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AP Precalculus Quiz

AP Precalculus Quiz: Change In Linear And Exponential Functions

Practice Change In Linear And Exponential Functions in AP Precalculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 18

0 of 18 answered

A car is worth V_0=\24{,}000.Lineardepreciation:. Linear depreciation: .Lineardepreciation:V_L(t)=24{,}000-1{,}800t,where, where ,wheretisyearsandis years andisyearsand$1{,}800/yearislost.Exponentialdepreciation:/year is lost. Exponential depreciation: /yearislost.Exponentialdepreciation:V_E(t)=24{,}000(0.90)^t$, where value drops 10% per year. Values are in dollars. Linear change subtracts the same amount yearly, while exponential change multiplies by the same factor yearly. In the context of this scenario, which model predicts a greater decrease after 3 years?

Select an answer to continue

What this quiz covers

This quiz focuses on Change In Linear And 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.

All questions

Question 1

A car is worth V_0=\24{,}000.Lineardepreciation:. Linear depreciation: .Lineardepreciation:V_L(t)=24{,}000-1{,}800t,where, where ,wheretisyearsandis years andisyearsand$1{,}800/yearislost.Exponentialdepreciation:/year is lost. Exponential depreciation: /yearislost.Exponentialdepreciation:V_E(t)=24{,}000(0.90)^t$, where value drops 10% per year. Values are in dollars. Linear change subtracts the same amount yearly, while exponential change multiplies by the same factor yearly. In the context of this scenario, which model predicts a greater decrease after 3 years?

  1. Linear model, because a constant percent always decreases more.
  2. Exponential model, because 10% of the current value compounds each year. (correct answer)
  3. Both models, because they start at the same initial value.
  4. Neither model, because depreciation cannot be modeled with functions.

Explanation: This question tests AP Precalculus understanding of changes in linear and exponential functions. Linear functions change at a constant rate, represented by a straight line, while exponential functions change at a rate proportional to their current value, creating a curve. In the given scenario, we need to compare the depreciation after 3 years: linear loses 1,800×3=1,800×3=1,800×3=5,400, while exponential retains 90% each year, resulting in 24,000×(0.90)3=24,000×(0.90)³=24,000×(0.90)3=17,496, a loss of 6,504.ChoiceBiscorrectbecausetheexponentialmodelpredictsagreaterdecrease(6,504. Choice B is correct because the exponential model predicts a greater decrease (6,504.ChoiceBiscorrectbecausetheexponentialmodelpredictsagreaterdecrease(6,504) compared to the linear model ($5,400) after 3 years. Choice A is incorrect because it makes the false claim that constant percent always decreases more, which depends on the specific values and time frame. Encourage students to calculate actual values at specific time points to compare models. Emphasize that exponential decay can exceed linear decay when the percentage is significant enough.

Question 2

A car is purchased for V_0=\30{,}000.Lineardepreciation:. Linear depreciation: .Lineardepreciation:V_L(t)=30{,}000-2{,}500t,where, where ,wheretisyearsandis years andisyearsand$2{,}500/yearislost.Exponentialdepreciation:/year is lost. Exponential depreciation: /yearislost.Exponentialdepreciation:V_E(t)=30{,}000(0.88)^t$, where value drops 12% per year. Values are in dollars. Linear change subtracts the same dollar amount yearly, while exponential change subtracts a percent of the current value yearly. Based on the scenario described, which function best describes the change in the exponential depreciation model?

  1. V(t)=30,000−2,500tV(t)=30{,}000-2{,}500tV(t)=30,000−2,500t
  2. V(t)=30,000(0.88)tV(t)=30{,}000(0.88)^tV(t)=30,000(0.88)t (correct answer)
  3. V(t)=2,500(0.88)tV(t)=2{,}500(0.88)^tV(t)=2,500(0.88)t
  4. V(t)=30,000(1.12)tV(t)=30{,}000(1.12)^tV(t)=30,000(1.12)t

Explanation: This question tests AP Precalculus understanding of changes in linear and exponential functions. Linear functions change at a constant rate, represented by a straight line, while exponential functions change at a rate proportional to their current value, creating a curve. In the given scenario, the exponential depreciation model loses 12% of value each year, which means multiplying by (1-0.12)=0.88 annually. Choice B is correct because V(t)=30,000(0.88)^t represents starting with $30,000 and retaining 88% (losing 12%) of the value each year. Choice D is incorrect because it uses 1.12, which would represent 12% growth rather than 12% depreciation. Encourage students to recognize that depreciation by x% means multiplying by (1-x/100). Use car value examples to demonstrate how percentage-based depreciation creates exponential decay.

