Answer:
1) Neither
2) Exponential
3) Exponential
4) Linear
Explanation:
1) x-values
y-values
0
0
1
1
2
4
3
9
4
16
Therefore, we have, the relationship of the function given as_f(x) = x² Which is neither a linear function nor an exponential function
2) x-values
y-values
0
1/3
1
1
2
3
3
9
4
27
The relationship between 'x', and 'y' is given as follows;
f(x) = 3⁽ˣ⁻¹⁾
Therefore, the relationship between 'x', and 'y', is an exponential relationship, and the function is an exponential function
3) x-values
y-values
0
5
1
5/2
2
5/4
3
5/8
4
5/16
The relationship between 'x', and 'y' is given as follows;
f(x) = 5×(1/2)ˣ
Therefore, the relationship between 'x', and 'y', is an exponential relationship, and the function is an exponential function
4) x-values
y-values
0
4.5
1
4
2
3.5
3
3
4
2.5
The given data for the y-values has a constant first common difference of 4 - 4.5 = -0.5, therefore, the relationship between the 'x', and 'y' values is a linear relationship.