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Sagot :
To determine which of the given tables shows a linear function, we need to check if the points in each table follow a consistent rate of change or slope. A linear function can be represented by the equation [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] represents the slope and [tex]\( b \)[/tex] represents the y-intercept. In other words, if we calculate the slope between each pair of points in the table and the slope remains constant, then the table represents a linear function.
### Analyzing the tables:
#### Table 1:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -4 & 8 \\ \hline -1 & 2 \\ \hline 1 & 2 \\ \hline 2 & 4 \\ \hline 3 & 6 \\ \hline \end{array} \][/tex]
To determine if this table shows a linear function, we calculate the slope between consecutive points:
1. Between [tex]\((-4, 8)\)[/tex] and [tex]\((-1, 2)\)[/tex]:
[tex]\[ \text{slope} = \frac{2 - 8}{-1 - (-4)} = \frac{-6}{3} = -2 \][/tex]
2. Between [tex]\((-1, 2)\)[/tex] and [tex]\( (1, 2)\)[/tex]:
[tex]\[ \text{slope} = \frac{2 - 2}{1 - (-1)} = \frac{0}{2} = 0 \][/tex]
3. Between [tex]\( (1, 2)\)[/tex] and [tex]\( (2, 4)\)[/tex]:
[tex]\[ \text{slope} = \frac{4 - 2}{2 - 1} = \frac{2}{1} = 2 \][/tex]
4. Between [tex]\( (2, 4)\)[/tex] and [tex]\( (3, 6)\)[/tex]:
[tex]\[ \text{slope} = \frac{6 - 4}{3 - 2} = \frac{2}{1} = 2 \][/tex]
The slopes are not consistent (we have [tex]\(-2\)[/tex], [tex]\(0\)[/tex], and [tex]\(2\)[/tex]), so Table 1 does not show a linear function.
#### Table 2:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -6 & -16 \\ \hline -2 & -8 \\ \hline 0 & -4 \\ \hline 1 & -2 \\ \hline 3 & 2 \\ \hline \end{array} \][/tex]
1. Between [tex]\((-6, -16)\)[/tex] and [tex]\((-2, -8)\)[/tex]:
[tex]\[ \text{slope} = \frac{-8 - (-16)}{-2 - (-6)} = \frac{8}{4} = 2 \][/tex]
2. Between [tex]\((-2, -8)\)[/tex] and [tex]\( (0, -4)\)[/tex]:
[tex]\[ \text{slope} = \frac{-4 - (-8)}{0 - (-2)} = \frac{4}{2} = 2 \][/tex]
3. Between [tex]\( (0, -4)\)[/tex] and [tex]\( (1, -2)\)[/tex]:
[tex]\[ \text{slope} = \frac{-2 - (-4)}{1 - 0} = \frac{2}{1} = 2 \][/tex]
4. Between [tex]\( (1, -2)\)[/tex] and [tex]\( (3, 2)\)[/tex]:
[tex]\[ \text{slope} = \frac{2 - (-2)}{3 - 1} = \frac{4}{2} = 2 \][/tex]
The slopes are consistent ([tex]\(2\)[/tex] between all consecutive points), so Table 2 does show a linear function.
#### Table 3:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & -2 \\ \hline \end{array} \][/tex]
This table only contains a single point, which trivially represents a linear function (with slope undefined due to lack of another point).
### Conclusion:
Among the provided tables, Table 2 shows a linear function because the slope between all consecutive points is consistent.
### Analyzing the tables:
#### Table 1:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -4 & 8 \\ \hline -1 & 2 \\ \hline 1 & 2 \\ \hline 2 & 4 \\ \hline 3 & 6 \\ \hline \end{array} \][/tex]
To determine if this table shows a linear function, we calculate the slope between consecutive points:
1. Between [tex]\((-4, 8)\)[/tex] and [tex]\((-1, 2)\)[/tex]:
[tex]\[ \text{slope} = \frac{2 - 8}{-1 - (-4)} = \frac{-6}{3} = -2 \][/tex]
2. Between [tex]\((-1, 2)\)[/tex] and [tex]\( (1, 2)\)[/tex]:
[tex]\[ \text{slope} = \frac{2 - 2}{1 - (-1)} = \frac{0}{2} = 0 \][/tex]
3. Between [tex]\( (1, 2)\)[/tex] and [tex]\( (2, 4)\)[/tex]:
[tex]\[ \text{slope} = \frac{4 - 2}{2 - 1} = \frac{2}{1} = 2 \][/tex]
4. Between [tex]\( (2, 4)\)[/tex] and [tex]\( (3, 6)\)[/tex]:
[tex]\[ \text{slope} = \frac{6 - 4}{3 - 2} = \frac{2}{1} = 2 \][/tex]
The slopes are not consistent (we have [tex]\(-2\)[/tex], [tex]\(0\)[/tex], and [tex]\(2\)[/tex]), so Table 1 does not show a linear function.
#### Table 2:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -6 & -16 \\ \hline -2 & -8 \\ \hline 0 & -4 \\ \hline 1 & -2 \\ \hline 3 & 2 \\ \hline \end{array} \][/tex]
1. Between [tex]\((-6, -16)\)[/tex] and [tex]\((-2, -8)\)[/tex]:
[tex]\[ \text{slope} = \frac{-8 - (-16)}{-2 - (-6)} = \frac{8}{4} = 2 \][/tex]
2. Between [tex]\((-2, -8)\)[/tex] and [tex]\( (0, -4)\)[/tex]:
[tex]\[ \text{slope} = \frac{-4 - (-8)}{0 - (-2)} = \frac{4}{2} = 2 \][/tex]
3. Between [tex]\( (0, -4)\)[/tex] and [tex]\( (1, -2)\)[/tex]:
[tex]\[ \text{slope} = \frac{-2 - (-4)}{1 - 0} = \frac{2}{1} = 2 \][/tex]
4. Between [tex]\( (1, -2)\)[/tex] and [tex]\( (3, 2)\)[/tex]:
[tex]\[ \text{slope} = \frac{2 - (-2)}{3 - 1} = \frac{4}{2} = 2 \][/tex]
The slopes are consistent ([tex]\(2\)[/tex] between all consecutive points), so Table 2 does show a linear function.
#### Table 3:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & -2 \\ \hline \end{array} \][/tex]
This table only contains a single point, which trivially represents a linear function (with slope undefined due to lack of another point).
### Conclusion:
Among the provided tables, Table 2 shows a linear function because the slope between all consecutive points is consistent.
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