To determine which table represents a linear function, recall that a linear function has the property that changes in [tex]\( y \)[/tex] are proportional to changes in [tex]\( x \)[/tex]. This means the rate of change (or the slope) between any two points should be constant.
Let's go through each table and compute the slope between consecutive points.
1. First Table:
[tex]\[
\begin{tabular}{|c|c|}
\hline
$x$ & $y$ \\
\hline
1 & \frac{1}{2} \\
\hline
2 & 1 \\
\hline
3 & 1\frac{1}{2} \\
\hline
4 & 2 \\
\hline
\end{tabular}
\][/tex]
Compute the slopes:
[tex]\[
\text{slope between (1, 0.5) and (2, 1)} = \frac{1 - 0.5}{2 - 1} = \frac{0.5}{1} = 0.5
\][/tex]
[tex]\[
\text{slope between (2, 1) and (3, 1.5)} = \frac{1.5 - 1}{3 - 2} = \frac{0.5}{1} = 0.5
\][/tex]
[tex]\[
\text{slope between (3, 1.5) and (4, 2)} = \frac{2 - 1.5}{4 - 3} = \frac{0.5}{1} = 0.5
\][/tex]
Since all the slopes are the same, the first table represents a linear function.
2. Second Table:
[tex]\[
\begin{tabular}{|c|c|}
\hline
$x$ & $y$ \\
\hline
1 & 1 \\
\hline
2 & \frac{1}{2} \\
\hline
3 & \frac{1}{3} \\
\hline
4 & \frac{1}{4} \\
\hline
\end{tabular}
\][/tex]
Compute the slopes:
[tex]\[
\text{slope between (1, 1) and (2, 0.5)} = \frac{0.5 - 1}{2 - 1} = \frac{-0.5}{1} = -0.5
\][/tex]
[tex]\[
\text{slope between (2, 0.5) and (3, 0.333)} \approx \frac{0.333 - 0.5}{3 - 2} \approx \frac{-0.167}{1} = -0.167
\][/tex]
[tex]\[
\text{slope between (3, 0.333) and (4, 0.25)} \approx \frac{0.25 - 0.333}{4 - 3} \approx \frac{-0.083}{1} = -0.083
\][/tex]
The slopes are not the same, so the second table does not represent a linear function.
3. Third Table:
[tex]\[
\begin{tabular}{|c|c|}
\hline
$x$ & $y$ \\
\hline
1 & 7 \\
\hline
2 & 9 \\
\hline
3 & 13 \\
\hline
4 & 21 \\
\hline
\end{tabular}
\][/tex]
Compute the slopes:
[tex]\[
\text{slope between (1, 7) and (2, 9)} = \frac{9 - 7}{2 - 1} = \frac{2}{1} = 2
\][/tex]
[tex]\[
\text{slope between (2, 9) and (3, 13)} = \frac{13 - 9}{3 - 2} = \frac{4}{1} = 4
\][/tex]
[tex]\[
\text{slope between (3, 13) and (4, 21)} = \frac{21 - 13}{4 - 3} = \frac{8}{1} = 8
\][/tex]
The slopes are not the same, so the third table does not represent a linear function.
4. Fourth Table:
[tex]\[
\begin{tabular}{|c|c|}
\hline
$x$ & $y$ \\
\hline
1 & 0 \\
\hline
2 & 6 \\
\hline
3 & 16 \\
\hline
4 & 30 \\
\hline
\end{tabular}
\][/tex]
Compute the slopes:
[tex]\[
\text{slope between (1, 0) and (2, 6)} = \frac{6 - 0}{2 - 1} = \frac{6}{1} = 6
\][/tex]
[tex]\[
\text{slope between (2, 6) and (3, 16)} = \frac{16 - 6}{3 - 2} = \frac{10}{1} = 10
\][/tex]
[tex]\[
\text{slope between (3, 16) and (4, 30)} = \frac{30 - 16}{4 - 3} = \frac{14}{1} = 14
\][/tex]
The slopes are not the same, so the fourth table does not represent a linear function.
Since the first table is the only one with consistent slopes between all points, the first table represents a linear function.
