4. 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). [ \begin{align*} x^{3}&=(x - 1)(x^{2}+x - 6)+(5x - 6)\ y&=\frac{x^{3}}{x^{2}+x - 6}=x - 1+\frac{5x - 6}{x^{2}+x - 6} \end{align*} ] As (x\rightarrow\pm\infty), (\frac{5x - 6}{x^{2}+x - 6}\rightarrow0). So the slant asymptote is (y=x - 1).

Answer:

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