Answer
$y = 2x-3~~$ is a slant asymptote.
Work Step by Step
$y = \frac{2x^3-5x^2+3x}{x^2-x-2}$
We can express $2x^3-5x^2+3x$ in terms of $x^2-x-2$:
$2x^3-5x^2+3x = (x^2-x-2)(2x-3)+(4x-6)$
Then:
$\frac{2x^3-5x^2+3x}{x^2-x-2} = (2x-3)+\frac{4x-6}{x^2-x-2}$
We can see that $~~y = 2x-3~~$ is a good candidate for a slant asymptote.
We can evaluate the limit as $x \to -\infty$:
$\lim\limits_{x \to -\infty} [\frac{2x^3-5x^2+3x}{x^2-x-2}-(2x-3)]$
$= \lim\limits_{x \to -\infty} [\frac{2x^3-5x^2+3x}{x^2-x-2}-\frac{(2x-3)(x^2-x-2)}{x^2-x-2}]$
$= \lim\limits_{x \to -\infty} (\frac{2x^3-5x^2+3x}{x^2-x-2}-\frac{2x^3-5x^2-x+6}{x^2-x-2})$
$= \lim\limits_{x \to -\infty} \frac{4x-6}{x^2-x-2}$
$= \lim\limits_{x \to -\infty} \frac{4x/x^2-6/x^2}{x^2/x^2-x/x^2-2/x^2}$
$= \lim\limits_{x \to -\infty} \frac{4/x-6/x^2}{1-1/x-2/x^2}$
$= \frac{0-0}{1-0-0}$
$= 0$
We can evaluate the limit as $x \to \infty$:
$\lim\limits_{x \to \infty} [\frac{2x^3-5x^2+3x}{x^2-x-2}-(2x-3)]$
$= \lim\limits_{x \to \infty} [\frac{2x^3-5x^2+3x}{x^2-x-2}-\frac{(2x-3)(x^2-x-2)}{x^2-x-2}]$
$= \lim\limits_{x \to \infty} (\frac{2x^3-5x^2+3x}{x^2-x-2}-\frac{2x^3-5x^2-x+6}{x^2-x-2})$
$= \lim\limits_{x \to \infty} \frac{4x-6}{x^2-x-2}$
$= \lim\limits_{x \to \infty} \frac{4x/x^2-6/x^2}{x^2/x^2-x/x^2-2/x^2}$
$= \lim\limits_{x \to \infty} \frac{4/x-6/x^2}{1-1/x-2/x^2}$
$= \frac{0-0}{1-0-0}$
$= 0$
Therefore, $~~y = 2x-3~~$ is a slant asymptote.