How do you find vertical, horizontal and oblique asymptotes for (3x^2 + 4)/(x+1)3x2+4x+1?

1 Answer
Nov 7, 2016

The vertical asymptote is x=-1x=1
The oblique asymptote is y=3x-3y=3x3

Explanation:

As you canot divide by 00, the vertical asymptote is x=-1x=1
As the degree of the numerator is >> than the degree of the dedenominator, so we expect an oblique asymptote:
Therefore we do a lond division
3x^23x2color(white)(aaaaaa)aaaaaa+4+4x+1x+1
3x^2+3x3x2+3xcolor(white)(aaaa)aaaa#3x-3#
color(white)(aa)aa0+3x+40+3x+4color(white)(a)a
color(white)(aaaaa)aaaaa+0-3+03color(white)(a)a
color(white)(aaaaaaaaa)aaaaaaaaa+7+7color(white)(a)a

So, (3x^2+4)/(x+1)=3x-3+7/(x+1)3x2+4x+1=3x3+7x+1
So the oblique asymptote is y=3x-3y=3x3
lim_(n rarr +-oo) (3x^2+4)/(x+1)=lim_(n rarr +-oo) 3x=+-oo
So there is no horizontal asymptote
graph{(y-(3x^2)/(x+1))(y-3x+3)=0 [-43.83, 38.34, -24.63, 16.5]}