Answer :
To determine which tables display linear functions, we need to verify if the rate of change (or slope) between consecutive points in each table is consistent. A table represents a linear function if the slope between each pair of consecutive points is the same.
Here is a step-by-step analysis of each table:
### Table 1
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -2 & 1.5 \\ \hline -1 & 0 \\ \hline 0 & -1.5 \\ \hline 1 & -3 \\ \hline \end{array} \][/tex]
1. Slope between [tex]\((-2, 1.5)\)[/tex] and [tex]\((-1, 0)\)[/tex]:
[tex]\[ \frac{0 - 1.5}{-1 - (-2)} = \frac{-1.5}{1} = -1.5 \][/tex]
2. Slope between [tex]\((-1, 0)\)[/tex] and [tex]\((0, -1.5)\)[/tex]:
[tex]\[ \frac{-1.5 - 0}{0 - (-1)} = \frac{-1.5}{1} = -1.5 \][/tex]
3. Slope between [tex]\((0, -1.5)\)[/tex] and [tex]\((1, -3)\)[/tex]:
[tex]\[ \frac{-3 - (-1.5)}{1 - 0} = \frac{-1.5}{1} = -1.5 \][/tex]
Since the slopes are consistent, Table 1 represents a linear function.
### Table 2
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -1 & 0 \\ \hline 0 & -2 \\ \hline 1 & -1 \\ \hline 2 & -3 \\ \hline \end{array} \][/tex]
1. Slope between [tex]\((-1, 0)\)[/tex] and [tex]\((0, -2)\)[/tex]:
[tex]\[ \frac{-2 - 0}{0 - (-1)} = \frac{-2}{1} = -2 \][/tex]
2. Slope between [tex]\((0, -2)\)[/tex] and [tex]\((1, -1)\)[/tex]:
[tex]\[ \frac{-1 - (-2)}{1 - 0} = \frac{1}{1} = 1 \][/tex]
Since the slopes are not consistent, Table 2 does not represent a linear function.
### Table 3
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 4 & -2.5 \\ \hline 5 & -5.5 \\ \hline 6 & -7.5 \\ \hline 7 & -10 \\ \hline \end{array} \][/tex]
1. Slope between [tex]\((4, -2.5)\)[/tex] and [tex]\((5, -5.5)\)[/tex]:
[tex]\[ \frac{-5.5 - (-2.5)}{5 - 4} = \frac{-3}{1} = -3 \][/tex]
2. Slope between [tex]\((5, -5.5)\)[/tex] and [tex]\((6, -7.5)\)[/tex]:
[tex]\[ \frac{-7.5 - (-5.5)}{6 - 5} = \frac{-2}{1} = -2 \][/tex]
3. Slope between [tex]\((6, -7.5)\)[/tex] and [tex]\((7, -10)\)[/tex]:
[tex]\[ \frac{-10 - (-7.5)}{7 - 6} = \frac{-2.5}{1} = -2.5 \][/tex]
Since the slopes are not consistent, Table 3 does not represent a linear function.
### Table 4
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -3 & 6 \\ \hline -4 & 7 \\ \hline -5 & 8 \\ \hline -6 & 9 \\ \hline \end{array} \][/tex]
1. Slope between [tex]\((-3, 6)\)[/tex] and [tex]\((-4, 7)\)[/tex]:
[tex]\[ \frac{7 - 6}{-4 - (-3)} = \frac{1}{-1} = -1 \][/tex]
2. Slope between [tex]\((-4, 7)\)[/tex] and [tex]\((-5, 8)\)[/tex]:
[tex]\[ \frac{8 - 7}{-5 - (-4)} = \frac{1}{-1} = -1 \][/tex]
3. Slope between [tex]\((-5, 8)\)[/tex] and [tex]\((-6, 9)\)[/tex]:
[tex]\[ \frac{9 - 8}{-6 - (-5)} = \frac{1}{-1} = -1 \][/tex]
Since the slopes are consistent, Table 4 represents a linear function.
