How do you find all the asymptotes for function #f(x) = (x+4)/(3x^2+5x-2)#?

1 Answer
Nov 12, 2016

The vertical asymptotes are #x=1/3# and #x=-2#
The vertical asymptote is #y=0#
There are no slant asymptotes

Explanation:

Let's factorise the denominator #3x^2+5x-2 =(3x-1)(x+2)#

As we cannot divide by #0#, the vertical asymptotes are #x=1/3# and #x=-2#

The degree of the numerator is #<# the degree of the denominator, so we don't have a slant asymptote.

#lim_(x->-oo)f(x)=lim_(x->-oo)x/(3x^2)=lim_(x->-oo)1/(3x)=0^(-)#

#lim_(x->+oo)f(x)=lim_(x->+oo)x/(3x^2)=lim_(x->+oo)1/(3x)=0^(+)#

So #y=0# is a horizontal asymptote
graph{(x+4)/(3x^2+5x-2) [-11.05, 6.73, -6.2, 2.69]}