Question 3

A town has P0=50,000P_0=50{,}000P0​=50,000 people. Linear model: PL(t)=50,000+800tP_L(t)=50{,}000+800tPL​(t)=50,000+800t, where ttt is years and 800 people/year is constant. Exponential model: PE(t)=50,000(1.016)tP_E(t)=50{,}000(1.016)^tPE​(t)=50,000(1.016)t, where 1.6% per year compounds. Both models measure population in people. The linear change adds the same number each year, while the exponential change adds a percentage of the current population. Using the information provided, how does the rate of change differ between the linear and exponential models?

  1. Both models increase by 800 people each year.
  2. Linear adds 800 people/year; exponential increases by 1.6% of the current population each year. (correct answer)
  3. Linear increases by 1.6% each year; exponential adds 800 people/year.
  4. Both models increase by a constant percent each year.

Explanation: This question tests AP Precalculus understanding of changes in linear and exponential functions. Linear functions change at a constant rate, represented by a straight line, while exponential functions change at a rate proportional to their current value, creating a curve. In the given scenario, the linear model adds 800 people each year (constant amount), while the exponential model grows by 1.6% of the current population annually (constant percentage). Choice B is correct because it accurately identifies that linear adds a fixed number (800 people/year) while exponential increases by a fixed percentage (1.6%) of the current population. Choice A is incorrect because it claims both models add 800 people yearly, missing that exponential growth depends on the current population size. Encourage students to identify whether the rate of change is a constant amount (linear) or constant percentage (exponential). Use population growth examples to illustrate how exponential growth accelerates over time as the base increases.

Question 4

A town begins with P0=60,000P_0=60{,}000P0​=60,000 people. Linear model: PL(t)=60,000+900tP_L(t)=60{,}000+900tPL​(t)=60,000+900t, where ttt is years and 900 people/year is constant. Exponential model: PE(t)=60,000(1.012)tP_E(t)=60{,}000(1.012)^tPE​(t)=60,000(1.012)t, where 1.2% per year compounds. Population is measured in people. Linear change is additive and constant, while exponential change is multiplicative and depends on current size. Based on the scenario described, what assumptions are made in the linear model vs the exponential model?

  1. Linear assumes the same number is added yearly; exponential assumes the same percent is applied yearly. (correct answer)
  2. Linear assumes the same percent is applied yearly; exponential assumes the same number is added yearly.
  3. Both assume the population decreases each year by a constant amount.
  4. Both assume the population change is unrelated to time.

Explanation: This question tests AP Precalculus understanding of changes in linear and exponential functions. Linear functions change at a constant rate, represented by a straight line, while exponential functions change at a rate proportional to their current value, creating a curve. In the given scenario, the linear model assumes a constant addition of 900 people yearly, while the exponential model assumes growth by 1.2% of the current population annually. Choice A is correct because it accurately states that linear assumes the same number (900) is added yearly while exponential assumes the same percent (1.2%) is applied yearly. Choice B is incorrect because it reverses these assumptions, confusing which model uses absolute versus relative change. Encourage students to identify the fundamental difference: linear models assume change independent of current size, while exponential models assume change proportional to current size. Use population dynamics to illustrate why exponential models often better represent natural growth.

Question 5

A lab culture begins with N0=5,000N_0=5{,}000N0​=5,000 bacteria. Linear model: NL(t)=5000+300tN_L(t)=5000+300tNL​(t)=5000+300t, where ttt is hours and 300 bacteria/hour is constant. Exponential model: NE(t)=5000(1.06)tN_E(t)=5000(1.06)^tNE​(t)=5000(1.06)t, where 6% per hour compounds. Counts are in bacteria. Linear growth adds a fixed number, while exponential growth scales by a fixed percent. Using the information provided, what assumptions are made in the linear model vs the exponential model?

  1. Linear assumes a constant increase; exponential assumes a constant percent increase. (correct answer)
  2. Linear assumes a constant percent increase; exponential assumes a constant increase.
  3. Both assume the increase per hour is proportional to time only.
  4. Both assume the culture decreases by the same amount each hour.

