Izanxeeidn
Answered

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Which table represents a linear function?

[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
0 & 1 \\
\hline
1 & 2 \\
\hline
2 & 4 \\
\hline
3 & 8 \\
\hline
\end{array}
\][/tex]

[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
0 & 0 \\
\hline
1 & 1 \\
\hline
2 & 3 \\
\hline
3 & 6 \\
\hline
\end{array}
\][/tex]

[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
0 & 0 \\
\hline
1 & 1 \\
\hline
2 & 0 \\
\hline
3 & 1 \\
\hline
\end{array}
\][/tex]

[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
0 & 1 \\
\hline
1 & 3 \\
\hline
2 & 5 \\
\hline
3 & 7 \\
\hline
\end{array}
\][/tex]

Sagot :

GinnyAnsweridn
Let's determine which of the given tables represents a linear function. A linear function has a constant rate of change, or slope, between any two points.

Table 1:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & 1 \\ \hline 1 & 2 \\ \hline 2 & 4 \\ \hline 3 & 8 \\ \hline \end{array} \][/tex]
To be linear:
- The change in [tex]\( y \)[/tex] divided by the change in [tex]\( x \)[/tex] should be constant.
- From [tex]\( x = 0 \)[/tex] to [tex]\( x = 1 \)[/tex]: [tex]\(\frac{2-1}{1-0} = 1\)[/tex]
- From [tex]\( x = 1 \)[/tex] to [tex]\( x = 2 \)[/tex]: [tex]\(\frac{4-2}{2-1} = 2\)[/tex]
- From [tex]\( x = 2 \)[/tex] to [tex]\( x = 3 \)[/tex]: [tex]\(\frac{8-4}{3-2} = 4\)[/tex]

The slope is not constant (changes), so Table 1 does not represent a linear function.

Table 2:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & 0 \\ \hline 1 & 1 \\ \hline 2 & 3 \\ \hline 3 & 6 \\ \hline \end{array} \][/tex]
- From [tex]\( x = 0 \)[/tex] to [tex]\( x = 1 \)[/tex]: [tex]\(\frac{1-0}{1-0} = 1\)[/tex]
- From [tex]\( x = 1 \)[/tex] to [tex]\( x = 2 \)[/tex]: [tex]\(\frac{3-1}{2-1} = 2\)[/tex]
- From [tex]\( x = 2 \)[/tex] to [tex]\( x = 3 \)[/tex]: [tex]\(\frac{6-3}{3-2} = 3\)[/tex]

The slope is not constant, so Table 2 does not represent a linear function.

Table 3:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & 0 \\ \hline 1 & 1 \\ \hline 2 & 0 \\ \hline 3 & 1 \\ \hline \end{array} \][/tex]
- From [tex]\( x = 0 \)[/tex] to [tex]\( x = 1 \)[/tex]: [tex]\(\frac{1-0}{1-0} = 1\)[/tex]
- From [tex]\( x = 1 \)[/tex] to [tex]\( x = 2 \)[/tex]: [tex]\(\frac{0-1}{2-1} = -1\)[/tex]
- From [tex]\( x = 2 \)[/tex] to [tex]\( x = 3 \)[/tex]: [tex]\(\frac{1-0}{3-2} = 1\)[/tex]

The slope is not constant, so Table 3 does not represent a linear function.

Table 4:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & 1 \\ \hline 1 & 3 \\ \hline 2 & 5 \\ \hline 3 & 7 \\ \hline \end{array} \][/tex]
- From [tex]\( x = 0 \)[/tex] to [tex]\( x = 1 \)[/tex]: [tex]\(\frac{3-1}{1-0} = 2\)[/tex]
- From [tex]\( x = 1 \)[/tex] to [tex]\( x = 2 \)[/tex]: [tex]\(\frac{5-3}{2-1} = 2\)[/tex]
- From [tex]\( x = 2 \)[/tex] to [tex]\( x = 3 \)[/tex]: [tex]\(\frac{7-5}{3-2} = 2\)[/tex]

The slope is constant and equals 2, so Table 4 does represent a linear function.

Thus, the table that represents a linear function is:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & 1 \\ \hline 1 & 3 \\ \hline 2 & 5 \\ \hline 3 & 7 \\ \hline \end{array} \][/tex]
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