Rates of Change and Behavior of Graphs

16 min read

Quick Reference

Field	Value
Textbook	OpenStax Precalculus 2e
Chapter	Ch 1: Functions
Section	1.3 Rates of Change and Behavior of Graphs
Subsection	1.3 Rates of Change and Behavior of Graphs
Pages	Page numbers pending faculty verification.
Course	MATH141

Try This First

A car trip covers $120$ miles. The trip starts at $1{:}00$ and the odometer reads as follows.

Clock time	Miles traveled
1:00	0
2:00	70
3:00	90
4:00	120

Before reading any formula, answer three questions out loud or on paper.

During which one-hour stretch did the car cover the most ground?
Was the car always moving forward, or did any hour show no progress?
If someone asked for “the speed of the trip,” would a single number describe how the speed changed from hour to hour?

What you probably noticed (click after you have tried it)

The first hour covered $70$ miles, the second covered $20$ miles, and the third covered $30$ miles. The car moved forward the whole time (the miles never went down), but it did not move at a steady speed. A single number, such as “the trip averaged $40$ miles per hour,” hides the fact that the first hour was much faster than the rest. The quantity that answers “how fast over this stretch” is the average rate of change, and the words “covered the most ground” and “always moving forward” are about whether the function is increasing or decreasing. Both ideas come next.

How Output Changes With Input

Picture filling a tall, narrow vase with water at a steady pour. The water level rises quickly because a small amount of water raises the level a lot. Now switch to a wide bucket: the same steady pour raises the level slowly. The pour rate did not change, but the height responds at a different rate depending on the container.

Average rate of change measures exactly this response: as the input moves from one value to another, how much does the output move, per unit of input. The behavior of a graph (where it rises, where it falls, where it turns around) is the picture of that response across the whole domain.

Three connected ideas follow.

Average rate of change turns “how the output changed over a stretch” into one number.
Increasing and decreasing intervals describe where the output rises or falls as the input grows.
Local extrema mark the turning points where rising switches to falling, or the reverse.

Prerequisite Hub

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    subgraph Builds_On["Builds on"]
        A["Functions and<br/>Function Notation"]
        B["Linear Equations<br/>(slope)"]
    end

    subgraph ThisSkill["This Skill"]
        C["Rates of Change and<br/>Behavior of Graphs"]
    end

    subgraph Unlocks["Unlocks"]
        D["Function<br/>Composition"]
    end

    A --> C
    B --> C
    C --> D

    style C fill:#d1fae5,stroke:#a565f0,stroke-width:3px

    click A "../ch1-sec1/functions-and-function-notation.html"
    click C "rates-of-change-and-behavior-of-graphs.html"
    click D "../ch1-sec4/function-composition.html"

Builds on (master these first):

m141-functions-and-function-notation: evaluating $f(x_1)$ and $f(x_2)$ is the first move in every rate-of-change problem.
linear-equations: the average rate of change between two points is the slope of the line through those two points.

Unlocks (this skill prepares you for):

function-composition: once a single function’s behavior is clear, the next step chains functions together.

There are no cross-course prerequisites for this skill.

Before You Start: Prerequisite Check

Can you do these? (Click to reveal self-test)

Function evaluation: If $f(x) = x^2 - 1$, find $f(2)$ and $f(5)$.

Check

$f(2) = 4 - 1 = 3$ and $f(5) = 25 - 1 = 24$.
Slope of a line: What is the slope of the line through $(1, 4)$ and $(3, 10)$?

Check

$\dfrac{10 - 4}{3 - 1} = \dfrac{6}{2} = 3$.
Reading a graph: A graph rises from left to right on the interval from $x = 0$ to $x = 4$. As $x$ grows from $0$ to $4$, do the output values get larger or smaller?

Check

Larger. A graph that rises from left to right has increasing output values.

If you struggled: review function notation and the slope of a line before continuing. The average rate of change uses both ideas at once.

Average Rate of Change

Definition

Average rate of change. The average rate of change between two input values is the total change of the function values (output values) divided by the change in the input values: $\dfrac{f(x_2) - f(x_1)}{x_2 - x_1}$.

Cite: OpenStax Precalculus 2e, Section 1.3 Rates of Change and Behavior of Graphs.

