To complete the table of values for the function [tex]\( y = x^2 + x \)[/tex], we will substitute each value of [tex]\( x \)[/tex] into the function to find the corresponding [tex]\( y \)[/tex]-values.
The [tex]\( x \)[/tex]-values given are [tex]\(-2\)[/tex], [tex]\(-1\)[/tex], [tex]\(0\)[/tex], [tex]\(1\)[/tex], and [tex]\(2\)[/tex].
1. For [tex]\( x = -2 \)[/tex]:
[tex]\[
y = (-2)^2 + (-2) = 4 - 2 = 2
\][/tex]
This matches the given value in the table.
2. For [tex]\( x = -1 \)[/tex]:
[tex]\[
y = (-1)^2 + (-1) = 1 - 1 = 0
\][/tex]
Therefore, [tex]\( A = 0 \)[/tex].
3. For [tex]\( x = 0 \)[/tex]:
[tex]\[
y = (0)^2 + 0 = 0 + 0 = 0
\][/tex]
Therefore, [tex]\( B = 0 \)[/tex].
4. For [tex]\( x = 1 \)[/tex]:
[tex]\[
y = (1)^2 + 1 = 1 + 1 = 2
\][/tex]
This matches the given value in the table.
5. For [tex]\( x = 2 \)[/tex]:
[tex]\[
y = (2)^2 + 2 = 4 + 2 = 6
\][/tex]
Therefore, [tex]\( C = 6 \)[/tex].
So, the completed table is:
[tex]\[
\begin{tabular}{c||c|c|c|c|c}
$x$ & -2 & -1 & 0 & 1 & 2 \\
\hline
$y$ & 2 & 0 & 0 & 2 & 6 \\
\end{tabular}
\][/tex]
Thus, the numbers replacing [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex] are [tex]\( 0 \)[/tex], [tex]\( 0 \)[/tex], and [tex]\( 6 \)[/tex] respectively.
