Question

1. which table represents a linear function? a. \

$$\begin{tabular}{|c|c|} \\hline x & y \\\\ \\hline 1 & 1 \\\\ \\hline 2 & 3 \\\\ \\hline 3 & 1 \\\\ \\hline 4 & 3 \\\\ \\hline \\end{tabular}$$

b. \

$$\begin{tabular}{|c|c|} \\hline x & y \\\\ \\hline -2 & 1 \\\\ \\hline -1 & 2 \\\\ \\hline 0 & 3 \\\\ \\hline 1 & 4 \\\\ \\hline \\end{tabular}$$

c. \

$$\begin{tabular}{|c|c|} \\hline x & y \\\\ \\hline 9 & -8 \\\\ \\hline 8 & -4 \\\\ \\hline 7 & -2 \\\\ \\hline 6 & -1 \\\\ \\hline \\end{tabular}$$

d. \

$$\begin{tabular}{|c|c|} \\hline x & y \\\\ \\hline 2 & 1 \\\\ \\hline 3 & 2 \\\\ \\hline 4 & 4 \\\\ \\hline 5 & 5 \\\\ \\hline \\end{tabular}$$

Explanation:

Step1: Recall linear function property

A linear function has a constant rate of change (slope), meaning the difference in \( y \)-values (\( \Delta y \)) over the difference in \( x \)-values (\( \Delta x \)) is constant for consecutive \( x \)-values.

Step2: Analyze Option A

For \( x = 1 \) to \( x = 2 \): \( \Delta x = 2 - 1 = 1 \), \( \Delta y = 3 - 1 = 2 \).
For \( x = 2 \) to \( x = 3 \): \( \Delta x = 3 - 2 = 1 \), \( \Delta y = 1 - 3 = -2 \).
Slopes are \( 2 \) and \( -2 \) (not constant). So A is not linear.

Step3: Analyze Option B

For \( x = -2 \) to \( x = -1 \): \( \Delta x = -1 - (-2) = 1 \), \( \Delta y = 2 - 1 = 1 \).
For \( x = -1 \) to \( x = 0 \): \( \Delta x = 0 - (-1) = 1 \), \( \Delta y = 3 - 2 = 1 \).
For \( x = 0 \) to \( x = 1 \): \( \Delta x = 1 - 0 = 1 \), \( \Delta y = 4 - 3 = 1 \).
Slope \( \frac{\Delta y}{\Delta x} = 1 \) (constant). So B is linear.

Step4: Analyze Option C (for confirmation)

For \( x = 9 \) to \( x = 8 \): \( \Delta x = 8 - 9 = -1 \), \( \Delta y = -4 - (-8) = 4 \). Slope \( \frac{4}{-1} = -4 \).
For \( x = 8 \) to \( x = 7 \): \( \Delta x = 7 - 8 = -1 \), \( \Delta y = -2 - (-4) = 2 \). Slope \( \frac{2}{-1} = -2 \).
Slopes are \( -4 \) and \( -2 \) (not constant). So C is not linear.

Step5: Analyze Option D (for confirmation)

For \( x = 2 \) to \( x = 3 \): \( \Delta x = 3 - 2 = 1 \), \( \Delta y = 2 - 1 = 1 \). Slope \( 1 \).
For \( x = 3 \) to \( x = 4 \): \( \Delta x = 4 - 3 = 1 \), \( \Delta y = 4 - 2 = 2 \). Slope \( 2 \).
Slopes are \( 1 \) and \( 2 \) (not constant). So D is not linear.

Answer:

B. \(

$$\begin{array}{|c|c|} \hline x & y \\ \hline -2 & 1 \\ \hline -1 & 2 \\ \hline 0 & 3 \\ \hline 1 & 4 \\ \hline \end{array}$$

\)