Answer:
No.
Step-by-step explanation:
The points from the table need to have a constant slope for the function to be linear.
Slope is rise over run.
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]
For (1,1) and (2,2):
[tex]m=\frac{2-1}{2-1}=\frac{1}{1}=\boxed{1}[/tex]
The slope is 1.
For (3,6) and (4,24):
[tex]m=\frac{24-6}{4-3}=\frac{18}{1}=\boxed{18}[/tex]
The slope is 18.
1 ≠ 18
Since the slopes of the points are no consistent, the table represents a nonlinear function.
Therefore,
The table does NOT represent a linear function.
