Answer :
To determine which table represents a function, we need to check each table for the definition of a function. A function is such that for every [tex]\( x \)[/tex]-value, there must be exactly one [tex]\( y \)[/tex]-value. In other words, each [tex]\( x \)[/tex]-value should map to a unique [tex]\( y \)[/tex]-value.
Let's review each table one by one:
### 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:
- [tex]\( x = -3 \)[/tex] maps to [tex]\( y = -1 \)[/tex]
- [tex]\( x = 0 \)[/tex] maps to [tex]\( y = 0 \)[/tex]
- [tex]\( x = -2 \)[/tex] maps to [tex]\( y = -1 \)[/tex]
- [tex]\( x = 8 \)[/tex] maps to [tex]\( y = 1 \)[/tex]
Each [tex]\( x \)[/tex]-value is unique and maps to only one [tex]\( y \)[/tex]-value. Thus, this table represents a function.
### 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:
- [tex]\( x = -5 \)[/tex] maps to [tex]\( y = -5 \)[/tex]
- [tex]\( x = 0 \)[/tex] maps to [tex]\( y = 0 \)[/tex]
- [tex]\( x = -5 \)[/tex] also maps to [tex]\( y = 5 \)[/tex]
- [tex]\( x = 6 \)[/tex] maps to [tex]\( y = -6 \)[/tex]
The [tex]\( x \)[/tex]-value [tex]\( -5 \)[/tex] maps to two different [tex]\( y \)[/tex]-values (-5 and 5). Thus, this table 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]
In this table:
- [tex]\( x = -4 \)[/tex] maps to [tex]\( y = 8 \)[/tex]
- [tex]\( x = -2 \)[/tex] maps to [tex]\( y = 2 \)[/tex]
- [tex]\( x = -2 \)[/tex] also maps to [tex]\( y = 4 \)[/tex]
- [tex]\( x = 0 \)[/tex] maps to [tex]\( y = 2 \)[/tex]
The [tex]\( x \)[/tex]-value [tex]\( -2 \)[/tex] maps to two different [tex]\( y \)[/tex]-values (2 and 4). Thus, this table 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]
In this table:
- [tex]\( x = -4 \)[/tex] maps to [tex]\( y = 2 \)[/tex]
- [tex]\( x = 3 \)[/tex] maps to [tex]\( y = 5 \)[/tex]
- [tex]\( x = 1 \)[/tex] maps to [tex]\( y = 3 \)[/tex]
- [tex]\( x = -4 \)[/tex] also maps to [tex]\( y = 0 \)[/tex]
The [tex]\( x \)[/tex]-value [tex]\( -4 \)[/tex] maps to two different [tex]\( y \)[/tex]-values (2 and 0). Thus, this table does not represent a function.
### Conclusion:
Only Table 1 represents a function.
Let's review each table one by one:
### 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:
- [tex]\( x = -3 \)[/tex] maps to [tex]\( y = -1 \)[/tex]
- [tex]\( x = 0 \)[/tex] maps to [tex]\( y = 0 \)[/tex]
- [tex]\( x = -2 \)[/tex] maps to [tex]\( y = -1 \)[/tex]
- [tex]\( x = 8 \)[/tex] maps to [tex]\( y = 1 \)[/tex]
Each [tex]\( x \)[/tex]-value is unique and maps to only one [tex]\( y \)[/tex]-value. Thus, this table represents a function.
### 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:
- [tex]\( x = -5 \)[/tex] maps to [tex]\( y = -5 \)[/tex]
- [tex]\( x = 0 \)[/tex] maps to [tex]\( y = 0 \)[/tex]
- [tex]\( x = -5 \)[/tex] also maps to [tex]\( y = 5 \)[/tex]
- [tex]\( x = 6 \)[/tex] maps to [tex]\( y = -6 \)[/tex]
The [tex]\( x \)[/tex]-value [tex]\( -5 \)[/tex] maps to two different [tex]\( y \)[/tex]-values (-5 and 5). Thus, this table 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]
In this table:
- [tex]\( x = -4 \)[/tex] maps to [tex]\( y = 8 \)[/tex]
- [tex]\( x = -2 \)[/tex] maps to [tex]\( y = 2 \)[/tex]
- [tex]\( x = -2 \)[/tex] also maps to [tex]\( y = 4 \)[/tex]
- [tex]\( x = 0 \)[/tex] maps to [tex]\( y = 2 \)[/tex]
The [tex]\( x \)[/tex]-value [tex]\( -2 \)[/tex] maps to two different [tex]\( y \)[/tex]-values (2 and 4). Thus, this table 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]
In this table:
- [tex]\( x = -4 \)[/tex] maps to [tex]\( y = 2 \)[/tex]
- [tex]\( x = 3 \)[/tex] maps to [tex]\( y = 5 \)[/tex]
- [tex]\( x = 1 \)[/tex] maps to [tex]\( y = 3 \)[/tex]
- [tex]\( x = -4 \)[/tex] also maps to [tex]\( y = 0 \)[/tex]
The [tex]\( x \)[/tex]-value [tex]\( -4 \)[/tex] maps to two different [tex]\( y \)[/tex]-values (2 and 0). Thus, this table does not represent a function.
### Conclusion:
Only Table 1 represents a function.
