AP Precalculus Quiz: Exponential Functions

What this quiz covers

This quiz focuses on Exponential Functions, 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. $f(x) = 4x + 3$
2. $f(x) = x^4$
3. $f(x) = 3(4)^x$ (correct answer)
4. $f(x) = 4(3)^x$

Explanation: The property described is the defining characteristic of an exponential function where the base is the constant multiplicative factor. For an exponential function $f(x) = ab^x$, we have $f(x+1) = ab^{x+1} = ab^x \cdot b = b \cdot f(x)$. Since the output is multiplied by 4 for every unit increase in $x$, the base $b$ must be 4. The function $f(x) = 3(4)^x$ fits this form.

Question 2

1. The graph has a horizontal asymptote at $y=5$, which it approaches as $x \to -\infty$.
2. The graph has a horizontal asymptote at $y=5$, which it approaches as $x \to \infty$.
3. The graph has a horizontal asymptote at $y=0$, which it approaches as $x \to -\infty$.
4. The graph has a horizontal asymptote at $y=0$, which it approaches as $x \to \infty$. (correct answer)

Explanation: The function $f(x) = 5(0.2)^x$ is an exponential decay function because its base, $0.2$, is between 0 and 1. For an exponential decay function of this form, the graph approaches the x-axis as $x$ becomes very large. Therefore, the line $y=0$ is the horizontal asymptote, and the function approaches it as $x \to \infty$.

Question 3

1. The graph of $g$ is a reflection of the graph of $f$ across the y-axis.
2. The graph of $g$ is a reflection of the graph of $f$ across the x-axis. (correct answer)
3. The graph of $g$ is a vertical shift of the graph of $f$ by -4 units.
4. The graph of $g$ is a horizontal shift of the graph of $f$.

Explanation: The function $g(x)$ can be written as $g(x) = -f(x)$. A transformation of the form $y=-f(x)$ reflects the graph of $y=f(x)$ across the x-axis. Each y-coordinate on the graph of $f$ is replaced by its opposite on the graph of $g$.

Question 4

1. $a=7$ and $b>1$
2. $a=7$ and $0<b<1$ (correct answer)
3. $a<0$ and $b>1$
4. $a>0$ and $b<0$

Explanation: The y-intercept occurs at $x=0$. So, $f(0) = ab^0 = a(1) = a$. Since the y-intercept is $(0, 7)$, we have $a=7$. An exponential function is decreasing when either $a>0$ and $0<b<1$ (decay) or $a<0$ and $b>1$ (reflected growth). Since $a=7$ (which is positive), the function must be an exponential decay, which requires $0<b<1$.

Question 5

1. The function $h$ is always increasing. (correct answer)
2. The function $h$ is always decreasing.
3. The function $h$ is increasing for $x>0$ and decreasing for $x<0$.
4. The function $h$ is decreasing for $x>0$ and increasing for $x<0$.

Explanation: For an exponential function $h(x) = ab^x$ with $a > 0$, the function is always increasing if the base $b > 1$. In this case, $a=4$ (which is positive) and $b=1.5$ (which is greater than 1). Therefore, the function $h$ is always increasing across its domain of all real numbers.

Question 6

1. A decrease of 1.8% per year
2. An increase of 18% per year
3. An increase of 1.8% per year (correct answer)
4. An increase of 0.18% per year

Explanation: This question tests understanding of exponential functions in AP Precalculus, specifically interpreting and calculating exponential growth or decay. Exponential functions model situations where quantities grow or shrink at constant percentage rates. They are represented by y = a * b^x, where a is the initial value, b is the growth/decay factor, and x is time. In this scenario, a town has 150,000 residents and the model is P(t) = 150000 * (1.018)^t. The growth factor of 1.018 represents 100% + 1.8% = 101.8%, indicating growth of 1.8% per year. Choice C is correct because the growth factor 1.018 corresponds to an increase of 1.8% per year (since 1.018 = 1 + 0.018). Choice B is incorrect because it interprets the decimal 0.018 as 18%, a common error when students forget to convert decimals to percentages by multiplying by 100. To help students: Emphasize that growth factor = 1 + growth rate, so growth rate = growth factor - 1. Practice converting between growth factors and percentage rates, stressing that 0.018 = 1.8%, not 18%.