In words, this is “change in output divided by change in input.” The Greek letter $\Delta$ (read “delta”) is shorthand for “the change in,” so the average rate of change is often written

$$\frac{\Delta y}{\Delta x} = \frac{f(x_2) - f(x_1)}{x_2 - x_1}.$$

The units are output units per one input unit: miles per hour, dollars per year, degrees per minute.

Two Connected Representations

The same average rate of change appears in a table and on a graph. The table below records a function $f$, and the graph reads the slope of the straight line joining two of its points.

$x$	$f(x)$
1	5
3	11
6	14

Between $x = 1$ and $x = 3$, the average rate of change is $\dfrac{11 - 5}{3 - 1} = \dfrac{6}{2} = 3$. On a graph, this is the slope of the line segment connecting $(1, 5)$ to $(3, 11)$. The line through two points of a graph is called a secant line, and its slope is the average rate of change.

Translate between the two: the number $3$ means the output rises $3$ units for every $1$ unit the input grows, over that stretch. Steeper secant line, larger average rate of change.

Worked Example 1: Average Rate of Change of a Quadratic

Problem. For $f(x) = x^2 + 2x$, find the average rate of change between $x = 1$ and $x = 4$.

Predict first. The graph of $f(x) = x^2 + 2x$ rises faster as $x$ grows, so the secant from $x = 1$ to $x = 4$ should be fairly steep. Predict a positive number larger than the slope of an ordinary line.

Compute.

$$f(1) = 1^2 + 2(1) = 3, \qquad f(4) = 4^2 + 2(4) = 24.$$

$$\frac{f(4) - f(1)}{4 - 1} = \frac{24 - 3}{4 - 1} = \frac{21}{3} = 7.$$

Compare. The average rate of change is $7$. The prediction of a steep positive value holds. Over the stretch from $x = 1$ to $x = 4$, the output rises $7$ units for every $1$ unit of input on average, even though the function does not rise at a constant rate within that stretch.

Worked Example 2: A Negative Average Rate of Change

Problem. A drone descends. Its height in meters is $h(t) = 100 - 5t^2$ for $t$ in seconds. Find the average rate of change of height between $t = 1$ and $t = 3$.

Predict first. The drone is falling, so the height output decreases as time grows. Predict a negative average rate of change.

Compute.

$$h(1) = 100 - 5(1)^2 = 95, \qquad h(3) = 100 - 5(3)^2 = 100 - 45 = 55.$$

$$\frac{h(3) - h(1)}{3 - 1} = \frac{55 - 95}{3 - 1} = \frac{-40}{2} = -20.$$

Compare. The average rate of change is $-20$ meters per second. The negative sign matches the prediction: the drone loses $20$ meters of height per second on average over that two-second window. The sign of an average rate of change records direction, and the size records steepness.

Worked Example 3: Rate of Change From a Table

Problem. A plant’s height is recorded weekly.

Week $w$	Height $H$ (cm)
0	4
2	9
5	18

Find the average rate of change in height from week $2$ to week $5$.

Predict first. The plant grew from $9$ cm to $18$ cm, so the rate is positive. The height roughly doubled over three weeks, so expect a few centimeters per week.

Compute.

$$\frac{H(5) - H(2)}{5 - 2} = \frac{18 - 9}{5 - 2} = \frac{9}{3} = 3.$$

Compare. The plant grew an average of $3$ cm per week between week $2$ and week $5$. The positive few-centimeters-per-week prediction holds. Notice that the data come straight from the table, not from a formula. Average rate of change works the same way whether the function is given by a rule, a graph, or a table.

Increasing and Decreasing Intervals

Definition

Increasing and decreasing functions. A function $f$ is increasing on an open interval if $f(b) > f(a)$ for every two input values $a$ and $b$ in the interval where $b > a$. A function $f$ is decreasing on an open interval if $f(b) < f(a)$ for every two input values $a$ and $b$ in the interval where $b > a$.

Cite: OpenStax Precalculus 2e, Section 1.3 Rates of Change and Behavior of Graphs.

