How do you find vertical, horizontal and oblique asymptotes for #y= x/((x+3)(x-4)#?

1 Answer
May 3, 2017

Answer:

#"vertical asymptotes at " x=-3" and " x=4#

#"horizontal asymptote at " y=0#

Explanation:

The denominator of y cannot be zero as this would make y undefined. Equating the denominator yo 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+3)(x-4)=0#

#rArrx=-3" and " x=4" are the asymptotes"#

#"Horizontal asymptotes occur as"#

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

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

#y=(x/x^2)/(x^2/x^2-x/x^2-12/x^2)=(1/x)/(1-1/x-12/x^2)#

as #xto+-oo,yto0/(1-0-0)#

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