Willis analyzed the following table to determine if the function it represents is linear or non-linear. First he found the diffe

His error lies in using the differences of 6, 10, and 14 instead of utilizing the differences of 7, 17, and 31. In a linear function, the y values must exhibit a constant rate of change rather than just relying on the differences.

He determined the y values for different x values as follows: 1, 7, 17, and 31. Willis then calculated the differences in the y values and observed they were rising by 4, concluding that the function had a constant rate of change, thus categorizing it as linear. However, his method was flawed. To establish the linearity of the function, one must assess if the slope of the graph yields a consistent rate of change, described by: where () denotes distinct interpolating points and their corresponding values. Hence, Willis needed to compute and confirm its equality for various pairs of points to substantiate that the function is indeed linear.