Read left to right along the graph. Where the curve goes uphill, the function is increasing. Where it goes downhill, the function is decreasing. The definition makes “uphill” precise: a larger input always gives a larger output on an increasing interval.

Reading Behavior From a Graph

Consider $f(x) = x^3 - 3x$. A small table reveals the turning behavior.

$x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$
$f(x)$	$-18$	$-2$	$2$	$0$	$-2$	$2$	$18$

Track the outputs from left to right. From $x = -3$ up to about $x = -1$, the outputs rise ($-18$ to $-2$ to $2$): increasing. From about $x = -1$ to $x = 1$, the outputs fall ($2$ to $0$ to $-2$): decreasing. From about $x = 1$ to $x = 3$, the outputs rise again ($-2$ to $2$ to $18$): increasing. The function increases, then decreases, then increases. The exact turning inputs for this cubic are $x = -1$ and $x = 1$, where the rising and falling switch.

One way to see this is the table of values above. Another way to see this is the sign of the average rate of change over short steps: a positive average rate of change over a small interval signals increasing behavior there, and a negative one signals decreasing behavior. Which view do you find clearer, and why?

Local Extrema

Definition

Local extrema. A function $f$ has a local maximum at a point $b$ in an open interval $(a, c)$ if $f(b)$ is greater than or equal to $f(x)$ for every $x$ (with $x$ not equal to $b$) in the interval. The function $f$ has a local minimum at $b$ in $(a, c)$ if $f(b)$ is less than or equal to $f(x)$ for every such $x$.

Cite: OpenStax Precalculus 2e, Section 1.3 Rates of Change and Behavior of Graphs.

A local maximum is a peak: the output there is at least as large as every output nearby. A local minimum is a valley: the output there is at least as small as every output nearby. The word local means “compared to the immediate neighborhood,” not “the highest or lowest point on the whole graph.”

Turning Points Connect the Three Ideas

A local maximum is exactly where the graph stops increasing and starts decreasing. A local minimum is exactly where the graph stops decreasing and starts increasing. For $f(x) = x^3 - 3x$ above, the switch from increasing to decreasing near $x = -1$ is a local maximum, and the switch from decreasing to increasing near $x = 1$ is a local minimum. Reading the increasing and decreasing intervals locates the extrema for free.

Worked Example 4: Locate Behavior and Extrema From a Table

Problem. A function $g$ is sampled below. State where $g$ is increasing, where it is decreasing, and identify any local maximum or minimum from the sample.

$x$	$0$	$1$	$2$	$3$	$4$
$g(x)$	$1$	$4$	$5$	$2$	$0$

Predict first. The outputs climb $1, 4, 5$ then drop $5, 2, 0$. Predict increasing early, decreasing late, with a peak near $x = 2$.

Reason. From $x = 0$ to $x = 2$, the outputs rise ($1$ to $4$ to $5$), so $g$ is increasing on that stretch. From $x = 2$ to $x = 4$, the outputs fall ($5$ to $2$ to $0$), so $g$ is decreasing on that stretch. The output at $x = 2$ is at least as large as its neighbors in the sample, so the sample shows a local maximum at $x = 2$ with value $5$.

Compare. The prediction of a peak near $x = 2$ holds. The sample shows no local minimum in the interior, because the outputs only fall after the peak. A table can locate behavior and extrema, with the caution that values between the sampled inputs are unknown.

Common Misconceptions

Common misconception

the average rate of change is the average (arithmetic mean) of the two output values. Averaging the outputs ignores the input change entirely. For $f(x) = x^2$ between $x = 2$ and $x = 6$, the mean of the outputs $4$ and $36$ is $20$, but the average rate of change is $\dfrac{36 - 4}{6 - 2} = \dfrac{32}{4} = 8$. The first number has units of output; the second has units of output per input. Average rate of change is a slope, not an average of heights.

Common misconception

a function has one fixed rate of change. A rate of change is a varying quantity, not a single property of the function. For $f(x) = x^2$, the average rate of change from $x = 0$ to $x = 1$ is $1$, while from $x = 3$ to $x = 4$ it is $7$. The same function changes at different rates over different stretches. Asking for “the rate of change” without naming an interval has no single answer for a curved graph.

