\begin{tabular}{|c|c|c|c|c|c|}
\hline
[tex]$x$[/tex] & -2 & -1 & 0 & 1 & 2 \\
[tex]$f(x)$[/tex] & & & & & \\
\hline
\end{tabular}

Hint: [tex]$f(x)=2^x$[/tex] or [tex]$f(x)=1(2)^x$[/tex]

[tex]$a = 2, b = 2, y$[/tex]-intercept = 1

End Behavior:

As [tex]$x \rightarrow \infty, y \rightarrow \square$[/tex]

As [tex]$x \rightarrow -\infty, y \rightarrow \square$[/tex]



Answer :

Let's consider the function [tex]\( f(x) = 2^x \)[/tex]. We need to evaluate [tex]\( f(x) \)[/tex] at specific [tex]\( x \)[/tex]-values and analyze its behavior.

1. First, we'll calculate [tex]\( f(x) \)[/tex] for each given [tex]\( x \)[/tex]-value:

- For [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = 2^{-2} = \frac{1}{2^2} = \frac{1}{4} = 0.25 \][/tex]
- For [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = 2^{-1} = \frac{1}{2} = 0.5 \][/tex]
- For [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 2^0 = 1 \][/tex]
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 2^1 = 2 \][/tex]
- For [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = 2^2 = 4 \][/tex]

2. Now, we will fill in the table with the computed values:

[tex]\[ \begin{tabular}{|c|c|cc|c|c|} \hline $x$ & -2 & -1 & 0 & 1 & 2 \\ \hline $f(x)$ & 0.25 & 0.5 & 1 & 2 & 4 \\ \hline \end{tabular} \][/tex]

3. Next, we analyze the end behavior of the function.

[tex]\[ \text{As } x \to \infty, f(x) = 2^x \to \infty \][/tex]

[tex]\[ \text{As } x \to -\infty, f(x) = 2^x \to 0 \][/tex]

Therefore, the end behavior can be summarized as:
[tex]\[ \text{As } x \to \infty, y \to \infty \][/tex]

[tex]\[ \text{As } x \to -\infty, y \to 0 \][/tex]

So we can complete the end behavior statement as follows:

End Behavior:
[tex]\[ \text{As } x \rightarrow \infty, y \rightarrow \infty \\ \][/tex]
[tex]\[ \text{And as } x \rightarrow -\infty, y \rightarrow 0 \][/tex]

In summary, we have evaluated the function [tex]\( f(x) = 2^x \)[/tex] at the given points, completed the table, and described the function's end behavior.