Let's analyze the given function step by step. The function is:

$f(x) = \frac{2(x-3)}{5(x-3)(x+2)^2}$

1. Horizontal Asymptotes:

To determine horizontal asymptotes, we compare the degrees of the numerator and the denominator:

• Degree of the numerator: The numerator $2(x-3)$ is a linear term, so the degree is 1.
• Degree of the denominator: The denominator $5(x-3)(x+2)^2$ contains a quadratic term $(x+2)^2$, so the degree is 3.

Since the degree of the numerator is smaller than the degree of the denominator, the horizontal asymptote is at $y = 0$.

Conclusion: The graph has a horizontal asymptote at $y = 0$.

2. Vertical Asymptotes:

Vertical asymptotes occur where the denominator equals zero, provided the numerator doesn't cancel the factor. The denominator is:

$5(x-3)(x+2)^2$

Setting the denominator equal to zero gives us:

$(x-3)(x+2)^2 = 0$

• $x - 3 = 0$ gives $x = 3$
• $(x+2)^2 = 0$ gives $x = -2$

Now, check for cancellations: the factor $(x - 3)$ appears in both the numerator and denominator, so it cancels out. Therefore, $x = 3$ is a hole, not a vertical asymptote.

Thus, the only vertical asymptote is at $x = -2$.

Conclusion: There is 1 vertical asymptote at $x = -2$.

3. Holes:

Holes occur where a factor cancels from both the numerator and the denominator. In this case, the factor $(x - 3)$ cancels, so there is a hole at $x = 3$.

Conclusion: There is a hole at $x = 3$.

Summary:

• Horizontal Asymptotes: 1 (at $y = 0$)
• Holes: 1 (at $x = 3$)
• Vertical Asymptotes: 1 (at $x = -2$)

Let me know if you'd like further explanation or more details! Here are some related questions to expand on:

1. How do you determine the slant asymptote for a rational function?
2. What happens to the behavior of the function as $x \to -2$?
3. Can there be more than one hole in a rational function?
4. How do you find the y-intercept of this function?
5. What is the domain of this function?

Tip: Always check for common factors between the numerator and denominator to identify holes in the graph.