Common misconception

a high point on the graph means a large rate of change there. Height and steepness are different quantities. At the top of a hill (a local maximum), the graph is momentarily level, so the rate of change is small even though the height is large. A point where $f$ is large is not the same as a point where $f$ is rising fast. Read height off the vertical axis; read rate off the steepness of the secant line.

Practice Problems

Level 1 Average Rate From a Formula

For $f(x) = 3x + 2$, find the average rate of change between $x = 1$ and $x = 5$.

Thought Process

Evaluate $f$ at both inputs, subtract the outputs, subtract the inputs, then divide. Because $f$ is a line, expect the slope of that line.

Show Answer

$f(1) = 3(1) + 2 = 5$ and $f(5) = 3(5) + 2 = 17$.

$$\frac{f(5) - f(1)}{5 - 1} = \frac{17 - 5}{5 - 1} = \frac{12}{4} = 3.$$

The average rate of change is $3$. For any straight line the average rate of change equals its slope, so the answer matches the coefficient of $x$.

Level 2 Reading Increasing and Decreasing

A function $f$ has the values below. State the interval where $f$ is increasing and the interval where $f$ is decreasing.

$x$	$0$	$1$	$2$	$3$	$4$
$f(x)$	$2$	$6$	$7$	$5$	$1$

Thought Process

Walk the outputs left to right. While the outputs rise, the function is increasing. Once they start to fall, it is decreasing. Find where the rising switches to falling.

Show Answer

The outputs rise on $2, 6, 7$ (from $x = 0$ to $x = 2$), so $f$ is increasing on the interval from $x = 0$ to $x = 2$.

The outputs fall on $7, 5, 1$ (from $x = 2$ to $x = 4$), so $f$ is decreasing on the interval from $x = 2$ to $x = 4$.

The switch at $x = 2$ marks a local maximum with value $7$.

Level 3 Average Rate of a Quadratic

For $f(x) = x^2 - 4x$, find the average rate of change between $x = 0$ and $x = 3$. Explain what the sign of your answer says about the graph over that stretch.

Thought Process

Predict the sign first: the vertex of $f(x) = x^2 - 4x$ sits at $x = 2$, so part of the stretch falls and part rises. Compute the two outputs and take the slope of the secant line, then read the sign.

Show Answer

$f(0) = 0^2 - 4(0) = 0$ and $f(3) = 3^2 - 4(3) = 9 - 12 = -3$.

$$\frac{f(3) - f(0)}{3 - 0} = \frac{-3 - 0}{3 - 0} = \frac{-3}{3} = -1.$$

The average rate of change is $-1$. The negative sign means that, taken as a whole, the output is lower at $x = 3$ than at $x = 0$, so the secant line over this stretch goes downhill. The function actually falls then rises inside the interval (the vertex is at $x = 2$), yet the net change over the whole stretch is downward.

Level 4 Diagnosing a Student Claim

A student computes the average rate of change of $f(x) = x^2$ between $x = 2$ and $x = 6$ and writes: “The outputs are $4$ and $36$, so the average rate of change is $\dfrac{4 + 36}{2} = 20$.”

Which statement best identifies the error?

(A) Nothing is wrong; $20$ is correct.

(B) The student averaged the two output values instead of dividing the change in output by the change in input.

(C) The student should have used $x = 2$ and $x = 6$ as the outputs.

(D) The student should have multiplied the outputs.

Thought Process

Average rate of change is change in output over change in input, a slope. Compare that to what the student actually did. The student combined the outputs with no reference to the inputs $2$ and $6$.

Show Answer

Answer: (B)

The correct computation is

$$\frac{f(6) - f(2)}{6 - 2} = \frac{36 - 4}{6 - 2} = \frac{32}{4} = 8.$$

(A) Incorrect: $20$ is the mean of the heights, not a rate.
(B) Correct: the student averaged the outputs and never used the input change of $4$.
(C) Incorrect: $2$ and $6$ are inputs, not outputs; using them as outputs would be a second error.
(D) Incorrect: average rate of change never multiplies the outputs.

The mean of two heights has units of output. The average rate of change has units of output per input. They are different quantities.

Level 5 Behavior, Extrema, and Steepest Stretch

A function $p$ is sampled at equal steps.