Explanation: This question tests AP Precalculus understanding of changes in linear and exponential functions. Linear functions change at a constant rate, represented by a straight line, while exponential functions change at a rate proportional to their current value, creating a curve. In the given scenario, the linear model assumes bacteria increase by a fixed number (300) each hour, while the exponential model assumes bacteria increase by a fixed percentage (6%) of the current population. Choice A is correct because it accurately states that linear assumes constant increase (300 bacteria/hour) while exponential assumes constant percent increase (6% per hour). Choice B is incorrect because it reverses the assumptions, confusing which model uses constant amounts versus constant percentages. Encourage students to identify the key difference: linear adds the same amount regardless of current size, while exponential adds more as the population grows. Use biological growth contexts to illustrate why exponential models often better represent population dynamics.

Question 6

A town begins with P0=48,000P_0=48{,}000P0​=48,000 people at t=0t=0t=0 years. One plan adds 1,2001{,}2001,200 people/year: PL(t)=48,000+1200tP_L(t)=48{,}000+1200tPL​(t)=48,000+1200t. Another plan grows by 2.5%2.5\%2.5% yearly: PE(t)=48,000(1.025)tP_E(t)=48{,}000(1.025)^tPE​(t)=48,000(1.025)t. Variables: ttt in years, PPP in people. The linear model’s increase stays constant, while the exponential model’s increase rises as PPP rises. Based on the scenario described, which model predicts a greater change after 20 years?

  1. Linear model, because constant addition always beats percent growth over long times.
  2. Exponential model, because the yearly increase becomes larger as the population base grows. (correct answer)
  3. Both models, because 2.5% of 48,000 equals 1,200 every year.
  4. Neither model, because population cannot be measured in people.

Explanation: This question tests AP Precalculus understanding of changes in linear and exponential functions. Linear functions change at a constant rate, represented by a straight line, while exponential functions change at a rate proportional to their current value, creating a curve. The linear model adds exactly 1,200 people per year, while the exponential model multiplies the population by 1.025 each year, meaning the actual number added increases as the population grows. Choice B is correct because after 20 years, the exponential model yields 48,000(1.025)20≈78,66348,000(1.025)^{20} ≈ 78,66348,000(1.025)20≈78,663 (change of 30,663), while the linear model yields 48,000+1,200(20)=72,00048,000 + 1,200(20) = 72,00048,000+1,200(20)=72,000 (change of 24,000). Choice C is incorrect because it assumes 2.5% of 48,000 remains constant, failing to recognize that in exponential growth, the base amount changes each year. Students should understand that exponential growth compounds, making the yearly increase larger over time. Use long-term projections to demonstrate how exponential models eventually dominate linear ones.

Question 7

A town has P0=20,000P_0=20{,}000P0​=20,000 residents at t=0t=0t=0 years. A linear model adds 600600600 residents/year: PL(t)=20,000+600tP_L(t)=20{,}000+600tPL​(t)=20,000+600t. An exponential model grows by 3%3\%3% per year: PE(t)=20,000(1.03)tP_E(t)=20{,}000(1.03)^tPE​(t)=20,000(1.03)t. Variables: ttt in years, PPP in residents. The linear rate of change is constant, while the exponential rate increases as the population grows. Using the information provided, how does the rate of change differ between the linear and exponential models?

  1. Linear adds 600 residents each year; exponential adds a constant 3% of the current population each year. (correct answer)
  2. Linear adds 3% each year; exponential adds 600 residents each year.
  3. Both models add the same number of residents each year because they start at 20,000.
  4. Exponential adds a constant number each year; linear adds an increasing percent each year.

Explanation: This question tests AP Precalculus understanding of changes in linear and exponential functions. Linear functions change at a constant rate, represented by a straight line, while exponential functions change at a rate proportional to their current value, creating a curve. In the given scenario, the linear model adds exactly 600 residents each year regardless of population size, while the exponential model adds 3% of the current population, meaning the actual number added increases as the population grows. Choice A is correct because it accurately identifies that linear adds a fixed 600 residents annually while exponential adds 3% of the current population each year. Choice B is incorrect because it reverses the models, suggesting linear uses percentage growth when it actually uses constant addition. Encourage students to identify whether the rate of change is constant in absolute terms (linear) or proportional to the current value (exponential). Use population growth examples to show how exponential models compound while linear models maintain steady increases.

