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Sagot :
To determine which table represents a linear function, we need to check if the rate of change (or slope) between consecutive points is constant for each table.
Table 1:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & -2 \\ \hline 2 & -6 \\ \hline 3 & -2 \\ \hline 4 & -6 \\ \hline \end{array} \][/tex]
Let's calculate the rate of change between each pair of points:
1. Between [tex]\( x = 1 \)[/tex] and [tex]\( x = 2 \)[/tex]:
[tex]\[ \frac{-6 - (-2)}{2 - 1} = \frac{-4}{1} = -4 \][/tex]
2. Between [tex]\( x = 2 \)[/tex] and [tex]\( x = 3 \)[/tex]:
[tex]\[ \frac{-2 - (-6)}{3 - 2} = \frac{4}{1} = 4 \][/tex]
3. Between [tex]\( x = 3 \)[/tex] and [tex]\( x = 4 \)[/tex]:
[tex]\[ \frac{-6 - (-2)}{4 - 3} = \frac{-4}{1} = -4 \][/tex]
The rate of change is not constant; thus, Table 1 does not represent a linear function.
Table 2:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & -2 \\ \hline 2 & -5 \\ \hline 3 & -9 \\ \hline 4 & -14 \\ \hline \end{array} \][/tex]
Let's calculate the rate of change between each pair of points:
1. Between [tex]\( x = 1 \)[/tex] and [tex]\( x = 2 \)[/tex]:
[tex]\[ \frac{-5 - (-2)}{2 - 1} = \frac{-3}{1} = -3 \][/tex]
2. Between [tex]\( x = 2 \)[/tex] and [tex]\( x = 3 \)[/tex]:
[tex]\[ \frac{-9 - (-5)}{3 - 2} = \frac{-4}{1} = -4 \][/tex]
3. Between [tex]\( x = 3 \)[/tex] and [tex]\( x = 4 \)[/tex]:
[tex]\[ \frac{-14 - (-9)}{4 - 3} = \frac{-5}{1} = -5 \][/tex]
The rate of change is not constant; thus, Table 2 does not represent a linear function.
Table 3:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & -2 \\ \hline 2 & -10 \\ \hline 3 & -18 \\ \hline 4 & -26 \\ \hline \end{array} \][/tex]
Let's calculate the rate of change between each pair of points:
1. Between [tex]\( x = 1 \)[/tex] and [tex]\( x = 2 \)[/tex]:
[tex]\[ \frac{-10 - (-2)}{2 - 1} = \frac{-8}{1} = -8 \][/tex]
2. Between [tex]\( x = 2 \)[/tex] and [tex]\( x = 3 \)[/tex]:
[tex]\[ \frac{-18 - (-10)}{3 - 2} = \frac{-8}{1} = -8 \][/tex]
3. Between [tex]\( x = 3 \)[/tex] and [tex]\( x = 4 \)[/tex]:
[tex]\[ \frac{-26 - (-18)}{4 - 3} = \frac{-8}{1} = -8 \][/tex]
The rate of change is constant; thus, Table 3 represents a linear function.
Table 4:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & -2 \\ \hline 2 & -4 \\ \hline \end{array} \][/tex]
Let's calculate the rate of change between the two points:
1. Between [tex]\( x = 1 \)[/tex] and [tex]\( x = 2 \)[/tex]:
[tex]\[ \frac{-4 - (-2)}{2 - 1} = \frac{-2}{1} = -2 \][/tex]
With only two points, the rate of change is constant by definition; thus, Table 4 represents a linear function.
In conclusion, the tables that represent a linear function are Table 3 and Table 4.
Table 1:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & -2 \\ \hline 2 & -6 \\ \hline 3 & -2 \\ \hline 4 & -6 \\ \hline \end{array} \][/tex]
Let's calculate the rate of change between each pair of points:
1. Between [tex]\( x = 1 \)[/tex] and [tex]\( x = 2 \)[/tex]:
[tex]\[ \frac{-6 - (-2)}{2 - 1} = \frac{-4}{1} = -4 \][/tex]
2. Between [tex]\( x = 2 \)[/tex] and [tex]\( x = 3 \)[/tex]:
[tex]\[ \frac{-2 - (-6)}{3 - 2} = \frac{4}{1} = 4 \][/tex]
3. Between [tex]\( x = 3 \)[/tex] and [tex]\( x = 4 \)[/tex]:
[tex]\[ \frac{-6 - (-2)}{4 - 3} = \frac{-4}{1} = -4 \][/tex]
The rate of change is not constant; thus, Table 1 does not represent a linear function.
Table 2:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & -2 \\ \hline 2 & -5 \\ \hline 3 & -9 \\ \hline 4 & -14 \\ \hline \end{array} \][/tex]
Let's calculate the rate of change between each pair of points:
1. Between [tex]\( x = 1 \)[/tex] and [tex]\( x = 2 \)[/tex]:
[tex]\[ \frac{-5 - (-2)}{2 - 1} = \frac{-3}{1} = -3 \][/tex]
2. Between [tex]\( x = 2 \)[/tex] and [tex]\( x = 3 \)[/tex]:
[tex]\[ \frac{-9 - (-5)}{3 - 2} = \frac{-4}{1} = -4 \][/tex]
3. Between [tex]\( x = 3 \)[/tex] and [tex]\( x = 4 \)[/tex]:
[tex]\[ \frac{-14 - (-9)}{4 - 3} = \frac{-5}{1} = -5 \][/tex]
The rate of change is not constant; thus, Table 2 does not represent a linear function.
Table 3:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & -2 \\ \hline 2 & -10 \\ \hline 3 & -18 \\ \hline 4 & -26 \\ \hline \end{array} \][/tex]
Let's calculate the rate of change between each pair of points:
1. Between [tex]\( x = 1 \)[/tex] and [tex]\( x = 2 \)[/tex]:
[tex]\[ \frac{-10 - (-2)}{2 - 1} = \frac{-8}{1} = -8 \][/tex]
2. Between [tex]\( x = 2 \)[/tex] and [tex]\( x = 3 \)[/tex]:
[tex]\[ \frac{-18 - (-10)}{3 - 2} = \frac{-8}{1} = -8 \][/tex]
3. Between [tex]\( x = 3 \)[/tex] and [tex]\( x = 4 \)[/tex]:
[tex]\[ \frac{-26 - (-18)}{4 - 3} = \frac{-8}{1} = -8 \][/tex]
The rate of change is constant; thus, Table 3 represents a linear function.
Table 4:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & -2 \\ \hline 2 & -4 \\ \hline \end{array} \][/tex]
Let's calculate the rate of change between the two points:
1. Between [tex]\( x = 1 \)[/tex] and [tex]\( x = 2 \)[/tex]:
[tex]\[ \frac{-4 - (-2)}{2 - 1} = \frac{-2}{1} = -2 \][/tex]
With only two points, the rate of change is constant by definition; thus, Table 4 represents a linear function.
In conclusion, the tables that represent a linear function are Table 3 and Table 4.
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