Certainly! Let's fill in the table step-by-step by substituting each value of [tex]\( x \)[/tex] into the given quadratic function [tex]\( y = 2x^2 - 3x + 1 \)[/tex].
We start with the general formula:
[tex]\[ y = 2x^2 - 3x + 1 \][/tex]
1. For [tex]\( x = 1 \)[/tex]:
[tex]\[ y = 2(1)^2 - 3(1) + 1 = 2 - 3 + 1 = 0 \][/tex]
So, when [tex]\( x = 1 \)[/tex], [tex]\( y = 0 \)[/tex].
2. For [tex]\( x = 0 \)[/tex]:
[tex]\[ y = 2(0)^2 - 3(0) + 1 = 0 - 0 + 1 = 1 \][/tex]
So, when [tex]\( x = 0 \)[/tex], [tex]\( y = 1 \)[/tex].
3. For [tex]\( x = -1 \)[/tex]:
[tex]\[ y = 2(-1)^2 - 3(-1) + 1 = 2(1) + 3 + 1 = 2 + 3 + 1 = 6 \][/tex]
So, when [tex]\( x = -1 \)[/tex], [tex]\( y = 6 \)[/tex].
4. For [tex]\( x = 2 \)[/tex]:
[tex]\[ y = 2(2)^2 - 3(2) + 1 = 2(4) - 6 + 1 = 8 - 6 + 1 = 3 \][/tex]
So, when [tex]\( x = 2 \)[/tex], [tex]\( y = 3 \)[/tex].
Now, filling in these values into the table, we get:
[tex]\[
\begin{tabular}{|c|c|}
\hline
x & y \\
\hline
1 & 0 \\
\hline
0 & 1 \\
\hline
-1 & 6 \\
\hline
2 & 3 \\
\hline
\end{tabular}
\][/tex]
So the completed table is:
[tex]\[
\begin{tabular}{|c|c|}
\hline
x & y \\
\hline
1 & 0 \\
\hline
0 & 1 \\
\hline
-1 & 6 \\
\hline
2 & 3 \\
\hline
\end{tabular}
\][/tex]
