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We are given the rational function \( f(x) = \frac{x^k(x - 1)(x + 3)}{x^2 + 2x - 5} \) and asked to determine for which values of \( k \), the graph of \( f(x) \) has a horizontal asymptote at \( y = 0 \).

### Step-by-Step Solution:

To find the horizontal asymptote of a rational function, we need to analyze the degrees of the polynomials in the numerator and denominator:

1. **Degree of the numerator**:
   The numerator is \( x^k(x - 1)(x + 3) \). The highest power of \( x \) in the numerator will be determined by multiplying the factors:
   \[
   x^k \cdot x \cdot x = x^{k + 2}.
   \]
   Therefore, the degree of the numerator is \( k + 2 \).

2. **Degree of the denominator**:
   The denominator is \( x^2 + 2x - 5 \), a quadratic polynomial. The highest power of \( x \) in the denominator is \( x^2 \), so the degree of the denominator is 2.

### Horizontal Asymptote Rule:
For a rational function \( \frac{P(x)}{Q(x)} \) where \( P(x) \) and \( Q(x) \) are polynomials:
- If the degree of \( P(x) \) is less than the degree of \( Q(x) \), the horizontal asymptote is at \( y = 0 \).
- If the degree of \( P(x) \) is equal to the degree of \( Q(x) \), the horizontal asymptote is at \( y = \frac{\text{leading coefficient of } P(x)}{\text{leading coefficient of } Q(x)} \).
- If the degree of \( P(x) \) is greater than the degree of \( Q(x) \), there is no horizontal asymptote (the function grows without bound).

### Applying the Rule:
- The degree of the denominator is 2.
- The degree of the numerator is \( k + 2 \).

For a horizontal asymptote at \( y = 0 \), we need the degree of the numerator to be **less than** the degree of the denominator:
\[
k + 2 < 2.
\]
Solving this inequality:
\[
k < 0.
\]

Since \( k \) is given to be a positive integer, the inequality \( k < 0 \) has **no solution**. Thus, there is no valid value of \( k \) that would give a horizontal asymptote at \( y = 0 \).

However, considering the context of the question and available answer choices, it seems we may be looking for the case where the degree of the numerator equals the degree of the denominator. For this condition:
\[
k + 2 = 2,
\]
which simplifies to:
\[
k = 0.
\]
Since the problem specifies \( k \) is a positive integer, there seems to be an error in the setup, as no valid positive \( k \) can satisfy this condition for \( y = 0 \).

Would you like me to verify this logic or proceed with related questions on rational functions and horizontal asymptotes?

The rational function f(x) = (x^k(x-1)(x+3)) / (x^2 + 2x - 5) is given. For which values of k will the graph of f have a horizontal asymptote at y = 0?

Learn how to find the horizontal asymptote for the rational function f(x) = (x^k(x-1)(x+3)) / (x^2 + 2x - 5). This problem covers key concepts of rational functions, degree of polynomials, and the horizontal asymptote theorem. Suitable for high school students.

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