Scheitelpunkt berechnen

$\to \backslash ;\backslash rightarrow\backslash ;$ $\mathrm{S}(2\mathrm{\mid }-2)\backslash mathrm\; S\backslash left(2\backslash ;\backslash vert-2\backslash right)$

Achsenschnittpunkte berechnen

Schnittpunkte mit der x-Achse

$f\left(x\right)=x^2-4x+2f\backslash left(x\backslash right)=x^2-4x+2$

$\to \backslash ;\backslash rightarrow\backslash ;$ 2 Lösungen:

$x_1=\frac{4+\sqrt{8}}{2}=2+\sqrt{2}x\_1=\backslash frac\{4+\backslash sqrt8\}2=2+\backslash sqrt2$

$\to \backslash ;\backslash rightarrow\backslash ;$ ${\mathrm{P}}_{\mathrm{x}1}(2+\sqrt{2}\mathrm{\mid }0)\{\backslash mathrm\; P\}\_\{\backslash mathrm\; x1\}\backslash left(2+\backslash sqrt2\backslash vert0\backslash right)$

$x_2=\frac{4-\sqrt{8}}{2}=2-\sqrt{2}\approx \mathrm{0,586}x\_2=\backslash frac\{4-\backslash sqrt8\}2=2-\backslash sqrt2\backslash approx0\{,\}586$

$\to \backslash ;\backslash rightarrow\backslash ;$ ${\mathrm{P}}_{\mathrm{x}2}(2-\sqrt{2}\mathrm{\mid }0)\{\backslash mathrm\; P\}\_\{\backslash mathrm\; x2\}\backslash left(2-\backslash sqrt2\backslash vert0\backslash right)$

Schnittpunkt mit der y-Achse

$f\left(x\right)=x^2-4x+2f\backslash left(x\backslash right)=x^2-4x+2$

Für x gleich 0 setzen.

$f\left(0\right)=0-0+2=2f\backslash left(0\backslash right)=0-0+2=2$

$\to \backslash ;\backslash rightarrow\backslash ;$ ${\mathrm{P}}_{\mathrm{y}}\left(0\mathrm{\mid }2\right)\{\backslash mathrm\; P\}\_\backslash mathrm\; y\backslash left(0\backslash vert2\backslash right)$

Verschiebung

Normalparabel: $f\left(x\right)=x^{2}f\backslash left(x\backslash right)=x^2$

Die verschiedenen Verschiebungen lassen sich an der Scheitelform ablesen:

$f\left(x\right)={(x-2)}^2-2f\backslash left(x\backslash right)=\backslash left(x-2\backslash right)^2-2$

$\to \backslash ;\backslash rightarrow\backslash ;$ Verschiebung um 2 LE nach rechts.

$\to \backslash ;\backslash rightarrow\backslash ;$ Verschiebung um 2 LE nach unten.

$\to \backslash ;\backslash rightarrow\backslash ;$ $\mathrm{S}(-2\mathrm{\mid }-2)\backslash mathrm\; S\backslash left(-2\backslash ;\backslash vert-2\backslash right)$

Schnittpunkte mit der x-Achse

$\to \backslash ;\backslash rightarrow\backslash ;$ 2 Lösungen

$x_1=\frac{-4+\sqrt{8}}{2}=-2+\sqrt{2}\approx -\mathrm{0,586}x\_1=\backslash frac\{-4+\backslash sqrt8\}2=-2+\backslash sqrt2\backslash approx-0\{,\}586$

$\to \backslash ;\backslash rightarrow\backslash ;$ ${\mathrm{P}}_{\mathrm{x}1}(-2+\sqrt{2}\mathrm{\mid }0)\{\backslash mathrm\; P\}\_\{\backslash mathrm\; x1\}\backslash left(-2+\backslash sqrt2\backslash vert0\backslash right)$

$x_2=\frac{-4-\sqrt{8}}{2}=-2-\sqrt{2}\approx -\mathrm{3,414}x\_2=\backslash frac\{-4-\backslash sqrt8\}2=-2-\backslash sqrt2\backslash approx-3\{,\}414$

${\mathrm{P}}_{\mathrm{x}2}(-2-\sqrt{2}\mathrm{\mid }0)\{\backslash mathrm\; P\}\_\{\backslash mathrm\; x2\}\backslash left(-2-\backslash sqrt2\backslash vert0\backslash right)$

Schnittpunkt mit der y-Achse

$f\left(x\right)=x^2+4x+2f\backslash left(x\backslash right)=x^2+4x+2$

Für x gleich 0 setzen.

