Absolutely, let's complete the table step-by-step by evaluating the function [tex]\( f(x) = 3x^2 - 5 \)[/tex] for each given [tex]\( x \)[/tex] value.
1. For [tex]\( x = -2 \)[/tex]:
- Substitute [tex]\( x = -2 \)[/tex] into the function:
[tex]\[
f(-2) = 3(-2)^2 - 5
\][/tex]
- Calculate the square of [tex]\(-2\)[/tex]:
[tex]\[
(-2)^2 = 4
\][/tex]
- Multiply by 3:
[tex]\[
3 \cdot 4 = 12
\][/tex]
- Subtract 5:
[tex]\[
12 - 5 = 7
\][/tex]
- So, [tex]\( f(-2) = 7 \)[/tex].
2. For [tex]\( x = -1 \)[/tex]:
- Substitute [tex]\( x = -1 \)[/tex] into the function:
[tex]\[
f(-1) = 3(-1)^2 - 5
\][/tex]
- Calculate the square of [tex]\(-1\)[/tex]:
[tex]\[
(-1)^2 = 1
\][/tex]
- Multiply by 3:
[tex]\[
3 \cdot 1 = 3
\][/tex]
- Subtract 5:
[tex]\[
3 - 5 = -2
\][/tex]
- So, [tex]\( f(-1) = -2 \)[/tex].
3. For [tex]\( x = 0 \)[/tex]:
- Substitute [tex]\( x = 0 \)[/tex] into the function:
[tex]\[
f(0) = 3 \cdot 0^2 - 5
\][/tex]
- Calculate the square of 0:
[tex]\[
0^2 = 0
\][/tex]
- Multiply by 3:
[tex]\[
3 \cdot 0 = 0
\][/tex]
- Subtract 5:
[tex]\[
0 - 5 = -5
\][/tex]
- So, [tex]\( f(0) = -5 \)[/tex].
4. For [tex]\( x = 1 \)[/tex]:
- Substitute [tex]\( x = 1 \)[/tex] into the function:
[tex]\[
f(1) = 3 \cdot 1^2 - 5
\][/tex]
- Calculate the square of 1:
[tex]\[
1^2 = 1
\][/tex]
- Multiply by 3:
[tex]\[
3 \cdot 1 = 3
\][/tex]
- Subtract 5:
[tex]\[
3 - 5 = -2
\][/tex]
- So, [tex]\( f(1) = -2 \)[/tex].
Now, we can complete the table:
[tex]\[
\begin{tabular}{|c|c|}
\hline
$x$ & $f(x)$ \\
\hline
-2 & 7 \\
\hline
-1 & -2 \\
\hline
0 & -5 \\
\hline
1 & -2 \\
\hline
\end{tabular}
\][/tex]
