## Calculus: Early Transcendentals 8th Edition

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