$f\left(0\right)=0+0+2=2f\backslash left(0\backslash right)=0+0+2=2$

$\to \backslash ;\backslash rightarrow\backslash ;$ ${\mathrm{P}}_{\mathrm{y}}\left(0\mathrm{\mid }2\right)\{\backslash mathrm\; P\}\_\backslash mathrm\; y\backslash left(0\backslash vert2\backslash right)$

Normalparabel: $f\left(x\right)=x^{2}f\backslash left(x\backslash right)=x^2$

Die verschiedenen Verschiebungen lassen sich an der Scheitelform ablesen.

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

$\to \backslash ;\backslash rightarrow\backslash ;$ Verschiebung um 2 LE nach links.

$\to \backslash ;\backslash rightarrow\backslash ;$ Verschiebung um 2 LE nach unten.

$\to \backslash ;\backslash rightarrow\backslash ;$ $\mathrm{S}(-2\mathrm{\mid }7)\backslash mathrm\; S\backslash left(-2\backslash ;\backslash vert\backslash ;7\backslash right)$

Schnittpunkte mit der x-Achse

$\to \backslash ;\backslash rightarrow\backslash ;$ 2 Lösungen

$x_1=\frac{4+\sqrt{28}}{2\cdot (-1)}=-2-\sqrt{7}\approx -\mathrm{4,64}x\_1=\backslash frac\{4+\backslash sqrt\{28\}\}\{2\backslash cdot\backslash left(-1\backslash right)\}=-2-\backslash sqrt7\backslash approx-4\{,\}64$

$\to \backslash ;\backslash rightarrow\backslash ;$ ${\mathrm{P}}_{\mathrm{x}1}(-2-\sqrt{7}\mathrm{\mid }0)\{\backslash mathrm\; P\}\_\{\backslash mathrm\; x1\}\backslash left(-2-\backslash sqrt7\backslash vert0\backslash right)$

$x_2=\frac{-4-\sqrt{28}}{2\cdot (-1)}=-2+\sqrt{7}\approx \mathrm{0,65}x\_2=\backslash frac\{-4-\backslash sqrt\{28\}\}\{2\backslash cdot\backslash left(-1\backslash right)\}=-2+\backslash sqrt7\backslash approx0\{,\}65$

$\to \backslash ;\backslash rightarrow\backslash ;$ ${\mathrm{P}}_{\mathrm{x}2}(-2+\sqrt{7}\mathrm{\mid }0)\{\backslash mathrm\; P\}\_\{\backslash mathrm\; x2\}\backslash left(-2+\backslash sqrt7\backslash vert0\backslash right)$

Schnittpunkt mit der y-Achse

$f\left(x\right)=-x^2-4x+3f\backslash left(x\backslash right)=-x^2-4x+3$

Für x gleich 0 setzen.

$f\left(0\right)=0-0+3=3f\backslash left(0\backslash right)=0-0+3=3$

$f\left(0\right)=3f\backslash left(0\backslash right)=3$

$\to \backslash ;\backslash rightarrow\backslash ;$ ${\mathrm{P}}_{\mathrm{y}}\left(0\mathrm{\mid }3\right)\{\backslash mathrm\; P\}\_\backslash mathrm\; y\backslash left(0\backslash vert3\backslash right)$

Normalparabel: $f\left(x\right)=x^{2}f\backslash left(x\backslash right)=x^2$

Die verschiedenen Verschiebunge n lassen sich an der Scheitelform ablesen.

$f\left(x\right)=-{(x+2)}^2+7f\backslash left(x\backslash right)=-\backslash left(x+2\backslash right)^2+7$

$\to \backslash ;\backslash rightarrow\backslash ;$ Nach unten geöffnet.

$\to \backslash ;\backslash rightarrow\backslash ;$ Verschiebung um 2 LE nach links.

$\to \backslash ;\backslash rightarrow\backslash ;$ Verschiebung um 7 LE nach oben.