Question 8

A car is worth V_0=\24{,}000atpurchase( at purchase (atpurchase(t=0years).Alineardepreciationmodelsubtractsyears). A linear depreciation model subtractsyears).Alineardepreciationmodelsubtracts$2{,}000peryear:per year:peryear:V_L(t)=24{,}000-2000t.Anexponentialdepreciationmodelloses. An exponential depreciation model loses .Anexponentialdepreciationmodelloses12%peryear:per year:peryear:V_E(t)=24{,}000(0.88)^t.Variables:. Variables: .Variables:tinyears,in years,inyears,V$ in dollars. Linear change is a constant dollar drop, while exponential change is a constant percent drop. Based on the scenario described, which function best describes the change in value if the car loses the same percent each year?

  1. VL(t)=24,000−2000tV_L(t)=24{,}000-2000tVL​(t)=24,000−2000t, because subtracting is always percent loss.
  2. VE(t)=24,000(0.88)tV_E(t)=24{,}000(0.88)^tVE​(t)=24,000(0.88)t, because the value is multiplied by the same factor each year. (correct answer)
  3. VL(t)=24,000−2000tV_L(t)=24{,}000-2000tVL​(t)=24,000−2000t, because linear models always fit depreciation best.
  4. VE(t)=24,000(1.12)tV_E(t)=24{,}000(1.12)^tVE​(t)=24,000(1.12)t, because depreciation means the factor is greater than 1.

Explanation: This question tests AP Precalculus understanding of changes in linear and exponential functions. Linear functions change at a constant rate, represented by a straight line, while exponential functions change at a rate proportional to their current value, creating a curve. The scenario describes a car that loses the same percent of its value each year, which is characteristic of exponential decay where the value is multiplied by a constant factor less than 1. Choice B is correct because VE(t)=24,000(0.88)tV_E(t)=24,000(0.88)^tVE​(t)=24,000(0.88)t represents losing 12% each year (keeping 88%), which matches the constant percent loss described in the question. Choice D is incorrect because it uses a factor greater than 1 (1.12), which would represent growth rather than depreciation. Help students recognize that exponential decay uses factors between 0 and 1, while growth uses factors greater than 1. Practice converting between percent decrease and decay factors (100% - 12% = 88% = 0.88).

Question 9

An investment begins at A_0=\10{,}000whenwhenwhent=0years.Alinearreturnmodeladdsyears. A linear return model addsyears.Alinearreturnmodeladds$600/year:/year: /year:A_L(t)=10{,}000+600t.Acompoundmodelgrowsby. A compound model grows by .Acompoundmodelgrowsby5%/year:/year: /year:A_E(t)=10{,}000(1.05)^t.Variables:. Variables: .Variables:tinyearsandin years andinyearsandA$ in dollars. Linear growth has a constant dollar rate, while exponential growth has a constant percent rate. Based on the scenario described, how does the rate of change differ between the linear and exponential models?

  1. Linear has a constant \600$/year increase; exponential has a constant 5% increase on the current balance. (correct answer)
  2. Linear has a constant 5% increase; exponential has a constant \600$/year increase.
  3. Both have constant rates because both are functions of time.
  4. Exponential has a constant dollar increase; linear has an increasing percent increase.

Explanation: This question tests AP Precalculus understanding of changes in linear and exponential functions. Linear functions change at a constant rate, represented by a straight line, while exponential functions change at a rate proportional to their current value, creating a curve. The linear model adds exactly 600eachyearregardlessoftheaccountbalance,representingaconstantdollarrateofchange,whiletheexponentialmodelmultipliesthebalanceby1.05eachyear,representingaconstant5600 each year regardless of the account balance, representing a constant dollar rate of change, while the exponential model multiplies the balance by 1.05 each year, representing a constant 5% rate of change. Choice A is correct because it accurately identifies that linear has a constant 600eachyearregardlessoftheaccountbalance,representingaconstantdollarrateofchange,whiletheexponentialmodelmultipliesthebalanceby1.05eachyear,representingaconstant5600/year increase while exponential has a constant 5% increase on the current balance. Choice B is incorrect because it reverses these characteristics, suggesting linear uses percentage growth when it actually uses constant dollar addition. Help students distinguish between absolute change (dollars per year) and relative change (percent per year). Use investment examples to show how compound interest creates increasing dollar gains while maintaining a constant percentage rate.

