Answer :
To determine if the given relation is a function, we need to check if each input \( x \) has exactly one output \( y \). In other words, every \( x \) value in the set of ordered pairs should be paired with one and only one \( y \) value.
Let’s examine the pairs provided:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 4 \\ \hline -1 & -2 \\ \hline 3 & 10 \\ \hline 5 & 16 \\ \hline \end{array} \][/tex]
Now, let's list the \( x \) values:
- \( x = 1 \)
- \( x = -1 \)
- \( x = 3 \)
- \( x = 5 \)
Since all \( x \) values are unique and appear only once in the table, there is no case where a single \( x \) value is mapped to more than one \( y \) value.
Thus, each \( x \) value has exactly one corresponding \( y \) value, which meets the criterion for the relation to be a function.
Therefore, the given relation is a function.
The answer is:
Yes
Let’s examine the pairs provided:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 4 \\ \hline -1 & -2 \\ \hline 3 & 10 \\ \hline 5 & 16 \\ \hline \end{array} \][/tex]
Now, let's list the \( x \) values:
- \( x = 1 \)
- \( x = -1 \)
- \( x = 3 \)
- \( x = 5 \)
Since all \( x \) values are unique and appear only once in the table, there is no case where a single \( x \) value is mapped to more than one \( y \) value.
Thus, each \( x \) value has exactly one corresponding \( y \) value, which meets the criterion for the relation to be a function.
Therefore, the given relation is a function.
The answer is:
Yes
