Answer :
To determine which table represents a linear function, we need to check the consistency of the differences between the successive pairs of points in each table. A linear function will have constant differences between its [tex]$y$[/tex]-values when the [tex]$x$[/tex]-values are incremented consistently. Let's analyze each table:
### Table 1
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & \frac{1}{2} \\ \hline 2 & 1 \\ \hline 3 & 1 \frac{1}{2} \\ \hline 4 & 2 \\ \hline \end{array} \][/tex]
Calculate the differences between successive [tex]$y$[/tex] values:
- Between [tex]$(1, \frac{1}{2})$[/tex] and [tex]$(2, 1)$[/tex]: [tex]\(\frac{1 - \frac{1}{2}}{2-1} = \frac{\frac{1}{2}}{1} = \frac{1}{2}\)[/tex]
- Between [tex]$(2, 1)$[/tex] and [tex]$(3, 1.5)$[/tex]: [tex]\(\frac{1.5 - 1}{3-2} = \frac{0.5}{1} = \frac{1}{2}\)[/tex]
- Between [tex]$(3, 1.5)$[/tex] and [tex]$(4, 2)$[/tex]: [tex]\(\frac{2 - 1.5}{4-3} = \frac{0.5}{1} = \frac{1}{2}\)[/tex]
Since the differences are consistent, Table 1 represents a linear function.
### Table 2
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 1 \\ \hline 2 & \frac{1}{2} \\ \hline 3 & \frac{1}{3} \\ \hline 4 & \frac{1}{4} \\ \hline \end{array} \][/tex]
Calculate the differences between successive [tex]$y$[/tex] values:
- Between [tex]$(1, 1)$[/tex] and [tex]$(2, \frac{1}{2})$[/tex]: [tex]\(\frac{\frac{1}{2} - 1}{2-1} = \frac{-\frac{1}{2}}{1} = -\frac{1}{2}\)[/tex]
- Between [tex]$(2, \frac{1}{2})$[/tex] and [tex]$(3, \frac{1}{3})$[/tex]: [tex]\(\frac{\frac{1}{3} - \frac{1}{2}}{3-2} = \frac{-\frac{1}{6}}{1} = -\frac{1}{6}\)[/tex]
- Between [tex]$(3, \frac{1}{3})$[/tex] and [tex]$(4, \frac{1}{4})$[/tex]: [tex]\(\frac{\frac{1}{4} - \frac{1}{3}}{4-3} = \frac{-\frac{1}{12}}{1} = -\frac{1}{12}\)[/tex]
Since the differences are not consistent, Table 2 does not represent a linear function.
### Table 3
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 7 \\ \hline 2 & 9 \\ \hline 3 & 13 \\ \hline 4 & 21 \\ \hline \end{array} \][/tex]
Calculate the differences between successive [tex]$y$[/tex] values:
- Between [tex]$(1, 7)$[/tex] and [tex]$(2, 9)$[/tex]: [tex]\(\frac{9 - 7}{2-1} = \frac{2}{1} = 2\)[/tex]
- Between [tex]$(2, 9)$[/tex] and [tex]$(3, 13)$[/tex]: [tex]\(\frac{13 - 9}{3-2} = \frac{4}{1} = 4\)[/tex]
- Between [tex]$(3, 13)$[/tex] and [tex]$(4, 21)$[/tex]: [tex]\(\frac{21 - 13}{4-3} = \frac{8}{1} = 8\)[/tex]
Since the differences are not consistent, Table 3 does not represent a linear function.
### Table 4
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 0 \\ \hline 2 & 6 \\ \hline 3 & 16 \\ \hline 4 & 30 \\ \hline \end{array} \][/tex]
Calculate the differences between successive [tex]$y$[/tex] values:
- Between [tex]$(1, 0)$[/tex] and [tex]$(2, 6)$[/tex]: [tex]\(\frac{6 - 0}{2-1} = \frac{6}{1} = 6\)[/tex]
- Between [tex]$(2, 6)$[/tex] and [tex]$(3, 16)$[/tex]: [tex]\(\frac{16 - 6}{3-2} = \frac{10}{1} = 10\)[/tex]
- Between [tex]$(3, 16)$[/tex] and [tex]$(4, 30)$[/tex]: [tex]\(\frac{30 - 16}{4-3} = \frac{14}{1} = 14\)[/tex]
Since the differences are not consistent, Table 4 does not represent a linear function.
