Linear Functions
1. Constant Rate of Change
Learning outcomes
- I can define rate of change.
- I can calculate the rate of change from tables.
- I can identify constant rates of change.
- I can compare different rates of change.
- I can interpret rates of change in context.
Constant Rate of Change
A rate of change describes how one quantity changes compared with another quantity.
Rates of change appear everywhere in mathematics, science, and everyday life. They can describe:
- how far a car travels each hour
- how much money is earned each day
- how quickly water fills a tank
- how temperature changes with time
- how a plant grows each week
- how much fuel a vehicle uses
- how the cost of an item changes with quantity
A constant rate of change occurs when one quantity changes by the same amount for every equal change in another quantity.
For example, if a cyclist travels 15 km every hour, the distance increases at a constant rate of:
15 km/h
What Is a Rate?
A rate compares two quantities that have different units.
Examples include:
- 60 kilometres per hour
- $18 per hour
- 5 litres per minute
- 3 centimetres per year
- 20 words per minute
- 4°C per hour
The word per is important.
It tells us that one quantity is being compared with another.
For example:
60 km per hour
means:
60 kilometres for every 1 hour
What Is Rate of Change?
A rate of change tells us how much the output changes when the input changes.
The basic relationship is:
rate of change = change in output / change in input
Using x and y:
rate of change = change in y / change in x
or:
rate of change = Δy/Δx
The Greek letter Δ (delta) means change in.
So:
Δy = change in y
and:
Δx = change in x
Therefore:
rate of change = Δy/Δx
Calculating Change
To calculate a change, subtract the earlier value from the later value.
change = final value − initial value
For example, if distance increases from 20 km to 50 km:
change in distance = 50 − 20
= 30 km
If time increases from 2 hours to 4 hours:
change in time = 4 − 2
= 2 hours
Therefore:
rate of change = 30 km / 2 h
= 15 km/h
Rate of Change and Slope
Rate of change is closely connected to the slope of a graph.
For a straight line:
slope = change in y / change in x
Therefore:
slope = rate of change
You may also see the slope formula written as:
m = (y₂ − y₁)/(x₂ − x₁)
where:
- (x₁, y₁) is one point
- (x₂, y₂) is another point
- m is the slope
Another common way to remember this is:
slope = rise/run
where:
rise = vertical change
and:
run = horizontal change
Worked Example: Rate of Change
A car travels:
120 km in 2 hours
and:
300 km in 5 hours.
Calculate its rate of change in distance.
Change in distance:
300 − 120 = 180 km
Change in time:
5 − 2 = 3 h
Rate of change:
180/3
= 60 km/h
The car's distance is increasing at:
60 km per hour
Positive Rates of Change
A rate of change is positive when the output increases as the input increases.
For example:
| Time |
Distance (km) |
|---|---|
| 0 | 0 |
| 1 | 20 |
| 2 | 40 |
| 3 | 60 |
| 4 | 80 |
Each hour, the distance increases by:
20 km
Therefore:
rate of change = +20 km/h
The positive sign indicates that distance is increasing.
On a graph, a positive rate of change produces a line that rises from left to right.
Negative Rates of Change
Rates of change can also be negative.
Consider the temperature of a cooling drink:
| Time (min) | Temperature (°C) |
|---|---|
| 0 | 80 |
| 5 | 70 |
| 10 | 60 |
| 15 | 50 |
| 20 | 40 |
Every 5 minutes, temperature changes by:
50 − 60 = −10°C
The rate of change is:
−10/5
= −2°C/min
The negative sign tells us that temperature is decreasing.
On a graph, a negative rate of change produces a line that falls from left to right.
Zero Rate of Change
A rate of change can also be zero.
Consider:
| Time (min) | Water Level (cm) |
|---|---|
| 0 | 50 |
| 2 | 50 |
| 4 | 50 |
| 6 | 50 |
The water level does not change.
