Question

find the vertical and horizontal asymptopes of (x^2+3x)/(x^2+x-6)

Answer 1

well, vertical asympotes occur at the value of x which makes the denominator 0

Answer 2

so, it would behovve you to factor that denomiantor

Answer 3

Then for the horizzontal asymptotes, in your case, would be the ratio of the leading coeffiecents, since the degree of the numerator is equal to the degree of the denomiator

Answer 4

\[\frac{x^2+3x}{x^2+x-6}=\frac{x(x+3)}{(x-2)(x+3)}\]\[\mbox{horizontal:}\]\[\lim_{x\rightarrow\infty}\frac{x^2+3x}{x^2+x-6}=\lim_{x\rightarrow\infty}\frac{1+3/x}{1+1/x-6/x^2}=1\quad(y=1)\]\[\mbox{vertical:}\]\[\lim_{x\rightarrow 2}\frac{x(x+3)}{(x-2)(x+3)}=\infty\quad(x=2)\]\[\lim_{x\rightarrow -3}\frac{x(x+3)}{(x-2)(x+3)}=\frac{3}{5}\quad\mbox{(x=-3 isn't vertical asympote)}\]