How do you find the vertical, horizontal or slant asymptotes for #f(x) = (3x^2 + 4)/(x+1)#?

1 Answer
Jun 29, 2017

Answer:

Vertical asymptote at #x= -1#, Slant asymptote: #y=3x-3# ,
Horizontal asymptote: Absent

Explanation:

#f(x) =(3x^2+4)/(x+1)#
Vertical asymptote : Denominator is #0 :. x+1 = 0 or x = -1#

Vertical asymptote at #x= -1#

Degree i.e maximum power of #x# in numerator is #1# more than denominator, so there is no horizontal asymptote.

Slant asymptote: Degree i.e maximum power of #x# in numerator is #1# more than denominator , so there is slant asymptote.

By long division of (f(x) we get dividend as #y=3x-3# and remainder #7 :. y = 3x-3# is the slant asymptote.

graph{(3x^2+4)/(x+1) [-80, 80, -40, 40]} [Ans]