Question

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

Explanation:

Step1: Simplify the function

First, factor the denominator \(x^{2}+x - 6=(x + 3)(x - 2)\). The function is \(y=\frac{x^{3}}{(x + 3)(x - 2)}\). The vertical asymptotes occur where the denominator is zero (and the numerator is non - zero). Set \((x + 3)(x - 2)=0\), so \(x=-3\) and \(x = 2\) are vertical asymptotes.

Step2: Check for horizontal asymptote

For horizontal asymptotes, compare the degrees of the numerator \(n = 3\) and denominator \(m=2\). Since \(n>m\), there is no horizontal asymptote.

Step3: Find the slant asymptote

Use polynomial long division. Divide \(x^{3}\) by \(x^{2}+x - 6\).

$$
LATEXBLOCK0
$$

As \(x
ightarrow\pm\infty\), \(\frac{5x - 6}{x^{2}+x - 6}
ightarrow0\). So the slant asymptote is \(y=x - 1\).

Answer:

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