To determine the type of function represented by the data in the table and to find the best description, we need to analyze the differences between successive values.
Given the table:
[tex]\[
\begin{tabular}{|c|c|}
\hline
$x$ & $y$ \\
\hline
-2 & 8 \\
\hline
-1 & 2 \\
\hline
0 & 0 \\
\hline
1 & 2 \\
\hline
2 & 8 \\
\hline
\end{tabular}
\][/tex]
First, we calculate the first differences. The first differences are found by subtracting each [tex]$y$[/tex] value from the next [tex]$y$[/tex] value:
[tex]\[
\begin{align*}
\Delta y_1 & = y(-1) - y(-2) = 2 - 8 = -6 \\
\Delta y_2 & = y(0) - y(-1) = 0 - 2 = -2 \\
\Delta y_3 & = y(1) - y(0) = 2 - 0 = 2 \\
\Delta y_4 & = y(2) - y(1) = 8 - 2 = 6 \\
\end{align*}
\][/tex]
So, the first differences are:
[tex]\[
[-6, -2, 2, 6]
\][/tex]
Next, we calculate the second differences. The second differences are found by subtracting each first difference from the next first difference:
[tex]\[
\begin{align*}
\Delta^2 y_1 & = \Delta y_2 - \Delta y_1 = (-2) - (-6) = 4 \\
\Delta^2 y_2 & = \Delta y_3 - \Delta y_2 = 2 - (-2) = 4 \\
\Delta^2 y_3 & = \Delta y_4 - \Delta y_3 = 6 - 2 = 4 \\
\end{align*}
\][/tex]
So, the second differences are:
[tex]\[
[4, 4, 4]
\][/tex]
Since the second differences are constant, this tells us that the function is quadratic. The common second difference of [tex]\(4\)[/tex] further validates this.
Thus, the best description of the function represented by the data in the table is:
quadratic with a common second difference of 4.
