How do you find the vertical, horizontal or slant asymptotes for #x/(x^2+ 5x +6)#?

1 Answer
Jun 11, 2016

Answer:

vertical asymptotes x = -3 , x = -2
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+5x+6=0rArr(x+2)(x+3)=0#

#rArrx=-3,x=-2" are the asymptotes"#

Horizontal asymptotes occur as

#lim_(xto+-oo),f(x)toc" (a constant)"#

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

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

as #xto+-oo,f(x)to0/(1+0+0)#

#rArry=0" is the asymptote"#

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 1 , denominator-degree 2 ) Hence there are no slant asymptotes.
graph{x/(x^2+5x+6) [-10, 10, -5, 5]}