The function $F(x) = 5(4)^x - 2$ is an exponential function. To describe its end behavior, we examine what happens to $F(x)$ as $x \to \infty$ and $x \to -\infty$.

As $x \to \infty$:

• The term $(4)^x$ grows very rapidly because $4^x$ is an exponential term with base 4.
• Therefore, as $x \to \infty$, $4^x \to \infty$.
• As a result, $F(x) = 5(4^x) - 2$ will also increase rapidly and approach infinity.

Thus, as $x \to \infty$, $F(x) \to \infty$.

As $x \to -\infty$:

• The term $(4)^x$ approaches 0 because $4^x \to 0$ as $x \to -\infty$.
• Therefore, $F(x) = 5(4^x) - 2$ will approach $5(0) - 2 = -2$.

Thus, as $x \to -\infty$, $F(x) \to -2$.

Conclusion:

• As $x \to \infty$, $F(x) \to \infty$.
• As $x \to -\infty$, $F(x) \to -2$.

The graph of the function rises sharply as $x$ increases and approaches the horizontal asymptote $y = -2$ as $x$ decreases.

Would you like further details or have any other questions?

Here are 5 related questions:

1. How does the base of an exponential function affect its growth?
2. What is the significance of the horizontal asymptote in this function?
3. How would the function change if the constant $-2$ were different?
4. Can you determine the y-intercept of this function?
5. How would the graph of the function change if the base were less than 1?

Tip: When analyzing end behavior, always focus on what happens to the largest terms as $x \to \infty$ or $x \to -\infty$.