which table represents a nonlinear function?

Explanation:

Step1: Recall the property of a linear function

For a linear function, the rate of change (slope) (m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}) is constant.

Step2: Calculate the slopes for the first table

For the first table: When (x = 0,y=-1.5); (x = 1,y = 1.5). Slope (m_{1}=\frac{1.5-\left(-1.5\right)}{1 - 0}=\frac{1.5 + 1.5}{1}=3) When (x = 1,y = 1.5); (x = 2,y = 3). Slope (m_{2}=\frac{3 - 1.5}{2 - 1}=\frac{1.5}{1}=3) When (x = 2,y = 3); (x = 3,y = 4.5). Slope (m_{3}=\frac{4.5-3}{3 - 2}=\frac{1.5}{1}=3)

Step3: Calculate the slopes for the second table

For the second table: When (x = 0,y=-19); (x = 1,y=-11). Slope (m_{1}=\frac{-11-\left(-19\right)}{1 - 0}=\frac{-11 + 19}{1}=8) When (x = 1,y=-11); (x = 2,y=-3). Slope (m_{2}=\frac{-3-\left(-11\right)}{2 - 1}=\frac{-3 + 11}{1}=8) When (x = 2,y=-3); (x = 3,y = 5). Slope (m_{3}=\frac{5-\left(-3\right)}{3 - 2}=\frac{5 + 3}{1}=8)

Step4: Calculate the slopes for the third table

For the third table: When (x = 0,y = 15); (x = 1,y = 12). Slope (m_{1}=\frac{12-15}{1 - 0}=\frac{-3}{1}=-3) When (x = 1,y = 12); (x = 2,y = 9). Slope (m_{2}=\frac{9 - 12}{2 - 1}=\frac{-3}{1}=-3) When (x = 2,y = 9); (x = 3,y = 6). Slope (m_{3}=\frac{6 - 9}{3 - 2}=\frac{-3}{1}=-3)

Answer:

None of the tables represent a nonlinear function. All of them have a constant rate of change (slope).