Which of the following tables represents a linear function?

x	1	1	1	1	1
y	−3	−2	−1	0	1

x	−4	−2	0	2	4
y	4	2	0	2	4

x	−5	−3	−1	1	3
y	negative one half	1	2	7 over 2	5

x	−6	−4	−2	0	2
y	5	13 over 3	11 over 3	3	7 over 3

To determine which table represents a linear function, we need to check if the relationship between \( x \) and \( y \) can be expressed in the form \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept. This requires that the differences between consecutive \( y \)-values are constant when the differences between consecutive \( x \)-values are constant.

Let's analyze each table:

### Table 1
\[
\begin{array}{c|ccccc}
x & 1 & 1 & 1 & 1 & 1 \\
\hline
y & -3 & -2 & -1 & 0 & 1 \\
\end{array}
\]
In this table, \( x \) is constant, so this cannot be a function. This table does not represent a linear function.

### Table 2
\[
\begin{array}{c|ccccc}
x & -4 & -2 & 0 & 2 & 4 \\
\hline
y & 4 & 2 & 0 & 2 & 4 \\
\end{array}
\]

Calculate the differences:
- \( \Delta y = 2 - 4 = -2 \)
- \( \Delta y = 0 - 2 = -2 \)
- \( \Delta y = 2 - 0 = 2 \)
- \( \Delta y = 4 - 2 = 2 \)

The differences between \( y \)-values are not consistent (they alternate between -2 and 2), so this table does not represent a linear function.

### Table 3
\[
\begin{array}{c|ccccc}
x & -5 & -3 & -1 & 1 & 3 \\
\hline
y & -\frac{1}{2} & 1 & 2 & \frac{7}{2} & 5 \\
\end{array}
\]

Calculate the differences:
- \( \Delta y = 1 - \left(-\frac{1}{2}\right) = 1 + \frac{1}{2} = \frac{3}{2} \)
- \( \Delta y = 2 - 1 = 1 \)
- \( \Delta y = \frac{7}{2} - 2 = \frac{7}{2} - \frac{4}{2} = \frac{3}{2} \)
- \( \Delta y = 5 - \frac{7}{2} = \frac{10}{2} - \frac{7}{2} = \frac{3}{2} \)

The differences are \( \frac{3}{2} \), 1, \( \frac{3}{2} \), \( \frac{3}{2} \), so this table does not represent a linear function as the differences between \( y \)-values are not consistent.

### Table 4
\[
\begin{array}{c|ccccc}
x & -6 & -4 & -2 & 0 & 2 \\
\hline
y & 5 & \frac{13}{3} & \frac{11}{3} & 3 & \frac{7}{3} \\
\end{array}
\]

Calculate the differences:
- \( \Delta y = \frac{13}{3} - 5 = \frac{13}{3} - \frac{15}{3} = -\frac{2}{3} \)
- \( \Delta y = \frac{11}{3} - \frac{13}{3} = -\frac{2}{3} \)
- \( \Delta y = 3 - \frac{11}{3} = \frac{9}{3} - \frac{11}{3} = -\frac{2}{3} \)
- \( \Delta y = \frac{7}{3} - 3 = \frac{7}{3} - \frac{9}{3} = -\frac{2}{3} \)

The differences between \( y \)-values are consistent, all being \( -\frac{2}{3} \). Therefore, this table represents a linear function with a slope of \( -\frac{2}{3} \).

So, the table that represents a linear function is:

\[
\begin{array}{c|ccccc}
x & -6 & -4 & -2 & 0 & 2 \\
\hline
y & 5 & \frac{13}{3} & \frac{11}{3} & 3 & \frac{7}{3} \\
\end{array}
\]

Would you like more details or have any questions?

Here are 5 questions you might want to ask next:
1. How do you determine the slope of a linear function from a table?
2. Can you explain how to convert a sequence of points into the equation of a linear function?
3. How do you identify a non-linear function from a table of values?
4. What are some real-world examples of linear functions?
5. How can you graph a linear function given a table of values?

**Tip:** To check if a function is linear, ensure the differences between the \( y \)-values are consistent when the differences between the \( x \)-values are consistent.

Learn how to identify linear functions from tables of values. Analyze different examples to understand the consistency of differences between x and y values. Suitable for students in Grades 7-9.

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