How do you find the vertical, horizontal and slant asymptotes of: #y = (x^2 + 2x - 3)/( x^2 - 5x - 6) #?

1 Answer
Jun 10, 2016

Answer:

vertical asymptotes x = -1 , x = 6
horizontal asymptote y = 1

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^2-5x-6)=0rArr(x-6)(x+1)=0#

#rArrx=-1,x=6" are the asymptotes"#

Horizontal asymptotes occur as

#lim_(xto+-oo),ytoc" (a constant)"#

divide terms on numerator/denominator by #x^2#

#(x^2/x^2+(2x)/x^2-3/x^2)/(x^2/x^2-(5x)/x^2-6/x^2)=(1+2/x-3/x^2)/(1-5/x-6/x^2)#

as #xto+-oo,yto(1+0-0)/(1-0-0)#

#rArry=1" is the asymptote"#

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both of degree 2 ) Hence there are no slant asymptotes.
graph{(x^2+2x-3)/(x^2-5x-6) [-10, 10, -5, 5]}