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. $a=7$ 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 3

1. The function $k$ represents exponential growth with a base of 2.
2. The function $k$ represents exponential growth with a base of -2.
3. The function $k$ represents exponential decay with a base of $1/2$. (correct answer)
4. The function $k$ represents exponential decay with a base of -2.

Explanation: Using the properties of exponents, the function can be rewritten as $k(t) = 50(2^{-1})^t = 50(1/2)^t$. This function is in the form $ab^t$ with $a=50$ and $b=1/2$. Since the base $b$ satisfies $0 < b < 1$, the function represents exponential decay.

Question 4

1. Domain: all real numbers; Range: all real numbers
2. Domain: $x > 0$; Range: $y > 0$
3. Domain: all real numbers; Range: $y > 0$ (correct answer)
4. Domain: all real numbers; Range: $y > 6$

Explanation: The domain of any exponential function of the form $ab^x$ is the set of all real numbers. The base $1/4$ raised to any real power $x$ is always a positive number. Multiplying by a positive constant $a=6$ results in an output that is always positive. Therefore, the range is all positive real numbers, or $y > 0$.

Question 5

1. The domain is all real numbers because the expression $50(2)^t$ is defined for all real $t$.
2. The domain is $t \ge 0$ because time since the start of the experiment cannot be negative. (correct answer)
3. The domain is all integers because the population is counted at discrete time intervals.
4. The domain is $t > 0$ because the population only exists after the experiment has started.

Explanation: While the mathematical domain of the function $P(t) = 50(2)^t$ is all real numbers, the context of the problem restricts the domain. The variable $t$ represents time in hours since the start of an experiment, so negative values of $t$ are not meaningful. The time $t=0$ represents the initial measurement, so it should be included in the domain. Therefore, the contextual domain is $t \ge 0$.

Question 6

1. $f$ must be a linear function with a slope of 5.
2. $f$ must be a quadratic function with a vertex at $y=5$.
3. $f$ must be an exponential function with a horizontal asymptote at $y=5$. (correct answer)
4. $f$ must be an exponential function with a horizontal asymptote at $y=-5$.

Explanation: The property that output values change by a constant multiplicative factor for equal-length input-value intervals defines an exponential function. Therefore, $g(x)$ must be an exponential function of the form $g(x) = ab^x$. Since $g(x) = f(x) - 5$, we can write $f(x) = g(x) + 5$, which means $f(x) = ab^x + 5$. This is an exponential function vertically shifted up by 5 units. The horizontal asymptote of $g(x)=ab^x$ is $y=0$, so the horizontal asymptote of $f(x)$ is $y=5$.

Question 7

1. The graph is always increasing or always decreasing, and such functions cannot have inflection points.
2. The graph's rate of change is always positive, and a positive rate of change prevents a change in concavity.
3. The graph is always concave up, so there is no change in concavity from up to down or down to up. (correct answer)
4. The graph has a horizontal asymptote, and functions with asymptotes do not have inflection points.

Explanation: A point of inflection is a point on a graph where the concavity changes (from concave up to concave down, or vice versa). For an exponential function $f(x) = ab^x$ with $a>0$, the graph is always concave up over its entire domain. Since there is no change in concavity, there are no points of inflection.

Question 8

1. 4
2. 8 (correct answer)
3. 16
4. 63

Explanation: When you encounter questions about average rates of change for exponential functions, you're dealing with the slope of secant lines over specific intervals. The key insight is that exponential functions have a special property: their average rates of change scale predictably. Let's calculate each average rate of change using the formula $\frac{f(b) - f(a)}{b - a}$. For $R_1$ over $[1, 3]$:

• $f(1) = 3(2)^1 = 6$
• $f(3) = 3(2)^3 = 24$
• $R_1 = \frac{24 - 6}{3 - 1} = \frac{18}{2} = 9$

• $f(4) = 3(2)^4 = 48$
• $f(6) = 3(2)^6 = 192$
• $R_2 = \frac{192 - 48}{6 - 4} = \frac{144}{2} = 72$

Question 9

1. The initial amount of the substance is 96 grams, and the amount decreases by 150% each year.
2. The initial amount of the substance is 150 grams, and the amount decreases by 96% each year.
3. The initial amount of the substance is 150 grams, and the amount decreases by 4% each year. (correct answer)
4. The amount of the substance after 1 year is 150 grams, and it decreases by 4% each year.

Explanation: When you encounter exponential decay models, you need to identify two key components: the initial value and the decay factor. The general form is $A(t) = A_0 \cdot r^t$, where $A_0$ is the initial amount and $r$ is the factor by which the quantity changes each time period. In $A(t) = 150(0.96)^t$, the initial amount is 150 grams because when $t = 0$, we get $A(0) = 150(0.96)^0 = 150(1) = 150$. The decay factor is 0.96, which means 96% of the substance remains after each year. If 96% remains, then 4% decays each year (since $100\% - 96\% = 4\%$). Choice A incorrectly identifies 96 as the initial amount and claims a 150% decrease, which is mathematically impossible since you can't lose more than 100% of something. Choice B correctly identifies the initial amount as 150 grams but misinterprets the decay factor—it states the substance decreases by 96% each year, which would mean only 4% remains, corresponding to a factor of 0.04, not 0.96. Choice D incorrectly suggests that 150 grams is the amount after one year rather than the initial amount. The correct answer is C: 150 grams initial amount with a 4% annual decrease. Strategy tip: For exponential functions $A(t) = A_0 \cdot r^t$, always evaluate at $t = 0$ to find the initial value, and remember that if $r < 1$, the percent decrease equals $(1 - r) \times 100\%$.

