Constant Rate of Change: Meaning, Formula and Examples

A constant rate of change means y changes by the same amount for each equal increase in x. Its value is the slope: m = (y₂ − y₁)/(x₂ − x₁), where x₂ ≠ x₁. For a linear function y = mx + b, m is constant. A positive rate means y increases, a negative rate means y decreases, and a zero rate means y stays the same.

How to find a constant rate of change

1. Identify the input and output. Usually x is the input and y is the output. Note their units.
2. Choose two points with different x-values. Label them (x₁, y₁) and (x₂, y₂).
3. Subtract in the same order. Calculate y₂ − y₁ and x₂ − x₁.
4. Divide: m = (y₂ − y₁)/(x₂ − x₁). State the result in y-units per x-unit.
5. Check other intervals when available. Matching rates across the table support a linear pattern. One pair alone gives an average rate; it does not prove the rate is constant everywhere.

For a graph of a linear function, choose two readable points on the line and calculate rise/run. A horizontal line has rate 0. A vertical line has an undefined slope because its change in x is 0; it is not a function of x.

Finding the rate from a table with unequal intervals

Suppose a table lists these points:

From x = 1 to x = 3, the rate is (13 − 7)/(3 − 1) = 6/2 = 3. From x = 3 to x = 6, it is (22 − 13)/(6 − 3) = 9/3 = 3. The y-differences, 6 and 9, differ because the x-intervals differ. Compare the ratios of changes, not just the y-differences.

These points fit y = 3x + 4. If the problem states that the relationship is linear, its rate is 3 throughout. A few table values alone cannot establish what happens between or beyond the listed points.

Constant rate of change examples

A positive rate: distance traveled

At a steady speed, a vehicle travels 60 miles in 2 hours. Its distance changes from 0 to 60 miles while time changes from 0 to 2 hours, so m = (60 − 0)/(2 − 0) = 30 miles per hour. Without the steady-speed assumption, this calculation gives the average speed for those 2 hours.

A negative rate: water leaving a tank

A tank has 48 liters after 2 minutes and 30 liters after 5 minutes. If it drains at a constant rate, m = (30 − 48)/(5 − 2) = −18/3 = −6 liters per minute. The negative sign means the amount remaining decreases; the tank loses 6 liters each minute while this model applies.

A zero rate: an unchanged value

The points (2, 9) and (5, 9) give m = (9 − 9)/(5 − 2) = 0. A constant function y = 9 has a horizontal graph. A zero rate is different from an undefined rate.

Is constant rate of change the same as constant of proportionality?

A proportional relationship has the form y = kx. Its graph passes through (0, 0), and its constant rate of change is k. A linear function y = mx + b can have a constant rate even when b ≠ 0; then it is not proportional.

For example, y = 3x + 4 has constant rate 3, but y/x is not constant: at x = 1, y/x = 7; at x = 3, y/x = 13/3. Use Δy/Δx for the rate of change. Use y/x for a constant of proportionality only when the relationship is proportional and x ≠ 0.

A changing rate can produce a curve. For y = x², the points (1, 1), (2, 4) and (3, 9) give rates 3 and 5 across equal x-intervals, so the rate is not constant.

Practice: calculate and interpret each rate

1. A linear function passes through (−2, 5) and (4, 17). Find its rate.
2. A table contains (0, 8), (2, 14) and (5, 23). Do the listed intervals have the same rate?
3. A tank drains steadily from 80 liters to 56 liters in 6 minutes. Find the signed rate of change.
4. Find the rate for y = −2x + 7. Is this relationship proportional?
5. A table contains (1, 2), (2, 4) and (3, 8). Is its rate constant?

Continue with graphing linear equations, or use the Algebra 1 learning center to choose a related lesson or practice resource.

References: OpenStax, linear functions and direct variation. The numerical examples and practice on this page are original.