How do you find vertical, horizontal and oblique asymptotes for #y=(x^3-x^2-10)/(3x^2-4x)#?

1 Answer
Binayaka C.
May 22, 2017

Vertical asymtotes are # x=0 , x = 4/3#. No H.A.
Oblique asymptote is #y = 1/3x+1/9#

Explanation:

#y= (x^3-x^2-10)/(3x^2-4x) = (x^3-x^2-10)/(x(3x-4)) #

Vertical asymtotes are # x=0 # , #3x-4=0 or x = 4/3#

Since degree of numerator is greater than degree of numerator there is no horizontal asymtote.

By long division we can find oblique asymptote as #y = 1/3x+1/9#

graph{(x^3-x^2-10)/(3x^2-4x) [-40, 40, -20, 20]} [Ans]

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