Question 7

1. $V(t)=12000\cdot(1.35)^t$
2. $V(t)=12000\cdot(1.035)^t$ (correct answer)
3. $V(t)=12000+0.035t$
4. $V(t)=12000\cdot(0.965)^t$

Explanation: This question tests understanding of exponential functions in AP Precalculus, specifically interpreting and calculating exponential growth or decay. Exponential functions model situations where quantities grow or shrink at constant percentage rates. They are represented by y = a * b^x, where a is the initial value, b is the growth/decay factor, and x is time. In this scenario, a $12,000 account increases 3.5$12,000, and the growth factor is 1.035 (representing 100% + 3.5% = 103.5%). Choice B is correct because V(t) = 12000 * (1.035)^t accurately models the account growing at 3.5% annually through compound interest. Choice D is incorrect because it uses 0.965 as the base, which would represent decay of 3.5%, not growth, a common error when students confuse growth and decay factors. To help students: Emphasize that growth rates require adding to 100% (1 + rate), while decay rates require subtracting from 100% (1 - rate). Practice identifying whether a scenario involves growth or decay before writing the equation.

Question 8

1. Solve $20000\cdot(1.042)^t=25000$ (correct answer)
2. Solve $20000+0.042t=25000$
3. Solve $20000\cdot(1.42)^t=25000$
4. Solve $20000\cdot(0.958)^t=25000$

Explanation: This question tests understanding of exponential functions in AP Precalculus, specifically interpreting and calculating exponential growth or decay. Exponential functions model situations where quantities grow or shrink at constant percentage rates. They are represented by y = a * b^x, where a is the initial value, b is the growth/decay factor, and x is time. In this scenario, a $20,000 certificate of deposit grows 4.2$25,000, we need to solve 20000 * (1.042)^t = 25000. Choice A is correct because it sets up the proper exponential equation with the correct growth factor of 1.042 to solve for the time needed. Choice B is incorrect because it assumes linear growth by adding 0.042t, failing to recognize that interest compounds exponentially. To help students: Emphasize that compound interest problems require exponential models, not linear ones. Practice setting up and solving exponential equations using logarithms to find the time variable.

Question 9

1. $120\cdot(1.12)^{10}$
2. $120\cdot(0.88)^{10}$ (correct answer)
3. $120-0.12\cdot 10$
4. $120\cdot(0.988)^{10}$

Explanation: This question tests understanding of exponential functions in AP Precalculus, specifically interpreting and calculating exponential growth or decay. Exponential functions model situations where quantities grow or shrink at constant percentage rates. They are represented by y = a * b^x, where a is the initial value, b is the growth/decay factor, and x is time. In this scenario, the lab has 120 grams of a substance that decays 12% per day, giving us m(t) = 120 * (0.88)^t. The initial value is 120 grams, and the decay factor is 0.88 (representing 100% - 12% = 88%). Choice B is correct because it accurately applies the exponential formula m(10) = 120 * (0.88)^10 to determine the mass after 10 days. Choice A is incorrect because it uses 1.12 as the base, treating decay as growth, a common error when students don't recognize that decay requires subtracting from 100%. To help students: Emphasize that decay means the remaining percentage (100% - decay rate), not adding the decay rate. Practice distinguishing between growth factors (greater than 1) and decay factors (between 0 and 1).

Question 10

1. $2500\cdot(1.09)^{12}$ (correct answer)
2. $2500\cdot(9)^{12}$
3. $2500+0.09\cdot 12$
4. $2500\cdot(1.9)^{12}$

Explanation: This question tests understanding of exponential functions in AP Precalculus, specifically interpreting and calculating exponential growth or decay. Exponential functions model situations where quantities grow or shrink at constant percentage rates. They are represented by y = a * b^x, where a is the initial value, b is the growth/decay factor, and x is time. In this scenario, a bacteria colony starts with 2,500 bacteria and grows 9% each hour, giving us B(t) = 2500 * (1.09)^t. The initial value is 2,500, and the growth factor is 1.09 (representing 100% + 9% = 109%). Choice A is correct because it accurately applies the exponential formula B(12) = 2500 * (1.09)^12 to determine the population after 12 hours. Choice B is incorrect because it uses 9 as the base instead of 1.09, a common error when students use the percentage directly without converting to a growth factor. To help students: Emphasize that percentage growth requires converting to a growth factor by adding 1. Practice problems involving bacterial growth to reinforce the exponential nature of population growth.

Question 11

1. $48000(1.03)^5$ (correct answer)
2. $48000+0.03(5)$
3. $48000(3)^5$
4. $48000(1.3)^5$

Explanation: This question tests understanding of exponential functions in AP Precalculus, specifically interpreting and calculating exponential growth or decay. Exponential functions model situations where quantities grow or shrink at constant percentage rates. They are represented by y = a * b^x, where a is the initial value, b is the growth/decay factor, and x is time. In this scenario, the town starts with 48,000 people and grows at 3% annually, meaning the growth factor is 1.03 (100% + 3% = 103% = 1.03). Choice A is correct because it accurately applies the exponential formula P(t) = 48000(1.03)^t with t = 5 years, giving us 48000(1.03)^5. Choice B is incorrect because it assumes linear growth by adding 0.03 × 5, which would only add 15% total rather than compounding the growth each year. To help students: Emphasize that percentage growth means multiplying by (1 + rate) each period, not adding the rate. Practice converting percentage rates to growth factors and stress the difference between linear and exponential growth.

