The given function is $f(x) = 2(5)^x - 4$, and you are asked to describe the long-run behavior, meaning what happens to $f(x)$ as $x \to \infty$ (positive infinity) and $x \to -\infty$ (negative infinity).

Long-run behavior as $x \to \infty$:

1. When $x \to \infty$, the term $5^x$ grows extremely large because the base of the exponent (5) is greater than 1. This causes $2(5)^x$ to also grow without bound.
2. Therefore, as $x \to \infty$, the term $2(5)^x$ dominates, and the function $f(x) \to \infty$.

$\boxed{\lim_{x \to \infty} f(x) = \infty}$

Long-run behavior as $x \to -\infty$:

1. When $x \to -\infty$, the term $5^x$ approaches 0 because raising 5 to a negative power results in very small values (approaching zero).
2. So, $2(5)^x$ approaches zero as well. The constant term $-4$ remains.
3. Therefore, as $x \to -\infty$, $f(x) \to -4$.

$\boxed{\lim_{x \to -\infty} f(x) = -4}$

Would you like further details or have any questions?

Related Questions:

1. What is the y-intercept of $f(x) = 2(5)^x - 4$?
2. How would the long-run behavior change if the base of the exponent was less than 1?
3. How does the value of the constant term $-4$ affect the graph of the function?
4. Can you find the horizontal asymptote of the function $f(x) = 2(5)^x - 4$?
5. What is the derivative of $f(x) = 2(5)^x - 4$, and what does it tell us about the function’s behavior?

Tip:

When analyzing exponential functions, the base of the exponential term is critical in determining whether the function grows or decays as $x \to \infty$ or $x \to -\infty$.