Answer
$y = 4x+2~~$ is the slant asymptote.
Work Step by Step
$y = \frac{4x^3-10x^2-11x+1}{x^2-3x}$
We can express $4x^3-10x^2-11x+1$ in terms of $x^2-3x$:
$4x^3-10x^2-11x+1 = (x^2-3x)(4x+2)+(-5x+1)$
Then:
$\frac{4x^3-10x^2-11x+1}{x^2-3x} = (4x+2)+\frac{-5x+1}{x^2-3x}$
We can see that $~~y = 4x+2~~$ is a good candidate for the slant asymptote.
We can evaluate the limit as $x \to -\infty$:
$\lim\limits_{x \to -\infty} [\frac{4x^3-10x^2-11x+1}{x^2-3x}-(4x+2)]$
$= \lim\limits_{x \to -\infty} [\frac{4x^3-10x^2-11x+1}{x^2-3x}-\frac{(4x+2)(x^2-3x)}{x^2-3x}]$
$= \lim\limits_{x \to -\infty} (\frac{4x^3-10x^2-11x+1}{x^2-3x}-\frac{4x^3-10x^2-6x}{x^2-3x})$
$= \lim\limits_{x \to -\infty} \frac{-5x+1}{x^2-3x}$
$= \lim\limits_{x \to -\infty} \frac{-5x/x^2+1/x^2}{x^2/x^2-3x/x^2}$
$= \lim\limits_{x \to -\infty} \frac{-5/x+1/x^2}{1-3/x}$
$= \frac{0+0}{1-0}$
$= 0$
We can evaluate the limit as $x \to \infty$:
$\lim\limits_{x \to \infty} [\frac{4x^3-10x^2-11x+1}{x^2-3x}-(4x+2)]$
$= \lim\limits_{x \to \infty} [\frac{4x^3-10x^2-11x+1}{x^2-3x}-\frac{(4x+2)(x^2-3x)}{x^2-3x}]$
$= \lim\limits_{x \to \infty} (\frac{4x^3-10x^2-11x+1}{x^2-3x}-\frac{4x^3-10x^2-6x}{x^2-3x})$
$= \lim\limits_{x \to \infty} \frac{-5x+1}{x^2-3x}$
$= \lim\limits_{x \to \infty} \frac{-5x/x^2+1/x^2}{x^2/x^2-3x/x^2}$
$= \lim\limits_{x \to \infty} \frac{-5/x+1/x^2}{1-3/x}$
$= \frac{0+0}{1-0}$
$= 0$
Therefore, $~~y = 4x+2~~$ is the slant asymptote.