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

1 Answer
Nov 7, 2016

Answer:

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

Explanation:

As you canot divide by #0#, the vertical asymptote is #x=-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^2##color(white)(aaaaaa)##+4##∣##x+1#
#3x^2+3x##color(white)(aaaa)######3x-3#
#color(white)(aa)##0+3x+4##color(white)(a)#
#color(white)(aaaaa)##+0-3##color(white)(a)#
#color(white)(aaaaaaaaa)##+7##color(white)(a)#

So, #(3x^2+4)/(x+1)=3x-3+7/(x+1)#
So the oblique asymptote is #y=3x-3#
# 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]}