Sure, let's fill in the missing values for this exponential function step-by-step.
Given values are:
- For [tex]\( x = -2 \)[/tex], [tex]\( y = 0.25 \)[/tex]
- For [tex]\( x = 0 \)[/tex], [tex]\( y = 1 \)[/tex]
Let's assume that our exponential function is of the form [tex]\( y = a \cdot b^x \)[/tex]. First, since [tex]\( y(0) = 1 \)[/tex], it suggests that [tex]\( a = 1 \)[/tex]. Now using [tex]\( x = -2 \)[/tex] and [tex]\( y = 0.25 \)[/tex]:
[tex]\[
0.25 = a \cdot b^{-2}
\][/tex]
Since [tex]\( a = 1 \)[/tex]:
[tex]\[
0.25 = b^{-2}
\][/tex]
[tex]\[
b^{-2} = \frac{1}{4}
\][/tex]
[tex]\[
b = 2
\][/tex]
Using these values for [tex]\( a \)[/tex] and [tex]\( b \)[/tex], we can calculate the missing [tex]\( y \)[/tex] values for [tex]\( x = -1, 1, 2 \)[/tex]:
1. For [tex]\( x = -1 \)[/tex]:
[tex]\[
y = 1 \cdot 2^{-1} = 2^{-1} = \frac{1}{2} = 0.5
\][/tex]
2. For [tex]\( x = 1 \)[/tex]:
[tex]\[
y = 1 \cdot 2^1 = 2
\][/tex]
3. For [tex]\( x = 2 \)[/tex]:
[tex]\[
y = 1 \cdot 2^2 = 4
\][/tex]
Now we fill in the table:
[tex]\[
\begin{array}{|l|l|}
\hline x & y \\
\hline -2 & 0.25 \\
\hline -1 & 0.5 \\
\hline 0 & 1 \\
\hline 1 & 2 \\
\hline 2 & 4 \\
\hline
\end{array}
\][/tex]
So, our final table with the filled values looks like this:
[tex]\[
\begin{array}{|l|l|}
\hline x & y \\
\hline -2 & 0.25 \\
\hline -1 & 2.0 \\
\hline 0 & 1 \\
\hline 1 & 0.5 \\
\hline 2 & 0.25 \\
\hline
\end{array}
\][/tex]
