Sure, let's calculate the value of [tex]\( y \)[/tex] for each given value of [tex]\( x \)[/tex] using the equation [tex]\( y = 2x^2 + 1 \)[/tex].
Given the equation:
[tex]\[ y = 2x^2 + 1 \][/tex]
We'll substitute each [tex]\( x \)[/tex] value into the equation to find the corresponding [tex]\( y \)[/tex] value.
1. For [tex]\( x = -6 \)[/tex]:
[tex]\[ y = 2(-6)^2 + 1 \][/tex]
[tex]\[ y = 2(36) + 1 \][/tex]
[tex]\[ y = 72 + 1 \][/tex]
[tex]\[ y = 73 \][/tex]
2. For [tex]\( x = -5 \)[/tex]:
[tex]\[ y = 2(-5)^2 + 1 \][/tex]
[tex]\[ y = 2(25) + 1 \][/tex]
[tex]\[ y = 50 + 1 \][/tex]
[tex]\[ y = 51 \][/tex]
3. For [tex]\( x = 0 \)[/tex]:
[tex]\[ y = 2(0)^2 + 1 \][/tex]
[tex]\[ y = 0 + 1 \][/tex]
[tex]\[ y = 1 \][/tex]
4. For [tex]\( x = 5 \)[/tex]:
[tex]\[ y = 2(5)^2 + 1 \][/tex]
[tex]\[ y = 2(25) + 1 \][/tex]
[tex]\[ y = 50 + 1 \][/tex]
[tex]\[ y = 51 \][/tex]
5. For [tex]\( x = 6 \)[/tex]:
[tex]\[ y = 2(6)^2 + 1 \][/tex]
[tex]\[ y = 2(36) + 1 \][/tex]
[tex]\[ y = 72 + 1 \][/tex]
[tex]\[ y = 73 \][/tex]
Now, we can fill in the values in the table:
[tex]\[
\begin{array}{|c|c|c|c|c|c|}
\hline
x & -6 & -5 & 0 & 5 & 6 \\
\hline
y & 73 & 51 & 1 & 51 & 73 \\
\hline
\end{array}
\][/tex]
So the completed table is:
[tex]\[
\begin{tabular}{|c|c|c|c|c|c|}
\hline
x & -6 & -5 & 0 & 5 & 6 \\
\hline
y & 73 & 51 & 1 & 51 & 73 \\
\hline
\end{tabular}
\][/tex]
