To determine which table could be a partial set of values for a linear function, we need to check if the relationship between \( x \) and \( y \) is linear. For a function to be linear, the change in \( y \) must be proportional to the change in \( x \). This means that the ratio \( \frac{\Delta y}{\Delta x} \) should be constant.
Let's analyze each option:
A.
x y
0 0.0
1 0.5
2 2.0
3 4.5
- Change from \( x = 0 \) to \( x = 1 \): \( \Delta y = 0.5 - 0.0 = 0.5 \)
- Change from \( x = 1 \) to \( x = 2 \): \( \Delta y = 2.0 - 0.5 = 1.5 \)
- Change from \( x = 2 \) to \( x = 3 \): \( \Delta y = 4.5 - 2.0 = 2.5 \)
The changes are not constant, so this is not linear.
B.
x y
0 0.0
1 0.5
2 1.0
3 1.5
- Change from \( x = 0 \) to \( x = 1 \): \( \Delta y = 0.5 - 0.0 = 0.5 \)
- Change from \( x = 1 \) to \( x = 2 \): \( \Delta y = 1.0 - 0.5 = 0.5 \)
- Change from \( x = 2 \) to \( x = 3 \): \( \Delta y = 1.5 - 1.0 = 0.5 \)
The changes are constant (\( \Delta y = 0.5 \)), so this is a linear function.
C.
x y
0 0
1 1
2 4
3 9
- Change from \( x = 0 \) to \( x = 1 \): \( \Delta y = 1 - 0 = 1 \)
- Change from \( x = 1 \) to \( x = 2 \): \( \Delta y = 4 - 1 = 3 \)
- Change from \( x = 2 \) to \( x = 3 \): \( \Delta y = 9 - 4 = 5 \)
The changes are not constant, so this is not linear.
D.
x y
0 0
1 1
2 8
3 27
- Change from \( x = 0 \) to \( x = 1 \): \( \Delta y = 1 - 0 = 1 \)
- Change from \( x = 1 \) to \( x = 2 \): \( \Delta y = 8 - 1 = 7 \)
- Change from \( x = 2 \) to \( x = 3 \): \( \Delta y = 27 - 8 = 19 \)
The changes are not constant, so this is not linear.
Conclusion: The only table that represents a linear function is B.