Change in water level:
50 − 50 = 0 cm
Therefore:
rate of change = 0 cm/min
A zero rate of change appears as a horizontal line on a graph.
What Makes a Rate Constant?
A rate is constant when the same change occurs for every equal interval.
Consider:
| Time (s) | Distance (m) |
|---|---|
| 0 | 0 |
| 2 | 10 |
| 4 | 20 |
| 6 | 30 |
| 8 | 40 |
Every time increases by:
2 s
Distance increases by:
10 m
Therefore:
rate of change = 10/2
= 5 m/s
The same rate occurs throughout the table.
Therefore, the relationship has a:
constant rate of change of 5 m/s
Identifying Constant Rate from a Table
Consider:
| x | y |
|---|---|
| 1 | 7 |
| 2 | 11 |
| 3 | 15 |
| 4 | 19 |
| 5 | 23 |
Look at the changes.
x increases by:
+1 each time
y increases by:
+4 each time
Therefore:
rate of change = 4/1
= 4
Since this rate is the same throughout the table, the rate of change is constant.
Be Careful When x Does Not Increase by 1
Students sometimes look only at the changes in y.
That can give the wrong answer.
Consider:
| x | y |
|---|---|
| 0 | 5 |
| 3 | 11 |
| 6 | 17 |
| 9 | 23 |
y increases by:
6
But x increases by:
3
Therefore:
rate of change = 6/3
= 2
The rate is 2, not 6.
Always calculate:
change in y / change in x
Unequal Intervals Can Still Have a Constant Rate
The x-values do not need to be equally spaced.
Consider:
| Time |
Distance (km) |
|---|---|
| 0 | 0 |
| 1 | 12 |
| 3 | 36 |
| 7 | 84 |
From 0 to 1:
rate = 12/1 = 12 km/h
From 1 to 3:
rate = (36 − 12)/(3 − 1)
= 24/2
= 12 km/h
From 3 to 7:
rate = (84 − 36)/(7 − 3)
= 48/4
= 12 km/h
The intervals are different sizes, but the rate remains:
12 km/h
Therefore, the rate of change is constant.
Identifying a Non-Constant Rate
Consider:
| Time (s) | Distance (m) |
|---|---|
| 0 | 0 |
| 1 | 4 |
| 2 | 10 |
| 3 | 18 |
| 4 | 28 |
Calculate the changes in distance:
0 → 4 = +4
4 → 10 = +6
10 → 18 = +8
18 → 28 = +10
Since time increases by 1 second each time, the rates are:
4 m/s
6 m/s
8 m/s
10 m/s
The rate is changing.
Therefore:
the rate of change is not constant.
This could describe an object that is speeding up.
Constant Rate and Straight-Line Graphs
A constant rate of change produces a straight-line graph.
Why?
Because every equal horizontal change produces the same vertical change.
For example, suppose:
rate of change = 3
For every increase of 1 in x:
y increases by 3.
The pattern is consistent.
This produces a straight line.
Linear Relationships
A relationship with a constant rate of change is called a linear relationship.
Linear relationships can often be written as:
y = mx + b
where:
- y = output
- x = input
- m = rate of change or slope
- b = initial value or y-intercept
The value m tells us how much y changes for every increase of 1 in x.
Example: Interpreting y = mx + b
Consider:
y = 4x + 7
The coefficient of x is:
m = 4
Therefore, the rate of change is:
4
The value:
b = 7
means that when x = 0:
y = 7
So the relationship begins at 7 and increases by 4 for every increase of 1 in x.
Rate of Change Is Not the Starting Value
This distinction is important.
Consider:
y = 5x + 20
The rate of change is:
5
The starting value is:
20
These numbers have different meanings.
If this equation represented a taxi fare:
20 might represent the starting fee.
5 might represent the additional cost per kilometre.
