Question

Which table could be a partial set of values for a linear function?

A. x y
0 0.0
1 0.5
2 2.0
3 4.5

B. x y
0 0.0
1 0.5
2 1.0
3 1.5

C. x y
0 0
1 1
2 4
3 9

D. x y
0 0
1 1
2 8
3 27

Answer 1

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**.