To complete the table with the [tex]\( y \)[/tex]-values for the equation [tex]\( y = x^2 + 1 \)[/tex], we will plug in each [tex]\( x \)[/tex]-value into the equation and solve for [tex]\( y \)[/tex].
1. For [tex]\( x = -2 \)[/tex]:
[tex]\[
y = (-2)^2 + 1 = 4 + 1 = 5
\][/tex]
The output [tex]\( y \)[/tex] for [tex]\( x = -2 \)[/tex] is 5.
2. For [tex]\( x = -1 \)[/tex]:
[tex]\[
y = (-1)^2 + 1 = 1 + 1 = 2
\][/tex]
The output [tex]\( y \)[/tex] for [tex]\( x = -1 \)[/tex] is 2.
3. For [tex]\( x = 0 \)[/tex]:
[tex]\[
y = 0^2 + 1 = 0 + 1 = 1
\][/tex]
The output [tex]\( y \)[/tex] for [tex]\( x = 0 \)[/tex] is 1.
4. For [tex]\( x = 1 \)[/tex]:
[tex]\[
y = 1^2 + 1 = 1 + 1 = 2
\][/tex]
The output [tex]\( y \)[/tex] for [tex]\( x = 1 \)[/tex] is 2.
5. For [tex]\( x = 2 \)[/tex]:
[tex]\[
y = 2^2 + 1 = 4 + 1 = 5
\][/tex]
The output [tex]\( y \)[/tex] for [tex]\( x = 2 \)[/tex] is 5.
Now, we can fill in the table with the calculated [tex]\( y \)[/tex]-values.
[tex]\[
\begin{tabular}{|c|c|}
\hline Input $(x)$ & Output $(y)$ \\
\hline
-2 & 5 \\
\hline
-1 & 2 \\
\hline
0 & 1 \\
\hline
1 & 2 \\
\hline
2 & 5 \\
\hline
\end{tabular}
\][/tex]
Thus, the completed table is:
[tex]\[
\begin{tabular}{|c|c|}
\hline Input $(x)$ & Output $(y)$ \\
\hline
-2 & 5 \\
\hline
-1 & 2 \\
\hline
0 & 1 \\
\hline
1 & 2 \\
\hline
2 & 5 \\
\hline
\end{tabular}
\][/tex]