Question 10

1. $A_n = 64\sqrt{3} \cdot (0.5625)^{n-1}$ (correct answer)
2. $A_n = 64\sqrt{3} \cdot (0.75)^{n-1}$
3. $A_n = 4\sqrt{3} \cdot (0.75)^{n-1}$
4. $A_n = 36\sqrt{3} \cdot (0.5625)^{n-1}$

Explanation: This problem combines geometric sequences with area formulas, so you need to track how both the side lengths and areas change as the sequence progresses. Start by finding the pattern for side lengths. The first triangle has side length 16, and each subsequent triangle has a side length that's 75% of the previous one. This gives you the sequence: $s_1 = 16$, $s_2 = 16(0.75)$, $s_3 = 16(0.75)^2$, so generally $s_n = 16(0.75)^{n-1}$. Now apply the area formula. Since $A = \frac{\sqrt{3}}{4}s^2$, you have: $A_n = \frac{\sqrt{3}}{4}[16(0.75)^{n-1}]^2 = \frac{\sqrt{3}}{4} \cdot 256 \cdot (0.75)^{2(n-1)} = 64\sqrt{3} \cdot (0.75)^{2n-2}$ The key insight is that $(0.75)^2 = 0.5625$, so $A_n = 64\sqrt{3} \cdot (0.5625)^{n-1}$. Choice A is correct because it properly squares the ratio when going from side length to area. Choice B uses the original ratio (0.75) instead of its square, missing that area scales with the square of linear dimensions. Choice C has the wrong coefficient—it uses 4 instead of 64, likely from forgetting to square the initial side length. Choice D has an incorrect coefficient of 36, possibly from computational errors in squaring 16. Study tip: When geometric sequences involve area or volume, remember that the ratio gets raised to the 2nd or 3rd power respectively. Always square the linear scaling factor for area problems.

Question 11

1. The population of Town A is always increasing by a greater number of people each year than the population of Town B.
2. The populations of the two towns will never be equal because they have different initial values and different growth rates.
3. The population of Town B is growing at a slower percentage rate than Town A, but its population will eventually exceed Town A's population.
4. The population of Town B will eventually exceed the population of Town A, and its annual increase in population will also eventually exceed that of Town A. (correct answer)

Explanation: When you encounter exponential growth problems comparing two populations, you need to analyze both the growth rates and how the functions behave over time, not just their initial values. Let's examine what happens to these populations. Town A starts with 2000 people and grows at 3% annually, while Town B starts with 1000 people but grows at 5% annually. Although Town B starts smaller, its higher growth rate means it will eventually catch up and surpass Town A's population. To see when populations become equal, you'd solve $2000(1.03)^t = 1000(1.05)^t$. This equation has a solution (around t = 14 years), after which Town B's population exceeds Town A's. Moreover, since Town B has the higher growth rate, its annual population increase (the derivative of the exponential function) will also eventually exceed Town A's annual increase. This confirms answer D is correct. Answer A is wrong because it assumes Town A always has a larger annual increase, but exponential functions with higher growth rates eventually dominate. Answer B incorrectly claims the populations will never be equal—different initial values and growth rates don't prevent intersection when one growth rate exceeds the other. Answer C correctly identifies that Town B will eventually exceed Town A despite its slower percentage rate, but this contradicts the given information since 5% > 3%. Strategy tip: In exponential growth comparisons, the function with the larger base (growth rate) will always eventually dominate, regardless of initial values. Focus on the exponents, not just the coefficients.

Question 12

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 13

1. The graph of $f$ is always concave up. (correct answer)
2. The graph of $f$ is always concave down.
3. The graph of $f$ is concave up for $x<0$ and concave down for $x>0$.
4. The graph of $f$ has a point of inflection at $x=0$.

Explanation: The graph of an exponential function of the form $f(x) = ab^x$ with a positive initial value ($a>0$) is always concave up. This is true for both exponential growth ($b>1$) and exponential decay ($0<b<1$). Since $a=2$ is positive, the graph of $f$ is always concave up.

Question 14

1. $\lim_{x\to\infty} p(x) = \infty$ and $\lim_{x\to-\infty} p(x) = 0$. (correct answer)
2. $\lim_{x\to\infty} p(x) = 0$ and $\lim_{x\to-\infty} p(x) = \infty$.
3. $\lim_{x\to\infty} p(x) = \infty$ and $\lim_{x\to-\infty} p(x) = -\infty$.
4. $\lim_{x\to\infty} p(x) = 0.5$ and $\lim_{x\to-\infty} p(x) = 0$.

Explanation: The function $p(x)$ represents exponential growth because the base $b=4$ is greater than 1. As $x$ increases without bound ($x \to \infty$), $4^x$ increases without bound, so $p(x) \to \infty$. As $x$ decreases without bound ($x \to -\infty$), $4^x$ approaches 0, so $p(x)$ approaches $0.5 \cdot 0 = 0$.

Question 15

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 16

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 17

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 18

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% per year. The initial value is $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 19

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% annually, giving us A(t) = 20000 * (1.042)^t. To find when it reaches $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 20

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).