\begin{tabular}{|c|c|}\hline\( x y \) \\ \hline 0 & 0 \\ \hline 1 & 0 \\ \hline 2 & 1 \\ \hline 3 & 3 \\ \hline \\ \hline funcionio \( y \) \\ \hline\end{tabular}

It looks like you're trying to create a table that shows the values of \( x \) and \( y \), along with a function that describes the relationship between them. Based on the values you've provided, we can analyze the data: \[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & 0 \\ 1 & 0 \\ 2 & 1 \\ 3 & 3 \\ \hline \end{array} \] From the table, we can observe the following pairs of \( (x, y) \): - When \( x = 0 \), \( y = 0 \) - When \( x = 1 \), \( y = 0 \) - When \( x = 2 \), \( y = 1 \) - When \( x = 3 \), \( y = 3 \) To find a function \( y = f(x) \) that fits these points, we can look for a pattern. 1. For \( x = 0 \) and \( x = 1 \), \( y \) remains 0. 2. For \( x = 2 \), \( y \) increases to 1. 3. For \( x = 3 \), \( y \) increases to 3. This suggests that \( y \) might be related to \( x \) in a non-linear way. One possible function that fits these points is: \[ y = \frac{(x-1)(x-2)}{2} \] This function gives: - For \( x = 0 \): \( y = \frac{(0-1)(0-2)}{2} = 0 \) - For \( x = 1 \): \( y = \frac{(1-1)(1-2)}{2} = 0 \) - For \( x = 2 \): \( y = \frac{(2-1)(2-2)}{2} = 1 \) - For \( x = 3 \): \( y = \frac{(3-1)(3-2)}{2} = 3 \) Thus, the function \( y = \frac{(x-1)(x-2)}{2} \) fits the data points provided. If you have any specific questions or need further assistance, feel free to ask!