How do you find the asymptotes for #f(x)=(2x+4)/(x^2-3x-4)#?

1 Answer
Sep 4, 2016

Answer:

vertical asymptotes at x = -1 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 values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

solve: #x^2-3x-4=0rArr(x+1)(x-4)=0#

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

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

#rArry=0" is the asymptote"#
graph{(2x+4)/(x^2-3x-4) [-10, 10, -5, 5]}