$\to \backslash ;\backslash rightarrow\backslash ;$ $\mathrm{S}\left(4\mathrm{\mid }7\right)\backslash mathrm\; S\backslash left(4\backslash ;\backslash vert\backslash ;7\backslash right)$

Schnittpunkte mit der x-Achse

$\to \backslash ;\backslash rightarrow\backslash ;$ 2 Lösungen

$x_1=\frac{-8+\sqrt{28}}{2\cdot (-1)}=4-\sqrt{7}\approx \mathrm{1,35}x\_1=\backslash frac\{-8+\backslash sqrt\{28\}\}\{2\backslash cdot\backslash left(-1\backslash right)\}=4-\backslash sqrt7\backslash approx1\{,\}35$

$\to \backslash ;\backslash rightarrow\backslash ;$ ${\mathrm{P}}_{\mathrm{x}1}(4-\sqrt{7}\mathrm{\mid }0)\{\backslash mathrm\; P\}\_\{\backslash mathrm\; x1\}\backslash left(4-\backslash sqrt7\backslash vert0\backslash right)$

$x_2=\frac{-8-\sqrt{28}}{2\cdot (-1)}=4+\sqrt{7}\approx \mathrm{6,65}x\_2=\backslash frac\{-8-\backslash sqrt\{28\}\}\{2\backslash cdot\backslash left(-1\backslash right)\}=4+\backslash sqrt7\backslash approx6\{,\}65$

$\to \backslash ;\backslash rightarrow\backslash ;$ ${\mathrm{P}}_{\mathrm{x}2}(4+\sqrt{7}\mathrm{\mid }0)\{\backslash mathrm\; P\}\_\{\backslash mathrm\; x2\}\backslash left(4+\backslash sqrt7\backslash vert0\backslash right)$

Schnittpunkt mit der y-Achse

$f\left(x\right)=-x^2+8x-9f\backslash left(x\backslash right)=-x^2+8x-9$

Für x gleich 0 setzen.

$f\left(0\right)=0+0-9=-9f\backslash left(0\backslash right)=0+0-9=-9$

$\to \backslash ;\backslash rightarrow\backslash ;$ ${\mathrm{P}}_{\mathrm{y}}(0\mathrm{\mid }-9)\{\backslash mathrm\; P\}\_\backslash mathrm\; y\backslash left(0\backslash vert-9\backslash right)$

Normalparabel: $f\left(x\right)=x^{2}f\backslash left(x\backslash right)=x^2$

Die verschiedenen Verschiebungen lassen sich an der Scheitelform ablesen.

$f\left(x\right)=-{(x-4)}^2+7f\backslash left(x\backslash right)=-\backslash left(x-\backslash textstyle4\backslash right)^2+7$

$\to \backslash ;\backslash rightarrow\backslash ;$ Nach unten geöffnet.

$\to \backslash ;\backslash rightarrow\backslash ;$ Verschiebung um 4 LE nach rechts.

$\to \backslash ;\backslash rightarrow\backslash ;$ Verschiebung um 7 LE nach oben.

$\to \backslash ;\backslash rightarrow\backslash ;$ $\mathrm{S}(4\mathrm{\mid }-3)\backslash mathrm\; S\backslash left(4\backslash ;\backslash vert\backslash ;-3\backslash right)$

Schnittpunkte mit der x-Achse

$\to \backslash ;\backslash rightarrow\backslash ;$ 2 Lösungen

$x_1=\frac{4+\sqrt{6}}{2\cdot {\displaystyle \frac{1}{2}}}=4+\sqrt{6}\approx \mathrm{6,45}x\_1=\backslash frac\{4+\backslash sqrt6\}\{2\backslash cdot\backslash displaystyle\backslash frac12\}=4+\backslash sqrt6\backslash approx6\{,\}45$

$\to \backslash ;\backslash rightarrow\backslash ;$ ${\mathrm{P}}_{\mathrm{x}1}(4+\sqrt{6}\mathrm{\mid }0)\{\backslash mathrm\; P\}\_\{\backslash mathrm\; x1\}\backslash left(4+\backslash sqrt6\backslash vert0\backslash right)$

$x_2=\frac{4-\sqrt{6}}{2\cdot {\displaystyle \frac{1}{2}}}=4-\sqrt{6}\approx \mathrm{1,55}x\_2=\backslash frac\{4-\backslash sqrt6\}\{2\backslash cdot\backslash displaystyle\backslash frac12\}=4-\backslash sqrt6\backslash approx1\{,\}55$

$\to \backslash ;\backslash rightarrow\backslash ;$ ${\mathrm{P}}_{\mathrm{x}2}(4-\sqrt{6}\mathrm{\mid }0)\{\backslash mathrm\; P\}\_\{\backslash mathrm\; x2\}\backslash left(4-\backslash sqrt6\backslash vert0\backslash right)$

Schnittpunkt mit der y-Achse

$f\left(x\right)=\frac{1}{2}{x}^2-4x+5f\backslash left(x\backslash right)=\backslash frac12x^2-4x+5$

Für x gleich 0 setzen.