Question 10

An investment starts at V_0=\2{,}000.Linearplan:. Linear plan: .Linearplan:V_L(t)=2000+120t,where, where ,wheretisyearsandis years andisyearsand$120/yearisadded.Compoundplan:/year is added. Compound plan: /yearisadded.Compoundplan:V_E(t)=2000(1.05)^t$, where 5% per year compounds. Values are in dollars. Linear growth changes by a constant amount, while exponential growth changes by a constant percent. Based on the scenario described, which function best describes the change in the compound plan’s value?

  1. V(t)=2000+120tV(t)=2000+120tV(t)=2000+120t
  2. V(t)=2000(1.05)tV(t)=2000(1.05)^tV(t)=2000(1.05)t (correct answer)
  3. V(t)=120(1.05)tV(t)=120(1.05)^tV(t)=120(1.05)t
  4. V(t)=2000−120tV(t)=2000-120tV(t)=2000−120t

Explanation: This question tests AP Precalculus understanding of changes in linear and exponential functions. Linear functions change at a constant rate, represented by a straight line, while exponential functions change at a rate proportional to their current value, creating a curve. In the given scenario, the compound plan grows by 5% annually, which means multiplying the current value by 1.05 each year, illustrating exponential growth. Choice B is correct because V(t)=2000(1.05)^t represents starting with 2,000andmultiplyingby1.05eachyear,whichmodels52,000 and multiplying by 1.05 each year, which models 5% compound growth. Choice A is incorrect because it represents linear growth (adding 2,000andmultiplyingby1.05eachyear,whichmodels5120 yearly) rather than the compound growth described. Encourage students to recognize that compound interest means multiplying by (1 + rate) each period. Use financial contexts to demonstrate how exponential models capture percentage-based growth.

Question 11

A cup of coffee is T0=90∘CT_0=90^\circ\text{C}T0​=90∘C at t=0t=0t=0 minutes in a 20∘C20^\circ\text{C}20∘C room. A linear model cools by 1.5∘C1.5^\circ\text{C}1.5∘C/min: TL(t)=90−1.5tT_L(t)=90-1.5tTL​(t)=90−1.5t. An exponential model follows Newton-style cooling: TE(t)=20+70(0.92)tT_E(t)=20+70(0.92)^tTE​(t)=20+70(0.92)t. Variables: ttt in minutes, TTT in ∘C^\circ\text{C}∘C. Linear cooling drops by the same number of degrees each minute, while exponential cooling slows as it approaches room temperature. Using the information provided, what assumptions are made in the linear model vs the exponential model?

  1. Linear assumes constant temperature ratio; exponential assumes constant degree decrease per minute.
  2. Linear assumes constant degree decrease per minute; exponential assumes a constant percent of the remaining difference each minute. (correct answer)
  3. Both assume the coffee reaches 0∘C0^\circ\text{C}0∘C after a fixed time.
  4. Both assume the coffee warms up because the room is cooler.

Explanation: This question tests AP Precalculus understanding of changes in linear and exponential functions. Linear functions change at a constant rate, represented by a straight line, while exponential functions change at a rate proportional to their current value, creating a curve. The linear cooling model assumes the coffee loses a constant 1.5°C every minute regardless of its current temperature, while the exponential model assumes the coffee loses a constant percentage of the temperature difference from room temperature each minute. Choice B is correct because it accurately describes that linear assumes constant degree decrease per minute while exponential assumes a constant percent of the remaining difference between coffee and room temperature. Choice A is incorrect because it reverses these assumptions, incorrectly stating that linear uses ratios. Teach students to recognize Newton's Law of Cooling as an exponential model where the rate of cooling is proportional to the temperature difference. Use physical examples to show why cooling slows as objects approach ambient temperature.

Question 12

A car’s value is V_0=\30{,}000atatatt=0years.Lineardepreciationsubtractsyears. Linear depreciation subtractsyears.Lineardepreciationsubtracts$3{,}500/year:/year: /year:V_L(t)=30{,}000-3500t.Exponentialdepreciationmultipliesby. Exponential depreciation multiplies by .Exponentialdepreciationmultipliesby0.85yearly:yearly:yearly:V_E(t)=30{,}000(0.85)^t.Variables:. Variables: .Variables:tinyears,in years,inyears,V$ in dollars. Linear change is a constant dollar decrease, while exponential change is a constant percent decrease. Using the information provided, how does the rate of change differ between the linear and exponential models?

