Change in linear and exponential functions — AP Precalculus Study Guide

For: AP Precalculus candidates sitting AP Precalculus.

Covers: Average rate of change for linear and exponential functions, constant absolute vs proportional relative change, the difference between absolute and relative change, and comparing growth rates of linear vs exponential functions over varying intervals.

You should already know: Slope calculation for linear functions, general form of exponential functions, basic difference quotient for average rate of change.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Precalculus style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Change in linear and exponential functions?

Change, in the context of functions, describes how the output value $f(x)$ shifts as the input $x$ increases by a fixed increment. For linear and exponential functions, change follows highly predictable patterns that clearly distinguish the two function types, forming the core of this foundational topic. Per the AP Precalculus CED, this is topic 2.1 in Unit 2 (Exponential and Logarithmic Functions), which contributes 19-26% of total exam score; this specific topic accounts for roughly 3-5% of the total AP exam score, and appears on both the multiple-choice (MCQ) and free-response (FRQ) sections. It is common to see this topic tested as a standalone MCQ, or as the opening part of a multi-part FRQ that asks you to model a real-world scenario and compare growth patterns.

Standard notation for this topic uses $\Delta x=h$ to represent a fixed, constant increment in the input $x$, and $\Delta f=f(x+h)-f(x)$ to represent the corresponding change in output. Common synonyms you will see on the exam include absolute change, relative change, proportional change, and average rate of change over a fixed interval.

2. Constant Change in Linear Functions

Linear functions have the general form $f(x)=mx+b$, where $m$ is the slope (constant rate of change) and $b$ is the $y$-intercept (output when $x=0$). The defining characteristic of a linear function, which separates it from all other function types on the exam, is that over any fixed interval of input $\Delta x=h$, the absolute change in output $\Delta f$ is constant, regardless of the starting value of $x$. We can confirm this with a simple derivation: $$ f(x+h) - f(x) = [m(x+h) + b] - [mx + b] = mh $$ The result $mh$ has no dependence on the starting input $x$, so it is identical for any interval of length $h$. The average rate of change over any interval is $\frac{\Delta f}{\Delta x}=\frac{mh}{h}=m$, which is also constant, equal to the slope of the line. Intuitively, linear change means every 1-unit increase in $x$ adds the same fixed amount to $f(x)$, for example, hourly wages where each extra hour adds the same fixed dollar amount to total pay.

Worked Example

Problem: The value of a bicycle depreciates linearly over time. After 2 years of ownership, the value is $420; after 5 years, the value is $180. What is the constant annual change in the value of the bicycle, and what will its value be after 7 years?

1. Define variables: let $t$ = time in years, $V(t)$ = value of the bicycle in dollars. We have two points: $(t_1,V_1)=(2,420)$ and $(t_2,V_2)=(5,180)$.
2. Calculate the change in input $\Delta t=5-2=3$ years, and the corresponding change in output $\Delta V=180-420=-240$ dollars.
3. Since change is constant for linear functions, the annual change is $\frac{\Delta V}{\Delta t}=\frac{-240}{3}=-80$ dollars per year. The negative sign confirms the value decreases each year.
4. To find the value after 7 years, add two more years of constant change to the value at 5 years: $V(7)=V(5)+2(-80)=180-160=20$ dollars.

Exam tip: When asked for change, always include the correct sign for decreasing functions. AP exam graders dock points for missing the sign in contextual problems.

3. Proportional Change in Exponential Functions

Exponential functions have the general form $f(x)=a{b}^x$, where $a\ne 0$ is the initial value (output when $x=0$) and $b>0,b\ne 1$ is the base (growth/decay factor per 1-unit increase in $x$). The defining characteristic of an exponential function is that over any fixed interval of input $\Delta x=h$, the relative (proportional) change in output is constant, regardless of the starting input $x$. We can derive this property similarly: $$ f(x+h) = ab^{x+h} = ab^x b^h = f(x) b^h $$ The ratio of outputs after the interval is $\frac{f(x+h)}{f(x)}=b^h$, which is constant and independent of $x$. The relative change is then: $$ \text{Relative change} = \frac{f(x+h) - f(x)}{f(x)} = b^h - 1 $$ This is also constant for any starting $x$. Intuitively, exponential change means every 1-unit increase in $x$ increases $f(x)$ by a fixed percentage of its current value, for example, savings account interest where the balance grows by a fixed percentage each year, so the absolute dollar increase grows over time.