Worked Example: Taxi Fare
Suppose a taxi charges:
$4 starting fee
plus:
$2.50 per kilometre
The total cost can be represented by:
C = 2.5d + 4
where:
- C = cost
- d = distance travelled
The rate of change is:
$2.50/km
This means:
for every additional kilometre travelled, the fare increases by $2.50.
The $4 is the initial cost, not the rate of change.
Comparing Rates of Change
Rates allow us to compare situations.
Suppose two cyclists travel at constant speeds.
Cyclist A
Travels 36 km in 2 hours.
Rate:
36/2 = 18 km/h
Cyclist B
Travels 50 km in 2 hours.
Rate:
50/2 = 25 km/h
Since:
25 km/h > 18 km/h
Cyclist B has the greater rate of change in distance.
Comparing Rates from Tables
Consider two machines filling containers.
Machine A
| Time (min) | Volume (L) |
|---|---|
| 0 | 0 |
| 2 | 16 |
| 4 | 32 |
Rate:
16/2 = 8 L/min
Machine B
| Time (min) | Volume (L) |
|---|---|
| 0 | 0 |
| 3 | 30 |
| 6 | 60 |
Rate:
30/3 = 10 L/min
Therefore, Machine B fills at a greater rate.
Comparing Positive Rates
Suppose two lines have rates:
Line A: +2
Line B: +6
Both are increasing.
But Line B increases more rapidly.
On a graph, Line B would generally appear steeper.
For positive slopes:
larger slope → steeper increase
Comparing Negative Rates
Suppose:
Line A: −2
Line B: −7
Both are decreasing.
But Line B decreases more rapidly because its rate has the greater magnitude.
For every increase of 1 in x:
Line A decreases by 2.
Line B decreases by 7.
Therefore, Line B has the steeper downward slope.
Units Matter
A rate of change should normally include appropriate units.
Suppose:
distance is measured in kilometres
and:
time is measured in hours.
Then:
rate of change = kilometres/hours
= km/h
If volume is measured in litres and time in minutes:
L/min
If temperature is measured in degrees Celsius and time in seconds:
°C/s
Units tell us what the rate actually means.
Units Come from the Axes
If a graph has:
y-axis: Distance (m)
and:
x-axis: Time (s)
then the rate of change has units:
m/s
because:
rate = change in distance / change in time
If the graph has:
y-axis: Cost ($)
and:
x-axis: Number of Tickets
then the rate has units:
$/ticket
Always examine what each variable represents.
Rate of Change from Two Points
Suppose a line passes through:
(2, 7)
and:
(6, 19)
Use:
m = (y₂ − y₁)/(x₂ − x₁)
Substitute:
m = (19 − 7)/(6 − 2)
= 12/4
= 3
Therefore, the rate of change is:
3
This means y increases by 3 for every increase of 1 in x.
The Order of Subtraction
When using:
m = (y₂ − y₁)/(x₂ − x₁)
you must subtract in the same order.
Correct:
m = (19 − 7)/(6 − 2)
You could also reverse both:
m = (7 − 19)/(2 − 6)
= −12/−4
= 3
Both work.
But you cannot reverse only one.
Incorrect:
(19 − 7)/(2 − 6)
= 12/−4
= −3
The numerator and denominator must use the points in the same order.
Rate of Change from a Graph
To calculate rate from a straight-line graph:
- Choose two clear points on the line.
- Find the vertical change.
- Find the horizontal change.
- Divide.
rate of change = vertical change / horizontal change
or:
rate of change = rise/run
Whenever possible, choose points that lie exactly on grid intersections.
This reduces measurement error.
Interpreting a Rate in Context
Calculating a number is only part of the task.
We should also explain what the number means.
Suppose:
rate = 12 L/min
A strong interpretation is:
The volume increases by 12 litres for every minute that passes.
Suppose:
rate = −3°C/h
Interpretation:
The temperature decreases by 3°C every hour.
Suppose:
rate = $8/day
Interpretation:
The cost increases by $8 for each additional day.
