How do you find the vertical, horizontal or slant asymptotes for #f(x)=(2x^3+3x^2+x)/(6x^2+x-1)#?

1 Answer
Oct 28, 2016

Answer:

The vertical asymptote is #x=1/3#
The slant asymptote is #y=x/3+4/9#

Explanation:

Let's start by fatorising the numerator and the denominator
#2x^3+3x^2+x=x(2x^2+3x+1)=x(2x+1)(x+1)#
and #6x^2+x-1=(3x-1)(2x+1)#

So #f(x)=(2x^3+3x^2+x)/(6x^2+x-1)=(xcancel(2x+1)(x+1))/((3x-1)cancel(2x+1))=(x(x+1))/((3x-1))#

As we cannot divide by #0#, so the vertical asymptote is #x=1/3#

As the degree of the numerator is greater than the degree of the denominator, we would expect a slant asymptote. So we make a long division.
#x^2+x##color(white)(aaaa)##∣##3x-1#
#x^2-x/3##color(white)(aaa)##∣##x/3+4/9#
#0+(4x)/3#
#color(white)(aaaa)##(4x)/3-4/9#
#color(white)(aaaaaa)##0+4/9#

So #f(x)=x/3+4/9+(4/9)/(3x-1)#
The slant asymptote is #y=x/3+4/9#
There is no horizontal asymptote as limit of #f(x)# as #x->+-oo# is #+-oo#
graph{(x^2+x)/(3x-1) [-7.02, 7.024, -3.51, 3.51]}