To solve this problem, we need to determine the finite differences and assess the type of function that the data models. Let's proceed step-by-step.
### Step 1: Calculate the First Differences
The first differences are computed by subtracting each [tex]\( y \)[/tex]-value from the subsequent [tex]\( y \)[/tex]-value in the table.
1. From [tex]\( y_1 = 21 \)[/tex] to [tex]\( y_2 = 11 \)[/tex]:
[tex]\[
11 - 21 = -10
\][/tex]
2. From [tex]\( y_2 = 11 \)[/tex] to [tex]\( y_3 = 5 \)[/tex]:
[tex]\[
5 - 11 = -6
\][/tex]
3. From [tex]\( y_3 = 5 \)[/tex] to [tex]\( y_4 = 3 \)[/tex]:
[tex]\[
3 - 5 = -2
\][/tex]
4. From [tex]\( y_4 = 3 \)[/tex] to [tex]\( y_5 = 5 \)[/tex]:
[tex]\[
5 - 3 = 2
\][/tex]
So, the first differences are:
[tex]\[
[-10, -6, -2, 2]
\][/tex]
### Step 2: Calculate the Second Differences
The second differences are calculated by subtracting each first difference from the subsequent first difference.
1. From [tex]\(-10\)[/tex] to [tex]\(-6\)[/tex]:
[tex]\[
-6 - (-10) = -6 + 10 = 4
\][/tex]
2. From [tex]\(-6\)[/tex] to [tex]\(-2\)[/tex]:
[tex]\[
-2 - (-6) = -2 + 6 = 4
\][/tex]
3. From [tex]\(-2\)[/tex] to [tex]\(2\)[/tex]:
[tex]\[
2 - (-2) = 2 + 2 = 4
\][/tex]
So, the second differences are:
[tex]\[
[4, 4, 4]
\][/tex]
### Step 3: Organize the Results in the Table
Now we can fill in the table with our computed differences:
\begin{tabular}{|c|c|c|c|}
\hline
[tex]$x$[/tex] & [tex]$y$[/tex] & \begin{tabular}{l}[tex]$1^{\text {st }}$[/tex] \\ difference\end{tabular} & \begin{tabular}{l}[tex]$2^{\text {nd }}$[/tex] \\ difference\end{tabular} \\
\hline-1 & 21 & -10 & \\
\hline 0 & 11 & -6 & 4 \\
\hline 1 & 5 & -2 & 4 \\
\hline 2 & 3 & 2 & 4 \\
\hline 3 & 5 & & \\
\hline
\end{tabular}
### Step 4: Analyze the Function Type
The second differences are constant and equal to 4. When the second differences are constant, this indicates that the function modeled by the data is a quadratic function.
[tex]\[
\boxed{\text{The table models a quadratic function because the second differences are constant.}}
\][/tex]
