How do you find vertical, horizontal and oblique asymptotes for #(x^2+x-6)/(x^2-x-6)#?

1 Answer
Jul 28, 2018

Answer:

#"vertical asymptotes at "x=-2" and "x=3#
#"horizontal asymptote at "y=1#

Explanation:

#"let "f(x)=(x^2+x-6)/(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=-2" and "x=3" 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^2/x^2+x/x^2-6/x^2)/(x^2/x^2-x/x^2-6/x^2)=(1+1/x-6/x^2)/(1-1/x-6/x^2)#

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

#y=1" 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^2+x-6)/(x^2-x-6) [-10, 10, -5, 5]}