Positive, Negative, and Zero Rates
The sign of a rate gives useful information.
| Rate | Meaning | Graph |
|---|---|---|
| Positive | Output increases | Rises left to right |
| Negative | Output decreases | Falls left to right |
| Zero | Output remains constant | Horizontal |
The sign tells us the direction of change.
The size tells us how quickly the change occurs.
Real-World Example: Water Tank
Suppose a tank contains:
| Time (min) | Water (L) |
|---|---|
| 0 | 50 |
| 2 | 80 |
| 4 | 110 |
| 6 | 140 |
| 8 | 170 |
Every 2 minutes:
water increases by 30 L
Rate:
30/2
= 15 L/min
Interpretation:
The amount of water in the tank increases by 15 litres every minute.
Notice that the tank begins with 50 L.
The starting amount and rate are different quantities.
Real-World Example: Draining a Tank
Now suppose:
| Time (min) | Water (L) |
|---|---|
| 0 | 200 |
| 5 | 175 |
| 10 | 150 |
| 15 | 125 |
Every 5 minutes:
water changes by:
175 − 200 = −25 L
Rate:
−25/5
= −5 L/min
Interpretation:
The amount of water decreases by 5 litres every minute.
The negative sign is meaningful.
Real-World Example: Temperature
Suppose a freezer cools from:
20°C at 0 minutes
to:
−10°C at 15 minutes.
Change in temperature:
−10 − 20
= −30°C
Change in time:
15 − 0
= 15 min
Rate:
−30/15
= −2°C/min
Interpretation:
The temperature decreases by 2°C each minute.
Real-World Example: Earnings
A student earns $72 for 6 hours of work.
Rate:
72/6
= $12/h
If the hourly rate remains the same, then:
2 hours → $24
4 hours → $48
6 hours → $72
8 hours → $96
This is a constant rate of change.
Real-World Example: Fuel Consumption
Suppose a vehicle uses 24 L of fuel while travelling 300 km.
A useful rate is:
24/300
= 0.08 L/km
This means the vehicle uses an average of:
0.08 litres per kilometre
We could also express the relationship differently:
300/24
= 12.5 km/L
This means:
12.5 kilometres per litre
Both are rates, but they answer different questions.
The Direction of a Rate Matters
Consider:
60 km/h
This compares:
distance / time
But:
1/60 h/km
compares:
time / distance
These are related, but they mean different things.
Always ask:
What quantity is changing compared with what other quantity?
The order matters.
Constant Rate and Proportional Relationships
Some constant-rate relationships are also proportional relationships.
Example:
d = 5t
If t = 0:
d = 0
The graph passes through:
(0, 0)
This is both:
- linear
- proportional
But consider:
C = 5t + 20
This still has a constant rate of change of 5.
However, when t = 0:
C = 20
Therefore, the graph does not pass through the origin.
It is linear but not proportional.
So:
Every proportional relationship has a constant rate of change.
But:
Not every constant-rate linear relationship is proportional.
Rate of Change and Calculus
Constant rate of change provides an important foundation for calculus.
For a straight line, the rate of change is the same everywhere.
For a curved graph, the rate may change continuously.
This leads to an important question:
How can we determine the rate of change at one exact point on a curve?
Calculus answers this using the derivative.
So there is an important progression:
constant rate → average rate → instantaneous rate → derivative
Constant Rate Versus Changing Rate
Consider a car travelling at exactly:
20 m/s
Its position increases by:
20 m every second.
The position-time graph is a straight line.
The rate of change is constant.
Now imagine a car accelerating.
It might travel:
5 m during one second
10 m during the next
15 m during the next
Its position is changing at an increasing rate.
The position-time graph becomes curved.
This distinction becomes extremely important in calculus.
