Answer :
To determine whether the given table represents a function, we need to check if each input (\( x \)) corresponds to exactly one output (\( y \)). In other words, for the table to represent a function, there must not be any repeated \( x \)-values with different \( y \)-values.
Let's examine the given table:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -1 & 2 \\ \hline 3 & 1 \\ \hline -1 & -2 \\ \hline 0 & -3 \\ \hline \end{array} \][/tex]
1. First, we list all the \( x \)-values and their corresponding \( y \)-values:
- \( x = -1 \) maps to \( y = 2 \)
- \( x = 3 \) maps to \( y = 1 \)
- \( x = -1 \) maps to \( y = -2 \)
- \( x = 0 \) maps to \( y = -3 \)
2. Next, we need to check for repeated \( x \)-values:
- The value \( x = -1 \) appears twice.
- For \( x = -1 \), the first \( y \)-value is 2, and the second \( y \)-value is -2.
3. Since the \( x \)-value \(-1\) maps to two different \( y \)-values (2 and -2), this violates the definition of a function where each input must correspond to exactly one output.
4. Therefore, due to the repeated \( x \)-value with different \( y \)-values, the table does not represent a function.
The answer is:
The above table is not a function.
Let's examine the given table:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -1 & 2 \\ \hline 3 & 1 \\ \hline -1 & -2 \\ \hline 0 & -3 \\ \hline \end{array} \][/tex]
1. First, we list all the \( x \)-values and their corresponding \( y \)-values:
- \( x = -1 \) maps to \( y = 2 \)
- \( x = 3 \) maps to \( y = 1 \)
- \( x = -1 \) maps to \( y = -2 \)
- \( x = 0 \) maps to \( y = -3 \)
2. Next, we need to check for repeated \( x \)-values:
- The value \( x = -1 \) appears twice.
- For \( x = -1 \), the first \( y \)-value is 2, and the second \( y \)-value is -2.
3. Since the \( x \)-value \(-1\) maps to two different \( y \)-values (2 and -2), this violates the definition of a function where each input must correspond to exactly one output.
4. Therefore, due to the repeated \( x \)-value with different \( y \)-values, the table does not represent a function.
The answer is:
The above table is not a function.
