Question

How do you find the vertical, horizontal or slant asymptotes for `(3+2x^2+5x^3) /( 3x^3-8x)`?

Answer 1

Vertical asymptotes are `x=0`, `x=-sqrt(8/3)` and `x=sqrt(8/3)` and horizontal asymptote is `y=5/3`.

Explanation:

To find all the asymptotes for function `y=(3+2x^2+5x^3)/(3x^3-8x)`, let us first start with vertical asymptotes, which are given by putting denominator equal to zero or `(3x^3-8x)=0`.

One factor is obviously `x` as `3x^3-8x=3x(x^2-8/3)`.

Other factors are obviously `(x+sqrt(8/3))(x-sqrt(8/3))` and hence

Vertical asymptotes are `x=0`, `x=-sqrt(8/3)` and `x=sqrt(8/3)`.

As the highest degree of numerator `5x^3` and denominator `3x^3` are equal, we have a horizontal asymptote `y=5/3`.