To solve the given problem, where [tex]\( f(x) = -3x \)[/tex], we need to calculate the value of [tex]\( f(x) \)[/tex] for different values of [tex]\( x \)[/tex]. Let's fill in the table step by step:
1. Calculate [tex]\( f(x) \)[/tex] for [tex]\( x = -2 \)[/tex]:
[tex]\[
f(-2) = -3 \times (-2) = 6
\][/tex]
So, [tex]\( f(-2) = 6 \)[/tex].
2. Calculate [tex]\( f(x) \)[/tex] for [tex]\( x = -1 \)[/tex]:
[tex]\[
f(-1) = -3 \times (-1) = 3
\][/tex]
So, [tex]\( f(-1) = 3 \)[/tex].
3. Calculate [tex]\( f(x) \)[/tex] for [tex]\( x = 0 \)[/tex]:
[tex]\[
f(0) = -3 \times 0 = 0
\][/tex]
So, [tex]\( f(0) = 0 \)[/tex].
4. Calculate [tex]\( f(x) \)[/tex] for [tex]\( x = 1 \)[/tex]:
[tex]\[
f(1) = -3 \times 1 = -3
\][/tex]
So, [tex]\( f(1) = -3 \)[/tex].
5. Calculate [tex]\( f(x) \)[/tex] for [tex]\( x = 2 \)[/tex]:
[tex]\[
f(2) = -3 \times 2 = -6
\][/tex]
So, [tex]\( f(2) = -6 \)[/tex].
Now let's fill in the table with these calculations:
[tex]\[
\begin{array}{|c|c|}
\hline
x & f(x) \\
\hline
-2 & 6 \\
\hline
-1 & 3 \\
\hline
0 & 0 \\
\hline
1 & -3 \\
\hline
2 & -6 \\
\hline
\end{array}
\][/tex]
This gives us the final values for [tex]\( f(x) \)[/tex] for each [tex]\( x \)[/tex] as follows:
- [tex]\( f(-2) = 6 \)[/tex]
- [tex]\( f(-1) = 3 \)[/tex]
- [tex]\( f(0) = 0 \)[/tex]
- [tex]\( f(1) = -3 \)[/tex]
- [tex]\( f(2) = -6 \)[/tex]