Question 12

1. $120-0.12(7)$
2. $120(0.12)^7$
3. $120(0.88)^7$ (correct answer)
4. $120(1.12)^7$

Explanation: This question tests understanding of exponential functions in AP Precalculus, specifically interpreting and calculating exponential growth or decay. Exponential functions model situations where quantities grow or shrink at constant percentage rates. They are represented by y = a * b^x, where a is the initial value, b is the growth/decay factor, and x is time. In this scenario, the lab starts with 120 grams of chemical that decays 12% daily, meaning 88% remains each day, so the decay factor is 0.88 (100% - 12% = 88% = 0.88). Choice C is correct because it accurately applies the exponential formula m(t) = 120(0.88)^t with t = 7 days, giving us 120(0.88)^7. Choice B is incorrect because it uses 0.12 as the base, which represents the amount lost rather than the amount remaining. To help students: Emphasize that for decay problems, the base is (1 - decay rate), not the decay rate itself. Practice identifying whether to subtract from or add to 1 when converting percentage changes to growth/decay factors.

Question 13

1. Solve $3200\cdot(1.04)^t=4000$ (correct answer)
2. Solve $3200+0.04t=4000$
3. Solve $3200\cdot(1.4)^t=4000$
4. Solve $3200\cdot(0.96)^t=4000$

Explanation: This question tests understanding of exponential functions in AP Precalculus, specifically interpreting and calculating exponential growth or decay. Exponential functions model situations where quantities grow or shrink at constant percentage rates. They are represented by y = a * b^x, where a is the initial value, b is the growth/decay factor, and x is time. In this scenario, the wildlife reserve starts with 3,200 deer and grows 4% annually, giving us N(t) = 3200 * (1.04)^t. To find when the population reaches 4,000, we need to solve 3200 * (1.04)^t = 4000. Choice A is correct because it sets up the proper exponential equation to solve for the time when the population reaches 4,000 deer. Choice B is incorrect because it assumes linear growth by adding 0.04t, a common error when students don't recognize the compound nature of percentage growth. To help students: Emphasize that solving exponential equations requires setting the exponential expression equal to the target value. Practice using logarithms to solve for the exponent and reinforce the difference between linear and exponential models.

Question 14

1. $A(t)=2500+0.06t$
2. $A(t)=2500(1.6)^t$
3. $A(t)=2500(1.06)^t$ (correct answer)
4. $A(t)=2500(6)^t$

Explanation: This question tests understanding of exponential functions in AP Precalculus, specifically interpreting and calculating exponential growth or decay. Exponential functions model situations where quantities grow or shrink at constant percentage rates. They are represented by y = a * b^x, where a is the initial value, b is the growth/decay factor, and x is time. In this scenario, the account starts with $2,500 and earns 6% interest annually, so the growth factor is 1.06 (100% + 6% = 106% = 1.06). Choice C is correct because it accurately represents the exponential formula A(t) = 2500(1.06)^t, where 2500 is the initial amount and 1.06 is the annual growth factor. Choice A is incorrect because it represents linear growth, adding 0.06t instead of compounding the interest. To help students: Emphasize that compound interest means the balance is multiplied by the growth factor each year. Practice converting interest rates to growth factors by adding 1 to the decimal rate.

Question 15

1. $260(1.05)^5$
2. $260(1.05)^6$ (correct answer)
3. $260+0.05(6)$
4. $260(5)^6$

Explanation: This question tests understanding of exponential functions in AP Precalculus, specifically interpreting and calculating exponential growth or decay. Exponential functions model situations where quantities grow or shrink at constant percentage rates. They are represented by y = a * b^x, where a is the initial value, b is the growth/decay factor, and x is time. In this scenario, the tank starts with 260 fish and increases by 5% each month, so the growth factor is 1.05 (100% + 5% = 105% = 1.05). Choice B is correct because it accurately applies the exponential formula N(t) = 260(1.05)^t with t = 6 months, giving us 260(1.05)^6. Choice A is incorrect because it uses an exponent of 5 instead of 6, calculating the population after 5 months rather than 6 months. To help students: Emphasize careful reading to identify the correct time value. Practice substituting the given time value into the exponential formula and double-checking that the exponent matches the time period asked for.