Worked Example: Table
Determine whether the rate of change is constant.
| x | y |
|---|---|
| 2 | 9 |
| 4 | 15 |
| 6 | 21 |
| 8 | 27 |
From the first pair to the second:
Δy = 15 − 9 = 6
Δx = 4 − 2 = 2
Rate:
6/2 = 3
Next interval:
(21 − 15)/(6 − 4)
= 6/2
= 3
Next:
(27 − 21)/(8 − 6)
= 6/2
= 3
The rate is always 3.
Therefore:
the relationship has a constant rate of change.
Worked Example: Non-Constant Rate
Consider:
| x | y |
|---|---|
| 1 | 2 |
| 2 | 5 |
| 3 | 10 |
| 4 | 17 |
Rates:
(5 − 2)/(2 − 1) = 3
(10 − 5)/(3 − 2) = 5
(17 − 10)/(4 − 3) = 7
Since:
3 ≠ 5 ≠ 7
the rate of change is not constant.
Worked Example: Comparing Two Rates
Company A charges:
$30 for 3 hours
Company B charges:
$56 for 8 hours
Company A:
30/3 = $10/h
Company B:
56/8 = $7/h
Therefore:
Company A's cost increases at a greater rate.
Notice that this statement compares the rates, not necessarily the total cost in every possible situation. Starting fees could change the total-price comparison.
Common Mistakes
Using y/x instead of change in y/change in x
The rate of change between two points is:
Δy/Δx
not automatically:
y/x
Looking only at changes in y
You must also consider how much x changed.
Assuming x always increases by 1
Always check the table.
Forgetting units
A rate such as 12 is much less informative than:
12 km/h
Confusing starting value with rate
In:
y = 4x + 20
the rate is 4, not 20.
Ignoring negative signs
A rate of:
−5°C/h
means the temperature is decreasing.
Thinking a larger negative number means faster decrease
Compare magnitude.
−8 represents a faster decrease than −3 because the quantity changes by 8 units rather than 3 units per interval.
Subtracting coordinates in different orders
Keep the subtraction order consistent.
Assuming every straight line passes through the origin
A straight line can have a nonzero starting value.
Assuming every linear relationship is proportional
A proportional relationship must pass through the origin.
Assuming any changing quantity has a constant rate
Many real systems have changing rates.
Key Terms
Rate — A comparison of two quantities with different units.
Rate of change — The amount one quantity changes relative to another.
Constant rate of change — A rate that remains the same throughout a relationship.
Variable rate of change — A rate that changes over the interval being considered.
Input — The independent quantity, commonly represented by x.
Output — The dependent quantity, commonly represented by y.
Change — The difference between a final value and an initial value.
Delta (Δ) — A symbol meaning "change in."
Δx — Change in the x-variable.
Δy — Change in the y-variable.
Slope — The rate of change of a straight line.
Rise — The vertical change between two points.
Run — The horizontal change between two points.
Linear relationship — A relationship with a constant rate of change that produces a straight-line graph.
Positive rate of change — A rate where the output increases as the input increases.
Negative rate of change — A rate where the output decreases as the input increases.
Zero rate of change — A rate where the output remains unchanged.
Initial value — The output value when the input is zero.
y-intercept — The point where a graph crosses the y-axis.
Proportional relationship — A constant-rate relationship that passes through the origin.
Unit rate — A rate expressed for one unit of another quantity.
Independent variable — A variable used as the input of a relationship.
Dependent variable — A variable whose value depends on the independent variable.
magnitude — The size or absolute value of a quantity.
Derivative — In calculus, a quantity describing instantaneous rate of change.
Instantaneous rate of change — The rate at one particular point or instant.
Key Takeaways
- A rate compares two quantities.
- A rate of change describes how one quantity changes relative to another.
- The basic formula is:
rate of change = change in output / change in input
- In x and y notation:
rate of change = Δy/Δx
- For a straight line, rate of change is the same as slope.
- Slope can be calculated using:
m = (y₂ − y₁)/(x₂ − x₁)
- A constant rate of change means the same rate occurs throughout the relationship.
