$f\left(\mathrm{x}\right)=-\frac{1}{2}{x}^2+2x+1f\backslash left(\backslash mathrm\{x\}\backslash right)=-\backslash frac\{1\}\{2\}x^2+2x+1$

Distributivgesetz anwenden.  $-\frac{1}{2}-\backslash frac12$ ausklammern.

$=-\frac{1}{2}(x^2-4x-2)==-\backslash frac\{1\}\{2\}\backslash left(x^2-4x-2\backslash right)=$

Quadratische Ergänzung.

$=-\frac{1}{2}(x^2-4x+{\left(\frac{4}{2}\right)}^2-{\left(\frac{4}{2}\right)}^2-2)==-\backslash frac\{1\}\{2\}\backslash left(x^2-4x+\backslash left(\backslash frac\{4\}\{2\}\backslash right)^2-\backslash left(\backslash frac\{4\}\{2\}\backslash right)^2-2\backslash right)=$

Schreibe in eine Binomische Formel um.

$=-\frac{1}{2}({(x-\frac{4}{2})}^2-{\left(\frac{4}{2}\right)}^2-2)==-\backslash frac\{1\}\{2\}\backslash left(\backslash left(x-\backslash frac\{4\}\{2\}\backslash right)^2-\backslash left(\backslash frac\{4\}\{2\}\backslash right)^2-2\backslash right)=$

$=-\frac{1}{2}({(x-2)}^2-2^2-2)==-\backslash frac12\backslash left(\backslash left(x-\backslash textstyle2\backslash right)^2-\{\backslash textstyle2\}^2-2\backslash right)=$

$=-\frac{1}{2}({(x-2)}^2-4-2)==-\backslash frac12\backslash left(\backslash left(x-\backslash textstyle2\backslash right)^2-\{\backslash textstyle4\}-2\backslash right)=$

$=-\frac{1}{2}({(x-2)}^2-6)==-\backslash frac12\backslash left(\backslash left(x-\backslash textstyle2\backslash right)^2-6\backslash right)=$

Wende das Distributivgesetz an.

$=-\frac{1}{2}{(x-2)}^2-6\cdot (-\frac{1}{2})==-\backslash frac12\backslash left(x-\backslash textstyle2\backslash right)^2-6\backslash cdot\backslash left(-\backslash frac12\backslash right)=$

$=-\frac{1}{2}{(x-2)}^2+3=-\backslash frac12\backslash left(x-\backslash textstyle2\backslash right)^2\backslash textstyle+\backslash textstyle3$

Lies den Scheitelpunkt ab.

$\to \backslash rightarrow\backslash ;$ Scheitelpunkt: $\mathrm{S}\left(2\mid 3\right)\backslash mathrm\; S\backslash left(2\backslash ;\backslash left|\backslash ;3\backslash right.\backslash right)$

Schnittpunkte mit der x-Achse bestimmen: Nullstelle berechnen

$f\left(\mathrm{x}\right)=-\frac{1}{2}{x}^2+2x+1f\backslash left(\backslash mathrm\{x\}\backslash right)=-\backslash frac\{1\}\{2\}x^2+2x+1$

Wende die Mitternachtsformel an. Dazu erst die Diskriminante D berechnen.

$\mathrm{D}=2^2-4\cdot (-\frac{1}{2})\cdot 1=\backslash mathrm\{D\}=2^2-4\backslash cdot\; \backslash left(-\backslash frac\{1\}\{2\}\backslash right)\backslash cdot\; 1=$

$=4-4\cdot (-\frac{1}{2})==4-4\backslash cdot\; \backslash left(-\backslash frac\{1\}\{2\}\backslash right)=$

$=4+2=6=4+2=6$

→ 2 Lösungen. Wende die Mitternachtsformel an.

$x_1=\frac{-2+\sqrt{D}}{2\cdot (-\mathrm{0,5})}=2-\sqrt{6}x\_1=\backslash frac\{-2+\backslash sqrt\{D\}\}\{2\backslash cdot\; (-0\{,\}5)\}=2-\backslash sqrt\{6\}$

$\to \text{}{P}_{x_1}(2-\sqrt{6}\mathrm{\mid }0)Px1\text{}(2-6\text{}\mid 0)\to \; P\_\{x\_1\}\backslash left(2-\backslash sqrt6\backslash vert\backslash ;0\backslash right)Px1(2-6\mid 0)$

$x_2=\frac{-2-\sqrt{D}}{2\cdot (-\mathrm{0,5})}=2+\sqrt{6}x\_2=\backslash frac\{-2-\backslash sqrt\; D\}\{2\backslash cdot(-0\{,\}5)\}=\backslash textstyle2\backslash textstyle+\backslash textstyle\backslash sqrt6$

$\to \backslash rightarrow\backslash ;$ $P_{x_1}(2+\sqrt{6}\mathrm{\mid }0)P\_\{x\_1\}\backslash left(2+\backslash sqrt6\backslash vert\backslash ;0\backslash right)$

Schnittpunkt mit der y-Achse

$f\left(\mathrm{x}\right)=-\frac{1}{2}{x}^2+2x+1f\backslash left(\backslash mathrm\{x\}\backslash right)=-\backslash frac\{1\}\{2\}x^2+2x+1$

Setze x gleich 0.

