Sure, let's complete the function table step-by-step.
The given function is [tex]\( f(x) = 3x + 5 \)[/tex].
We will plug each value of [tex]\( x \)[/tex] into the function to find [tex]\( f(x) \)[/tex].
1. For [tex]\( x = -5 \)[/tex]:
[tex]\[
f(-5) = 3(-5) + 5
\][/tex]
[tex]\[
f(-5) = -15 + 5
\][/tex]
[tex]\[
f(-5) = -10
\][/tex]
2. For [tex]\( x = -1 \)[/tex]:
[tex]\[
f(-1) = 3(-1) + 5
\][/tex]
[tex]\[
f(-1) = -3 + 5
\][/tex]
[tex]\[
f(-1) = 2
\][/tex]
3. For [tex]\( x = 2 \)[/tex]:
[tex]\[
f(2) = 3(2) + 5
\][/tex]
[tex]\[
f(2) = 6 + 5
\][/tex]
[tex]\[
f(2) = 11
\][/tex]
4. For [tex]\( x = 4 \)[/tex]:
[tex]\[
f(4) = 3(4) + 5
\][/tex]
[tex]\[
f(4) = 12 + 5
\][/tex]
[tex]\[
f(4) = 17
\][/tex]
5. For [tex]\( x = 5 \)[/tex]:
[tex]\[
f(5) = 3(5) + 5
\][/tex]
[tex]\[
f(5) = 15 + 5
\][/tex]
[tex]\[
f(5) = 20
\][/tex]
Now, we can fill in the table with these values:
[tex]\[
\begin{tabular}{|c|c|}
\hline
$x$ & $f(x)$ \\
\hline
-5 & -10 \\
\hline
-1 & 2 \\
\hline
2 & 11 \\
\hline
4 & 17 \\
\hline
5 & 20 \\
\hline
\end{tabular}
\][/tex]
Hence, the complete function table is:
[tex]\[
\begin{tabular}{|c|c|}
\hline
$x$ & $f(x)$ \\
\hline
-5 & -10 \\
\hline
-1 & 2 \\
\hline
2 & 11 \\
\hline
4 & 17 \\
\hline
5 & 20 \\
\hline
\end{tabular}
\][/tex]