Based on the analysis above, the tables that display linear functions are:
- Table 1
- Table 4
Thus, the final answer is:
[tex]\[ \boxed{\{1, 4\}} \][/tex]
Here is a step-by-step analysis of each table:
### Table 1
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -2 & 1.5 \\ \hline -1 & 0 \\ \hline 0 & -1.5 \\ \hline 1 & -3 \\ \hline \end{array} \][/tex]
1. Slope between [tex]\((-2, 1.5)\)[/tex] and [tex]\((-1, 0)\)[/tex]:
[tex]\[ \frac{0 - 1.5}{-1 - (-2)} = \frac{-1.5}{1} = -1.5 \][/tex]
2. Slope between [tex]\((-1, 0)\)[/tex] and [tex]\((0, -1.5)\)[/tex]:
[tex]\[ \frac{-1.5 - 0}{0 - (-1)} = \frac{-1.5}{1} = -1.5 \][/tex]
3. Slope between [tex]\((0, -1.5)\)[/tex] and [tex]\((1, -3)\)[/tex]:
[tex]\[ \frac{-3 - (-1.5)}{1 - 0} = \frac{-1.5}{1} = -1.5 \][/tex]
Since the slopes are consistent, Table 1 represents a linear function.
### Table 2
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -1 & 0 \\ \hline 0 & -2 \\ \hline 1 & -1 \\ \hline 2 & -3 \\ \hline \end{array} \][/tex]
1. Slope between [tex]\((-1, 0)\)[/tex] and [tex]\((0, -2)\)[/tex]:
[tex]\[ \frac{-2 - 0}{0 - (-1)} = \frac{-2}{1} = -2 \][/tex]
2. Slope between [tex]\((0, -2)\)[/tex] and [tex]\((1, -1)\)[/tex]:
[tex]\[ \frac{-1 - (-2)}{1 - 0} = \frac{1}{1} = 1 \][/tex]
Since the slopes are not consistent, Table 2 does not represent a linear function.
### Table 3
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 4 & -2.5 \\ \hline 5 & -5.5 \\ \hline 6 & -7.5 \\ \hline 7 & -10 \\ \hline \end{array} \][/tex]
1. Slope between [tex]\((4, -2.5)\)[/tex] and [tex]\((5, -5.5)\)[/tex]:
[tex]\[ \frac{-5.5 - (-2.5)}{5 - 4} = \frac{-3}{1} = -3 \][/tex]
2. Slope between [tex]\((5, -5.5)\)[/tex] and [tex]\((6, -7.5)\)[/tex]:
[tex]\[ \frac{-7.5 - (-5.5)}{6 - 5} = \frac{-2}{1} = -2 \][/tex]
3. Slope between [tex]\((6, -7.5)\)[/tex] and [tex]\((7, -10)\)[/tex]:
[tex]\[ \frac{-10 - (-7.5)}{7 - 6} = \frac{-2.5}{1} = -2.5 \][/tex]
Since the slopes are not consistent, Table 3 does not represent a linear function.
### Table 4
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -3 & 6 \\ \hline -4 & 7 \\ \hline -5 & 8 \\ \hline -6 & 9 \\ \hline \end{array} \][/tex]
1. Slope between [tex]\((-3, 6)\)[/tex] and [tex]\((-4, 7)\)[/tex]:
[tex]\[ \frac{7 - 6}{-4 - (-3)} = \frac{1}{-1} = -1 \][/tex]
2. Slope between [tex]\((-4, 7)\)[/tex] and [tex]\((-5, 8)\)[/tex]:
[tex]\[ \frac{8 - 7}{-5 - (-4)} = \frac{1}{-1} = -1 \][/tex]
3. Slope between [tex]\((-5, 8)\)[/tex] and [tex]\((-6, 9)\)[/tex]:
[tex]\[ \frac{9 - 8}{-6 - (-5)} = \frac{1}{-1} = -1 \][/tex]
Since the slopes are consistent, Table 4 represents a linear function.
Based on the analysis above, the tables that display linear functions are:
- Table 1
- Table 4
Thus, the final answer is:
[tex]\[ \boxed{\{1, 4\}} \][/tex]
