Answer :
To determine which of the given tables represents a linear function, let's examine each table individually and check if the relationship between [tex]\( x \)[/tex] and [tex]\( y \)[/tex] can be expressed in the form [tex]\( y = mx + b \)[/tex], for some constants [tex]\( m \)[/tex] and [tex]\( b \)[/tex].
### Table 1
[tex]\[ \begin{array}{|l|l|l|l|l|l|} \hline x & -2 & -1 & 0 & 1 & 2 \\ \hline y & 4 & 1 & -2 & -5 & -8 \\ \hline \end{array} \][/tex]
Calculate the differences in [tex]\( y \)[/tex] and [tex]\( x \)[/tex]:
[tex]\[ \begin{align*} \Delta y_1 &= y(-1) - y(-2) = 1 - 4 = -3, & \Delta x_1 &= -1 - (-2) = 1 \\ \Delta y_2 &= y(0) - y(-1) = -2 - 1 = -3, & \Delta x_2 &= 0 - (-1) = 1 \\ \Delta y_3 &= y(1) - y(0) = -5 - (-2) = -3, & \Delta x_3 &= 1 - 0 = 1 \\ \Delta y_4 &= y(2) - y(1) = -8 - (-5) = -3, & \Delta x_4 &= 2 - 1 = 1 \\ \end{align*} \][/tex]
Here, [tex]\(\Delta y / \Delta x\)[/tex] is constant:
[tex]\[ \frac{\Delta y}{\Delta x} = \frac{-3}{1} = -3 \][/tex]
Since the ratio [tex]\(\frac{\Delta y}{\Delta x}\)[/tex] is constant, the table represents a linear function with slope [tex]\( m = -3 \)[/tex].
### Table 2
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline x & -2 & -1 & 0 & 1 & 2 \\ \hline y & 4 & 1 & 0 & 1 & 4 \\ \hline \end{array} \][/tex]
Calculate the differences:
[tex]\[ \begin{align*} \Delta y_1 &= 1 - 4 = -3, & \Delta x_1 &= 1 \\ \Delta y_2 &= 0 - 1 = -1, & \Delta x_2 &= 1 \\ \Delta y_3 &= 1 - 0 = 1, & \Delta x_3 &= 1 \\ \Delta y_4 &= 4 - 1 = 3, & \Delta x_4 &= 1 \\ \end{align*} \][/tex]
Here, [tex]\(\Delta y / \Delta x\)[/tex] is not constant. The ratio changes:
[tex]\[ \frac{\Delta y_1}{\Delta x_1} = -3, \quad \frac{\Delta y_2}{\Delta x_2} = -1, \quad \frac{\Delta y_3}{\Delta x_3} = 1, \quad \frac{\Delta y_4}{\Delta x_4} = 3 \][/tex]
This table does not represent a linear function.
### Table 3
[tex]\[ \begin{array}{|l|l|l|l|l|l|} \hline x & 2 & 2 & 0 & 2 & 2 \\ \hline y & -2 & -1 & 0 & 1 & 2 \\ \hline \end{array} \][/tex]
This table contains repeated x-values (2) with different y-values, which violates the definition of a function. Therefore, it cannot represent a linear function.
### Table 4
[tex]\[ \begin{array}{|l|l|l|l|l|l|} \hline x & 0 & 1 & 2 & 3 & 4 \\ \hline y & -2 & 1 & 0 & 1 & -2 \\ \hline \end{array} \][/tex]
Calculate the differences:
[tex]\[ \begin{align*} \Delta y_1 &= 1 - (-2) = 3, & \Delta x_1 &= 1 \\ \Delta y_2 &= 0 - 1 = -1, & \Delta x_2 &= 1 \\ \Delta y_3 &= 1 - 0 = 1, & \Delta x_3 &= 1 \\ \Delta y_4 &= -2 - 1 = -3, & \Delta x_4 &= 1 \\ \end{align*} \][/tex]
Here, [tex]\(\Delta y / \Delta x\)[/tex] is not constant. The ratio changes:
[tex]\[ \frac{\Delta y_1}{\Delta x_1} = 3, \quad \frac{\Delta y_2}{\Delta x_2} = -1, \quad \frac{\Delta y_3}{\Delta x_3} = 1, \quad \frac{\Delta y_4}{\Delta x_4} = -3 \][/tex]
This table also does not represent a linear function.
