Change in linear and exponential functions

AP Precalculus· AP Precalculus CED — Exponential and Logarithmic Functions· 14 min read

1. Constant Change in Linear Functions★★☆☆☆⏱ 4 min

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Linear functions have the general form $f (x) = m x + b$ , 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 is that over any fixed interval of input $Δ x = h$ , the absolute change in output $Δ f$ is constant, regardless of the starting value of $x$ .

🔬 Derivation

Goal:

Prove absolute change is constant for all linear functions

Starting from:

General form $f (x) = m x + b$

1
Calculate the difference between output after and before the increment:
2
$f(x+h) - f(x) = [m(x+h) + b] - [mx + b]$
3
Simplify the expression:
4
$mh + mx + b - mx - b = mh$

Result:

The resulting change $mh$ does not depend on the starting input $x$ , so absolute change is constant for any interval of length $h$ .

📐 Worked Example

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 $Δ t = 5 - 2 = 3$ years, and the corresponding change in output $Δ V = 180 - 420 = - 240$ dollars.
3
Since change is constant for linear functions, the annual change is:
4
$\frac{\Delta V}{\Delta t} = \frac{-240}{3} = -80$
5
The negative sign confirms the value decreases each year. To find the value after 7 years, add two more years of constant change to the value at 5 years:
6
$V(7) = V(5) + 2(-80) = 180 - 160 = 20$
7
The value of the bicycle after 7 years is $20.

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.

2. Proportional Change in Exponential Functions★★★☆☆⏱ 4 min

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Exponential functions have the general form $f (x) = a b^{x}$ , where $a \neq = 0$ is the initial value (output when $x = 0$ ) and $b > 0, b \neq = 1$ 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 $Δ x = h$ , the relative (proportional) change in output is constant, regardless of the starting input $x$ .

🔬 Derivation

Goal:

Prove relative change is constant for all exponential functions

Starting from:

General form $f (x) = a b^{x}$

1
Rewrite $f (x + h)$ using exponent rules:
2
$f(x+h) = ab^{x+h} = ab^x b^h = f(x) b^h$
3
The ratio of outputs after and before the interval is:
4
$\frac{f(x+h)}{f(x)} = b^h$
5
Relative change is defined as:
6
$\text{Relative change} = \frac{f(x+h) - f(x)}{f(x)} = b^h - 1$

Result:

Relative change $b^{h} - 1$ does not depend on the starting input $x$ , so relative change is constant for any interval of length $h$ .

📐 Worked Example

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 $Δ t = 5 - 3 = 2$ hours. We write the ratio of populations:
3
$\frac{P(5)}{P(3)} = \frac{ab^5}{ab^3} = b^2$
4
Substitute the given population values and solve for $b$ (we take the positive root since growth factor is always positive):
5
$\frac{8000}{2000} = 4 = b^2 \implies b = 2$
6
The hourly relative change is $b - 1 = 2 - 1 = 1$ , or 100% per hour, which matches the given 2-hour 4x increase.

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.

3. Comparing Growth Rates of Linear and Exponential Functions★★★☆☆⏱ 3 min

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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 is that for any increasing linear function $f (x) = m x + b$ with $m > 0$ and any increasing exponential growth function $g (x) = a b^{x}$ with $a > 0$ and $b > 1$ , exponential growth will eventually outpace linear growth. However, over short time horizons, linear growth can often be larger than exponential growth, so you must always evaluate both functions for the given input.

📐 Worked Example

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:
2
$A(t) = 1000 + 200t \quad \text{(linear, constant absolute change)}$
3
$B(t) = 1000(1.1)^t \quad \text{(exponential, constant relative change)}$
4
We need the smallest positive integer $t$ where $B (t) > A (t)$ . Test $t = 20$ to narrow the range: $A (20) = 5000$ , $B (20) \approx 6727.5$ , so $B$ is larger at $t = 20$ .
5
Check $t = 15$ : $A (15) = 4000$ , $B (15) \approx 4177.2$ , so $B > A$ at $t = 15$ .
6
Confirm it is the first full year by checking $t = 14$ : $A (14) = 3800$ , $B (14) \approx 3797.5$ , 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.

4. Classifying Functions from Tabular Data★★★☆☆⏱ 3 min

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AP Precalculus exams often ask you to classify a function as linear or exponential from tabular data, using the defining change properties of each function type. To do this, check for constant absolute change over equal input intervals to confirm linearity, or constant relative change over equal input intervals to confirm exponentiality.

📐 Worked Example

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

$x$	0	2	4	6
$f (x)$	10	26	42	58
$g (x)$	10	15	22.5	33.75

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

1
Part (a): Check change over fixed $Δ x = 2$ intervals. For $f (x)$ , absolute change between consecutive points is $26 - 10 = 16$ , $42 - 26 = 16$ , $58 - 42 = 16$ . Constant absolute change over fixed intervals confirms $f (x)$ is linear. For $g (x)$ , relative change between consecutive points is $\frac{15 - 10}{10} = 0.5$ , $\frac{22.5 - 15}{15} = 0.5$ , $\frac{33.75 - 22.5}{22.5} = 0.5$ . Constant relative change confirms $g (x)$ is exponential.
2
Part (b): Write equations. For $f (x)$ , initial value $f (0) = 10$ , per-unit slope is $\frac{16}{2} = 8$ , so:
3
$f(x) = 10 + 8x$
4
For $g (x)$ , initial value $g (0) = 10$ , 2-unit growth factor is $1.5$ , so per-unit growth factor is $1. 5^{1/2}$ , and:
5
$g(x) = 10(1.5)^{x/2}$
6
Part (c): Calculate the difference:
7
$f(8) = 74, \quad g(8) = 50.625, \quad f(8) - g(8) = 23.375$

5. Common Pitfalls

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 $\frac{final - initial}{initial} = b^{h} - 1$ , 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 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 an $n$ -year interval, uses the total $n$ -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. Quick Reference Cheatsheet

Category	Formula/Property	Notes
Linear function general form	$f (x) = m x + b$	$m$ = constant slope/rate of change
Absolute change (linear, $Δ x = h$ )	$Δ f = mh$	Constant for any starting input $x$
Defining property: linear	Constant absolute change over fixed input intervals	True for all linear functions
Exponential function general form	$f (x) = a b^{x}$	$a$ = initial value, $b > 0, b \neq = 1$ = per-unit growth/decay factor
Relative change (exponential, $Δ x = h$ )	$\frac{f ( x + h ) - f ( x )}{f ( x )} = b^{h} - 1$	Constant for any starting input $x$
Defining property: exponential	Constant relative change over fixed input intervals	True for all exponential functions
Per-interval growth factor	$b = G^{1/ n}, G = \frac{final}{initial}$	Use for $n$ total intervals to find per-interval factor
Exponential vs linear growth rule	For $b > 1, m > 0$ , $a b^{x} > m x + b$ for all sufficiently large $x$	Exponential always overtakes linear eventually, but may not for small $x$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

All · AP Precalculus
Compare linear vs exponential change

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.