To determine the horizontal and vertical asymptotes of the function $f(x) = \frac{4x^2}{x^3 - 1}$, let's go through the process step by step.

1. Vertical Asymptotes:

Vertical asymptotes occur where the denominator is equal to zero (as long as the numerator isn't also zero at the same point). So, we need to solve for $x$ when the denominator is zero:

$x^3 - 1 = 0$

$x^3 = 1$

$x = 1$

Thus, there is a vertical asymptote at $x = 1$.

2. Horizontal Asymptote:

To find the horizontal asymptote, we compare the degrees of the polynomial in the numerator and the denominator:

• The degree of the numerator ($4x^2$) is 2.
• The degree of the denominator ($x^3 - 1$) is 3.

When the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is $y = 0$.

Thus, the horizontal asymptote is $y = 0$.

Summary:

• Vertical asymptote: $x = 1$
• Horizontal asymptote: $y = 0$

Would you like more details on how I arrived at these results, or do you have any specific questions?

Here are 5 related questions for further exploration:

1. How do you find oblique asymptotes?
2. What happens if the degree of the numerator is greater than the denominator?
3. Can a function have more than one vertical asymptote?
4. How does the behavior of the function change near the vertical asymptote?
5. Can rational functions have no asymptotes at all?

Tip: Always check for any common factors before determining asymptotes, as canceling them might change the function's behavior at certain points.