  1. Linear decreases by a constant \3{,}500$/year; exponential decreases by a constant 15% of the current value each year. (correct answer)
  2. Linear decreases by 15% each year; exponential decreases by \3{,}500$ each year.
  3. Both decrease by \3{,}500eachyearbecausebothstartateach year because both start ateachyearbecausebothstartat$30{,}000$.
  4. Exponential decreases by a constant \3{,}500$/year; linear decreases by a constant percent each year.

Explanation: This question tests AP Precalculus understanding of changes in linear and exponential functions. Linear functions change at a constant rate, represented by a straight line, while exponential functions change at a rate proportional to their current value, creating a curve. The linear depreciation model subtracts a fixed 3,500eachyearregardlessofthecar′scurrentvalue,whiletheexponentialmodelmultipliesby0.85,whichmeansthecarloses153,500 each year regardless of the car's current value, while the exponential model multiplies by 0.85, which means the car loses 15% of its current value each year. Choice A is correct because it accurately states that linear decreases by a constant 3,500eachyearregardlessofthecar′scurrentvalue,whiletheexponentialmodelmultipliesby0.85,whichmeansthecarloses153,500/year while exponential decreases by a constant 15% of the current value. Choice B is incorrect because it reverses the models, incorrectly claiming linear uses percentage depreciation. Students should recognize that linear depreciation results in equal dollar losses each year, while exponential depreciation results in decreasing dollar losses but constant percentage losses. Use depreciation schedules to illustrate how these models diverge over time.

Question 13

A metal object is T0=100∘CT_0=100^\circ\text{C}T0​=100∘C at t=0t=0t=0 minutes in a 25∘C25^\circ\text{C}25∘C room. A linear model cools by 2∘C2^\circ\text{C}2∘C/min: TL(t)=100−2tT_L(t)=100-2tTL​(t)=100−2t. An exponential model is TE(t)=25+75(0.90)tT_E(t)=25+75(0.90)^tTE​(t)=25+75(0.90)t. Variables: ttt in minutes, TTT in ∘C^\circ\text{C}∘C. Linear cooling keeps the same degree drop each minute, while exponential cooling drops quickly at first then slows near 25∘C25^\circ\text{C}25∘C. Based on the scenario described, which function best describes the change in temperature if the cooling slows as it approaches room temperature?

  1. TL(t)=100−2tT_L(t)=100-2tTL​(t)=100−2t, because slowing cooling means subtracting a constant each minute.
  2. TE(t)=25+75(0.90)tT_E(t)=25+75(0.90)^tTE​(t)=25+75(0.90)t, because the remaining difference shrinks by the same factor each minute. (correct answer)
  3. TL(t)=25+75tT_L(t)=25+75tTL​(t)=25+75t, because approaching room temperature requires a positive slope.
  4. TE(t)=25+75t0.90T_E(t)=25+75t^{0.90}TE​(t)=25+75t0.90, because exponential models always use fractional powers.

Explanation: This question tests AP Precalculus understanding of changes in linear and exponential functions. Linear functions change at a constant rate, represented by a straight line, while exponential functions change at a rate proportional to their current value, creating a curve. The question describes cooling that slows as temperature approaches room temperature, which is characteristic of exponential decay where the rate of change depends on the current temperature difference from ambient. Choice B is correct because TE(t)=25+75(0.90)tT_E(t)=25+75(0.90)^tTE​(t)=25+75(0.90)t models temperature approaching 25°C asymptotically, with the cooling rate decreasing as the temperature difference shrinks. Choice A is incorrect because linear cooling maintains a constant 2°C/min decrease, which would eventually make temperature go below room temperature rather than approaching it. Teach students that exponential models are appropriate when the rate of change is proportional to the distance from an equilibrium value. Use cooling curves to show how exponential models naturally approach but never cross asymptotic values.

Question 14

An account has A_0=\8{,}000atatatt=0years.Alinearplanaddsyears. A linear plan addsyears.Alinearplanadds$320eachyear:each year:eachyear:A_L(t)=8000+320t.Acompoundplangrowsby. A compound plan grows by .Acompoundplangrowsby3.8%peryear:per year:peryear:A_E(t)=8000(1.038)^t.Variables:. Variables: .Variables:tinyears,in years,inyears,Aindollars.Lineargrowthhasconstantslopeindollarsperyear,whileexponentialgrowthhasachangingdollarincrease.Usingtheinformationprovided,whichfunctionbestdescribesthechangeinaccountbalanceiftheyearlyincreaseisalwaysin dollars. Linear growth has constant slope in dollars per year, while exponential growth has a changing dollar increase. Using the information provided, which function best describes the change in account balance if the yearly increase is alwaysindollars.Lineargrowthhasconstantslopeindollarsperyear,whileexponentialgrowthhasachangingdollarincrease.Usingtheinformationprovided,whichfunctionbestdescribesthechangeinaccountbalanceiftheyearlyincreaseisalways$320$?

