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Sagot :
To determine which table represents a function, we need to check if each input [tex]\(x\)[/tex] has exactly one corresponding output [tex]\(y\)[/tex]. A function assigns only one output to each input.
Let's evaluate each table one by one:
1. Table [tex]\(W\) \[ \begin{array}{c|c} x & y \\ \hline -1 & 15 \\ -2 & 6 \\ 0 & 15 \\ 4 & 6 \\ \end{array} \] - \(x = -1 \rightarrow y = 15\)[/tex]
- [tex]\(x = -2 \rightarrow y = 6\)[/tex]
- [tex]\(x = 0 \rightarrow y = 15\)[/tex]
- [tex]\(x = 4 \rightarrow y = 6\)[/tex]
In this table, each [tex]\(x\)[/tex] has a unique [tex]\(y\)[/tex] value. This represents a function.
2. Table [tex]\(X\) \[ \begin{array}{c|c} x & y \\ \hline 0 & 4 \\ 6 & 15 \\ 0 & 6 \\ -2 & -1 \\ \end{array} \] - \(x = 0 \rightarrow y = 4\)[/tex]
- [tex]\(x = 6 \rightarrow y = 15\)[/tex]
- [tex]\(x = 0 \rightarrow y = 6\)[/tex]
- [tex]\(x = -2 \rightarrow y = -1\)[/tex]
In this table, [tex]\(x = 0\)[/tex] corresponds to two different [tex]\(y\)[/tex] values (4 and 6). This does not represent a function.
3. Table [tex]\(Y\) \[ \begin{array}{c|c} x & y \\ \hline 4 & 0 \\ -1 & -2 \\ 6 & -2 \\ 6 & 15 \\ \end{array} \] - \(x = 4 \rightarrow y = 0\)[/tex]
- [tex]\(x = -1 \rightarrow y = -2\)[/tex]
- [tex]\(x = 6 \rightarrow y = -2\)[/tex]
- [tex]\(x = 6 \rightarrow y = 15\)[/tex]
In this table, [tex]\(x = 6\)[/tex] corresponds to two different [tex]\(y\)[/tex] values (-2 and 15). This does not represent a function.
4. Table [tex]\(Z\) \[ \begin{array}{c|c} x & y \\ \hline 6 & -2 \\ 4 & 0 \\ 15 & -1 \\ 4 & 3 \\ \end{array} \] - \(x = 6 \rightarrow y = -2\)[/tex]
- [tex]\(x = 4 \rightarrow y = 0\)[/tex]
- [tex]\(x = 15 \rightarrow y = -1\)[/tex]
- [tex]\(x = 4 \rightarrow y = 3\)[/tex]
In this table, [tex]\(x = 4\)[/tex] corresponds to two different [tex]\(y\)[/tex] values (0 and 3). This does not represent a function.
From the evaluation, only Table [tex]\(W\)[/tex] meets the requirement that each input [tex]\(x\)[/tex] has exactly one corresponding output [tex]\(y\)[/tex].
Therefore, the correct answer is:
A. [tex]\(W\)[/tex]
Let's evaluate each table one by one:
1. Table [tex]\(W\) \[ \begin{array}{c|c} x & y \\ \hline -1 & 15 \\ -2 & 6 \\ 0 & 15 \\ 4 & 6 \\ \end{array} \] - \(x = -1 \rightarrow y = 15\)[/tex]
- [tex]\(x = -2 \rightarrow y = 6\)[/tex]
- [tex]\(x = 0 \rightarrow y = 15\)[/tex]
- [tex]\(x = 4 \rightarrow y = 6\)[/tex]
In this table, each [tex]\(x\)[/tex] has a unique [tex]\(y\)[/tex] value. This represents a function.
2. Table [tex]\(X\) \[ \begin{array}{c|c} x & y \\ \hline 0 & 4 \\ 6 & 15 \\ 0 & 6 \\ -2 & -1 \\ \end{array} \] - \(x = 0 \rightarrow y = 4\)[/tex]
- [tex]\(x = 6 \rightarrow y = 15\)[/tex]
- [tex]\(x = 0 \rightarrow y = 6\)[/tex]
- [tex]\(x = -2 \rightarrow y = -1\)[/tex]
In this table, [tex]\(x = 0\)[/tex] corresponds to two different [tex]\(y\)[/tex] values (4 and 6). This does not represent a function.
3. Table [tex]\(Y\) \[ \begin{array}{c|c} x & y \\ \hline 4 & 0 \\ -1 & -2 \\ 6 & -2 \\ 6 & 15 \\ \end{array} \] - \(x = 4 \rightarrow y = 0\)[/tex]
- [tex]\(x = -1 \rightarrow y = -2\)[/tex]
- [tex]\(x = 6 \rightarrow y = -2\)[/tex]
- [tex]\(x = 6 \rightarrow y = 15\)[/tex]
In this table, [tex]\(x = 6\)[/tex] corresponds to two different [tex]\(y\)[/tex] values (-2 and 15). This does not represent a function.
4. Table [tex]\(Z\) \[ \begin{array}{c|c} x & y \\ \hline 6 & -2 \\ 4 & 0 \\ 15 & -1 \\ 4 & 3 \\ \end{array} \] - \(x = 6 \rightarrow y = -2\)[/tex]
- [tex]\(x = 4 \rightarrow y = 0\)[/tex]
- [tex]\(x = 15 \rightarrow y = -1\)[/tex]
- [tex]\(x = 4 \rightarrow y = 3\)[/tex]
In this table, [tex]\(x = 4\)[/tex] corresponds to two different [tex]\(y\)[/tex] values (0 and 3). This does not represent a function.
From the evaluation, only Table [tex]\(W\)[/tex] meets the requirement that each input [tex]\(x\)[/tex] has exactly one corresponding output [tex]\(y\)[/tex].
Therefore, the correct answer is:
A. [tex]\(W\)[/tex]
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