D. Characteristics of Functions · Rates of Change

Average Rate of Change: The Slope of a Secant

D1.4 D1.7About 70 minutes

Learning goals

I can calculate the average rate of change of a function over an interval from an equation, a table or a graph.
I can explain that the average rate of change over [a, b] is the slope of the secant line through (a, f(a)) and (b, f(b)).
I can find the equation of a secant line.
I can interpret an average rate of change, with units and sign, in context.

Words to know

interval [a, b]
All the x-values from a to b, including a and b.

average rate of change
Over [a, b]: (f(b) − f(a)) ÷ (b − a). It describes the overall change per unit, ignoring what happens in between.

secant line
A line through two points on a curve. Its slope is the average rate of change between those points.

average velocity
The average rate of change of position (displacement) with respect to time.

The road trip question

Your family drives 240 km along Highway 401 to visit relatives, and the trip takes 3 hours. You stopped for a coffee, slowed down in traffic and sped up on open highway. Still, your average speed was 240 ÷ 3 = 80 km/h.

Average speed only cares about where you started and where you finished. That is the idea of an average rate of change. In this lesson you will calculate it for any function and see it as the slope of a line on the graph.

The formula

The average rate of change of a function f over the interval [a, b] is

average rate of change = (f(b) − f(a)) ÷ (b − a) = Δy ÷ Δx

On the graph, the points P(a, f(a)) and Q(b, f(b)) lie on the curve. The line through them is called a secant line. The formula above is exactly the slope of that line:

average rate of change over [a, b] = slope of the secant PQ

Remember from Grade 11: in the MCR3U unit Quadratic Functions, a secant was a line that crosses a parabola at two points. Now you will use secants on any kind of graph, and their slopes become rates.

The secant through P(1, 2) and Q(4, 11) on f(x) = x² − 2x + 3. Its slope, 3, is the average rate of change of f from x = 1 to x = 4.

Worked example

Example 1: From an equation

For f(x) = x² − 2x + 3, find the average rate of change on [1, 4], then find the equation of the secant line.

f(4) = 16 − 8 + 3 = 11 and f(1) = 1 − 2 + 3 = 2.
Average rate = (11 − 2) ÷ (4 − 1) = 9 ÷ 3 = 3.
Secant line through (1, 2) with slope 3: y − 2 = 3(x − 1), so y = 3x − 3 + 2 = 3x − 1.
Check with Q: 3(4) − 1 = 11 ✓.

Average rate of change 3; secant line y = 3x − 1.

Worked example

Example 2: Average velocity of a ball

A ball is thrown upward. Its height in metres after t seconds is h(t) = −4.9t² + 20t + 1.5. Find the average velocity (a) in the first second, (b) from t = 1 to t = 3, (c) from t = 2 to t = 4. Interpret each.

Values: h(0) = 1.5, h(1) = −4.9 + 20 + 1.5 = 16.6, h(2) = −19.6 + 40 + 1.5 = 21.9, h(3) = −44.1 + 60 + 1.5 = 17.4, h(4) = −78.4 + 80 + 1.5 = 3.1.
(a) (16.6 − 1.5) ÷ (1 − 0) = 15.1 m/s.
(b) (17.4 − 16.6) ÷ (3 − 1) = 0.8 ÷ 2 = 0.4 m/s.
(c) (3.1 − 21.9) ÷ (4 − 2) = −18.8 ÷ 2 = −9.4 m/s.

(a) 15.1 m/s: rising fast. (b) 0.4 m/s: the ball went up and came back down, ending just a little higher than it was at t = 1. (c) −9.4 m/s: on average it fell 9.4 m each second.

Average means "ignore the middle"

In part (b), the ball rose to its highest point and fell again, yet the average velocity was only 0.4 m/s. The average rate of change uses only the endpoints. It can hide a lot of action in between. That is why we will need the instantaneous rate of change, starting in the lesson Instantaneous Rate of Change: The Slope of a Tangent.

Two trips with the same start and finish have the same secant, so they have the same average speed (80 km/h), even though one trip had two stops.

Worked example

Example 3: From a table

A weather station recorded these temperatures on a February morning in Sudbury.

time: 6:00, 8:00, 10:00, 12:00, 14:00
T (°C): −14, −12, −7, −3, −1

(This is a sample data set for practice.) Find the average rate of change of temperature from 6:00 to 12:00 and from 10:00 to 14:00. When was it warming faster?

6:00 to 12:00: ΔT = −3 − (−14) = 11 °C and Δt = 6 h. Rate = 11 ÷ 6 ≈ 1.83 °C/h.
10:00 to 14:00: ΔT = −1 − (−7) = 6 °C and Δt = 4 h. Rate = 6 ÷ 4 = 1.5 °C/h.
Compare: 1.83 > 1.5.

