Let's determine the function [tex]\( f(x) \)[/tex] given the points and the missing [tex]\( f(x) \)[/tex] values. We'll assume the function is linear, which means it has the form [tex]\( f(x) = ax + b \)[/tex].
We are given:
[tex]\[
\begin{array}{|c|c|}
\hline$x$ & $f(x)$ \\
\hline-8 & 8 \\
\hline-4 & \text{Unknown} \\
\hline-2 & \text{Unknown} \\
\hline
\end{array}
\][/tex]
Given this, we need to find the coefficients [tex]\( a \)[/tex] and [tex]\( b \)[/tex] for the linear equation [tex]\( f(x) = ax + b \)[/tex].
1. Use the point [tex]\((-8, 8)\)[/tex] to write out the equation:
[tex]\[ 8 = a(-8) + b \][/tex]
Since this is linear interpolation and we need to determine the missing values of [tex]\( f(x) \)[/tex], we solve for the coefficients.
Looking at the given outputs:
2. Since the given point is [tex]\((-8, 8)\)[/tex], we use it to understand the function's behavior.
Since the function is linear and the pattern suggests no variation in [tex]\( f(x) \)[/tex], we can check what this implies.
Let’s compute the slope [tex]\( a \)[/tex]. With the given data, it can hint that the slope might be zero.
[tex]\[ a = 0 \][/tex]
Substituting [tex]\( a = 0 \)[/tex] back into the equation to find [tex]\( b \)[/tex]:
[tex]\[ 8 = 0(-8) + b \][/tex]
[tex]\[ b = 8 \][/tex]
Thus, the linear function is:
[tex]\[ f(x) = 0x + 8 \][/tex]
[tex]\[ f(x) = 8 \][/tex]
Now, let's find the missing [tex]\( f(x) \)[/tex] values:
- For [tex]\( x = -4 \)[/tex]:
[tex]\[ f(-4) = 0(-4) + 8 \][/tex]
[tex]\[ f(-4) = 8 \][/tex]
- For [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = 0(-2) + 8 \][/tex]
[tex]\[ f(-2) = 8 \][/tex]
So, the complete table is:
[tex]\[
\begin{array}{|c|c|}
\hline$x$ & $f(x)$ \\
\hline-8 & 8 \\
\hline-4 & 8 \\
\hline-2 & 8 \\
\hline
\end{array}
\][/tex]
The linear function that fits the given points is [tex]\( f(x) = 8 \)[/tex]. Therefore, for [tex]\( x = -4 \)[/tex] and [tex]\( x = -2 \)[/tex], the values of [tex]\( f(x) \)[/tex] are 8, making the final table:
[tex]\[
\begin{array}{|c|c|}
\hline$x$ & $f(x)$ \\
\hline-8 & 8 \\
\hline-4 & 8 \\
\hline-2 & 8 \\
\hline
\end{array}
\][/tex]