Worked Example

Problem: The population of a bacteria colony grows exponentially over time. After 3 hours, the population is 2000 bacteria; after 5 hours, the population is 8000 bacteria. What is the constant hourly relative change in the population?

1. Let $P(t)=a{b}^t$ be the population at time $t$ hours, where $b$ is the hourly growth factor. We use the fact that the ratio of populations separated by a fixed interval of time is constant for exponential functions.
2. The time difference between measurements is $\Delta t=5-3=2$ hours. We write the ratio of populations: $\frac{P(5)}{P(3)}=\frac{a{b}^5}{a{b}^3}=b^2$.
3. Substitute the given population values: $\frac{8000}{2000}=4=b^2$, so solve for $b$: since growth factor is positive, $b=2$.
4. The hourly relative change is $b-1=2-1=1$, or 100% per hour. This checks out: after 2 hours, the population is $2\times 2=4$ times its original size, matching the given values.

Exam tip: Always distinguish between growth factor $b$ and relative change. If the question asks for percentage change, you need to calculate $b-1$, not leave your answer as $b$. This is one of the most common point-losing mistakes on this topic.

4. Comparing Growth Rates of Linear and Exponential Functions

A common AP exam question asks you to compare the output or growth rate of a linear and an exponential function, or find when exponential growth first overtakes linear growth. The core rule for this comparison is: for any linear function $f(x)=mx+b$ with $m>0$ and any exponential growth function $g(x)=a{b}^x$ with $a>0$ and $b>1$, exponential growth will eventually outpace linear growth, no matter how large the slope of the linear function or how slow the exponential growth. However, over short time horizons, linear growth can often be larger than exponential growth, so you cannot rely on the general rule alone—you must evaluate both functions at the input in question. To find when exponential overtakes linear, you can test integer values (for questions asking for full years/periods) or set the functions equal and solve using logarithms.

Worked Example

Problem: Two investment accounts start with $1000. Account A grows linearly by $200 per year. Account B grows exponentially by 10% per year. After how many full years will the value of Account B first exceed the value of Account A?

1. Write the equation for each account after $t$ years: $A(t)=1000+200t$ (linear, constant absolute change), $B(t)=1000(1.1{)}^t$ (exponential, constant relative change). We need the smallest positive integer $t$ where $B(t)>A(t)$.
2. Narrow down the range by testing a mid-range value: at $t=20$, $A(20)=1000+200(20)=5000$, $B(20)=1000(1.1{)}^{20}\approx 6727.5$, so $B$ is larger at $t=20$.
3. Check lower values to find the first crossing: at $t=15$, $A(15)=1000+200(15)=4000$, $B(15)\approx 1000(4.1772)=4177.2$, so $B>A$ at $t=15$.
4. Confirm it is the first full year by checking $t=14$: $A(14)=1000+200(14)=3800$, $B(14)\approx 1000(3.7975)=3797.5$, so $B<A$ at $t=14$. Thus, the first full year where $B$ exceeds $A$ is $t=15$.

Exam tip: When asked for the first full period where exponential overtakes linear, always check the integer before your candidate value to confirm it is still smaller. AP MCQ distractors are often designed to match the candidate value without this check.

5. Common Pitfalls (and how to avoid them)

• Wrong move: When asked for relative change, reports the growth factor $b$ instead of $b-1$. Why: Students confuse the ratio of final to initial output with the proportional change from initial to final. Correct move: Whenever a question asks for relative or percentage change, always compute $\frac{\text{final}-\text{initial}}{\text{initial}}=b^h-1$, not just $b^h$.
• Wrong move: Assumes exponential growth is faster than linear growth for all positive $x$, so it is always larger for any input. Why: Students memorize the general rule that exponential outpaces linear and forget it only applies for sufficiently large input. Correct move: Always evaluate both functions at the input given in the question, do not rely on the general rule to answer comparison questions.
• Wrong move: Calculates absolute change for an exponential function over a fixed interval and concludes the function is not exponential because the absolute change is not constant. Why: Students mix up the defining properties of linear and exponential functions. Correct move: For any exponential function, check that relative change is constant over a fixed interval, not absolute change.
• Wrong move: When calculating annual growth rate for an exponential function over a 10-year interval, uses the 10-year relative change as the annual relative change. Why: Students forget that exponential growth compounds, so growth factors multiply across intervals. Correct move: If total growth factor over $n$ intervals is $G$, the per-interval growth factor is $G^{1/n}$, so per-interval relative change is $G^{1/n}-1$.
• Wrong move: For exponential decay with $0<b<1$, reports a positive relative change instead of a negative one. Why: Students focus on the magnitude of the change and forget the direction. Correct move: For a decreasing exponential function, always write the relative change as a negative value (or clearly state it is a percentage decrease with a positive magnitude) to match the question's prompt.