$f\left(0\right)=0-0+5=5f\backslash left(0\backslash right)=0-0+5=5$

$\to \backslash ;\backslash rightarrow\backslash ;$ ${\mathrm{P}}_{\mathrm{y}}\left(0\mathrm{\mid }5\right)\{\backslash mathrm\; P\}\_\backslash mathrm\; y\backslash left(0\backslash vert5\backslash right)$

Normalparabel: $f\left(x\right)=x^{2}f\backslash left(x\backslash right)=x^2$

Die verschiedenen Verschiebungen lassen sich an der Scheitelform ablesen.

$f\left(x\right)=\frac{1}{2}{(x-4)}^2-3f\backslash left(x\backslash right)=\backslash frac12\backslash left(x-\backslash textstyle4\backslash right)^2-3$

$\to \backslash ;\backslash rightarrow\backslash ;$ Gestaucht durch den Faktor $\frac{1}{2}\backslash frac12$.

$\to \backslash ;\backslash rightarrow\backslash ;$ Verschiebung um 4 LE nach rechts.

$\to \backslash ;\backslash rightarrow\backslash ;$ Verschiebung um 3 LE nach unten.

$\to \backslash ;\backslash rightarrow\backslash ;$ $\mathrm{S}(-2\mathrm{\mid }8)\backslash mathrm\; S\backslash left(-2\backslash ;\backslash vert\backslash ;8\backslash right)$

Schnittpunkte mit der x-Achse

$\to \backslash ;\backslash rightarrow\backslash ;$ ${\mathrm{P}}_{\mathrm{x}1}(-6\mathrm{\mid }0)\{\backslash mathrm\; P\}\_\{\backslash mathrm\; x1\}\backslash left(-6\backslash vert0\backslash right)$ ; ${\mathrm{P}}_{\mathrm{x}2}\left(2\mathrm{\mid }0\right)\{\backslash mathrm\; P\}\_\{\backslash mathrm\; x2\}\backslash left(2\backslash vert\backslash ;0\backslash right)$

Schnittpunkt mit der y-Achse

$f\left(x\right)=-\frac{1}{2}{x}^2-2x+6f\backslash left(x\backslash right)=-\backslash frac12x^2-2x+6$

Für x gleich 0 setzen.

$f\left(0\right)=0-0+6=6f\backslash left(0\backslash right)=0-0+6=6$

$\to \backslash ;\backslash rightarrow\backslash ;$ ${\mathrm{P}}_{\mathrm{y}}\left(0\mathrm{\mid }6\right)\{\backslash mathrm\; P\}\_\backslash mathrm\; y\backslash left(0\backslash vert6\backslash right)$

Normalparabel: $f\left(x\right)=x^{2}f\backslash left(x\backslash right)=x^2$

Die verschiedenen Verschiebungen lassen sich an der Scheitelform ablesen.

$=-\frac{1}{2}{(x+2)}^2+8=-\backslash frac12\backslash left(x+\backslash textstyle2\backslash right)^2+\backslash textstyle8$

$\to \backslash ;\backslash rightarrow\backslash ;$ Nach unten geöffnet.

$\to \backslash ;\backslash rightarrow\backslash ;$ Gestaucht durch den Faktor $\frac{1}{2}\backslash frac12$.

$\to \backslash ;\backslash rightarrow\backslash ;$ Verschiebung um 2 LE nach links.

$\to \backslash ;\backslash rightarrow\backslash ;$ Verschiebung um 8 LE nach oben.

$\to \backslash ;\backslash rightarrow\backslash ;$ $\mathrm{S}(1\mathrm{\mid }-\frac{7}{3})\backslash mathrm\; S\backslash left(1\backslash ;\backslash vert\backslash ;-\backslash frac73\backslash right)$

Schnittpunkte mit der x-Achse

$\to \backslash ;\backslash rightarrow\backslash ;$ 2 Lösungen

$x_1=\frac{{\displaystyle \frac{2}{3}}+\sqrt{{\displaystyle \frac{28}{9}}}}{2\cdot {\displaystyle \frac{1}{3}}}=1+\sqrt{7}\approx \mathrm{3,65}x\_1=\backslash frac\{\{\backslash displaystyle\backslash frac23\}+\backslash sqrt\{\backslash displaystyle\backslash frac\{28\}9\}\}\{2\backslash cdot\backslash displaystyle\backslash frac13\}=1+\backslash sqrt7\backslash approx3\{,\}65$

