To determine which of the given tables represents a function, we need to ensure that each [tex]\( x \)[/tex]-value is paired with exactly one [tex]\( y \)[/tex]-value. A relation is a function if each input value from the domain (all possible [tex]\( x \)[/tex]-values) is associated with one and only one output value from the range (all possible [tex]\( y \)[/tex]-values).
Let's examine each table:
### Table [tex]\( W \)[/tex]
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
-1 & 15 \\
\hline
-2 & 6 \\
\hline
0 & 15 \\
\hline
4 & 6 \\
\hline
\end{array}
\][/tex]
- The x-values are [tex]\(-1, -2, 0, 4\)[/tex], which are all unique.
- Each [tex]\( x \)[/tex]-value maps to exactly one [tex]\( y \)[/tex]-value.
Since each [tex]\( x \)[/tex]-value corresponds to one and only one [tex]\( y \)[/tex]-value, Table [tex]\( W \)[/tex] is a function.
### Table [tex]\( X \)[/tex]
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
0 & 4 \\
\hline
6 & 15 \\
\hline
0 & 6 \\
\hline
-2 & -1 \\
\hline
\end{array}
\][/tex]
- The x-values are [tex]\( 0, 6, 0, -2\)[/tex].
- The [tex]\( x = 0 \)[/tex] appears twice with different [tex]\( y \)[/tex]-values (4 and 6).
Since the [tex]\( x \)[/tex]-value [tex]\( 0 \)[/tex] maps to two different [tex]\( y \)[/tex]-values, Table [tex]\( X \)[/tex] is not a function.
### Table [tex]\( Y \)[/tex]
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
4 & 0 \\
\hline
-1 & -2 \\
\hline
6 & -2 \\
\hline
6 & 15 \\
\hline
\end{array}
\][/tex]
- The x-values are [tex]\( 4, -1, 6, 6 \)[/tex].
- The [tex]\( x = 6 \)[/tex] appears twice with different [tex]\( y \)[/tex]-values (-2 and 15).
Since the [tex]\( x \)[/tex]-value [tex]\( 6 \)[/tex] maps to two different [tex]\( y \)[/tex]-values, Table [tex]\( Y \)[/tex] is not a function.
### Table [tex]\( Z \)[/tex]
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
6 & -2 \\
\hline
4 & 0 \\
\hline
15 & -1 \\
\hline
4 & 3 \\
\hline
\end{array}
\][/tex]
- The x-values are [tex]\( 6, 4, 15, 4 \)[/tex].
- The [tex]\( x = 4 \)[/tex] appears twice with different [tex]\( y \)[/tex]-values (0 and 3).
Since the [tex]\( x \)[/tex]-value [tex]\( 4 \)[/tex] maps to two different [tex]\( y \)[/tex]-values, Table [tex]\( Z \)[/tex] is not a function.
Therefore, the correct answer is:
[tex]\[
\boxed{A. \, W}
\][/tex]
