Let's determine the values of [tex]\( A \)[/tex] and [tex]\( B \)[/tex] in the table for the equation [tex]\( y = x^2 + 6 \)[/tex].
We'll plug in the given [tex]\( x \)[/tex]-values into the equation [tex]\( y = x^2 + 6 \)[/tex], and compute the corresponding [tex]\( y \)[/tex]-values.
1. For [tex]\( x = -2 \)[/tex]:
[tex]\[
y = (-2)^2 + 6 = 4 + 6 = 10
\][/tex]
So, [tex]\( A \)[/tex] is 10.
2. For [tex]\( x = -1 \)[/tex]:
[tex]\[
y = (-1)^2 + 6 = 1 + 6 = 7
\][/tex]
3. For [tex]\( x = 0 \)[/tex]:
[tex]\[
y = 0^2 + 6 = 0 + 6 = 6
\][/tex]
4. For [tex]\( x = 1 \)[/tex]:
[tex]\[
y = 1^2 + 6 = 1 + 6 = 7
\][/tex]
5. For [tex]\( x = 2 \)[/tex]:
[tex]\[
y = 2^2 + 6 = 4 + 6 = 10
\][/tex]
So, [tex]\( B \)[/tex] is 10.
Based on these computations, the completed table of values for [tex]\( y = x^2 + 6 \)[/tex] is:
[tex]\[
\begin{tabular}{c||r|c|c|c|c}
$x$ & -2 & -1 & 0 & 1 & 2 \\
\hline
$y$ & 10 & 7 & 6 & 7 & 10 \\
\end{tabular}
\][/tex]
Therefore, the numbers that replace [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are both 10.