### Conclusion
Only Table 1 represents a linear function.
### Table 1
[tex]\[ \begin{array}{|l|l|l|l|l|l|} \hline x & -2 & -1 & 0 & 1 & 2 \\ \hline y & 4 & 1 & -2 & -5 & -8 \\ \hline \end{array} \][/tex]
Calculate the differences in [tex]\( y \)[/tex] and [tex]\( x \)[/tex]:
[tex]\[ \begin{align*} \Delta y_1 &= y(-1) - y(-2) = 1 - 4 = -3, & \Delta x_1 &= -1 - (-2) = 1 \\ \Delta y_2 &= y(0) - y(-1) = -2 - 1 = -3, & \Delta x_2 &= 0 - (-1) = 1 \\ \Delta y_3 &= y(1) - y(0) = -5 - (-2) = -3, & \Delta x_3 &= 1 - 0 = 1 \\ \Delta y_4 &= y(2) - y(1) = -8 - (-5) = -3, & \Delta x_4 &= 2 - 1 = 1 \\ \end{align*} \][/tex]
Here, [tex]\(\Delta y / \Delta x\)[/tex] is constant:
[tex]\[ \frac{\Delta y}{\Delta x} = \frac{-3}{1} = -3 \][/tex]
Since the ratio [tex]\(\frac{\Delta y}{\Delta x}\)[/tex] is constant, the table represents a linear function with slope [tex]\( m = -3 \)[/tex].
### Table 2
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline x & -2 & -1 & 0 & 1 & 2 \\ \hline y & 4 & 1 & 0 & 1 & 4 \\ \hline \end{array} \][/tex]
Calculate the differences:
[tex]\[ \begin{align*} \Delta y_1 &= 1 - 4 = -3, & \Delta x_1 &= 1 \\ \Delta y_2 &= 0 - 1 = -1, & \Delta x_2 &= 1 \\ \Delta y_3 &= 1 - 0 = 1, & \Delta x_3 &= 1 \\ \Delta y_4 &= 4 - 1 = 3, & \Delta x_4 &= 1 \\ \end{align*} \][/tex]
Here, [tex]\(\Delta y / \Delta x\)[/tex] is not constant. The ratio changes:
[tex]\[ \frac{\Delta y_1}{\Delta x_1} = -3, \quad \frac{\Delta y_2}{\Delta x_2} = -1, \quad \frac{\Delta y_3}{\Delta x_3} = 1, \quad \frac{\Delta y_4}{\Delta x_4} = 3 \][/tex]
This table does not represent a linear function.
### Table 3
[tex]\[ \begin{array}{|l|l|l|l|l|l|} \hline x & 2 & 2 & 0 & 2 & 2 \\ \hline y & -2 & -1 & 0 & 1 & 2 \\ \hline \end{array} \][/tex]
This table contains repeated x-values (2) with different y-values, which violates the definition of a function. Therefore, it cannot represent a linear function.
### Table 4
[tex]\[ \begin{array}{|l|l|l|l|l|l|} \hline x & 0 & 1 & 2 & 3 & 4 \\ \hline y & -2 & 1 & 0 & 1 & -2 \\ \hline \end{array} \][/tex]
Calculate the differences:
[tex]\[ \begin{align*} \Delta y_1 &= 1 - (-2) = 3, & \Delta x_1 &= 1 \\ \Delta y_2 &= 0 - 1 = -1, & \Delta x_2 &= 1 \\ \Delta y_3 &= 1 - 0 = 1, & \Delta x_3 &= 1 \\ \Delta y_4 &= -2 - 1 = -3, & \Delta x_4 &= 1 \\ \end{align*} \][/tex]
Here, [tex]\(\Delta y / \Delta x\)[/tex] is not constant. The ratio changes:
[tex]\[ \frac{\Delta y_1}{\Delta x_1} = 3, \quad \frac{\Delta y_2}{\Delta x_2} = -1, \quad \frac{\Delta y_3}{\Delta x_3} = 1, \quad \frac{\Delta y_4}{\Delta x_4} = -3 \][/tex]
This table also does not represent a linear function.
### Conclusion
Only Table 1 represents a linear function.
