Exponential Functions - AP Precalculus

A city has 310,000 residents and grows 1.5% yearly, modeled by $P(t)=310000(1.015)^t$ . Using the information, what is the population after 12 years?
Answer: $310000(1.015)^{12}$ 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.
The function $f$ is given by $f(x) = 15(0.82)^x$ . Which of the following statements best describes the function $f$ ?
Answer: $f$ models exponential decay because the base, $0.82$ , is between $0$ and $1$ . In an exponential function of the form $f(x) = ab^x$ , the function models exponential growth if the base $b > 1$ and exponential decay if $0 < b < 1$ . In this case, the base is $b = 0.82$ , which satisfies $0 < 0.82 < 1$ . Therefore, the function models exponential decay. The initial value $a$ does not determine whether the function models growth or decay.
A scientist models the decay of a radioactive substance. The amount of the substance remaining, $A(t)$ , in grams, after $t$ years is modeled by the function $A(t) = 150(0.96)^t$ . Which of the following statements is the best interpretation of the numbers in this function?
Answer: The initial amount of the substance is 150 grams, and the amount decreases by 4% each year. 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 $(1 - r) \times 100\%$ .
Which of the following describes the horizontal asymptote of the graph of the function $f(x) = 5(0.2)^x$ ?
Answer: The graph has a horizontal asymptote at $y=0$ , which it approaches as $x \to \infty$ . 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 $x \to \infty$ .
An exponential function $f$ has the property that for any increase of 1 in the input variable $x$ , the output $f(x)$ is multiplied by a factor of 4. Which of the following could be an expression for $f(x)$ ?
Answer: $f(x) = 3(4)^x$ 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.
The function $g$ is defined as $g(x) = f(x) - 5$ . It is known that the output values of $g(x)$ change by a constant multiplicative factor for equal-length input-value intervals. Which of the following function types must $f$ be?
Answer: $f$ must be an exponential function with a horizontal asymptote at $y=5$ . 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 $y=5$ .
The function $q$ is given by $q(x) = 10(1/3)^x$ . Which of the following correctly describes the end behavior of $q$ ?
Answer: $\lim_{x\to\infty} q(x) = 0$ and $\lim_{x\to-\infty} q(x) = \infty$ . The function $q(x)$ represents exponential decay because the base $b=1/3$ is between 0 and 1. As $x$ increases without bound ( $x \to \infty$ ), $(1/3)^x$ approaches 0, so $q(x)$ approaches $10 \cdot 0 = 0$ . As $x$ decreases without bound ( $x \to -\infty$ ), $(1/3)^x$ increases without bound, so $q(x)$ increases without bound ( $q(x) \to \infty$ ).
A device battery starts at 100% and loses 8% of its charge per hour, modeled by $C(t)=100(0.92)^t$ . Which equation represents the scenario described?
Answer: $C(t)=100(0.92)^t$ 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.
Consider the exponential function $f(x) = 3(2)^x$ . Which statement about the extrema of $f$ on the interval $(-\infty, \infty)$ is true?
Answer: The function $f$ has neither a global maximum nor a global minimum. The function $f(x) = 3(2)^x$ represents exponential growth. Its domain is all real numbers and its range is $(0, \infty)$ . As $x \to \infty$ , $f(x) \to \infty$ , so there is no global maximum. As $x \to -\infty$ , $f(x)$ approaches 0 but never reaches it, so there is no global minimum. Thus, the function has no global extrema on its domain.
A car's value is $\$18{,}500$ and depreciates 7% yearly, modeled by $V(t)=18500(0.93)^t$ . Using the information, what will the value be after 3 years?
Answer: $18500(0.93)^3$ 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.
A medicine dosage is 75 mg and decreases 20% each hour, modeled by $D(t)=75(0.80)^t$ . Using the information, what is the dosage after 4 hours?
Answer: $75(0.80)^4$ 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.
A fish tank contains 260 fish and increases 5% each month, modeled by $N(t)=260(1.05)^t$ . Using the information, what is the number of fish after 6 months?
Answer: $260(1.05)^6$ 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.
A research colony begins with 2,500 bacteria and grows 9% each hour, modeled by $B(t)=2500\cdot(1.09)^t$ . Using the information, what will the population be after 12 hours?
Answer: $2500\cdot(1.09)^{12}$ 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.
The function $k$ is given by $k(t) = 50(2)^{-t}$ . Which of the following statements is true?
Answer: The function $k$ represents exponential decay with a base of $1/2$ . 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.
A lab has 120 grams of a chemical that decays 12% each day, modeled by $m(t)=120(0.88)^t$ . Using the information, what is the mass after 7 days?
Answer: $120(0.88)^7$ 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.
A lab has 120 grams of a substance that decays 12% per day, modeled by $m(t)=120\cdot(0.88)^t$ . Using the information, what will the mass be after 10 days?
Answer: $120\cdot(0.88)^{10}$ 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).
An investment of $\$8{,}000$ grows 4% per year, modeled by $V(t)=8000(1.04)^t$ . Using the information, what will the value be after 9 years?
Answer: $8000(1.04)^9$ 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.
Consider the functions $f(x) = (1.5)^x$ and $g(x) = (2.5)^x$ . Which statement best compares the behavior of these two functions for $x > 0$ ?
Answer: Both functions are increasing, and $g(x)$ increases at a faster proportional rate than $f(x)$ . Both functions represent exponential growth because their bases (1.5 and 2.5) are greater than 1. For exponential growth, a larger base corresponds to a faster proportional rate of increase. Since $2.5 > 1.5$ , the function $g(x)$ increases at a faster rate than $f(x)$ for $x > 0$ .
A wildlife reserve begins with 3,200 deer and grows 4% each year, modeled by $N(t)=3200\cdot(1.04)^t$ . How long will it take for the population to reach 4,000 deer?
Answer: Solve $3200\cdot(1.04)^t=4000$ 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.
A $\$12{,}000$ account increases 3.5% per year, modeled by $V(t)=12000\cdot(1.035)^t$ . Which equation represents the scenario described?
Answer: $V(t)=12000\cdot(1.035)^t$ 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.

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