Question

Willis analyzed the following table to determine if the function it represents is linear or non-linear. First he found the differences in the y-values as 7 – 1 = 6, 17 – 7 = 10, and 31 – 17 = 14. Then he concluded that since the differences of 6, 10, and 14 are increasing by 4 each time, the function has a constant rate of change and is linear. What was Willis’s mistake?

x
y
1
1
2
7
3
17
4
31
He found the differences in the y-values as 7 – 1 = 6, 17 – 7 = 10, and 31 – 17 = 14.
He determined that the differences of 6, 10, and 14 are increasing by 4 each time.
He concluded that the function has a constant rate of change.
He reasoned that a function that has a constant rate of change is linear.

Answer 1

Willis confused “differences increasing by a constant amount” with a constant rate of change.

For a linear function the first differences (changes in y for equal changes in x) must be constant. Here the first differences are 6, 10, 14 — not equal, so the rate of change is not constant and the function is not linear. The fact that the first differences increase by the constant 4 means the second differences are constant, which indicates a quadratic function. In fact the points lie on y = 2x^2 − 1.