To determine the end behavior of the exponential function $f(x) = 3^{-x} + 2$ as $x \to \infty$, let's break it down:

1. Examine the behavior of $3^{-x}$ as $x \to \infty$:

   • The term $3^{-x}$ represents an exponential decay because the base (3) is greater than 1 and the exponent is negative.
   • As $x$ increases (i.e., as $x \to \infty$), $3^{-x}$ gets closer and closer to 0.

2. Combine this with the constant 2:

   • The function becomes: $f(x) = 3^{-x} + 2$.
   • As $x \to \infty$, $3^{-x}$ approaches 0, so the function approaches $0 + 2 = 2$.

Thus, the end behavior of $f(x)$ as $x \to \infty$ is:

$f(x) \to 2$

Answer: f(x) ➝ 2

Would you like further clarification on any part of this explanation? Here are a few related questions for deeper understanding:

1. What is the end behavior of the function as $x \to -\infty$?
2. How would the end behavior change if the function were $f(x) = 3^x + 2$?
3. How do exponential functions compare to polynomial functions in terms of end behavior?
4. Can you explain how the base of an exponential affects the decay or growth rate?
5. What is the impact of changing the constant term (the "+2") on the end behavior?

Tip: For exponential decay functions like $3^{-x}$, the function approaches zero as $x \to \infty$ but never actually reaches zero. It asymptotically approaches the horizontal line $y = 2$.