6. Practice Questions (AP Precalculus Style)

Question 1 (Multiple Choice)

Which of the following statements correctly describes the difference between a linear function $f(x)=mx+b$ with $m>0$ and an exponential function $g(x)=a{b}^x$ with $a>0,b>1$ over intervals of fixed length $\Delta x=1$?

A) Both the absolute change and relative change are constant for both functions. B) Absolute change is constant for $f(x)$, relative change is constant for $g(x)$. C) Absolute change is constant for $g(x)$, relative change is constant for $f(x)$. D) Neither function has constant absolute change over all fixed intervals.

Worked Solution: This question tests the core defining properties of linear and exponential change. By definition, a linear function has constant absolute change over any fixed input interval: $\Delta f=f(x+1)-f(x)=m(x+1)+b-(mx+b)=m$, which is constant for any $x$. An exponential function has constant relative change over any fixed input interval: $\frac{g(x+1)-g(x)}{g(x)}=\frac{a{b}^{x+1}-a{b}^x}{a{b}^x}=b-1$, which is also constant for any $x$. This matches option B. Correct answer: $\boxed{B}$.

Question 2 (Free Response)

The table below shows the value of two functions $f(x)$ and $g(x)$ at integer values of $x$:

(a) Classify each function as linear or exponential. Justify your answer using the properties of change for each function. (b) Write the explicit equation for each function. (c) Find the value of $f(8)-g(8)$.

Worked Solution: (a) For $f(x)$, calculate absolute change over intervals of $\Delta x=2$: $26-10=16$, $42-26=16$, $58-42=16$. Absolute change over fixed intervals is constant, so $f(x)$ is linear. For $g(x)$, calculate relative change over intervals of $\Delta x=2$: $\frac{15-10}{10}=0.5$, $\frac{22.5-15}{15}=0.5$, $\frac{33.75-22.5}{22.5}=0.5$. Relative change over fixed intervals is constant, so $g(x)$ is exponential. (b) For $f(x)$, the per-unit slope is $\frac{16}{2}=8$, with initial value $f(0)=10$, so $f(x)=10+8x$. For $g(x)$, the growth factor per 2 units is $1+0.5=1.5$, so per-unit growth factor is $1.5^{1/2}$, so $g(x)=10(1.5{)}^{x/2}$. (c) Calculate values: $f(8)=10+8(8)=74$, $g(8)=10(1.5{)}^{8/2}=10(1.5{)}^4=50.625$. So $f(8)-g(8)=74-50.625=23.375$.

Question 3 (Application / Real-World Style)

A city is comparing two proposed plans for increasing its park space over the next 20 years. Plan 1 adds 12 acres of park space each year, starting from 150 acres of existing park space. Plan 2 increases the total park space by 4% per year, starting from the same 150 acres. Calculate the total park space under each plan after 20 years, rounded to the nearest whole acre, and state which plan results in more park space after 20 years.

Worked Solution: Define $t$ = number of years from now, $P_1(t)$ = total park space (acres) under Plan 1, $P_2(t)$ = total park space (acres) under Plan 2. Plan 1 has constant annual absolute change, so it is linear: $P_1(t)=150+12t$. For $t=20$, $P_1(20)=150+12(20)=390$ acres. Plan 2 has constant annual relative change, so it is exponential: $P_2(t)=150(1.04{)}^t$. For $t=20$, $P_2(20)=150(1.04{)}^{20}\approx 150(2.1911)\approx 329$ acres, rounded to the nearest whole acre. After 20 years, Plan 1 results in more total park space, because the low exponential growth rate has not yet outpaced the constant annual absolute addition from Plan 1.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundational prerequisite for all remaining topics in Unit 2 of AP Precalculus. Next, you will use the constant proportional change property of exponential functions to model exponential growth and decay in real-world contexts, including compound interest, population growth, and radioactive decay. Without mastering the difference between constant absolute and constant relative change, you will not be able to correctly classify real-world scenarios, build accurate models, or earn full points on multi-part FRQ questions. This topic also feeds into the larger study of rates of change across the AP Precalculus course, preparing you for the study of logarithmic function properties and the comparison of all function types in later units.

Exponential growth and decay modeling Logarithms and their properties Modeling with exponential functions Comparing function types