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 |
A. He found the differences in the $y$-values as $7-1=6, 17-7=10$, and $31-17=14$.
B. He determined that the differences of 6, 10, and 14 are increasing by 4 each time.
C. He concluded that the function has a constant rate of change.
D. He reasoned that a function that has a constant rate of change is linear.
Answer (1)
The function is linear if the rate of change Δ x Δ y is constant.
Willis only checked the differences in y -values, which were increasing.
The rate of change between consecutive points is not constant (6, 10, 14).
Willis's mistake was assuming that a constant change in the differences of y -values implies linearity, without considering the change in x -values.
Explanation
Understanding the Problem Willis analyzed the differences in consecutive y -values of the table and found that they are increasing by a constant amount. He incorrectly concluded that the function is linear. Let's identify his mistake. A function is linear if the rate of change is constant. The rate of change is calculated as the change in y divided by the change in x .
Checking for Constant Rate of Change To determine if the function is linear, we need to check if the ratio of the difference in y -values to the difference in x -values is constant for consecutive points in the table. The x -values are 1, 2, 3, and 4, so the difference between consecutive x -values is always 1. Therefore, we only need to check if the difference in y -values is constant.
Calculating Rate of Change Let's calculate the rate of change between consecutive points:
Between (1, 1) and (2, 7): Rate of change = 2 − 1 7 − 1 = 1 6 = 6 .
Between (2, 7) and (3, 17): Rate of change = 3 − 2 17 − 7 = 1 10 = 10 .
Between (3, 17) and (4, 31): Rate of change = 4 − 3 31 − 17 = 1 14 = 14 .
Identifying the Mistake The rates of change are 6, 10, and 14. Since the rate of change is not constant, the function is not linear. Willis's mistake was assuming that a constant change in the differences of y -values implies linearity. He failed to consider the corresponding change in x -values. For a function to be linear, the rate of change ( Δ x Δ y ) must be constant.
Conclusion Therefore, Willis's mistake was that he didn't verify that the rate of change ( Δ x Δ y ) is constant. He only looked at the differences in y -values and incorrectly concluded that the function is linear because those differences increased by a constant amount.
Examples
Imagine you're tracking the distance a car travels over time. If the car is moving at a constant speed, the relationship between time and distance is linear. However, if the car is accelerating, the distance covered in each time interval increases, but the relationship is not linear because the speed is changing. Similarly, in this problem, Willis incorrectly assumed linearity based on the increasing differences in y -values, without considering the constant change in x -values.
