Certainly! Let's complete the table by calculating the value of [tex]\( y \)[/tex] for each given [tex]\( x \)[/tex] using the function [tex]\( y = 2x^2 - 1 \)[/tex].
1. For [tex]\( x = -2 \)[/tex]:
[tex]\[
y = 2(-2)^2 - 1
\][/tex]
[tex]\[
y = 2(4) - 1
\][/tex]
[tex]\[
y = 8 - 1
\][/tex]
[tex]\[
y = 7
\][/tex]
2. For [tex]\( x = -1 \)[/tex]:
[tex]\[
y = 2(-1)^2 - 1
\][/tex]
[tex]\[
y = 2(1) - 1
\][/tex]
[tex]\[
y = 2 - 1
\][/tex]
[tex]\[
y = 1
\][/tex]
3. For [tex]\( x = 0 \)[/tex]:
[tex]\[
y = 2(0)^2 - 1
\][/tex]
[tex]\[
y = 2(0) - 1
\][/tex]
[tex]\[
y = 0 - 1
\][/tex]
[tex]\[
y = -1
\][/tex]
4. For [tex]\( x = 1 \)[/tex]:
[tex]\[
y = 2(1)^2 - 1
\][/tex]
[tex]\[
y = 2(1) - 1
\][/tex]
[tex]\[
y = 2 - 1
\][/tex]
[tex]\[
y = 1
\][/tex]
5. For [tex]\( x = 2 \)[/tex]:
[tex]\[
y = 2(2)^2 - 1
\][/tex]
[tex]\[
y = 2(4) - 1
\][/tex]
[tex]\[
y = 8 - 1
\][/tex]
[tex]\[
y = 7
\][/tex]
Now, let's fill in the table with the computed [tex]\( y \)[/tex] values:
[tex]\[
\begin{tabular}{|l|l|l|l|l|l|}
\hline
$x$ & -2 & -1 & 0 & 1 & 2 \\
\hline
$y$ & 7 & 1 & -1 & 1 & 7 \\
\hline
\end{tabular}
\][/tex]
So the completed table is:
[tex]\[
\begin{tabular}{|l|l|l|l|l|l|}
\hline
$x$ & -2 & -1 & 0 & 1 & 2 \\
\hline
$y$ & 7 & 1 & -1 & 1 & 7 \\
\hline
\end{tabular}
\][/tex]
