How do you find the Vertical, Horizontal, and Oblique Asymptote given #f(x)=(2x^3+3x^2+x)/(6x^2+x-1)#?

1 Answer
Jun 24, 2016

Answer:

A vertical assymptote at #x = 1/3# and
a slant assymptote given by #y = x/3+4/9#

Explanation:

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

We have a vertical assymptote at #x = 1/3# and also we know

#1/3x(x+1) = (x-1/3)(ax+b)+c#

Equating power coefficients

#{ (b/3 - c=0), ( 1/3 + a/3 - b=0), (1/3 - a=0) :}#

and solving

#a = 1/3,b=4/9,c=4/27#

The assymptote is given by

#y = x/3+4/9#

Attached a figure with vertical and slant assymptotes

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