Answer: The third table represents a nonlinear function
x y
2 0
4 6
6 16
Step-by-step explanation:
We know that for a nonlinear function, the rate of change of y is not constant w.r.t x.
The rate of change of y w.r.t. x is given by :-
[tex]\dfrac{\text{Change in y}}{\text{Change in x}}[/tex]
For Table 1.
The rate of change of function is given by :_
[tex]\dfrac{1-(-9)}{4-2}=\dfrac{10}{2}=5[/tex]
[tex]\dfrac{11-1}{6-4}=\dfrac{10}{2}=5[/tex]
Thus , the rate of change is constant through out the function, hence it is not representing a nonlinear function.
For Table 2.
The rate of change of function is given by :_
[tex]\dfrac{-16-(-14)}{4-2}=\dfrac{-2}{2}=-1[/tex]
[tex]\dfrac{-18-(-16)}{6-4}=\dfrac{-2}{2}=-1[/tex]
Thus , the rate of change is constant through out the function, hence it is not representing a nonlinear function.
For Table 3.
The rate of change of function is given by :_
[tex]\dfrac{6-0}{4-2}=\dfrac{6}{2}=3[/tex]
[tex]\dfrac{16-6}{6-4}=\dfrac{10}{2}=5[/tex]
Thus , the rate of change is not constant through out the function, hence it is representing a nonlinear function.
For Table 4.
The rate of change of function is given by :_
[tex]\dfrac{-6-(-9)}{4-2}=\dfrac{3}{2}[/tex]
[tex]\dfrac{-3-(-6)}{6-4}=\dfrac{3}{2}[/tex]
Thus , the rate of change is constant through out the function, hence it is not representing a nonlinear function.
