# How do you find the vertical, horizontal or slant asymptotes for #(3x^2+2x-1)/(x^2-4) #?

##### 1 Answer

May 7, 2016

vertical asymptotes x = ± 2

horizontal asymptote y = 3

#### Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation/s set the denominator equal to zero.

solve :

#x^2-4=0 → (x-2)(x+2)=0 → x=±2#

#rArrx=-2" and " x=2" are the asymptotes "# Horizontal asymptotes occur as

#lim_(x to +- oo) , f(x) to 0# divide all terms on numerator/denominator by

#x^2#

#((3x^2)/x^2+(2x)/x^2-1/x^2)/(x^2/x^2-4/x^2)=(3+2/x-1/x^2)/(1-4/x^2)# as

#x to +- oo , y to (3+0-0)/(1-0)#

#rArry=3" is the asymptote "# Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both degree 2) hence there are no slant asymptotes.

graph{(3x^2+2x-1)/(x^2-4) [-10, 10, -5, 5]}