# How do you find the slant asymptote of f (x ) = ( 2x^3 + x^2) / (2x^2 - 3x + 3)?

Jun 28, 2016

$y = x + 2$

#### Explanation:

I must obtain an equation of the type

y=mx+n

First I find m by calculating:

$m = {\lim}_{x \rightarrow \infty} \frac{2 {x}^{3} + {x}^{2}}{2 {x}^{2} - 3 x + 3} \cdot \frac{1}{x}$

$m = {\lim}_{x \rightarrow \infty} \frac{2 {x}^{2} + x}{2 {x}^{2} - 3 x + 3}$

$m = 1$

the I calculate n:

$n = {\lim}_{x \rightarrow \infty} \left(\frac{2 {x}^{3} + {x}^{2}}{2 {x}^{2} - 3 x + 3} - x\right)$

$n = {\lim}_{x \rightarrow \infty} \frac{\cancel{2 {x}^{3}} + {x}^{2} \cancel{- 2 {x}^{3}} + 3 {x}^{2} - 3 x}{2 {x}^{2} - 3 x + 3}$

$n = {\lim}_{x \rightarrow \infty} \frac{4 {x}^{2} - 3 x}{2 {x}^{2} - 3 x + 3}$

$n = 2$

so the slant asymptote is

$y = x + 2$