How do you find the vertical, horizontal and slant asymptotes of: #(2x-1)/(x^2+3x)#?

1 Answer
Apr 30, 2016

Answer:

vertical asymptotes x = -3 , x = 0
horizontal asymptote y = 0

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+3x=0 → x(x+3)=0 → x = 0 , x = -3 #

#rArr x=-3" and " x=0" are the asymptotes "#

Horizontal asymptotes occur as #lim_(x to +- oo) , f(x) to 0#

If the degree of the numerator < degree of denominator , as is the case here numerator (degree 1) and denominator (degree 2) then the equation is always y = 0

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there are no slant asymptotes.
graph{(2x-1)/(x^2+3x) [-10, 10, -5, 5]}