y = \frac{- 5 x^{2} + 5}{x^{3}}

$D = \{x \in ℝ \ x = 0\}$
ungerade Symmetrie
$y = \frac{- 5 (x^{2} - 1)}{x^{3}} = - \frac{5 (x - 1) (x + 1)}{x^{3}}$
Nullstellen $x_{1} = - 1$ und $x_{2} = 1$
Polstelle $x_{3} = 0$
Ableitungen:
$y^{'} = \frac{- 10 x \cdot x^{3} - (- 5 x^{2} + 5) 3 x^{2}}{x^{6}}$
$= \frac{- 10 x^{2} + 15 x^{2} - 15}{x^{4}}$
$= \frac{5 (x^{2} - 3)}{x^{4}}$
$y^{″} = \frac{- 10 (x^{2} - 6)}{x^{5}}$
$y^{‴} = \frac{30 (x^{2} - 10)}{x^{6}}$

horizontale Tangente: $y^{'} = 0$
$x^{2} - 3 = 0 \Rightarrow x = \pm \sqrt{3}$
$y^{″} (\sqrt{3}) = \frac{10}{3} \sqrt{3} > 0$ Minimum
$y^{″} (- \sqrt{3}) = - \frac{10}{3} \sqrt{3} < 0$ Maximum
Wendepunkte:
$y^{″} = 0 \Rightarrow x^{2} - 6 = 0$
$x = \pm \sqrt{6}$
$y^{‴} (\sqrt{6}) = \frac{30 (6 - 10)}{6^{3}} < 0$
$y^{‴} (- \sqrt{6}) = - \frac{30 (6 - 10)}{6^{3}} > 0$
Asymptote: echt gebrochen rational $\Rightarrow x$ -Achse
$W$ ertebereich $- \infty < x < \infty$
.

Abbildung 2: $y = \frac{- 5 x^{2} + 5}{x^{3}}$