How do you find the asymptotes for # (2x^2)/((-3x-1)^2)#?

1 Answer
Sep 20, 2016

Answer:

vertical asymptote at #x=-1/3#
horizontal asymptote at #y=2/9#

Explanation:

The denominator of the function cannot be zero as this would make the function undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve: #(-3x-1)^2=0rArr-3x-1=0rArrx=-1/3#

#rArrx=-1/3" is the asymptote"#

Horizontal asymptotes occur as

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

#f(x)=(2x^2)/((-3x-1)^2)=(2x^2)/(9x^2+6x+1)#

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

#f(x)=((2x^2)/(x^2))/((9x^2)/x^2+(6x)/x^2+1/x^2)=2/(9+6/x+1/x^2)#

as #xto+-oo,f(x)to2/(9+0+0)#

#rArry=2/9" is the asymptote"#
graph{(2x^2)/(9x^2+6x+1) [-20, 20, -10, 10]}