How do you find the vertical, horizontal or slant asymptotes for #f(x) = (2x^2-5x-12)/(3x^2-11x-4 )#?

1 Answer
Oct 4, 2016

Answer:

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

Explanation:

The first step is to factorise and simplify f(x).

#f(x)=((2x+3)cancel((x-4)))/((3x+1)cancel((x-4)))=(2x+3)/(3x+1)#

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 this value then it is a vertical asymptote.

solve : #3x+1=0rArrx=-1/3" is the asymptote"#

Horizontal asymptotes occur as

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

divide numerator/denominator by x

#f(x)=((2x)/x+3/x)/((3x)/x+1/x)=(2+3/x)/(3+1/x)#

as #xto+-oo,f(x)to(2+0)/(3+0)#

#rArry=2/3" is the asymptote"#

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both of degree 1 ) Hence there are no slant asymptotes.
graph{(2x+3)/(3x+1) [-10, 10, -5, 5]}