  1. AE(t)=8000(1.038)tA_E(t)=8000(1.038)^tAE​(t)=8000(1.038)t, because compound interest adds the same dollars each year.
  2. AL(t)=8000+320tA_L(t)=8000+320tAL​(t)=8000+320t, because the amount increases by a constant \320$/year. (correct answer)
  3. AE(t)=8000(0.962)tA_E(t)=8000(0.962)^tAE​(t)=8000(0.962)t, because growth must use a factor less than 1.
  4. AL(t)=8000+320tA_L(t)=8000+320^tAL​(t)=8000+320t, because linear models use exponents on the rate.

Explanation: This question tests AP Precalculus understanding of changes in linear and exponential functions. Linear functions change at a constant rate, represented by a straight line, while exponential functions change at a rate proportional to their current value, creating a curve. The question asks which function describes a situation where the yearly increase is always 320,whichisaconstantadditivechangecharacteristicoflineargrowth.ChoiceBiscorrectbecause320, which is a constant additive change characteristic of linear growth. Choice B is correct because 320,whichisaconstantadditivechangecharacteristicoflineargrowth.ChoiceBiscorrectbecauseA_L(t)=8000+320taddsexactlyadds exactlyaddsexactly320 each year, matching the described constant yearly increase. Choice A is incorrect because it suggests compound interest (exponential model) adds the same dollars each year, when actually the dollar amount added increases each year in exponential growth. Help students distinguish between constant dollar increases (linear) and constant percentage increases (exponential). Emphasize reading problems carefully to identify whether the change is described in absolute or relative terms.

Question 15

A savings account starts at V_0=\1{,}500.Lineardepositplan:. Linear deposit plan: .Lineardepositplan:V_L(t)=1500+200t,where, where ,wheretisyearsandis years andisyearsand$200/yearisadded.Compoundplan:/year is added. Compound plan: /yearisadded.Compoundplan:V_E(t)=1500(1.04)^t$, where 4% per year compounds. Values are in dollars. Linear change is constant in dollars, while exponential change is constant in percent. Using the information provided, how does the rate of change differ between the linear and exponential models?

  1. Linear increases by a constant percent; exponential increases by a constant dollar amount.
  2. Both increase by \200$ each year, but exponential starts later.
  3. Linear increases by \200$/year; exponential increases by 4% of the current balance each year. (correct answer)
  4. Both increase by 4% each year because they start at \1{,}500$.

Explanation: This question tests AP Precalculus understanding of changes in linear and exponential functions. Linear functions change at a constant rate, represented by a straight line, while exponential functions change at a rate proportional to their current value, creating a curve. In the given scenario, the linear plan adds a fixed 200eachyear,whilethecompoundplangrowsby4200 each year, while the compound plan grows by 4% of the current balance annually. Choice C is correct because it accurately identifies that linear increases by 200eachyear,whilethecompoundplangrowsby4200/year (constant dollar amount) while exponential increases by 4% of the current balance each year (constant percentage). Choice A is incorrect because it reverses the characteristics, claiming linear uses constant percent and exponential uses constant dollars. Encourage students to distinguish between absolute change (dollars) and relative change (percent). Use savings account examples to show how compound interest accelerates growth over time.

Question 16

An investment starts at V_0=\5{,}000.Linearplan:. Linear plan: .Linearplan:V_L(t)=5000+300t,where, where ,wheretisyearsandis years andisyearsand$300/yearisadded.Exponentialplan:/year is added. Exponential plan: /yearisadded.Exponentialplan:V_E(t)=5000(1.03)^t$, where 3% per year compounds. Values are in dollars. Linear growth adds a constant amount, while exponential growth adds a percent of the current value. In the context of this scenario, which model predicts a greater change after 20 years?

  1. Linear model, because compounding makes the increase smaller each year.
  2. Exponential model, because a constant percent increase compounds over time. (correct answer)
  3. Both models, because they have the same initial value.
  4. Linear model, because 3% is always less than adding \300$.

