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

1 Answer
May 7, 2016

Answer:

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]}