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Average & Approximate Rates of Change (College Board AP® Precalculus): Study Guide

Written by: Roger B

Reviewed by: Mark Curtis

Updated on 8 April 2026

Average rate of change

What is the average rate of change?

The average rate of change of a function over an interval is
- the constant rate of change
  - that would produce the same change in output values as the function actually produces over that interval
- It is the ratio of the change in output values to the change in input values over the interval
For a function $f$ on the interval $[a, b]$ :
- Average rate of change $= \frac{f (b) - f (a)}{b - a}$
- This is sometimes written using delta notation:
  - Average rate of change $= \frac{change in y}{change in x} = \frac{Δ y}{Δ x}$
The average rate of change tells you how quickly the output is changing on average across the whole interval
- It does not tell you what the function is doing at every individual point within the interval
- E.g. a function might increase and decrease within the interval, but the average rate of change only reflects the overall change from start to finish

How does the average rate of change relate to a graph?

The average rate of change over the interval $[a, b]$ is the slope of the secant line from the point $(a, f (a))$ to the point $(b, f (b))$
- A secant line is a straight line connecting two points on the graph of a function

Graph of a function y=f(x) showing a curve between points (a,f(a)) and (b,f(b)), with a secant line and slope formula. Axes labelled x and y. — **Average rate of change as the slope of a secant line**

This can be interpreted geometrically:
- A positive average rate of change means the secant line slopes upward
  - the output increases overall across the interval
- A negative average rate of change means the secant line slopes downward
  - the output decreases overall across the interval
- An average rate of change of zero means the secant line is horizontal
  - the output values at the start and end of the interval are equal

How do I calculate an average rate of change?

Identify the two input values that define the interval, $a$ and $b$
- Then find the corresponding output values, $f (a)$ and $f (b)$
- Substitute into the formula:

Average rate of change $= \frac{f (b) - f (a)}{b - a}$

E.g. if $f (2) = 5$ and $f (6) = 17$ , then the average rate of change over $[2, 6]$ is $\frac{17 - 5}{6 - 2} = \frac{12}{4} = 3$
- This means that, on average, the output increases by 3 for every 1 unit increase in the input over this interval

Examiner Tips and Tricks

When a question asks you to find an average rate of change, always show the setup of the fraction $\frac{f (b) - f (a)}{b - a}$ with the values substituted in, before simplifying.

This is especially important in free response questions, where showing your working is required to earn full marks

Don't forget to include units if the question is set in a real-world context (e.g. "points per month", "gallons per hour").

Approximating & comparing rates of change

What is the rate of change at a point?

The rate of change of a function at a point describes how quickly the output values would change if the input values were to change at that specific point
- On a graph of the function, it is equal to the slope of the tangent line to the graph at the point
- This is different from the average rate of change, which describes the overall change across a whole interval
The rate of change at a point can be approximated using average rates of change over small intervals that contain the point
- The smaller the interval, the better the approximation

Graph with curve y = f(x), and a tangent line drawn at a point near a peak. Three secant lines are drawn between closer and closer points on either side of the point with the tangent. Axes labelled x and y . — **Improving rate of change approximations over smaller and smaller intervals**

E.g. to approximate the rate of change of $f$ at $x = 3$
- you could calculate the average rate of change over $[2.9, 3.1]$
- or over an even smaller interval like $[2.99, 3.01]$

How can I compare rates of change at two points?

The rates of change at two different points can be compared
- by calculating average rate of change approximations for each point
- over sufficiently small intervals containing each point
This allows you to make statements like
- "the function is changing more rapidly at $x = 2$ than at $x = 5$ "
- or "the function is increasing at $x = 3$ but decreasing at $x = 7$ "
The absolute values (magnitudes) of the average rates of change tell you which point has the faster rate of change
- An average rate of 8 is changing faster than an average rate of 6
- An average rate of -9 is changing faster than an average rate of -4
  - because $|- 9| = 9 > 4 = |- 4|$
The signs tell you the direction of change at each point
- A positive rate of change means an increasing function
- A negative rate of change means a decreasing function

Examiner Tips and Tricks

Make sure you understand the basic ideas of average rate of change and approximating the rate of change at a point.

