To determine which table represents a linear function, we need to check if the sets of values for [tex]\(x\)[/tex] and [tex]\(y\)[/tex] exhibit a constant rate of change (i.e., the slope between points is consistent). Let's analyze each table individually.
### Table 1
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
1 & 5 \\
\hline
2 & 9 \\
\hline
3 & 5 \\
\hline
4 & 9 \\
\hline
\end{array}
\][/tex]
Calculate the slopes between consecutive points:
- Between [tex]\((1, 5)\)[/tex] and [tex]\((2, 9)\)[/tex]:
[tex]\[
\text{slope} = \frac{9 - 5}{2 - 1} = \frac{4}{1} = 4
\][/tex]
- Between [tex]\((2, 9)\)[/tex] and [tex]\((3, 5)\)[/tex]:
[tex]\[
\text{slope} = \frac{5 - 9}{3 - 2} = \frac{-4}{1} = -4
\][/tex]
- Between [tex]\((3, 5)\)[/tex] and [tex]\((4, 9)\)[/tex]:
[tex]\[
\text{slope} = \frac{9 - 5}{4 - 3} = \frac{4}{1} = 4
\][/tex]
The slopes are not consistent, so this table does not represent a linear function.
### Table 2
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
1 & -5 \\
\hline
2 & 10 \\
\hline
3 & -15 \\
\hline
4 & 20 \\
\hline
\end{array}
\][/tex]
Calculate the slopes between consecutive points:
- Between [tex]\((1, -5)\)[/tex] and [tex]\((2, 10)\)[/tex]:
[tex]\[
\text{slope} = \frac{10 - (-5)}{2 - 1} = \frac{15}{1} = 15
\][/tex]
- Between [tex]\((2, 10)\)[/tex] and [tex]\((3, -15)\)[/tex]:
[tex]\[
\text{slope} = \frac{-15 - 10}{3 - 2} = \frac{-25}{1} = -25
\][/tex]
- Between [tex]\((3, -15)\)[/tex] and [tex]\((4, 20)\)[/tex]:
[tex]\[
\text{slope} = \frac{20 - (-15)}{4 - 3} = \frac{35}{1} = 35
\][/tex]
The slopes are not consistent, so this table does not represent a linear function.
### Table 3
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
1 & 5 \\
\hline
2 & 10 \\
\hline
3 & 20 \\
\hline
4 & 40 \\
\hline
\end{array}
\][/tex]
Calculate the slopes between consecutive points:
- Between [tex]\((1, 5)\)[/tex] and [tex]\((2, 10)\)[/tex]:
[tex]\[
\text{slope} = \frac{10 - 5}{2 - 1} = \frac{5}{1} = 5
\][/tex]
- Between [tex]\((2, 10)\)[/tex] and [tex]\((3, 20)\)[/tex]:
[tex]\[
\text{slope} = \frac{20 - 10}{3 - 2} = \frac{10}{1} = 10
\][/tex]
- Between [tex]\((3, 20)\)[/tex] and [tex]\((4, 40)\)[/tex]:
[tex]\[
\text{slope} = \frac{40 - 20}{4 - 3} = \frac{20}{1} = 20
\][/tex]
The slopes are not consistent, so this table does not represent a linear function.
### Table 4
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
1 & -5 \\
\hline
2 & 0 \\
\hline
\end{array}
\][/tex]
Calculate the slope between the two points:
- Between [tex]\((1, -5)\)[/tex] and [tex]\((2, 0)\)[/tex]:
[tex]\[
\text{slope} = \frac{0 - (-5)}{2 - 1} = \frac{5}{1} = 5
\][/tex]
With only two points, the consistency of the slope is maintained by default. Therefore, this table represents a linear function.
### Conclusion
The table that represents a linear function is the one with [tex]\( x \)[/tex] and [tex]\( y \)[/tex] values:
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
1 & -5 \\
\hline
2 & 0 \\
\hline
\end{array}
\][/tex]
So, Table 4 represents a linear function.
