Alright, let's go through the problem step by step.
### Step 1: Identify [tex]\(a\)[/tex] and [tex]\(b\)[/tex]
Given the exponential function [tex]\( f(x) = 3 \cdot 2^x \)[/tex], we can identify the coefficients [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
- [tex]\(a\)[/tex] is the coefficient in front of the exponential term, which is [tex]\(3\)[/tex].
- [tex]\(b\)[/tex] is the base of the exponent, which is [tex]\(2\)[/tex].
### Step 2: Calculate [tex]\( f(x) \)[/tex] for given [tex]\( x \)[/tex] values
We are given a table with specific [tex]\( x \)[/tex] values: [tex]\(-2, -1, 0, 1, 2\)[/tex]. We will plug these [tex]\( x \)[/tex] values into the function [tex]\( f(x) = 3 \cdot 2^x \)[/tex] to find their corresponding [tex]\( f(x) \)[/tex] values.
1. For [tex]\( x = -2 \)[/tex]:
[tex]\[
f(-2) = 3 \cdot 2^{-2} = 3 \cdot \frac{1}{4} = 0.75
\][/tex]
2. For [tex]\( x = -1 \)[/tex]:
[tex]\[
f(-1) = 3 \cdot 2^{-1} = 3 \cdot \frac{1}{2} = 1.5
\][/tex]
3. For [tex]\( x = 0 \)[/tex]:
[tex]\[
f(0) = 3 \cdot 2^0 = 3 \cdot 1 = 3
\][/tex]
4. For [tex]\( x = 1 \)[/tex]:
[tex]\[
f(1) = 3 \cdot 2^1 = 3 \cdot 2 = 6
\][/tex]
5. For [tex]\( x = 2 \)[/tex]:
[tex]\[
f(2) = 3 \cdot 2^2 = 3 \cdot 4 = 12
\][/tex]
Thus, the completed table is:
[tex]\[
\begin{tabular}{|c|c|c|c|c|c|}
\hline
$x$ & -2 & -1 & 0 & 1 & 2 \\
\hline
$f ( x )$ & 0.75 & 1.5 & 3 & 6 & 12 \\
\hline
\end{tabular}
\][/tex]
### Step 3: Identify the [tex]\( y \)[/tex]-intercept
The [tex]\( y \)[/tex]-intercept is the value of the function when [tex]\( x = 0 \)[/tex].
[tex]\[
f(0) = 3
\][/tex]
So, the [tex]\( y \)[/tex]-intercept is [tex]\( 3 \)[/tex].
### Step 4: Determine the end behavior
For the end behavior, we analyze what happens to [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches [tex]\( \infty \)[/tex] and [tex]\( -\infty \)[/tex]:
- As [tex]\( x \to \infty \)[/tex], the term [tex]\( 2^x \)[/tex] grows exponentially, and thus [tex]\( f(x) \to \infty \)[/tex]. Therefore, as [tex]\( x \to \infty \)[/tex], [tex]\( f(x) \to \infty \)[/tex].
- As [tex]\( x \to -\infty \)[/tex], the term [tex]\( 2^x \)[/tex] approaches [tex]\( 0 \)[/tex] because [tex]\( 2^x \)[/tex] is a fraction that gets smaller and smaller. So, [tex]\( f(x) \to 0 \)[/tex]. Therefore, as [tex]\( x \to -\infty \)[/tex], [tex]\( f(x) \to 0 \)[/tex].
### Summary
- [tex]\( a = 3 \)[/tex]
- [tex]\( b = 2 \)[/tex]
- [tex]\( y \)[/tex]-intercept: [tex]\( 3 \)[/tex]
- End behavior:
- As [tex]\( x \to \infty \)[/tex], [tex]\( f(x) \to \infty \)[/tex]
- As [tex]\( x \to -\infty \)[/tex], [tex]\( f(x) \to 0 \)[/tex]
The graph of the exponential function [tex]\( f(x) = 3 \cdot 2^x \)[/tex] will show the values calculated in the table, with the [tex]\( y \)[/tex]-intercept at [tex]\( 3 \)[/tex], growing rapidly as [tex]\( x \)[/tex] increases, and approaching [tex]\( 0 \)[/tex] as [tex]\( x \)[/tex] decreases.
