Sure, let's address the given question in detail:
### Question 8
a. Copy and complete the table for [tex]\( y = 3x^2 - 5x + 4 \)[/tex] for [tex]\(-3 \leq x \leq 4\)[/tex].
Given:
[tex]\[ y = 3x^2 - 5x + 4 \][/tex]
We will fill in the respective values for each term in the table.
Let's complete each cell in sequence:
- For [tex]\( x = -3 \)[/tex]:
[tex]\[
3(-3)^2 = 3 \cdot 9 = 27
\][/tex]
- For [tex]\( x = -2 \)[/tex]:
[tex]\[
3(-2)^2 = 3 \cdot 4 = 12
\][/tex]
- For [tex]\( x = -1 \)[/tex]:
[tex]\[
3(-1)^2 = 3 \cdot 1 = 3
\][/tex]
- For [tex]\( x = 0 \)[/tex]:
[tex]\[
3(0)^2 = 3 \cdot 0 = 0
\][/tex]
- For [tex]\( x = 1 \)[/tex]:
[tex]\[
3(1)^2 = 3 \cdot 1 = 3
\][/tex]
- For [tex]\( x = 2 \)[/tex]:
[tex]\[
3(2)^2 = 3 \cdot 4 = 12
\][/tex]
- For [tex]\( x = 3 \)[/tex]:
[tex]\[
3(3)^2 = 3 \cdot 9 = 27
\][/tex]
- For [tex]\( x = 4 \)[/tex]:
[tex]\[
3(4)^2 = 3 \cdot 16 = 48
\][/tex]
Now, let's complete the table:
[tex]\[
\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline
x & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 \\
\hline
3x^2 & 27 & 12 & 3 & 0 & 3 & 12 & 27 & 48 \\
\hline
\end{array}
\][/tex]
Therefore, the completed table for [tex]\( 3x^2 \)[/tex] is:
[tex]\[
\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline
x & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 \\
\hline
3x^2 & 27 & 12 & 3 & 0 & 3 & 12 & 27 & 48 \\
\hline
\end{array}
\][/tex]
If you have any further questions or need assistance with another part of the original problem, feel free to ask!
