How do you find the Vertical, Horizontal, and Oblique Asymptote given #x/(x^2+x-6)#?

1 Answer
Jun 19, 2018

Answer:

#"vertical asymptotes at "x--3" and "x=2#
#"horizontal asymptote at "y=0#

Explanation:

#"let "f(x)=x/(x^2+x-6)#

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

#"solve "x^2+x-6=0rArr(x+3)(x-2)=0#

#x=-3" and "x=2" are the asymptotes"#

#"horizontal asymptotes occur as"#

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

#"divide terms on numerator/denominator by the highest"#
#"power of "x" that is "x^2#

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

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

#y=0" is the asymptote"#

Oblique asymptotes occur when the degree of the numerator is greater than the degree of the denominator. This is not the case here hence there are no oblique asymptotes.
graph{x/(x^2+x-6) [-10, 10, -5, 5]}