Answer :
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.
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.
