Question

1. find the vertical, horizontal and slant asymptotes for $y = \frac{x^{3}}{x^{2}+x - 6}$

Explanation:

Step1: Simplify the denominator

Factor \(x^{2}+x - 6=(x + 3)(x-2)\)

Step2: Find the vertical asymptotes

Set the denominator equal to zero \((x + 3)(x - 2)=0\), so \(x=-3\) and \(x = 2\) are vertical asymptotes.

Step3: Determine the horizontal or slant asymptote

Since the degree of the numerator (\(n = 3\)) is one more than the degree of the denominator (\(m=2\)), we use polynomial long - division.
Divide \(x^{3}\) by \(x^{2}+x - 6\):

$$
LATEXBLOCK0
$$

As \(x\to\pm\infty\), \(\frac{7x - 6}{x^{2}+x - 6}\to0\). So the slant asymptote is \(y=x - 1\) and there is no horizontal asymptote.

Answer:

Vertical asymptotes: \(x=-3\) and \(x = 2\); Slant asymptote: \(y=x - 1\); No horizontal asymptote.