2.718) ≈ 12 * 0.6065, which equals 7.278 grams.