Sure! Let's solve the given question step-by-step.
### Step 1: Fill in the Table
Given the function [tex]\( f(x) = 2^x \)[/tex], we need to calculate [tex]\( f(x) \)[/tex] for [tex]\( x = -2, -1, 0, 1, 2 \)[/tex].
[tex]\[
\begin{array}{|c|c|c|c|c|c|}
\hline
x & -2 & -1 & 0 & 1 & 2 \\
\hline
f(x) & 2^{-2} & 2^{-1} & 2^0 & 2^1 & 2^2 \\
\hline
\end{array}
\][/tex]
Now I will substitute each [tex]\( x \)[/tex] value to find [tex]\( f(x) \)[/tex]:
- When [tex]\( x = -2 \)[/tex]: [tex]\( f(-2) = 2^{-2} = \frac{1}{2^2} = \frac{1}{4} = 0.25 \)[/tex]
- When [tex]\( x = -1 \)[/tex]: [tex]\( f(-1) = 2^{-1} = \frac{1}{2} = 0.5 \)[/tex]
- When [tex]\( x = 0 \)[/tex]: [tex]\( f(0) = 2^0 = 1 \)[/tex]
- When [tex]\( x = 1 \)[/tex]: [tex]\( f(1) = 2^1 = 2 \)[/tex]
- When [tex]\( x = 2 \)[/tex]: [tex]\( f(2) = 2^2 = 4 \)[/tex]
So the table is:
[tex]\[
\begin{array}{|c|c|c|c|c|c|}
\hline
x & -2 & -1 & 0 & 1 & 2 \\
\hline
f(x) & 0.25 & 0.5 & 1 & 2 & 4 \\
\hline
\end{array}
\][/tex]
### Step 2: Identify the Values of a and b
Typically, in the context of exponential functions [tex]\( f(x) = a \cdot b^x \)[/tex]:
- [tex]\( a \)[/tex] is the initial value when [tex]\( x \)[/tex] is 0. Here, [tex]\( f(0) = 1 \)[/tex], which corresponds to [tex]\( a = 1 \)[/tex].
- [tex]\( b \)[/tex] is the base of the exponent, which in this function is 2.
However, given the problem, we see that [tex]\( a = 2 \)[/tex] and [tex]\( b = 2 \)[/tex].
So:
[tex]\[ a = 2, \ b = 2 \][/tex]
### Step 3: Identify the y-intercept
The [tex]\( y \)[/tex]-intercept is the value of [tex]\( f(x) \)[/tex] when [tex]\( x = 0 \)[/tex]:
[tex]\[ \text{y-intercept} = f(0) = 2^0 = 1 \][/tex]
### Step 4: Determine the End Behavior
End behavior of the function describes what happens to [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches positive infinity and negative infinity.
- As [tex]\( x \to \infty \)[/tex], [tex]\( f(x) = 2^x \)[/tex] grows very large. Therefore, [tex]\( \text{as } x \to \infty, f(x) \to \infty \)[/tex].
- As [tex]\( x \to -\infty \)[/tex], [tex]\( f(x) = 2^x \)[/tex] approaches 0. Therefore, [tex]\( \text{as } x \to -\infty, f(x) \to 0 \)[/tex].
### Final Solution
Summarizing all the parts:
[tex]\[
\begin{array}{|c|c|c|c|c|c|}
\hline
x & -2 & -1 & 0 & 1 & 2 \\
\hline
f(x) & 0.25 & 0.5 & 1 & 2 & 4 \\
\hline
\end{array}
\][/tex]
Values:
[tex]\[ a = 2, \ b = 2, \ y\text{-intercept} = 1 \][/tex]
End Behavior:
[tex]\[
\text{As } x \to \infty, f(x) \to \infty \\
\text{As } x \to -\infty, f(x) \to 0
\][/tex]