$x$	$0$	$1$	$2$	$3$	$4$	$5$
$p(x)$	$1$	$7$	$9$	$9$	$4$	$0$

(a) State the increasing and decreasing intervals visible in the sample. (b) Identify any local maximum. (c) Over which one-unit step is the average rate of change steepest in size, and what is that value?

Thought Process

For (a) and (b), walk the outputs left to right and watch for rises, falls, and turning points. For (c), the steps are all one unit wide, so the average rate of change over each step is just the change in output. Find the step with the largest change in size, and keep its sign.

Show Answer

(a) The outputs rise on $1, 7, 9$ (from $x = 0$ to $x = 2$), so $p$ is increasing on that stretch. The outputs fall on $9, 4, 0$ (from $x = 3$ to $x = 5$), so $p$ is decreasing on that stretch. Between $x = 2$ and $x = 3$ the output holds at $9$ (neither strictly increasing nor strictly decreasing in the sample).

(b) The output reaches $9$ at both $x = 2$ and $x = 3$ and is at least as large there as its neighbors, so the sample shows a local maximum value of $9$ across $x = 2$ to $x = 3$.

(c) Because each step has width $1$, the average rate of change over a step equals the change in output:

Step	Change in output	Average rate of change
$0$ to $1$	$7 - 1 = 6$	$6$
$1$ to $2$	$9 - 7 = 2$	$2$
$2$ to $3$	$9 - 9 = 0$	$0$
$3$ to $4$	$4 - 9 = -5$	$-5$
$4$ to $5$	$0 - 4 = -4$	$-4$

The largest in size is $6$, over the step from $x = 0$ to $x = 1$. The steepest stretch is the first one, with an average rate of change of $6$. The output is not highest there, which shows again that height and steepness are different quantities.

Mastery Checklist

Novice (Level 1-2):

I can compute the average rate of change between two inputs from a formula, a table, or a graph.
I can state the units of an average rate of change in a context.
I can read increasing and decreasing intervals off a graph or a table.

Competent (Level 3-4):

I can explain what the sign of an average rate of change says about the graph over a stretch.
I can locate a local maximum or local minimum as a turning point between increasing and decreasing behavior.
I can spot the error when someone averages the outputs instead of dividing the change in output by the change in input.

Proficient (Level 5):

I can compare average rates of change across several stretches and identify the steepest.
I can explain why “the rate of change” of a curved graph has no single value without naming an interval.
I can argue why a high point on a graph need not be a steep point.

Mental Model

The road-trip odometer. Average rate of change is the trip’s “miles per hour over this stretch”: total distance covered divided by time elapsed, for whatever stretch you choose. Pick a different stretch and you can get a different speed. Increasing and decreasing intervals are the parts of the trip where the odometer climbs versus the (rare) parts where it would go down if the car backed up. A local maximum is the farthest point reached before turning back. The secant line is a ruler laid between two odometer readings; its steepness is the average speed.

Connections

Looking back:

m141-functions-and-function-notation supplies the evaluation $f(x_1)$ and $f(x_2)$ that every rate computation begins with.
linear-equations supplies the slope idea: the average rate of change is the slope of the secant line through two points.

Looking ahead:

function-composition chains functions together, and reading the behavior of a single function first makes the behavior of a composed function easier to predict.
In calculus, shrinking the interval of an average rate of change toward a single point produces the instantaneous rate of change, which is the derivative. The average rate of change over a stretch is the first face of that idea.

Real-world connections:

Average speed over a trip leg is an average rate of change of distance with respect to time.
A drone’s rate of climb or descent is an average rate of change of height with respect to time.
A plant’s weekly growth rate is an average rate of change of height with respect to time.

Resources

Textbook section. OpenStax Precalculus 2e, Section 1.3 Rates of Change and Behavior of Graphs.
OpenStax Precalculus 2e 1.3 Rates of Change and Behavior of Graphs. https://openstax.org/books/precalculus-2e/pages/1-3-rates-of-change-and-behavior-of-graphs
OpenStax Precalculus 2e Ch 1 Key Concepts. https://openstax.org/books/precalculus-2e/pages/1-key-concepts

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Last updated: 2026-06-16