Question 16

1. $8000(1.04)^9$ (correct answer)
2. $8000+0.04(9)$
3. $8000(4)^9$
4. $8000(1.4)^9$

Explanation: This question tests understanding of exponential functions in AP Precalculus, specifically interpreting and calculating exponential growth or decay. Exponential functions model situations where quantities grow or shrink at constant percentage rates. They are represented by y = a * b^x, where a is the initial value, b is the growth/decay factor, and x is time. In this scenario, the investment starts at $8,000 and grows at 4% per year, so the growth factor is 1.04 (100% + 4% = 104% = 1.04). Choice A is correct because it accurately applies the exponential formula V(t) = 8000(1.04)^t with t = 9 years, giving us 8000(1.04)^9. Choice B is incorrect because it represents linear growth by adding 0.04 × 9, which ignores the compounding effect of exponential growth. To help students: Emphasize that investment growth compounds, meaning each year's growth is based on the new total, not just the original amount. Use real-world examples to show how compound growth differs significantly from simple interest over time.

Question 17

1. $18500(0.93)^3$ (correct answer)
2. $18500-0.07(3)$
3. $18500(0.07)^3$
4. $18500(1.07)^3$

Explanation: This question tests understanding of exponential functions in AP Precalculus, specifically interpreting and calculating exponential growth or decay. Exponential functions model situations where quantities grow or shrink at constant percentage rates. They are represented by y = a * b^x, where a is the initial value, b is the growth/decay factor, and x is time. In this scenario, the car starts at $18,500 and depreciates 7% yearly, meaning 93% of its value remains each year, so the decay factor is 0.93 (100% - 7% = 93% = 0.93). Choice A is correct because it accurately applies the exponential formula V(t) = 18500(0.93)^t with t = 3 years, giving us 18500(0.93)^3. Choice D is incorrect because it uses 1.07 as the base, which would represent 7% growth rather than 7% depreciation. To help students: Emphasize that depreciation is a form of decay, so we subtract the rate from 1. Practice distinguishing between growth and decay scenarios to select the correct base for the exponential function.

Question 18

1. $310000(1.15)^{12}$
2. $310000(1.015)^{12}$ (correct answer)
3. $310000+0.015(12)$
4. $310000(1.015)^{11}$

Explanation: This question tests understanding of exponential functions in AP Precalculus, specifically interpreting and calculating exponential growth or decay. Exponential functions model situations where quantities grow or shrink at constant percentage rates. They are represented by y = a * b^x, where a is the initial value, b is the growth/decay factor, and x is time. In this scenario, the city starts with 310,000 residents and grows at 1.5% yearly, so the growth factor is 1.015 (100% + 1.5% = 101.5% = 1.015). Choice B is correct because it accurately applies the exponential formula P(t) = 310000(1.015)^t with t = 12 years, giving us 310000(1.015)^12. Choice A is incorrect because it uses 1.15 as the growth factor, which would represent 15% growth rather than 1.5% growth. To help students: Emphasize careful conversion of percentages to decimals - 1.5% = 0.015, not 0.15. Practice with various percentage values to build accuracy in converting rates to growth factors.

Question 19

1. $C(t)=100(1.08)^t$
2. $C(t)=100(0.92)^t$ (correct answer)
3. $C(t)=100-8t$
4. $C(t)=100(0.08)^t$

Explanation: This question tests understanding of exponential functions in AP Precalculus, specifically interpreting and calculating exponential growth or decay. Exponential functions model situations where quantities grow or shrink at constant percentage rates. They are represented by y = a * b^x, where a is the initial value, b is the growth/decay factor, and x is time. In this scenario, the battery starts at 100% and loses 8% per hour, meaning 92% remains each hour, so the decay factor is 0.92 (100% - 8% = 92% = 0.92). Choice B is correct because it accurately represents the exponential formula C(t) = 100(0.92)^t, where 100 is the initial charge and 0.92 is the hourly retention factor. Choice C is incorrect because it represents linear decay, subtracting 8t, which would mean losing exactly 8% of the original charge each hour rather than 8% of the remaining charge. To help students: Emphasize that percentage decay is multiplicative, not additive. Practice identifying the retention factor (1 - decay rate) for exponential decay problems.

Question 20

1. $75(1.20)^4$
2. $75(0.20)^4$
3. $75-0.20(4)$
4. $75(0.80)^4$ (correct answer)

Explanation: This question tests understanding of exponential functions in AP Precalculus, specifically interpreting and calculating exponential growth or decay. Exponential functions model situations where quantities grow or shrink at constant percentage rates. They are represented by y = a * b^x, where a is the initial value, b is the growth/decay factor, and x is time. In this scenario, the medicine starts at 75 mg and decreases by 20% each hour, meaning 80% remains each hour, so the decay factor is 0.80 (100% - 20% = 80% = 0.80). Choice D is correct because it accurately applies the exponential formula D(t) = 75(0.80)^t with t = 4 hours, giving us 75(0.80)^4. Choice B is incorrect because it uses 0.20 as the base, which represents the amount lost per hour rather than the amount remaining. To help students: Emphasize that in decay problems, we use the retention rate (1 - decay rate) as the base. Practice converting decay percentages to retention factors to avoid this common error.