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Sagot :
To determine which table represents a function, we need to ensure that for every [tex]\( x \)[/tex]-value, there is exactly one corresponding [tex]\( y \)[/tex]-value. This means that, in a function, an [tex]\( x \)[/tex]-value cannot map to multiple [tex]\( y \)[/tex]-values. Let's examine each table:
### Table 1:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -3 & -1 \\ \hline 0 & 0 \\ \hline -2 & -1 \\ \hline 8 & 1 \\ \hline \end{array} \][/tex]
- For [tex]\( x = -3 \)[/tex], [tex]\( y = -1 \)[/tex]
- For [tex]\( x = 0 \)[/tex], [tex]\( y = 0 \)[/tex]
- For [tex]\( x = -2 \)[/tex], [tex]\( y = -1 \)[/tex]
- For [tex]\( x = 8 \)[/tex], [tex]\( y = 1 \)[/tex]
Each [tex]\( x \)[/tex]-value maps to a unique [tex]\( y \)[/tex]-value.
### Table 2:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -5 & -5 \\ \hline 0 & 0 \\ \hline -5 & 5 \\ \hline 6 & -6 \\ \hline \end{array} \][/tex]
- For [tex]\( x = -5 \)[/tex], [tex]\( y = -5 \)[/tex]
- For [tex]\( x = 0 \)[/tex], [tex]\( y = 0 \)[/tex]
- For [tex]\( x = -5 \)[/tex], [tex]\( y = 5 \)[/tex] (conflict with previous [tex]\(-5\)[/tex])
The [tex]\( x \)[/tex]-value [tex]\(-5\)[/tex] maps to both [tex]\(-5\)[/tex] and [tex]\(5\)[/tex], which means it does not represent a function.
### Table 3:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -4 & 8 \\ \hline -2 & 2 \\ \hline -2 & 4 \\ \hline 0 & 2 \\ \hline \end{array} \][/tex]
- For [tex]\( x = -4 \)[/tex], [tex]\( y = 8 \)[/tex]
- For [tex]\( x = -2 \)[/tex], [tex]\( y = 2 \)[/tex]
- For [tex]\( x = -2 \)[/tex], [tex]\( y = 4 \)[/tex] (conflict with previous [tex]\(-2\)[/tex])
- For [tex]\( x = 0 \)[/tex], [tex]\( y = 2 \)[/tex]
The [tex]\( x \)[/tex]-value [tex]\(-2\)[/tex] maps to both [tex]\(2\)[/tex] and [tex]\(4\)[/tex], which means it does not represent a function.
### Table 4:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -4 & 2 \\ \hline 3 & 5 \\ \hline 1 & 3 \\ \hline -4 & 0 \\ \hline \end{array} \][/tex]
- For [tex]\( x = -4 \)[/tex], [tex]\( y = 2 \)[/tex]
- For [tex]\( x = 3 \)[/tex], [tex]\( y = 5 \)[/tex]
- For [tex]\( x = 1 \)[/tex], [tex]\( y = 3 \)[/tex]
- For [tex]\( x = -4 \)[/tex], [tex]\( y = 0 \)[/tex] (conflict with previous [tex]\(-4\)[/tex])
The [tex]\( x \)[/tex]-value [tex]\(-4\)[/tex] maps to both [tex]\(2\)[/tex] and [tex]\(0\)[/tex], which means it does not represent a function.
### Conclusion:
Only the first table satisfies the condition of a function, where each [tex]\( x \)[/tex]-value corresponds to exactly one [tex]\( y \)[/tex]-value.
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -3 & -1 \\ \hline 0 & 0 \\ \hline -2 & -1 \\ \hline 8 & 1 \\ \hline \end{array} \][/tex]
Therefore, Table 1 represents a function.
### Table 1:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -3 & -1 \\ \hline 0 & 0 \\ \hline -2 & -1 \\ \hline 8 & 1 \\ \hline \end{array} \][/tex]
- For [tex]\( x = -3 \)[/tex], [tex]\( y = -1 \)[/tex]
- For [tex]\( x = 0 \)[/tex], [tex]\( y = 0 \)[/tex]
- For [tex]\( x = -2 \)[/tex], [tex]\( y = -1 \)[/tex]
- For [tex]\( x = 8 \)[/tex], [tex]\( y = 1 \)[/tex]
Each [tex]\( x \)[/tex]-value maps to a unique [tex]\( y \)[/tex]-value.
### Table 2:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -5 & -5 \\ \hline 0 & 0 \\ \hline -5 & 5 \\ \hline 6 & -6 \\ \hline \end{array} \][/tex]
- For [tex]\( x = -5 \)[/tex], [tex]\( y = -5 \)[/tex]
- For [tex]\( x = 0 \)[/tex], [tex]\( y = 0 \)[/tex]
- For [tex]\( x = -5 \)[/tex], [tex]\( y = 5 \)[/tex] (conflict with previous [tex]\(-5\)[/tex])
The [tex]\( x \)[/tex]-value [tex]\(-5\)[/tex] maps to both [tex]\(-5\)[/tex] and [tex]\(5\)[/tex], which means it does not represent a function.
### Table 3:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -4 & 8 \\ \hline -2 & 2 \\ \hline -2 & 4 \\ \hline 0 & 2 \\ \hline \end{array} \][/tex]
- For [tex]\( x = -4 \)[/tex], [tex]\( y = 8 \)[/tex]
- For [tex]\( x = -2 \)[/tex], [tex]\( y = 2 \)[/tex]
- For [tex]\( x = -2 \)[/tex], [tex]\( y = 4 \)[/tex] (conflict with previous [tex]\(-2\)[/tex])
- For [tex]\( x = 0 \)[/tex], [tex]\( y = 2 \)[/tex]
The [tex]\( x \)[/tex]-value [tex]\(-2\)[/tex] maps to both [tex]\(2\)[/tex] and [tex]\(4\)[/tex], which means it does not represent a function.
### Table 4:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -4 & 2 \\ \hline 3 & 5 \\ \hline 1 & 3 \\ \hline -4 & 0 \\ \hline \end{array} \][/tex]
- For [tex]\( x = -4 \)[/tex], [tex]\( y = 2 \)[/tex]
- For [tex]\( x = 3 \)[/tex], [tex]\( y = 5 \)[/tex]
- For [tex]\( x = 1 \)[/tex], [tex]\( y = 3 \)[/tex]
- For [tex]\( x = -4 \)[/tex], [tex]\( y = 0 \)[/tex] (conflict with previous [tex]\(-4\)[/tex])
The [tex]\( x \)[/tex]-value [tex]\(-4\)[/tex] maps to both [tex]\(2\)[/tex] and [tex]\(0\)[/tex], which means it does not represent a function.
### Conclusion:
Only the first table satisfies the condition of a function, where each [tex]\( x \)[/tex]-value corresponds to exactly one [tex]\( y \)[/tex]-value.
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -3 & -1 \\ \hline 0 & 0 \\ \hline -2 & -1 \\ \hline 8 & 1 \\ \hline \end{array} \][/tex]
Therefore, Table 1 represents a function.
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