Answer :
Sure! Let's complete the table of values for the function [tex]\( y = x^2 - x \)[/tex].
We'll fill in the [tex]\( y \)[/tex] values for the given [tex]\( x \)[/tex] values step-by-step using the given results.
### Given Values:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline x & -3 & -2 & 1 & 0 & 1 & 2 & 3 \\ \hline y & 12 & 6 & & 0 & & 2 & 6 \\ \hline \end{array} \][/tex]
### Step-by-Step Solution:
#### For [tex]\( x = 1 \)[/tex]:
[tex]\[ (x, y) = (1, 0) \][/tex]
#### For [tex]\( x = 3 \)[/tex]:
[tex]\[ (x, y) = (3, 6) \][/tex]
#### For [tex]\( x = 2 \)[/tex]:
[tex]\[ (x, y) = (2, 2) \][/tex]
Now, plug these values into the table:
### Completed Table:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline x & -3 & -2 & 1 & 0 & 1 & 2 & 3 \\ \hline y & 12 & 6 & 0 & 0 & 0 & 2 & 6 \\ \hline \end{array} \][/tex]
The completed table of values for the function [tex]\( y = x^2 - x \)[/tex] is confirmed correct based on the values provided.
We'll fill in the [tex]\( y \)[/tex] values for the given [tex]\( x \)[/tex] values step-by-step using the given results.
### Given Values:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline x & -3 & -2 & 1 & 0 & 1 & 2 & 3 \\ \hline y & 12 & 6 & & 0 & & 2 & 6 \\ \hline \end{array} \][/tex]
### Step-by-Step Solution:
#### For [tex]\( x = 1 \)[/tex]:
[tex]\[ (x, y) = (1, 0) \][/tex]
#### For [tex]\( x = 3 \)[/tex]:
[tex]\[ (x, y) = (3, 6) \][/tex]
#### For [tex]\( x = 2 \)[/tex]:
[tex]\[ (x, y) = (2, 2) \][/tex]
Now, plug these values into the table:
### Completed Table:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline x & -3 & -2 & 1 & 0 & 1 & 2 & 3 \\ \hline y & 12 & 6 & 0 & 0 & 0 & 2 & 6 \\ \hline \end{array} \][/tex]
The completed table of values for the function [tex]\( y = x^2 - x \)[/tex] is confirmed correct based on the values provided.
