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Sagot :
To determine which tables represent [tex]\( y \)[/tex] as a function of [tex]\( x \)[/tex] and are one-to-one, we need to confirm two things:
1. Each [tex]\( x \)[/tex] value must map to exactly one [tex]\( y \)[/tex] value.
2. Each [tex]\( y \)[/tex] value must be unique (no [tex]\( y \)[/tex] value is repeated for different [tex]\( x \)[/tex] values).
Let's analyze each table step-by-step:
### Table 1:
[tex]\[ \begin{array}{|r|r|r|r|} \hline x & 5 & 8 & 12 \\ \hline y & 3 & 10 & 15 \\ \hline \end{array} \][/tex]
- Condition 1: [tex]\( y \)[/tex] is a function of [tex]\( x \)[/tex]: Each [tex]\( x \)[/tex] has a unique [tex]\( y \)[/tex].
- [tex]\( x = 5 \rightarrow y = 3 \)[/tex]
- [tex]\( x = 8 \rightarrow y = 10 \)[/tex]
- [tex]\( x = 12 \rightarrow y = 15 \)[/tex]
- Condition 2: One-to-one function: Each [tex]\( y \)[/tex] value is unique.
- [tex]\( y = 3 \)[/tex] for [tex]\( x = 5 \)[/tex]
- [tex]\( y = 10 \)[/tex] for [tex]\( x = 8 \)[/tex]
- [tex]\( y = 15 \)[/tex] for [tex]\( x = 12 \)[/tex]
Since both conditions are satisfied, Table 1 represents [tex]\( y \)[/tex] as a function of [tex]\( x \)[/tex] and is one-to-one.
### Table 2:
[tex]\[ \begin{array}{|r|r|r|r|} \hline x & 5 & 8 & 12 \\ \hline y & 3 & 10 & 10 \\ \hline \end{array} \][/tex]
- Condition 1: [tex]\( y \)[/tex] is a function of [tex]\( x \)[/tex]: Each [tex]\( x \)[/tex] has a unique [tex]\( y \)[/tex].
- [tex]\( x = 5 \rightarrow y = 3 \)[/tex]
- [tex]\( x = 8 \rightarrow y = 10 \)[/tex]
- [tex]\( x = 12 \rightarrow y = 10 \)[/tex]
- Condition 2: One-to-one function: Not satisfied because [tex]\( y \)[/tex] has repeated values.
- [tex]\( y = 3 \)[/tex] for [tex]\( x = 5 \)[/tex]
- [tex]\( y = 10 \)[/tex] for [tex]\( x = 8 \)[/tex]
- [tex]\( y = 10 \)[/tex] for [tex]\( x = 12 \)[/tex]
Since the second condition is not satisfied (repeated [tex]\( y \)[/tex] values), Table 2 does not represent [tex]\( y \)[/tex] as a one-to-one function of [tex]\( x \)[/tex].
### Table 3:
[tex]\[ \begin{array}{|r|r|r|r|} \hline x & 5 & 8 & 8 \\ \hline y & 3 & 10 & 15 \\ \hline \end{array} \][/tex]
- Condition 1: [tex]\( y \)[/tex] is a function of [tex]\( x \)[/tex]: Not satisfied because [tex]\( x = 8 \)[/tex] maps to two different [tex]\( y \)[/tex] values.
- [tex]\( x = 5 \rightarrow y = 3 \)[/tex]
- [tex]\( x = 8 \rightarrow y = 10 \)[/tex]
- [tex]\( x = 8 \rightarrow y = 15 \)[/tex]
Thus, Table 3 does not represent [tex]\( y \)[/tex] as a function of [tex]\( x \)[/tex], and it cannot be one-to-one.