Therefore, only Table 1 represents a linear function.
### Table 1
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & \frac{1}{2} \\ \hline 2 & 1 \\ \hline 3 & 1 \frac{1}{2} \\ \hline 4 & 2 \\ \hline \end{array} \][/tex]
Calculate the differences between successive [tex]$y$[/tex] values:
- Between [tex]$(1, \frac{1}{2})$[/tex] and [tex]$(2, 1)$[/tex]: [tex]\(\frac{1 - \frac{1}{2}}{2-1} = \frac{\frac{1}{2}}{1} = \frac{1}{2}\)[/tex]
- Between [tex]$(2, 1)$[/tex] and [tex]$(3, 1.5)$[/tex]: [tex]\(\frac{1.5 - 1}{3-2} = \frac{0.5}{1} = \frac{1}{2}\)[/tex]
- Between [tex]$(3, 1.5)$[/tex] and [tex]$(4, 2)$[/tex]: [tex]\(\frac{2 - 1.5}{4-3} = \frac{0.5}{1} = \frac{1}{2}\)[/tex]
Since the differences are consistent, Table 1 represents a linear function.
### Table 2
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 1 \\ \hline 2 & \frac{1}{2} \\ \hline 3 & \frac{1}{3} \\ \hline 4 & \frac{1}{4} \\ \hline \end{array} \][/tex]
Calculate the differences between successive [tex]$y$[/tex] values:
- Between [tex]$(1, 1)$[/tex] and [tex]$(2, \frac{1}{2})$[/tex]: [tex]\(\frac{\frac{1}{2} - 1}{2-1} = \frac{-\frac{1}{2}}{1} = -\frac{1}{2}\)[/tex]
- Between [tex]$(2, \frac{1}{2})$[/tex] and [tex]$(3, \frac{1}{3})$[/tex]: [tex]\(\frac{\frac{1}{3} - \frac{1}{2}}{3-2} = \frac{-\frac{1}{6}}{1} = -\frac{1}{6}\)[/tex]
- Between [tex]$(3, \frac{1}{3})$[/tex] and [tex]$(4, \frac{1}{4})$[/tex]: [tex]\(\frac{\frac{1}{4} - \frac{1}{3}}{4-3} = \frac{-\frac{1}{12}}{1} = -\frac{1}{12}\)[/tex]
Since the differences are not consistent, Table 2 does not represent a linear function.
### Table 3
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 7 \\ \hline 2 & 9 \\ \hline 3 & 13 \\ \hline 4 & 21 \\ \hline \end{array} \][/tex]
Calculate the differences between successive [tex]$y$[/tex] values:
- Between [tex]$(1, 7)$[/tex] and [tex]$(2, 9)$[/tex]: [tex]\(\frac{9 - 7}{2-1} = \frac{2}{1} = 2\)[/tex]
- Between [tex]$(2, 9)$[/tex] and [tex]$(3, 13)$[/tex]: [tex]\(\frac{13 - 9}{3-2} = \frac{4}{1} = 4\)[/tex]
- Between [tex]$(3, 13)$[/tex] and [tex]$(4, 21)$[/tex]: [tex]\(\frac{21 - 13}{4-3} = \frac{8}{1} = 8\)[/tex]
Since the differences are not consistent, Table 3 does not represent a linear function.
### Table 4
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 0 \\ \hline 2 & 6 \\ \hline 3 & 16 \\ \hline 4 & 30 \\ \hline \end{array} \][/tex]
Calculate the differences between successive [tex]$y$[/tex] values:
- Between [tex]$(1, 0)$[/tex] and [tex]$(2, 6)$[/tex]: [tex]\(\frac{6 - 0}{2-1} = \frac{6}{1} = 6\)[/tex]
- Between [tex]$(2, 6)$[/tex] and [tex]$(3, 16)$[/tex]: [tex]\(\frac{16 - 6}{3-2} = \frac{10}{1} = 10\)[/tex]
- Between [tex]$(3, 16)$[/tex] and [tex]$(4, 30)$[/tex]: [tex]\(\frac{30 - 16}{4-3} = \frac{14}{1} = 14\)[/tex]
Since the differences are not consistent, Table 4 does not represent a linear function.
Therefore, only Table 1 represents a linear function.
