Given the standard form of a quadratic function:

[tex] y = 2x^2 - 3x + 1 [/tex]

Complete the table:

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



Answer :

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]