Finde Definitionslücken, indem du schaust, wann der Nenner Null wird.

${\mathrm{x}}^2+\mathrm{a}=0\backslash mathrm\{x\}^2+\backslash mathrm\{a\}=0$ $\mid -\mathrm{a}\backslash left|-\backslash mathrm\; a\backslash right.$

${\mathrm{x}}^2=-\mathrm{a}\backslash mathrm\{x\}^2=-\backslash mathrm\{a\}$

Mache eine Fallunterscheidung für a.

$\mathrm{a}>0\backslash mathrm\{a\}>0$ :  Gleichung nie erfüllt

Bestimme den Definitionsbereich.

$\Rightarrow {\mathrm{D}}_{{\mathrm{f}}_{\mathrm{a}}}=\mathbb{R}\backslash Rightarrow\backslash ;\backslash ;\{\backslash mathrm\; D\}\_\{\{\backslash mathrm\; f\}\_\backslash mathrm\; a\}=\backslash mathbb\{R\}$

$\mathrm{a}\le 0:\backslash mathrm\{a\}\backslash le0:$  $\mathrm{x}=\pm \sqrt{-\mathrm{a}}\backslash mathrm\{x\}=\backslash pm\backslash sqrt\{-\backslash mathrm\{a\}\}$

Bestimme den Definitionsbereich.

$\Rightarrow {\mathrm{D}}_{{\mathrm{f}}_{\mathrm{a}}}=\mathbb{R}\mathrm{\backslash }\{\pm \sqrt{-\mathrm{a}}\}\backslash Rightarrow\backslash ;\backslash ;\{\backslash mathrm\; D\}\_\{\{\backslash mathrm\; f\}\_\backslash mathrm\; a\}=\backslash mathbb\{R\}\backslash ;\backslash backslash\backslash ;\backslash left\backslash \{\backslash pm\backslash sqrt\{-\backslash mathrm\; a\}\backslash right\backslash \}$

Nullstellen bestimmen

$-2{\mathrm{x}}^2+50=0\mid -50-2\backslash mathrm\{x\}^2+50=0\; \backslash left|-50\backslash right.$

$-2{\mathrm{x}}^2=-50-2\backslash mathrm\{x\}^2=-50$ $\mid :(-2)\backslash left|:\backslash left(-2\backslash right)\backslash right.$

${\mathrm{x}}^2=25\backslash mathrm\{x\}^2=25$ $\mid \sqrt{}\backslash left|\backslash sqrt\{\}\backslash right.$

$\mathrm{x}=\pm 5\backslash mathrm\{x\}=\backslash pm5$

Gib die beiden Nullstellen an.

$\Rightarrow {\mathrm{x}}_1=-5,{\mathrm{x}}_2=5\backslash Rightarrow\backslash ;\backslash ;\{\backslash mathrm\; x\}\_1=-5\backslash ;\backslash ;,\backslash ;\backslash ;\{\backslash mathrm\; x\}\_2=5$

Y-Achsenabschnitt bestimmen

Setze $00$ in die Funktion ein

${\mathrm{f}}_{\mathrm{a}}(0)=\frac{-2\cdot {\left(0\right)}^2+50}{{\left(0\right)}^2+\mathrm{a}}=\frac{50}{\mathrm{a}}\backslash mathrm\{f\}\_\{\backslash mathrm\{a\}\}(0)=\backslash frac\{-2\backslash cdot\; \backslash left(0\backslash right)^2+50\}\{\backslash left(0\backslash right)^2+\backslash mathrm\{a\}\}=\backslash frac\{50\}\{\backslash mathrm\{a\}\}$

Gib den Schnittpunkt an.

$\Rightarrow {\mathrm{Y}}_{\mathrm{a}}\left(0\mid \frac{50}{\mathrm{a}}\right)\backslash Rightarrow\backslash ;\backslash ;\{\backslash mathrm\; Y\}\_\backslash mathrm\; a\backslash left(\backslash left.0\backslash ;\backslash right|\backslash ;\backslash frac\{50\}\{\backslash mathrm\; a\}\backslash right)$

Berechne nun den Grenzwert

Bestimme nun den Grenzwert. Je nachdem von welcher Seite man sich $\sqrt{-a}\backslash sqrt\{-a\}$ nähert, erhält man entweder ein $++$ oder ein $--$ als Vorzeichen:

$\backslash overset\{\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\}\{\backslash rightarrow\backslash pm\backslash infty\}$

${\mathrm{f}}_{\mathrm{a}}(\mathrm{x})=\frac{-2{\mathrm{x}}^2+50}{{\mathrm{x}}^2+\mathrm{a}}\backslash mathrm\{f\}\_\{\backslash mathrm\{a\}\}(\backslash mathrm\{x\})=\backslash frac\{-2\backslash mathrm\{x\}^2+50\}\{\backslash mathrm\{x\}^2+\backslash mathrm\{a\}\}$

Setze die geforderten Werte für a ein und bestimme so die 3 Funktionsterme

${\mathrm{f}}_{-25}(\mathrm{x})=\frac{-2{\mathrm{x}}^2+50}{{\mathrm{x}}^2-25}\frac{-2({\mathrm{x}}^2-25)}{({\mathrm{x}}^2-25)}-2\backslash mathrm\{f\}\_\{-25\}(\backslash mathrm\{x\})=\backslash frac\{-2\backslash mathrm\{x\}^2+50\}\{\backslash mathrm\{x\}^2-25\}\backslash frac\{-2\backslash left(\backslash mathrm\{x\}^2-25\backslash right)\}\{\backslash left(\backslash mathrm\{x\}^2-25\backslash right)\}-2$

Beachte hier aber die Definitionslücken beim Einzeichnen!

${\mathrm{f}}_{-16}(\mathrm{x})=\frac{-2{\mathrm{x}}^2+50}{{\mathrm{x}}^2-16}\backslash mathrm\{f\}\_\{-16\}(\backslash mathrm\{x\})=\backslash frac\{-2\backslash mathrm\{x\}^2+50\}\{\backslash mathrm\{x\}^2-16\}$

${\mathrm{f}}_{25}(\mathrm{x})=\frac{-2{\mathrm{x}}^2+50}{{\mathrm{x}}^2+25}\backslash mathrm\{f\}\_\{25\}(\backslash mathrm\{x\})=\backslash frac\{-2\backslash mathrm\{x\}^2+50\}\{\backslash mathrm\{x\}^2+25\}$

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