To determine which of these tables represents a linear function, we need to check if the differences between each pair of consecutive points exhibit a constant rate of change, otherwise known as the slope. For a function to be linear, the slope between every pair of points should be the same.
Let's examine each table one by one:
1. First Table:
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
1 & 3 \\
\hline
2 & 6 \\
\hline
3 & 12 \\
\hline
4 & 24 \\
\hline
\end{array}
\][/tex]
- Slope from [tex]\((1,3)\)[/tex] to [tex]\((2,6)\)[/tex]: [tex]\(\frac{6-3}{2-1} = 3\)[/tex]
- Slope from [tex]\((2,6)\)[/tex] to [tex]\((3,12)\)[/tex]: [tex]\(\frac{12-6}{3-2} = 6\)[/tex]
- Slope from [tex]\((3,12)\)[/tex] to [tex]\((4,24)\)[/tex]: [tex]\(\frac{24-12}{4-3} = 12\)[/tex]
Since the slopes are not constant, this table does not represent a linear function.
2. Second Table:
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
1 & 2 \\
\hline
2 & 5 \\
\hline
3 & 9 \\
\hline
4 & 14 \\
\hline
\end{array}
\][/tex]
- Slope from [tex]\((1,2)\)[/tex] to [tex]\((2,5)\)[/tex]: [tex]\(\frac{5-2}{2-1} = 3\)[/tex]
- Slope from [tex]\((2,5)\)[/tex] to [tex]\((3,9)\)[/tex]: [tex]\(\frac{9-5}{3-2} = 4\)[/tex]
- Slope from [tex]\((3,9)\)[/tex] to [tex]\((4,14)\)[/tex]: [tex]\(\frac{14-9}{4-3} = 5\)[/tex]
Since the slopes are not constant, this table does not represent a linear function.
3. Third Table:
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
1 & -3 \\
\hline
2 & -5 \\
\hline
3 & -7 \\
\hline
4 & -9 \\
\hline
\end{array}
\][/tex]
- Slope from [tex]\((1, -3)\)[/tex] to [tex]\((2, -5)\)[/tex]: [tex]\(\frac{-5 + 3}{2 - 1} = -2\)[/tex]
- Slope from [tex]\((2, -5)\)[/tex] to [tex]\((3, -7)\)[/tex]: [tex]\(\frac{-7 + 5}{3 - 2} = -2\)[/tex]
- Slope from [tex]\((3, -7)\)[/tex] to [tex]\((4, -9)\)[/tex]: [tex]\(\frac{-9 + 7}{4 - 3} = -2\)[/tex]
The slopes are all the same, [tex]\(-2\)[/tex]. Thus, this table represents a linear function.
4. Fourth Table:
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
1 & -2 \\
\hline
2 & -4 \\
\hline
3 & -2 \\
\hline
4 & 0 \\
\hline
\end{array}
\][/tex]
- Slope from [tex]\((1, -2)\)[/tex] to [tex]\((2, -4)\)[/tex]: [tex]\(\frac{-4 +2}{2-1} = -2\)[/tex]
- Slope from [tex]\((2, -4)\)[/tex] to [tex]\((3, -2)\)[/tex]: [tex]\(\frac{-2 + 4}{3 - 2} = 2\)[/tex]
- Slope from [tex]\((3, -2)\)[/tex] to [tex]\((4, 0)\)[/tex]: [tex]\(\frac{0 + 2}{4-3} = 2\)[/tex]
Since the slopes are not constant, this table does not represent a linear function.
Therefore, the third table is the one that represents a linear function.