About 1.8 °C/h from 6:00 to 12:00 and 1.5 °C/h from 10:00 to 14:00. On average it warmed faster over the first interval.

Interpreting an average rate

A complete answer has three parts:

the number (with sign),
the units (y-units per x-unit),
a sentence in context that uses the words "on average".

For example: "From 10:00 to 14:00 the temperature rose on average 1.5 °C per hour." A negative value means the quantity decreased on average. A value of zero means the quantity ended where it started, not that it never changed.

Just for fun

The word secant comes from the Latin secare, "to cut", the same root as section. A secant line is basically a math knife: it slices through a curve at two points and reports how steep the cut was.

Hands-on activity

Cooling cup

You need: a mug, hot water from the tap (not boiling; ask a grown-up first and handle it carefully), a kitchen or candy thermometer, a timer, paper and a pencil.

Fill the mug with hot tap water and put in the thermometer. Wait 30 s and record the temperature at t = 0.
Record the temperature every 2 minutes for 20 minutes.
Make a table and plot temperature against time.
Calculate the average rate of change for [0, 4], [8, 12] and [16, 20]. Draw the three secants on your graph.
Calculate the average rate of change for [0, 20]. Is it the average of your three answers? Explain.

What to look for: all the rates are negative (the water cools), and their size shrinks as time goes on (decreasing at a decreasing rate). The average over [0, 20] depends only on the first and last readings, so it will generally not equal the average of the three shorter-interval rates.

Your turn

Find the average rate of change of f(x) = x² on [1, 3].

Find the average rate of change of f(x) = 3x² − x on [−1, 2].

Find the average rate of change of g(x) = 2ˣ on [0, 3].

For the ball h(t) = −4.9t² + 20t + 1.5, find the average velocity from t = 1 to t = 2.

Find the equation of the secant line of f(x) = x² − 2x + 3 through the points where x = 0 and x = 3.

A lake's water level was 245 cm on day 1, 230 cm on day 4 and 212 cm on day 10. Find the average rate of change from day 1 to day 10.

For the same ball, the average velocity over [0, 4] is (3.1 − 1.5) ÷ 4 = 0.4 m/s. Does this mean the ball hardly moved? Explain.

More practice: warm-up

Find the slope of the secant through (2, 5) and (6, 13).

Find the average rate of change of f(x) = 5x + 1 on [0, 10].

Find the average rate of change of f(x) = x² on [0, 2].

Find the average rate of change of f(x) = 10 − x² on [1, 3].

A train travels 360 km in 4.5 h. What is its average speed?

More practice: on your own

Find the average rate of change of f(x) = x³ on [−1, 2].

Find the average rate of change of f(x) = 1/x on [1, 4].

Find the average rate of change of f(x) = √x on [4, 9].

Find the equation of the secant of f(x) = −x² + 6x through x = 1 and x = 4.

A cyclist's distance is recorded: t = 0, 10, 25, 40 min and d = 0, 3.2, 7.0, 12.4 km. Find her average speed from t = 10 to t = 40 min in km/min and in km/h.

Challenge

Show that for f(x) = x², the average rate of change on [a, b] (a ≠ b) is a + b.

For f(x) = x² − 4x + 7, the average rate of change on [1, k] is 5. Find k (k ≠ 1).

A secant of y = x² passes through (1, 1) and has slope 6. Find the other point where it meets the parabola.

Check your understanding

Quiz

What is the average rate of change of f(x) = x² from x = 1 to x = 4?

The average rate of change of a function over [a, b] equals the slope of which line?

A ball's height is 1.5 m at t = 0 s and 16.6 m at t = 1 s. What is its average velocity over that second?

A town had 1500 people in 2020 and 1860 people in 2026. What was the average rate of change of its population?

Two drivers both start at the same place at noon and arrive at the same place at 3 p.m. One drives steadily; the other stops twice. Which is true?

What is the slope of the secant through (2, 4) and (5, 1)?

For parents and teachers

Big ideas (D1.4, D1.7): the average rate of change over [a, b] is (f(b) − f(a)) ÷ (b − a), and on a graph it is the slope of the secant line through the two endpoints. It uses only the endpoints.

How to help: On your next drive or transit trip, note the time and distance at the start and the end and have your teen work out the average speed. Ask: "Were we ever going faster than that? Slower?" This leads straight into the idea of an instantaneous rate.

Common difficulties:

Evaluating f(a) and f(b) incorrectly, especially with negatives (for example, −4.9(3)² is −44.1, not +44.1).
Dividing by b instead of by b − a.
Thinking an average rate of 0 means nothing happened.
Leaving out units or the sign when interpreting.