$f\left(0\right)=-\frac{1}{2}{0}^2+2\cdot 0+1=1f\backslash left(0\backslash right)=-\backslash frac\{1\}\{2\}0^2+2\backslash cdot\; 0+1=1$

$\to \backslash rightarrow\backslash ;$ $P_y\left(0\mathrm{\mid }1\right)P\_y\backslash left(0\backslash vert\backslash ;1\backslash right)$

$P\mathrm{‘}(-3\mathrm{\mid }-1)P`\backslash left(-3\backslash vert\backslash ;-1\backslash right)$

0 $\to \backslash rightarrow\backslash ;$ 3

Verschiebung um 3 Einheiten nach links.

-1 $\to \backslash rightarrow\backslash ;$ 1

Verschiebung um 2 Einheiten nach unten.

Um eine Parabel zu verschieben, muss die Scheitelform nach vorherig gemachten Angaben verändert werden.

$f\left(\mathrm{x}\right)=-\frac{1}{2}{(x-2)}^2+3f\backslash left(\backslash mathrm\; x\backslash right)=-\backslash frac12\backslash left(x-\backslash textstyle2\backslash right)^2\backslash textstyle+\backslash textstyle3$ $\to \backslash rightarrow\backslash ;$$g\left(\mathrm{x}\right)=-\frac{1}{2}{\left(x+1\right)}^2+1g\backslash left(\backslash mathrm\; x\backslash right)=-\backslash frac12\backslash left(x\backslash textstyle+\backslash textstyle1\backslash right)^2\backslash textstyle+\backslash textstyle1$

Multipliziere die Funktion aus. Binomische Formel.

$=-\frac{1}{2}(x^2+2x+1)+1==-\backslash frac12\{\backslash textstyle\backslash left(x^2+2x+1\backslash right)\}+1\backslash textstyle=$

$=-\frac{1}{2}{x}^2-x-\frac{1}{2}+1==-\backslash frac\{1\}\{2\}x^2-x-\backslash frac\{1\}\{2\}+1=$

$=-\frac{1}{2}{x}^2-x+\frac{1}{2}=-\backslash frac\{1\}\{2\}x^2-x+\backslash frac\{1\}\{2\}$

Ansatz: $f(x)=g(x)f(x)=g(x)$

$f\left(x\right)=-\frac{1}{2}{x}^2+2x+1f\backslash left(x\backslash right)=-\backslash frac\{1\}\{2\}x^2+2x+1$ ;

$g\left(x\right)=-\frac{1}{2}{x}^2-x+\frac{1}{2}g\backslash left(x\backslash right)=-\backslash frac\{1\}\{2\}x^2-x+\backslash frac\{1\}\{2\}$

$-\frac{1}{2}{x}^2+2x+1=-\frac{1}{2}{x}^2-x+\frac{1}{2}-\backslash frac\{1\}\{2\}x^2+2x+1=-\backslash frac\{1\}\{2\}x^2-x+\backslash frac\{1\}\{2\}$

Nach x auflösen.

$\iff 2x+1=-x+\frac{1}{2}\backslash Leftrightarrow\; 2x+1=-x+\backslash frac\{1\}\{2\}$

$\iff \backslash Leftrightarrow$ $3x=-\frac{1}{2}3x=-\backslash frac\{1\}\{2\}$

$\iff \backslash Leftrightarrow$ $x=-\frac{1}{6}x=-\backslash frac\{1\}\{6\}$

Setze den gefundenen x-Wert in eine der Anfangsgleichungen, um den zugehörigen y-Wert zu berechnen.

$f(-\frac{1}{6})=-\frac{1}{2}\cdot {(-\frac{1}{6})}^2+\frac{1}{6}+\frac{1}{2}=f\backslash left(-\backslash frac\{1\}\{6\}\backslash right)=-\backslash frac\{1\}\{2\}\backslash cdot\; \backslash left(-\backslash frac\{1\}\{6\}\backslash right)^2+\backslash frac\{1\}\{6\}+\backslash frac\{1\}\{2\}=$

$=-\frac{1}{2}\cdot \frac{1}{36}+\frac{1}{6}+\frac{1}{2}==-\backslash frac\{1\}\{2\}\backslash cdot\; \backslash frac\{1\}\{36\}+\backslash frac\{1\}\{6\}+\backslash frac\{1\}\{2\}=$

$=-\frac{1}{72}+\frac{1}{6}+\frac{1}{2}==-\backslash frac\{1\}\{72\}+\backslash frac\{1\}\{6\}+\backslash frac\{1\}\{2\}=$

Hauptnenner bilden und auf diesen erweitern . $\to \backslash rightarrow\backslash ;$$66$

$=-\frac{1}{72}+\frac{1}{6}+\frac{3}{6}==-\backslash frac\{1\}\{72\}+\backslash frac\{1\}\{6\}+\backslash frac\{3\}\{6\}=$

$=-\frac{1}{72}+\frac{4}{6}==-\backslash frac\{1\}\{72\}+\backslash frac\{4\}\{6\}=$

Hauptnenner bilden und auf diesen erweitern . $\to \backslash rightarrow\backslash ;$72

$=-\frac{1}{72}+\frac{48}{72}==-\backslash frac\{1\}\{72\}+\backslash frac\{48\}\{72\}=$

$=\frac{47}{72}=\backslash frac\{47\}\{72\}$

$\to \mathrm{S}(-\frac{1}{6}\mathrm{\mid }\frac{47}{72})\backslash rightarrow\backslash ;\; \backslash mathrm\; S\backslash left(-\backslash frac16\backslash vert\backslash ;\backslash frac\{47\}\{72\}\backslash right)$