The problem involves determining the value of $k$ for which the given rational function $f(x) = \frac{x^k(x-1)(x+3)}{x^5+2x-5}$ has a horizontal asymptote at $y = 0$.

Horizontal Asymptote Analysis:

The behavior of $f(x)$ for large values of $x$ depends on the degree of the numerator and the denominator:

1. Degree of the numerator:
   $x^k(x-1)(x+3)$ expands to $x^k \cdot (x^2 - x - 3)$, which has a leading term of $x^{k+2}$. Thus, the degree of the numerator is $k+2$.

2. Degree of the denominator:
   The denominator $x^5 + 2x - 5$ has a leading term $x^5$. Hence, the degree of the denominator is $5$.

3. Horizontal asymptote conditions:

   • If $\text{degree of numerator} < \text{degree of denominator}$, the horizontal asymptote is $y = 0$.
   • If $\text{degree of numerator} = \text{degree of denominator}$, the horizontal asymptote is determined by the ratio of leading coefficients.
   • If $\text{degree of numerator} > \text{degree of denominator}$, there is no horizontal asymptote.

Solving for $k$:

For $y = 0$ as a horizontal asymptote, the degree of the numerator must be less than the degree of the denominator: $k + 2 < 5 \implies k < 3.$ Since $k$ is a positive integer, the possible values are $k = 1$ or $k = 2$.

From the given choices, $k = 2$ (option A) is the correct answer.

Would you like additional details or explanations?
Here are 5 related questions to deepen your understanding:

1. How are horizontal asymptotes different from vertical asymptotes?
2. What happens to the graph when $k = 3$?
3. How do you calculate slant asymptotes if the degree of the numerator is exactly one more than the denominator?
4. Can horizontal asymptotes change if the function is transformed?
5. How can we apply limits at infinity to confirm horizontal asymptotes?

Tip: To identify asymptotes, focus on the dominant terms (highest-degree terms) of the numerator and denominator.