$\begin{array}{l} \text{Gesucht:}\\ \text{Definitions- und Wertebereich} \\ \text{Grenzwerte} \\ \text{Symmetrie} \\ \text{Nullstellen - Schnittpunkt mit der x-Achse} \\ \text{Ableitungen - Stammfunktion} \\ \text{Extremwerte - Monotonie} \\ \text{Wendepunkte - Krümmung} \\ \text{Funktion:}f\left(x\right)=\displaystyle \frac{ x^2+x-6}{ 5x-2} \ <br/> \bullet \text{Funktion/Faktorisieren} \\ f\left(x\right)=\displaystyle \frac{ x^2+x-6}{ 5x-2} \\ \text{Zaehler faktorisieren: } \\ x^2+x-6 = 0 \\ \\ \\ 1x^{2}+1x-6 =0 \\ x_{1/2}=\displaystyle\frac{-1 \pm\sqrt{1^{2}-4\cdot 1 \cdot \left(-6\right)}}{2\cdot1} \\ x_{1/2}=\displaystyle \frac{-1 \pm\sqrt{25}}{2} \\ x_{1/2}=\displaystyle \frac{-1 \pm5}{2} \\ x_{1}=\displaystyle \frac{-1 +5}{2} \qquad x_{2}=\displaystyle \frac{-1 -5}{2} \\ x_{1}=2 \qquad x_{2}=-3 \\ \underline{x_1=-3; \quad1\text{-fache Nullstelle}} \\\underline{x_2=2; \quad1\text{-fache Nullstelle}} \\ \\\text{Nenner faktorisieren:} \\ 5x-2 = 0 \\ \\ 5 x-2 =0 \qquad /+2 \\ 5 x= 2 \qquad /:5 \\ x=\displaystyle\frac{2}{5}\\ x=\frac{2}{5} \\ \underline{x_3=\frac{2}{5}; \quad1\text{-fache Nullstelle}} \\ \\ \text{Faktorisierter Term:}\\ f\left(x\right)=\displaystyle\frac{(x+3)(x-2)}{5(x-\frac{2}{5})} \\ \\ \bullet\text{Definitionsbereich:}\qquad \mathbb{D} = \mathbb{R}\setminus \left\{\frac{2}{5}\right\} \\ f\left(x\right)= \displaystyle \frac{ \frac{1}{5}x^2+\frac{1}{5}x-1\frac{1}{5}}{ x-\frac{2}{5}} \\ Polynomdivision:\\\small \begin{matrix} ( \frac{1}{5}x^2&+\frac{1}{5}x&-1\frac{1}{5}&):( x -\frac{2}{5} )= \frac{1}{5}x +\frac{7}{25} \\ \,-( \frac{1}{5}x^2&-\frac{2}{25}x) \\ \hline & \frac{7}{25}x&-1\frac{1}{5}&\\ &-( \frac{7}{25}x&-\frac{14}{125}) \\ \hline &&-1\frac{11}{125}&\\ \end{matrix} \\ \normalsize \\ \\ f(x)= \frac{1}{5}x+\frac{7}{25}+\frac{-1\frac{11}{125}}{ x-\frac{2}{5}} \\ \\ \bullet \text{1. Ableitungen und 2.Ableitung} \\f'\left(x\right)=\frac{( \frac{2}{5}x+\frac{1}{5})\cdot( x-\frac{2}{5})-( \frac{1}{5}x^2+\frac{1}{5}x-1\frac{1}{5})\cdot 1}{( x-\frac{2}{5})^2}\\ = \frac{( \frac{2}{5}x^2+\frac{1}{25}x-\frac{2}{25})-( \frac{1}{5}x^2+\frac{1}{5}x-1\frac{1}{5})}{( x-\frac{2}{5})^2}\\ = \frac{ \frac{1}{5}x^2-\frac{4}{25}x+1\frac{3}{25}}{( x-\frac{2}{5})^2}\\ = \frac{ \frac{1}{5}x^2-\frac{4}{25}x+1\frac{3}{25}}{( x-\frac{2}{5})^2}\\ f''\left(x\right)=\frac{( \frac{2}{5}x-\frac{4}{25})\cdot( x^2-\frac{4}{5}x+\frac{4}{25})-( \frac{1}{5}x^2-\frac{4}{25}x+1\frac{3}{25})\cdot( 2x-\frac{4}{5})}{( x^2-\frac{4}{5}x+\frac{4}{25})^2}\\ = \frac{( \frac{2}{5}x^3-\frac{12}{25}x^2+\frac{24}{125}x-0,0256)-( \frac{2}{5}x^3-\frac{12}{25}x^2+2\frac{46}{125}x-\frac{112}{125})}{( x^2-\frac{4}{5}x+\frac{4}{25})^2}\\ = \frac{-2\frac{22}{125}x+0,87}{( x^2-\frac{4}{5}x+\frac{4}{25})^2}\\ = \frac{-2\frac{22}{125}x+0,87}{( x^2-\frac{4}{5}x+\frac{4}{25})^2} \\ \\ \bullet \text{Nullstellen / Schnittpunkt mit der x-Achse:} \\Zaehler =0 \\ \frac{1}{5}x^2+\frac{1}{5}x-1\frac{1}{5} = 0 \\ \underline{x_4=-3; \quad1\text{-fache Nullstelle}} \\\underline{x_5=2; \quad1\text{-fache Nullstelle}} \\ \\ \\ \bullet \text{Vorzeichentabelle:} \\ \begin{array}{|c|c|c|c|c|c|c|c|c|c||} \hline & x < &-3&< x <&\frac{2}{5}&< x <&2&< x\\ \hline f(x)&-&0&+&0&-&0&+\\ \hline \end{array}\\ \\ \underline{\quad x \in ]-3;\frac{2}{5}[\quad \cup \quad]2;\infty[\quad f(x)>0 \quad \text{oberhalb der x-Achse}}\\ \\ \underline{\quad x \in ]-\infty;-3[\quad \cup \quad]\frac{2}{5};2[\quad f(x)<0 \quad \text{unterhalb der x-Achse}} \\ \\ \bullet\text{Grenzwerte und Asymtoten: } \\ \\ \lim\limits_{x \rightarrow \infty}{\displaystyle \frac{x^2( 1+x-\dfrac{6}{x^2}) }{x( 5-\dfrac{2}{x}) }}=\infty \\\lim\limits_{x \rightarrow -\infty}{\displaystyle \frac{x^2( 1+x-\dfrac{6}{x^2}) }{x( 5-\dfrac{2}{x}) }}=\infty \\ \\ \text{Schiefe Asymptote:} y= \frac{1}{5}x+\frac{7}{25} \\\lim\limits_{x \rightarrow \frac{2}{5}^+}{\displaystyle\frac{\frac{1}{5}(x+3)(x-2)}{(x-\frac{2}{5})}}=-\infty\\ \lim\limits_{x \rightarrow \frac{2}{5}^-}{\displaystyle\frac{\frac{1}{5}(x+3)(x-2)}{(x-\frac{2}{5})}}=\infty\\ \\ \text{Vertikale Asymptote (Polstelle): } x=\frac{2}{5}\\ \\ \\ \bullet \text{Extremwerte/Hochpunkte/Tiefpunkte:} \\f'(x)=\displaystyle \frac{ \frac{1}{5}x^2-\frac{4}{25}x+1\frac{3}{25}}{ x^2-\frac{4}{5}x+\frac{4}{25}} = 0 \\ \\ \frac{1}{5}x^{2}-\frac{4}{25}x+1\frac{3}{25} =0\\ x_{1/2}=\displaystyle\frac{+\frac{4}{25} \pm\sqrt{\left(-\frac{4}{25}\right)^{2}-4 \cdot \frac{1}{5} \cdot 1\frac{3}{25}}}{2\cdot\frac{1}{5}}\\ x_{1/2}=\displaystyle \frac{+\frac{4}{25} \pm\sqrt{-0,87}}{\frac{2}{5}}\\ \text{Diskriminante negativ keine Lösung} \\ \\ \, \, \\\bullet\text{Monotonie/ streng monoton steigend (sms)/streng monoton fallend (smf) } \\f'\left(x\right)=\displaystyle \frac{ \frac{1}{5}x^2-\frac{4}{25}x+1\frac{3}{25}}{ x^2-\frac{4}{5}x+\frac{4}{25}}\\ \,\text{Zaehler} =0 \\\, \\\text{Nullstellen des Nenners aus f(x) übernehmen} \\\underline{x_6=\frac{2}{5}; \quad1\text{-fache Nullstelle}} \\\, \\ \begin{array}{|c|c|c|c||} \hline & x < &\frac{2}{5}&< x\\ \hline f'(x)&+&0&+\\ \hline \end{array}\\ \\ \underline{\quad x \in ]-\infty;\frac{2}{5}[\quad \cup \quad]\frac{2}{5};\infty[\quad f'(x)>0 \quad \text{streng monoton steigend }} \\ \\ \bullet\text{Kruemmung} \\f''\left(x\right)=\displaystyle \frac{-2\frac{22}{125}x+0,87}{ x^4-1\frac{3}{5}x^3+\frac{24}{25}x^2-\frac{32}{125}x+0,0256}\\ \,Zaehler =0 \\\\ -2\frac{22}{125} x+0,87 =0 \qquad /-0,87 \\ -2\frac{22}{125} x= -0,87 \qquad /:\left(-2\frac{22}{125}\right) \\ x=\displaystyle\frac{-0,87}{-2\frac{22}{125}}\\ x=\frac{2}{5} \\ \underline{x_7=\frac{2}{5}; \quad1\text{-fache Nullstelle}} \\\, \\\text{Nullstelle des Nenners aus f(x) übernehmen} \\\underline{x_8=\frac{2}{5}; \quad1\text{-fache Nullstelle}} \\ \begin{array}{|c|c|c|c|c|c|c||} \hline & x < &\frac{2}{5}&< x <&\frac{2}{5}&< x\\ \hline f''(x)&+&0&+&0&-\\ \hline \end{array}\\ \\ \underline{\quad x \in ]-\infty;\frac{2}{5}[\quad \cup \quad]\frac{2}{5};\frac{2}{5}[\quad f''(x)>0 \quad \text{linksgekrümmt}}\\ \\ \underline{\quad x \in ]\frac{2}{5};\infty[\quad f''(x)<0 \quad \text{rechtsgekrümmt}} \\ Funktionsgraph und Wertetabelle \\ \end{array}$