### Conclusion
Only Table 1 represents [tex]\( y \)[/tex] as a function of [tex]\( x \)[/tex] and is also one-to-one. Therefore, the correct answer is:
[tex]\[ \boxed{ \begin{array}{|r|r|r|} \hline x & 5 & 8 & 12 \\ \hline y & 3 & 10 & 15 \\ \hline \end{array} } \][/tex]
1. Each [tex]\( x \)[/tex] value must map to exactly one [tex]\( y \)[/tex] value.
2. Each [tex]\( y \)[/tex] value must be unique (no [tex]\( y \)[/tex] value is repeated for different [tex]\( x \)[/tex] values).
Let's analyze each table step-by-step:
### Table 1:
[tex]\[ \begin{array}{|r|r|r|r|} \hline x & 5 & 8 & 12 \\ \hline y & 3 & 10 & 15 \\ \hline \end{array} \][/tex]
- Condition 1: [tex]\( y \)[/tex] is a function of [tex]\( x \)[/tex]: Each [tex]\( x \)[/tex] has a unique [tex]\( y \)[/tex].
- [tex]\( x = 5 \rightarrow y = 3 \)[/tex]
- [tex]\( x = 8 \rightarrow y = 10 \)[/tex]
- [tex]\( x = 12 \rightarrow y = 15 \)[/tex]
- Condition 2: One-to-one function: Each [tex]\( y \)[/tex] value is unique.
- [tex]\( y = 3 \)[/tex] for [tex]\( x = 5 \)[/tex]
- [tex]\( y = 10 \)[/tex] for [tex]\( x = 8 \)[/tex]
- [tex]\( y = 15 \)[/tex] for [tex]\( x = 12 \)[/tex]
Since both conditions are satisfied, Table 1 represents [tex]\( y \)[/tex] as a function of [tex]\( x \)[/tex] and is one-to-one.
### Table 2:
[tex]\[ \begin{array}{|r|r|r|r|} \hline x & 5 & 8 & 12 \\ \hline y & 3 & 10 & 10 \\ \hline \end{array} \][/tex]
- Condition 1: [tex]\( y \)[/tex] is a function of [tex]\( x \)[/tex]: Each [tex]\( x \)[/tex] has a unique [tex]\( y \)[/tex].
- [tex]\( x = 5 \rightarrow y = 3 \)[/tex]
- [tex]\( x = 8 \rightarrow y = 10 \)[/tex]
- [tex]\( x = 12 \rightarrow y = 10 \)[/tex]
- Condition 2: One-to-one function: Not satisfied because [tex]\( y \)[/tex] has repeated values.
- [tex]\( y = 3 \)[/tex] for [tex]\( x = 5 \)[/tex]
- [tex]\( y = 10 \)[/tex] for [tex]\( x = 8 \)[/tex]
- [tex]\( y = 10 \)[/tex] for [tex]\( x = 12 \)[/tex]
Since the second condition is not satisfied (repeated [tex]\( y \)[/tex] values), Table 2 does not represent [tex]\( y \)[/tex] as a one-to-one function of [tex]\( x \)[/tex].
### Table 3:
[tex]\[ \begin{array}{|r|r|r|r|} \hline x & 5 & 8 & 8 \\ \hline y & 3 & 10 & 15 \\ \hline \end{array} \][/tex]
- Condition 1: [tex]\( y \)[/tex] is a function of [tex]\( x \)[/tex]: Not satisfied because [tex]\( x = 8 \)[/tex] maps to two different [tex]\( y \)[/tex] values.
- [tex]\( x = 5 \rightarrow y = 3 \)[/tex]
- [tex]\( x = 8 \rightarrow y = 10 \)[/tex]
- [tex]\( x = 8 \rightarrow y = 15 \)[/tex]
Thus, Table 3 does not represent [tex]\( y \)[/tex] as a function of [tex]\( x \)[/tex], and it cannot be one-to-one.
### Conclusion
Only Table 1 represents [tex]\( y \)[/tex] as a function of [tex]\( x \)[/tex] and is also one-to-one. Therefore, the correct answer is:
[tex]\[ \boxed{ \begin{array}{|r|r|r|} \hline x & 5 & 8 & 12 \\ \hline y & 3 & 10 & 15 \\ \hline \end{array} } \][/tex]
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