Which table of values represents a linear function?
A
xx	yy
minus, 6−6	minus, 5−5
minus, 2−2	minus, 2−2
22	11
55	44
B
xx	yy
11	minus, 5−5
22	minus, 3−3
33	minus, 1−1
66	44
C
xx	yy
minus, 5−5	minus, 7−7
11	minus, 3−3
44	minus, 1−1
77	11
D
xx	yy
minus, 2−2	77
11	66
44	55
77	33
Answer

To determine which table of values represents a linear function, we need to check whether the change in the \( y \) values is consistent with the change in the \( x \) values. For a linear function, the difference between consecutive \( y \) values divided by the difference between consecutive \( x \) values (i.e., the slope) should be constant.

Let's examine each table:

**Table A:**

\[
\begin{array}{c|c}
x & y \\
\hline
-6 & -5 \\
-2 & -2 \\
2 & 1 \\
5 & 4 \\
\end{array}
\]

**Changes:**

\[
\text{slope between } (-6, -5) \text{ and } (-2, -2): \frac{-2 - (-5)}{-2 - (-6)} = \frac{3}{4}
\]

\[
\text{slope between } (-2, -2) \text{ and } (2, 1): \frac{1 - (-2)}{2 - (-2)} = \frac{3}{4}
\]

\[
\text{slope between } (2, 1) \text{ and } (5, 4): \frac{4 - 1}{5 - 2} = \frac{3}{3} = 1
\]

The slopes are not consistent, so **Table A is not linear**.

---

**Table B:**

\[
\begin{array}{c|c}
x & y \\
\hline
1 & -5 \\
2 & -3 \\
3 & -1 \\
6 & 4 \\
\end{array}
\]

**Changes:**

\[
\text{slope between } (1, -5) \text{ and } (2, -3): \frac{-3 - (-5)}{2 - 1} = \frac{2}{1} = 2
\]

\[
\text{slope between } (2, -3) \text{ and } (3, -1): \frac{-1 - (-3)}{3 - 2} = \frac{2}{1} = 2
\]

\[
\text{slope between } (3, -1) \text{ and } (6, 4): \frac{4 - (-1)}{6 - 3} = \frac{5}{3}
\]

The slopes are not consistent, so **Table B is not linear**.

---

**Table C:**

\[
\begin{array}{c|c}
x & y \\
\hline
-5 & -7 \\
1 & -3 \\
4 & -1 \\
7 & 1 \\
\end{array}
\]

**Changes:**

\[
\text{slope between } (-5, -7) \text{ and } (1, -3): \frac{-3 - (-7)}{1 - (-5)} = \frac{4}{6} = \frac{2}{3}
\]

\[
\text{slope between } (1, -3) \text{ and } (4, -1): \frac{-1 - (-3)}{4 - 1} = \frac{2}{3}
\]

\[
\text{slope between } (4, -1) \text{ and } (7, 1): \frac{1 - (-1)}{7 - 4} = \frac{2}{3}
\]

The slopes are consistent, so **Table C is linear**.

---

**Table D:**

\[
\begin{array}{c|c}
x & y \\
\hline
-2 & 7 \\
1 & 6 \\
4 & 5 \\
7 & 3 \\
\end{array}
\]

**Changes:**

\[
\text{slope between } (-2, 7) \text{ and } (1, 6): \frac{6 - 7}{1 - (-2)} = \frac{-1}{3} = -\frac{1}{3}
\]

\[
\text{slope between } (1, 6) \text{ and } (4, 5): \frac{5 - 6}{4 - 1} = \frac{-1}{3}
\]

\[
\text{slope between } (4, 5) \text{ and } (7, 3): \frac{3 - 5}{7 - 4} = \frac{-2}{3}
\]

The slopes are not consistent, so **Table D is not linear**.

---

**Conclusion:** **Table C** represents a linear function.

Would you like more details or have any questions?

---

**Related Questions:**

1. How do you determine the slope of a linear function from a table of values?
2. What is the importance of consistency in slope for identifying linear functions?
3. Can a function be linear if its slope is zero?
4. How would you graph a linear function given a table of values?
5. What are real-life examples of linear functions?

**Tip:** Consistent slopes between points in a table indicate that the function is linear.

Discover how to identify linear functions from tables of values with examples and explanations. Learn about slope consistency and its importance in determining linearity. Suitable for Grades 7-9 students.

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