Explanation: This question tests AP Precalculus understanding of changes in linear and exponential functions. Linear functions change at a constant rate, represented by a straight line, while exponential functions change at a rate proportional to their current value, creating a curve. In the given scenario, after 20 years: linear adds 300×20=6,000,whileexponentialgrowsto5000×(1.03)20≈6,000, while exponential grows to 5000×(1.03)²⁰≈6,000,whileexponentialgrowsto5000×(1.03)20≈9,031, an increase of about 4,031.ChoiceBiscorrectbecausetheexponentialmodel′s34,031. Choice B is correct because the exponential model's 3% compounding effect accumulates over 20 years to produce greater growth than the linear model's constant 4,031.ChoiceBiscorrectbecausetheexponentialmodel′s3300 yearly addition. Choice D is incorrect because it misunderstands that 3% of an increasing base eventually exceeds $300, especially over longer time periods. Encourage students to calculate long-term values to see how compound growth accelerates. Emphasize that time horizon matters when comparing linear and exponential models.

Question 17

A car’s value starts at V_0=\18{,}000.Lineardepreciation:. Linear depreciation: .Lineardepreciation:V_L(t)=18{,}000-1{,}200t,where, where ,wheretisyearsandis years andisyearsand$1{,}200/yearislost.Exponentialdepreciation:/year is lost. Exponential depreciation: /yearislost.Exponentialdepreciation:V_E(t)=18{,}000(0.93)^t$, where value drops 7% per year. Values are in dollars. Linear change has constant slope, while exponential change has a constant multiplier. Using the information provided, how does the rate of change differ between the linear and exponential models?

  1. Linear decreases by \1{,}200$/year; exponential decreases by 7% of the current value each year. (correct answer)
  2. Linear decreases by 7% each year; exponential decreases by \1{,}200$/year.
  3. Both decrease by \1{,}200eachyearbecausetheystartateach year because they start ateachyearbecausetheystartat$18{,}000$.
  4. Both decrease by a constant percent each year, so both are exponential.

Explanation: This question tests AP Precalculus understanding of changes in linear and exponential functions. Linear functions change at a constant rate, represented by a straight line, while exponential functions change at a rate proportional to their current value, creating a curve. In the given scenario, the linear model subtracts a fixed 1,200eachyear,whiletheexponentialmodelretains931,200 each year, while the exponential model retains 93% (loses 7%) of the current value annually. Choice A is correct because it accurately identifies that linear decreases by 1,200eachyear,whiletheexponentialmodelretains931,200/year (constant amount) while exponential decreases by 7% of the current value each year (constant percentage). Choice B is incorrect because it reverses the characteristics of each model, confusing absolute and relative change. Encourage students to recognize that linear depreciation removes the same dollar amount yearly, while exponential depreciation removes a percentage of the remaining value. Use depreciation contexts to show how exponential decay slows over time in absolute terms.

Question 18

A town’s population is modeled two ways starting at P0=40,000P_0=40{,}000P0​=40,000 people. Linear: PL(t)=40,000+500tP_L(t)=40{,}000+500tPL​(t)=40,000+500t, with ttt in years and 500 people/year constant. Exponential: PE(t)=40,000(1.02)tP_E(t)=40{,}000(1.02)^tPE​(t)=40,000(1.02)t, with 2% per year compounding. Population is measured in people. Linear adds the same number each year, while exponential adds a percentage of the current population. Based on the scenario described, in the context of population growth, which model predicts a greater change after 10 years?

  1. Linear model, because exponential growth always slows over time.
  2. Exponential model, because compounding 2% on a growing base can exceed adding 500/year. (correct answer)
  3. Both models, because they start at 40,000 people.
  4. Linear model, because percent change is the same as adding 500 each year.

Explanation: This question tests AP Precalculus understanding of changes in linear and exponential functions. Linear functions change at a constant rate, represented by a straight line, while exponential functions change at a rate proportional to their current value, creating a curve. In the given scenario, after 10 years: linear adds 500×10=5,000 people, while exponential grows to 40,000×(1.02)¹⁰≈48,760, an increase of about 8,760 people. Choice B is correct because the exponential model's 2% compounding on an increasing base eventually exceeds the linear model's constant addition of 500 people per year. Choice A is incorrect because it falsely claims exponential growth slows over time, when actually exponential growth accelerates as the base increases. Encourage students to calculate values at different time points to see how exponential growth eventually overtakes linear growth. Emphasize that even small percentage growth can compound significantly over time.