These ideas occur throughout the course in the context of different function types

Describing & interpreting rates of change

What do rates of change tell us about how quantities vary together?

Rates of change quantify how two quantities vary together
- The rate of change describes how a function's output responds as the input changes
A positive rate of change indicates that the function is increasing
- As one quantity increases, the other quantity also increases
- Or equally, if one quantity decreases, the other also decreases
  - The two quantities move in the same direction
- E.g. if the rate of change of distance with respect to time is positive
  - then the distance is increasing as time passes
A negative rate of change indicates that the function is decreasing
- As one quantity increases, the other quantity decreases
  - The two quantities move in opposite directions
- E.g. if the rate of change of a test score with respect to time since last studying the material is negative
  - then the score is decreasing as time passes
The average rate of change over equal-length input-value intervals can tell you about the concavity of the function
- If the average rates of change are increasing for all small-length intervals
  - then the graph of the function is concave up
- If the average rates of change are decreasing for all small-length intervals
  - then the graph of the function is concave down

Worked Example

A population of bacteria in a laboratory experiment is modeled by the function $P$ , where $P (t)$ is the number of bacteria, in thousands, at time $t$ hours after the start of the experiment. Selected values of $P (t)$ are given in the table below.

$t$ (hours)	0	2	5	8	12
$P (t)$ (thousands)	3	5	11	18	22

(a) Find the average rate of change of $P (t)$ , in thousands of bacteria per hour, from $t = 0$ to $t = 8$ hours. Show the computations that lead to your answer.

Answer:

(a)

Use the average rate of change formula

$\begin{array}{rcl} Average rate of change & = & \frac{P (8) - P (0)}{8 - 0} \\ = & \frac{18 - 3}{8 - 0} \\ = & \frac{15}{8} \\ = & 1.875 \end{array}$

Be sure to state units in the context of the question

The average rate of change from $t = 0$ to $t = 8$ is $1.875$ thousand bacteria per hour

(b) Is the average rate of change of $P (t)$ from $t = 0$ to $t = 5$ greater than or less than the average rate of change from $t = 5$ to $t = 12$ ? Show the work that supports your answer.

Answer:

Calculate the two average rates of change

From $t = 0$ to $t = 5$ :

$\frac{P (5) - P (0)}{5 - 0} = \frac{11 - 3}{5 - 0} = \frac{8}{5} = 1.6$

From $t = 5$ to $t = 12$ :

$\frac{P (12) - P (5)}{12 - 5} = \frac{22 - 11}{12 - 5} = \frac{11}{7} \approx 1.571$

Compare the two values found

The average rate of change from $t = 0$ to $t = 5$ ( $1.6$ thousand bacteria per hour) is greater than the average rate of change from $t = 5$ to $t = 12$ (approximately $1.571$ thousand bacteria per hour)

Worked Example

A business tracks the total number of online orders received since the start of the year. The total number of orders, in thousands, can be modeled by the function $N$ , where $N (t)$ is the total number of orders, in thousands, received by the end of month $t$ . At $t = 0$ (the start of the year), the business had received 2,000 orders. At $t = 6$ , the business had received 20,000 orders.

Use the given data to find the average rate of change of the total number of orders, in thousands per month, from $t = 0$ to $t = 6$ months. Show the computations that lead to your answer.

Answer:

Use the average rate of change formula with the figures from the question

Remember that $N$ represents thousands of orders
- so 2,000 orders at the start means $N (0) = 2$
- and 20,000 orders at $t = 6$ means $N (6) = 20$

$\frac{N (6) - N (0)}{6 - 0} = \frac{20 - 2}{6 - 0} = \frac{18}{6} = 3$

Answer in the context of the question

The average rate of change is 3 thousand orders per month

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Average & Approximate Rates of Change (College Board AP® Precalculus): Study Guide

Average rate of change

What is the average rate of change?

How does the average rate of change relate to a graph?

How do I calculate an average rate of change?

Examiner Tips and Tricks

Approximating & comparing rates of change

What is the rate of change at a point?

How can I compare rates of change at two points?

Examiner Tips and Tricks

Describing & interpreting rates of change

What do rates of change tell us about how quantities vary together?

Worked Example

Worked Example

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Author: Roger B

Reviewer: Mark Curtis