- Constant rates produce straight-line graphs.
- Relationships with constant rates of change are linear.
- A positive rate indicates an increasing quantity.
- A negative rate indicates a decreasing quantity.
- A zero rate indicates no change.
- The magnitude of a rate tells us how rapidly the quantity changes.
- Units are essential when interpreting rates.
- The units of a rate come from output units divided by input units.
- A table can show a constant rate even when the input intervals are unequal.
- Always compare Δy with Δx rather than looking only at changes in y.
- The starting value is different from the rate of change.
- In y = mx + b, m is the rate of change and b is the initial value.
- A proportional relationship has a constant rate and passes through the origin.
- A linear relationship can have a constant rate without being proportional.
- Real-world rates include speed, wages, flow rate, temperature change, growth, and cost per item.
- Comparing rates allows us to compare how quickly different quantities change.
- A changing rate usually produces a curved rather than straight graph.
- Constant rate of change provides an important foundation for understanding instantaneous rate of change in calculus.
A useful question to ask whenever you see a rate is:
"How much does the output change for each unit of input?"
Check Your Understanding
1. Define rate of change in your own words.
2. What does Δ mean?
3. Write the formula for rate of change using Δx and Δy.
4. A car travels 180 km in 3 hours. Calculate its rate.
5. Interpret your answer to Question 4 in a complete sentence.
6. A plant grows from 12 cm to 32 cm in 5 weeks. Calculate its average rate of growth.
7. A temperature decreases from 30°C to 18°C in 4 hours. Calculate the rate of change.
8. Why is the answer to Question 7 negative?
9. Determine the rate of change:
| x | y |
|---|---|
| 0 | 4 |
| 2 | 10 |
| 4 | 16 |
| 6 | 22 |
10. Is the rate in Question 9 constant? Explain.
11. Determine whether this relationship has a constant rate:
| x | y |
|---|---|
| 1 | 5 |
| 2 | 9 |
| 3 | 15 |
| 4 | 23 |
12. Explain how you know.
13. Find the rate of change between:
(3, 8)
and:
(7, 20)
14. A line has a rate of change of −4. What does the negative sign tell you?
15. What does a zero rate of change look like on a graph?
16. What does a constant rate of change look like on a graph?
17. For:
y = 7x + 3
identify:
a. the rate of change
b. the initial value
18. For:
C = 12t + 50
explain what 12 could mean if C represents cost in dollars and t represents time in hours.
19. Why is 50 not the rate of change in Question 18?
20. Machine A fills a tank at 8 L/min. Machine B fills at 12 L/min. Which has the greater rate of change?
21. Compare the rates:
−3°C/min
and:
−8°C/min
Which represents the more rapid decrease?
22. Determine whether this table shows a constant rate:
| x | y |
|---|---|
| 0 | 5 |
| 2 | 13 |
| 5 | 25 |
| 9 | 41 |
23. Explain why unequal x-intervals do not automatically mean the rate is non-constant.
24. A water tank contains 300 L and loses 15 L every minute. State the rate of change, including its sign and units.
25. Explain the difference between a rate of change and an initial value.
26. Explain the difference between a linear relationship and a proportional relationship.
27. Give three real-world examples of constant rates of change.
28. A cyclist's distance-time graph is a straight line with slope 6. If distance is measured in kilometres and time in hours, interpret the slope.
29. Explain why rate of change is an important idea for calculus.
30. Challenge: Two delivery companies use these pricing models:
Company A:
C = 3d + 12
Company B:
C = 4.5d + 5
where C is cost in dollars and d is distance in kilometres.
For each company:
a. identify the rate of change
b. identify the initial value
c. explain what each rate means
d. determine which company's price increases more quickly with distance
e. calculate the cost of a 10 km delivery
f. explain why comparing only the rates is not enough to determine which company is cheaper for every possible delivery distance.