Complete the table for the function [tex]y = 2x^2 - 1[/tex].

[tex]\[
\begin{tabular}{|c|c|c|c|c|c|}
\hline
$x$ & -2 & -1 & 0 & 1 & 2 \\
\hline
$y$ & & & & & \\
\hline
\end{tabular}
\][/tex]



Answer :

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]