The following table represents an exponential function. Fill in the missing blanks.

\begin{tabular}{|l|l|}
\hline
[tex]$x$[/tex] & [tex]$y$[/tex] \\
\hline
-2 & 0.25 \\
\hline
-1 & \\
\hline
0 & 1 \\
\hline
1 & \\
\hline
2 & \\
\hline
\end{tabular}



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]