Answer :
To determine whether the given set of ordered pairs forms a function, we need to apply the definition of a function from mathematics. A set of ordered pairs \((x, y)\) is a function if for every \(x\)-value there is exactly one \(y\)-value associated with it.
Let's analyze the provided set of ordered pairs step by step:
[tex]\[ \begin{array}{|r|r|} \hline x & y \\ \hline -1 & 2 \\ \hline 3 & 1 \\ \hline -1 & -2 \\ \hline 0 & -3 \\ \hline \end{array} \][/tex]
1. For \(x = -1\), the corresponding \(y\)-values are \(2\) and \(-2\). This means that the \(x\)-value \(-1\) maps to two different \(y\)-values.
2. For \(x = 3\), the corresponding \(y\)-value is \(1\). There is only one \(y\)-value for this \(x\)-value.
3. For \(x = 0\), the corresponding \(y\)-value is \(-3\). There is only one \(y\)-value for this \(x\)-value.
As per the definition of a function, each \(x\) should map to exactly one \(y\). However, in this case, the \(x\)-value \(-1\) maps to two different \(y\)-values (\(2\) and \(-2\)). This violates the definition of a function.
Therefore, based on the analysis, the given set of ordered pairs is not a function.
Let's analyze the provided set of ordered pairs step by step:
[tex]\[ \begin{array}{|r|r|} \hline x & y \\ \hline -1 & 2 \\ \hline 3 & 1 \\ \hline -1 & -2 \\ \hline 0 & -3 \\ \hline \end{array} \][/tex]
1. For \(x = -1\), the corresponding \(y\)-values are \(2\) and \(-2\). This means that the \(x\)-value \(-1\) maps to two different \(y\)-values.
2. For \(x = 3\), the corresponding \(y\)-value is \(1\). There is only one \(y\)-value for this \(x\)-value.
3. For \(x = 0\), the corresponding \(y\)-value is \(-3\). There is only one \(y\)-value for this \(x\)-value.
As per the definition of a function, each \(x\) should map to exactly one \(y\). However, in this case, the \(x\)-value \(-1\) maps to two different \(y\)-values (\(2\) and \(-2\)). This violates the definition of a function.
Therefore, based on the analysis, the given set of ordered pairs is not a function.
