Answer :
To determine which table represents a linear function, we need to check if the values in the table form a linear relationship. A linear function has the form \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept. This means that the difference in \( y \) values divided by the difference in \( x \) values (the slope) should be constant for all points in the table.
Let's analyze each table step by step:
### 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]
- From \( (1, \frac{1}{2}) \) to \( (2, 1) \):
[tex]\[ \text{slope} = \frac{1 - \frac{1}{2}}{2 - 1} = \frac{1}{2} \][/tex]
- From \( (2, 1) \) to \( (3, 1\frac{1}{2}) \):
[tex]\[ \text{slope} = \frac{1\frac{1}{2} - 1}{3 - 2} = \frac{\frac{3}{2} - 1}{1} = \frac{1}{2} \][/tex]
- From \( (3, 1\frac{1}{2}) \) to \( (4, 2) \):
[tex]\[ \text{slope} = \frac{2 - 1\frac{1}{2}}{4 - 3} = \frac{2 - \frac{3}{2}}{1} = \frac{1}{2} \][/tex]
All slopes are equal to \( \frac{1}{2} \), so this table 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]
- From \( (1, 1) \) to \( (2, \frac{1}{2}) \):
[tex]\[ \text{slope} = \frac{\frac{1}{2} - 1}{2 - 1} = -\frac{1}{2} \][/tex]
- From \( (2, \frac{1}{2}) \) to \( (3, \frac{1}{3}) \):
[tex]\[ \text{slope} = \frac{\frac{1}{3} - \frac{1}{2}}{3 - 2} = \frac{\frac{2}{6} - \frac{3}{6}}{1} = -\frac{1}{6} \][/tex]
- From \( (3, \frac{1}{3}) \) to \( (4, \frac{1}{4}) \):
[tex]\[ \text{slope} = \frac{\frac{1}{4} - \frac{1}{3}}{4 - 3} = \frac{\frac{3}{12} - \frac{4}{12}}{1} = -\frac{1}{12} \][/tex]
Slopes are not consistent, so this table does not represent a linear function.
### Table 3:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 7 \\ \hline 2 & 9 \\ \hline 3 & 13 \\ \hline 4 & 21 \\ \hline \end{array} \][/tex]
- From \( (1, 7) \) to \( (2, 9) \):
[tex]\[ \text{slope} = \frac{9 - 7}{2 - 1} = 2 \][/tex]
- From \( (2, 9) \) to \( (3, 13) \):
[tex]\[ \text{slope} = \frac{13 - 9}{3 - 2} = 4 \][/tex]
- From \( (3, 13) \) to \( (4, 21) \):
[tex]\[ \text{slope} = \frac{21 - 13}{4 - 3} = 8 \][/tex]
Slopes are not consistent, so this table does not represent a linear function.
### Table 4:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 0 \\ \hline 2 & 6 \\ \hline 3 & 16 \\ \hline 4 & 30 \\ \hline \end{array} \][/tex]
- From \( (1, 0) \) to \( (2, 6) \):
[tex]\[ \text{slope} = \frac{6 - 0}{2 - 1} = 6 \][/tex]
- From \( (2, 6) \) to \( (3, 16) \):
[tex]\[ \text{slope} = \frac{16 - 6}{3 - 2} = 10 \][/tex]
- From \( (3, 16) \) to \( (4, 30) \):
[tex]\[ \text{slope} = \frac{30 - 16}{4 - 3} = 14 \][/tex]
Slopes are not consistent, so this table does not represent a linear function.
Conclusion:
Table 1 represents a linear function because its slopes are consistent.
Let's analyze each table step by step:
### 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]
- From \( (1, \frac{1}{2}) \) to \( (2, 1) \):
[tex]\[ \text{slope} = \frac{1 - \frac{1}{2}}{2 - 1} = \frac{1}{2} \][/tex]
- From \( (2, 1) \) to \( (3, 1\frac{1}{2}) \):
[tex]\[ \text{slope} = \frac{1\frac{1}{2} - 1}{3 - 2} = \frac{\frac{3}{2} - 1}{1} = \frac{1}{2} \][/tex]
- From \( (3, 1\frac{1}{2}) \) to \( (4, 2) \):
[tex]\[ \text{slope} = \frac{2 - 1\frac{1}{2}}{4 - 3} = \frac{2 - \frac{3}{2}}{1} = \frac{1}{2} \][/tex]
All slopes are equal to \( \frac{1}{2} \), so this table 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]
- From \( (1, 1) \) to \( (2, \frac{1}{2}) \):
[tex]\[ \text{slope} = \frac{\frac{1}{2} - 1}{2 - 1} = -\frac{1}{2} \][/tex]
- From \( (2, \frac{1}{2}) \) to \( (3, \frac{1}{3}) \):
[tex]\[ \text{slope} = \frac{\frac{1}{3} - \frac{1}{2}}{3 - 2} = \frac{\frac{2}{6} - \frac{3}{6}}{1} = -\frac{1}{6} \][/tex]
- From \( (3, \frac{1}{3}) \) to \( (4, \frac{1}{4}) \):
[tex]\[ \text{slope} = \frac{\frac{1}{4} - \frac{1}{3}}{4 - 3} = \frac{\frac{3}{12} - \frac{4}{12}}{1} = -\frac{1}{12} \][/tex]
Slopes are not consistent, so this table does not represent a linear function.
### Table 3:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 7 \\ \hline 2 & 9 \\ \hline 3 & 13 \\ \hline 4 & 21 \\ \hline \end{array} \][/tex]
- From \( (1, 7) \) to \( (2, 9) \):
[tex]\[ \text{slope} = \frac{9 - 7}{2 - 1} = 2 \][/tex]
- From \( (2, 9) \) to \( (3, 13) \):
[tex]\[ \text{slope} = \frac{13 - 9}{3 - 2} = 4 \][/tex]
- From \( (3, 13) \) to \( (4, 21) \):
[tex]\[ \text{slope} = \frac{21 - 13}{4 - 3} = 8 \][/tex]
Slopes are not consistent, so this table does not represent a linear function.
### Table 4:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 0 \\ \hline 2 & 6 \\ \hline 3 & 16 \\ \hline 4 & 30 \\ \hline \end{array} \][/tex]
- From \( (1, 0) \) to \( (2, 6) \):
[tex]\[ \text{slope} = \frac{6 - 0}{2 - 1} = 6 \][/tex]
- From \( (2, 6) \) to \( (3, 16) \):
[tex]\[ \text{slope} = \frac{16 - 6}{3 - 2} = 10 \][/tex]
- From \( (3, 16) \) to \( (4, 30) \):
[tex]\[ \text{slope} = \frac{30 - 16}{4 - 3} = 14 \][/tex]
Slopes are not consistent, so this table does not represent a linear function.
Conclusion:
Table 1 represents a linear function because its slopes are consistent.
