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

Feb 27, 2016

Vertical asymptotes are $x = 0$, $x = - \sqrt{\frac{8}{3}}$ and $x = \sqrt{\frac{8}{3}}$ and horizontal asymptote is $y = \frac{5}{3}$.

#### Explanation:

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

One factor is obviously $x$ as $3 {x}^{3} - 8 x = 3 x \left({x}^{2} - \frac{8}{3}\right)$.

Other factors are obviously $\left(x + \sqrt{\frac{8}{3}}\right) \left(x - \sqrt{\frac{8}{3}}\right)$ and hence

Vertical asymptotes are $x = 0$, $x = - \sqrt{\frac{8}{3}}$ and $x = \sqrt{\frac{8}{3}}$.

As the highest degree of numerator $5 {x}^{3}$ and denominator $3 {x}^{3}$ are equal, we have a horizontal asymptote $y = \frac{5}{3}$.