Graph each exponential function. Identify [tex]\(a\)[/tex], [tex]\(b\)[/tex], the [tex]\(y\)[/tex]-intercept, and the end behavior of the graph.

1. [tex]\(f(x)=-3(3)^x\)[/tex]

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

[tex]\(a=\)[/tex] [tex]\(\qquad\)[/tex] [tex]\(b=\)[/tex] [tex]\(\qquad\)[/tex] [tex]\(y\)[/tex]-intercept [tex]\(=\)[/tex] [tex]\(\qquad\)[/tex]

End Behavior:
- As [tex]\(x \rightarrow +\infty, y \rightarrow\)[/tex] [tex]\(\qquad\)[/tex]
- As [tex]\(x \rightarrow -\infty, y \rightarrow\)[/tex] [tex]\(\qquad\)[/tex]



Answer :

Sure! Let's analyze the exponential function [tex]\( f(x) = -3(3^x) \)[/tex] step-by-step:

### 1. Identify [tex]\( a \)[/tex] and [tex]\( b \)[/tex]

The general form of an exponential function is [tex]\( f(x) = a(b^x) \)[/tex]. Here, we can directly see:

- [tex]\( a = -3 \)[/tex]
- [tex]\( b = 3 \)[/tex]

### 2. Determine the [tex]\( y \)[/tex]-intercept

The [tex]\( y \)[/tex]-intercept occurs where [tex]\( x = 0 \)[/tex]. To find this, substitute [tex]\( x = 0 \)[/tex] into the function:

[tex]\[ f(0) = -3 \cdot (3^0) = -3 \cdot 1 = -3 \][/tex]

Thus, the [tex]\( y \)[/tex]-intercept is [tex]\(-3\)[/tex].

### 3. Calculate [tex]\( f(x) \)[/tex] for specified [tex]\( x \)[/tex] values

We need to find the values of [tex]\( f(x) \)[/tex] for [tex]\( x = -2, -1, 0, 1, 2 \)[/tex]:

- For [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = -3 \cdot (3^{-2}) = -3 \cdot \frac{1}{3^2} = -3 \cdot \frac{1}{9} = -\frac{3}{9} = -0.3333 \][/tex]

- For [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = -3 \cdot (3^{-1}) = -3 \cdot \frac{1}{3} = -3 \cdot 0.3333 = -1 \][/tex]

- For [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -3 \cdot (3^0) = -3 \cdot 1 = -3 \][/tex]

- For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = -3 \cdot (3^1) = -3 \cdot 3 = -9 \][/tex]

- For [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = -3 \cdot (3^2) = -3 \cdot 9 = -27 \][/tex]

### 4. Determine the end behavior

Evaluate the behavior of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches positive and negative infinity:

- As [tex]\( x \to +\infty \)[/tex]: Since [tex]\( b = 3 > 1 \)[/tex] and the coefficient [tex]\( a = -3 \)[/tex] is negative, the function grows exponentially more negative. Graphically, this means the values of [tex]\( f(x) \)[/tex] become very large negative numbers.

So, as [tex]\( x \to +\infty \)[/tex], [tex]\( y \to -\infty \)[/tex].

- As [tex]\( x \to -\infty \)[/tex]: For very large negative values of [tex]\( x \)[/tex], the term [tex]\( 3^x \)[/tex] approaches 0 because any number raised to a negative power gets closer to zero. However, multiplied by [tex]\(-3\)[/tex], it becomes a very small negative number but still close to zero.

So, as [tex]\( x \to -\infty \)[/tex], [tex]\( y \to 0 \)[/tex].

### Summary

Here is the result:

[tex]\[ \begin{array}{|l|l|l|l|l|l|} \hline x & -2 & -1 & 0 & 1 & 2 \\ \hline f(x) & -0.3333 & -1 & -3 & -9 & -27 \\ \hline \end{array} \][/tex]

- [tex]\( a = -3 \)[/tex]
- [tex]\( b = 3 \)[/tex]
- [tex]\( y \)[/tex]-intercept = [tex]\(-3\)[/tex]
- End Behavior:
- As [tex]\( x \to +\infty \)[/tex], [tex]\( y \to -\infty \)[/tex]
- As [tex]\( x \to -\infty \)[/tex], [tex]\( y \to 0 \)[/tex]