How do you find the asymptotes for # f(x) = x/((x+3)(x-4))#?

1 Answer
Jul 20, 2018

Answer:

#"vertical asymptotes at "x=-3" and "x=4#
#"horizontal asymptote at "y=0#

Explanation:

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

#"solve "(x+3)(x-4)=0#

#rArrx=-3" and "x=4" 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^2-12/x^2)=(1/x)/(1-1/x-12/x^2)#

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

#y=0" is the asymptote"#
graph{x/((x+3)(x-4)) [-10, 10, -5, 5]}