Answer :
To determine which table represents a function, we need to examine if every [tex]\( x \)[/tex]-value maps to exactly one [tex]\( y \)[/tex]-value in each table. A relation is a function if no [tex]\( x \)[/tex]-value is repeated with different [tex]\( y \)[/tex]-values. Let's analyze 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]
In this table:
- Each [tex]\( x \)[/tex] value (-3, 0, -2, 8) is unique and maps to exactly one [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]
In this table:
- The [tex]\( x \)[/tex]-value -5 maps to two different [tex]\( y \)[/tex]-values (-5 and 5), which disqualifies it from being 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]
In this table:
- The [tex]\( x \)[/tex]-value -2 maps to two different [tex]\( y \)[/tex]-values (2 and 4), which disqualifies it from being 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]
In this table:
- The [tex]\( x \)[/tex]-value -4 maps to two different [tex]\( y \)[/tex]-values (2 and 0), which disqualifies it from being a function.
### Conclusion:
After examining all the tables, only Table 1 ensures that each [tex]\( x \)[/tex]-value maps to exactly one [tex]\( y \)[/tex]-value. 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]
In this table:
- Each [tex]\( x \)[/tex] value (-3, 0, -2, 8) is unique and maps to exactly one [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]
In this table:
- The [tex]\( x \)[/tex]-value -5 maps to two different [tex]\( y \)[/tex]-values (-5 and 5), which disqualifies it from being 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]
In this table:
- The [tex]\( x \)[/tex]-value -2 maps to two different [tex]\( y \)[/tex]-values (2 and 4), which disqualifies it from being 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]
In this table:
- The [tex]\( x \)[/tex]-value -4 maps to two different [tex]\( y \)[/tex]-values (2 and 0), which disqualifies it from being a function.
### Conclusion:
After examining all the tables, only Table 1 ensures that each [tex]\( x \)[/tex]-value maps to exactly one [tex]\( y \)[/tex]-value. Therefore, Table 1 represents a function.
