How do you find vertical, horizontal and oblique asymptotes for y=(x^3-x^2-10)/(3x^2-4x)y=x3−x2−103x2−4x?

1 Answer
Binayaka C.
May 22, 2017

Vertical asymtotes are x=0 , x = 4/3x=0,x=43. No H.A.
Oblique asymptote is y = 1/3x+1/9y=13x+19

Explanation:

y= (x^3-x^2-10)/(3x^2-4x) = (x^3-x^2-10)/(x(3x-4)) y=x3−x2−103x2−4x=x3−x2−10x(3x−4)

Vertical asymtotes are x=0 x=0 , 3x-4=0 or x = 4/33x−4=0orx=43

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+1/9y=13x+19

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

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