Answer :
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]
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]