Answer :
To determine which linear function represents a slope of [tex]\(\frac{1}{4}\)[/tex], we need to examine the slopes derived from the two data sets provided.
First, let's understand what slope represents. The slope [tex]\(m\)[/tex] between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] on a straight line is calculated using the formula:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
The two data sets given are:
1.
[tex]\[ \begin{tabular}{|c|c|} \hline $x$ & $y$ \\ \hline 3 & -11 \\ \hline 6 & 1 \\ \hline 9 & 13 \\ \hline 12 & 25 \\ \hline \end{tabular} \][/tex]
2.
[tex]\[ \begin{tabular}{|c|c|} \hline $x$ & $y$ \\ \hline -5 & 32 \\ \hline -1 & 24 \\ \hline 3 & 16 \\ \hline 7 & 8 \\ \hline \end{tabular} \][/tex]
Let's calculate the slopes for these points in both data sets.
For the first data set:
1. Between [tex]\((3, -11)\)[/tex] and [tex]\((6, 1)\)[/tex]:
[tex]\[ m = \frac{1 - (-11)}{6 - 3} = \frac{12}{3} = 4 \][/tex]
2. Between [tex]\((6, 1)\)[/tex] and [tex]\((9, 13)\)[/tex]:
[tex]\[ m = \frac{13 - 1}{9 - 6} = \frac{12}{3} = 4 \][/tex]
3. Between [tex]\((9, 13)\)[/tex] and [tex]\((12, 25)\)[/tex]:
[tex]\[ m = \frac{25 - 13}{12 - 9} = \frac{12}{3} = 4 \][/tex]
So, the slopes between consecutive points in the first data set are consistently 4.
For the second data set:
1. Between [tex]\((-5, 32)\)[/tex] and [tex]\((-1, 24)\)[/tex]:
[tex]\[ m = \frac{24 - 32}{-1 - (-5)} = \frac{-8}{4} = -2 \][/tex]
2. Between [tex]\((-1, 24)\)[/tex] and [tex]\((3, 16)\)[/tex]:
[tex]\[ m = \frac{16 - 24}{3 - (-1)} = \frac{-8}{4} = -2 \][/tex]
3. Between [tex]\((3, 16)\)[/tex] and [tex]\((7, 8)\)[/tex]:
[tex]\[ m = \frac{8 - 16}{7 - 3} = \frac{-8}{4} = -2 \][/tex]
So, the slopes between consecutive points in the second data set are consistently -2.
Comparing the slopes:
- The first data set has slopes of 4.
- The second data set has slopes of -2.
Neither of these consistently represents a slope of [tex]\(\frac{1}{4}\)[/tex]. There is no linear function in these data sets that has a slope of [tex]\(\frac{1}{4}\)[/tex]. The slopes are either 4 or -2, not [tex]\(\frac{1}{4}\)[/tex].
First, let's understand what slope represents. The slope [tex]\(m\)[/tex] between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] on a straight line is calculated using the formula:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
The two data sets given are:
1.
[tex]\[ \begin{tabular}{|c|c|} \hline $x$ & $y$ \\ \hline 3 & -11 \\ \hline 6 & 1 \\ \hline 9 & 13 \\ \hline 12 & 25 \\ \hline \end{tabular} \][/tex]
2.
[tex]\[ \begin{tabular}{|c|c|} \hline $x$ & $y$ \\ \hline -5 & 32 \\ \hline -1 & 24 \\ \hline 3 & 16 \\ \hline 7 & 8 \\ \hline \end{tabular} \][/tex]
Let's calculate the slopes for these points in both data sets.
For the first data set:
1. Between [tex]\((3, -11)\)[/tex] and [tex]\((6, 1)\)[/tex]:
[tex]\[ m = \frac{1 - (-11)}{6 - 3} = \frac{12}{3} = 4 \][/tex]
2. Between [tex]\((6, 1)\)[/tex] and [tex]\((9, 13)\)[/tex]:
[tex]\[ m = \frac{13 - 1}{9 - 6} = \frac{12}{3} = 4 \][/tex]
3. Between [tex]\((9, 13)\)[/tex] and [tex]\((12, 25)\)[/tex]:
[tex]\[ m = \frac{25 - 13}{12 - 9} = \frac{12}{3} = 4 \][/tex]
So, the slopes between consecutive points in the first data set are consistently 4.
For the second data set:
1. Between [tex]\((-5, 32)\)[/tex] and [tex]\((-1, 24)\)[/tex]:
[tex]\[ m = \frac{24 - 32}{-1 - (-5)} = \frac{-8}{4} = -2 \][/tex]
2. Between [tex]\((-1, 24)\)[/tex] and [tex]\((3, 16)\)[/tex]:
[tex]\[ m = \frac{16 - 24}{3 - (-1)} = \frac{-8}{4} = -2 \][/tex]
3. Between [tex]\((3, 16)\)[/tex] and [tex]\((7, 8)\)[/tex]:
[tex]\[ m = \frac{8 - 16}{7 - 3} = \frac{-8}{4} = -2 \][/tex]
So, the slopes between consecutive points in the second data set are consistently -2.
Comparing the slopes:
- The first data set has slopes of 4.
- The second data set has slopes of -2.
Neither of these consistently represents a slope of [tex]\(\frac{1}{4}\)[/tex]. There is no linear function in these data sets that has a slope of [tex]\(\frac{1}{4}\)[/tex]. The slopes are either 4 or -2, not [tex]\(\frac{1}{4}\)[/tex].
