To fill in the table using the function [tex]\( f(x) = 2 - 5x \)[/tex], we need to evaluate the function at each given value of [tex]\( x \)[/tex]. Let's walk through the process step-by-step for each value:
1. For [tex]\( x = -2 \)[/tex]:
[tex]\[
f(-2) = 2 - 5(-2) = 2 + 10 = 12
\][/tex]
So, when [tex]\( x = -2 \)[/tex], [tex]\( f(x) = 12 \)[/tex].
2. For [tex]\( x = -1 \)[/tex]:
[tex]\[
f(-1) = 2 - 5(-1) = 2 + 5 = 7
\][/tex]
So, when [tex]\( x = -1 \)[/tex], [tex]\( f(x) = 7 \)[/tex].
3. For [tex]\( x = 0 \)[/tex]:
[tex]\[
f(0) = 2 - 5(0) = 2 - 0 = 2
\][/tex]
So, when [tex]\( x = 0 \)[/tex], [tex]\( f(x) = 2 \)[/tex].
4. For [tex]\( x = 1 \)[/tex]:
[tex]\[
f(1) = 2 - 5(1) = 2 - 5 = -3
\][/tex]
So, when [tex]\( x = 1 \)[/tex], [tex]\( f(x) = -3 \)[/tex].
5. For [tex]\( x = 2 \)[/tex]:
[tex]\[
f(2) = 2 - 5(2) = 2 - 10 = -8
\][/tex]
So, when [tex]\( x = 2 \)[/tex], [tex]\( f(x) = -8 \)[/tex].
Now, we can fill in the table with the computed values:
[tex]\[
\begin{tabular}{|l|l|}
\hline
$x$ & $f(x)$ \\
\hline
-2 & 12 \\
\hline
-1 & 7 \\
\hline
0 & 2 \\
\hline
1 & -3 \\
\hline
2 & -8 \\
\hline
\end{tabular}
\][/tex]
This completes the table with the given function [tex]\( f(x) = 2 - 5x \)[/tex].
