The given function is:

$y = \frac{18(x+2)^2}{x^3}.$

We are tasked with finding the asymptotes of this function.

Step 1: Analyze vertical asymptotes

Vertical asymptotes occur when the denominator of the function equals zero, provided the numerator is nonzero at the same points. For this function, the denominator is $x^3$. Thus, the vertical asymptote is:

$x = 0.$

Step 2: Analyze horizontal or oblique asymptotes

To check for horizontal or oblique asymptotes, evaluate the behavior of the function as $x \to \infty$ or $x \to -\infty$.

1. Simplify the function for large $|x|$: $y = \frac{18(x+2)^2}{x^3}.$

   Expand $(x+2)^2 = x^2 + 4x + 4$, so: $y = \frac{18(x^2 + 4x + 4)}{x^3} = \frac{18x^2}{x^3} + \frac{72x}{x^3} + \frac{72}{x^3}.$

   Simplifying: $y = \frac{18}{x} + \frac{72}{x^2} + \frac{72}{x^3}.$

   As $x \to \infty$ or $x \to -\infty$, all terms vanish except $\frac{18}{x}$, leading to: $y \to 0.$

   Thus, the function has a horizontal asymptote at: $y = 0.$

Final Answer

The asymptotes of the function are:

1. Vertical asymptote: $x = 0$,
2. Horizontal asymptote: $y = 0$.

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Related Questions

1. How can asymptotes help sketch the graph of a function?
2. What is the difference between a horizontal and an oblique asymptote?
3. How do vertical asymptotes relate to the domain of a function?
4. What happens to the function $y$ near $x = 0$?
5. How can this function be approximated for very large $|x|$?

Tip

When finding asymptotes, always analyze the function as $x$ approaches infinity, negative infinity, and any points where the denominator is zero.