To analyze the problem from the image:

The function provided is:
$f(n) = -2(3)^n + 4$

We are tasked with describing the long-run behavior of $f(n)$, which involves understanding what happens to $f(n)$ as $n \to \infty$ (as $n$ becomes very large) and $n \to -\infty$ (as $n$ becomes very negative). Here's how to proceed:

1. Behavior as $n \to \infty$:

• The term $3^n$ grows exponentially as $n \to \infty$. Thus, $-2(3)^n$ grows negatively without bound (it approaches $-\infty$).
• Adding 4 does not significantly change this behavior, as the $-2(3)^n$ term dominates.
  $\text{As } n \to \infty, f(n) \to -\infty.$

2. Behavior as $n \to -\infty$:

• When $n \to -\infty$, $3^n = (3^{-1})^n = \left(\frac{1}{3}\right)^n$, which approaches 0 because powers of fractions decrease rapidly. Thus, $-2(3)^n \to 0$.
• Adding 4 means the result approaches: $\text{As } n \to -\infty, f(n) \to 4.$

Final Answer:

• As $n \to \infty$, $f(n) \to -\infty$.
• As $n \to -\infty$, $f(n) \to 4$.

Let me know if you'd like further clarification or a detailed breakdown! 😊

Related Questions:

1. What are the general rules for determining the end behavior of exponential functions?
2. How does changing the base of the exponential term affect the growth rate?
3. How does the coefficient $-2$ impact the function's range and direction?
4. What happens to $f(n)$ if we change $+4$ to $-4$?
5. How would the behavior of $f(n)$ change if the base $3^n$ were replaced by $(1/2)^n$?

Tip:

When analyzing the long-run behavior of functions, focus on the term with the highest growth rate as $n \to \infty$ or $n \to -\infty$, and consider how the sign and constants affect the outcome.