$\to \backslash ;\backslash rightarrow\backslash ;$ ${\mathrm{P}}_{\mathrm{x}1}(1+\sqrt{7}\mathrm{\mid }0)\{\backslash mathrm\; P\}\_\{\backslash mathrm\; x1\}\backslash left(1+\backslash sqrt7\backslash vert0\backslash right)$

$x_2=\frac{{\displaystyle \frac{2}{3}}-\sqrt{{\displaystyle \frac{28}{9}}}}{2\cdot {\displaystyle \frac{1}{3}}}=1-\sqrt{7}\approx -\mathrm{1,65}x\_2=\backslash frac\{\{\backslash displaystyle\backslash frac23\}-\backslash sqrt\{\backslash displaystyle\backslash frac\{28\}9\}\}\{2\backslash cdot\backslash displaystyle\backslash frac13\}=1-\backslash sqrt7\backslash approx-1\{,\}65$

$\to \backslash ;\backslash rightarrow\backslash ;$ ${\mathrm{P}}_{\mathrm{x}2}(1-\sqrt{7}\mathrm{\mid }0)\{\backslash mathrm\; P\}\_\{\backslash mathrm\; x2\}\backslash left(1-\backslash sqrt7\backslash vert0\backslash right)$

Schnittpunkt mit der y-Achse

$f\left(x\right)=\frac{1}{3}{x}^2-\frac{2}{3}x-2f\backslash left(x\backslash right)=\backslash frac13x^2-\backslash frac23x-2$

Für x gleich 0 setzen.

$f\left(0\right)=0-0-2=-2f\backslash left(0\backslash right)=0-0-2=-2$

$\to \backslash ;\backslash rightarrow\backslash ;$ ${\mathrm{P}}_{\mathrm{y}}(0\mathrm{\mid }-2)\{\backslash mathrm\; P\}\_\backslash mathrm\; y\backslash left(0\backslash vert-2\backslash right)$

Normalparabel: $f\left(x\right)=x^{2}f\backslash left(x\backslash right)=x^2$

Die verschiedenen Verschiebungen lassen sich an der Scheitelform ablesen.

$=\frac{1}{3}{(x-1)}^2-\frac{7}{3}=\backslash frac13\backslash left(x-\backslash textstyle1\backslash right)^2-\backslash frac73$

$\to \backslash ;\backslash rightarrow\backslash ;$ Gestaucht durch den Faktor $\frac{1}{3}\backslash frac13$.

$\to \backslash ;\backslash rightarrow\backslash ;$ Verschiebung um 1 LE nach rechts.

$\to \backslash ;\backslash rightarrow\backslash ;$ Verschiebung um  $\frac{7}{3}\backslash frac73$ LE nach unten.

$\to \backslash ;\backslash rightarrow\backslash ;$ $\mathrm{S}\left(\frac{9}{16}\mathrm{\mid }\frac{795}{128}\right)\backslash mathrm\; S\backslash left(\backslash frac9\{16\}\backslash ;\backslash vert\backslash ;\backslash frac\{795\}\{128\}\backslash right)$

Schnittpunkte mit der x-Achse

$\to \backslash ;\backslash rightarrow\backslash ;$ 2 Lösungen

$x_1=\frac{-{\displaystyle \frac{3}{4}}+\sqrt{{\displaystyle \frac{265}{16}}}}{2\cdot (-{\displaystyle \frac{2}{3}})}\approx -\mathrm{2,49}x\_1=\backslash frac\{-\{\backslash displaystyle\backslash frac34\}+\backslash sqrt\{\backslash displaystyle\backslash frac\{265\}\{16\}\}\}\{2\backslash cdot\backslash left(-\backslash displaystyle\backslash frac23\backslash right)\}\backslash approx-2\{,\}49$

$\to \backslash ;\backslash rightarrow\backslash ;$ ${\mathrm{P}}_{\mathrm{x}1}(-\mathrm{2,49}\mathrm{\mid }0)\{\backslash mathrm\; P\}\_\{\backslash mathrm\; x1\}\backslash left(-2\{,\}49\backslash vert0\backslash right)$

$x_2=\frac{-{\displaystyle \frac{3}{4}}-\sqrt{{\displaystyle \frac{265}{16}}}}{2\cdot (-{\displaystyle \frac{2}{3}})}\approx \mathrm{3,61}x\_2=\backslash frac\{-\{\backslash displaystyle\backslash frac34\}-\backslash sqrt\{\backslash displaystyle\backslash frac\{265\}\{16\}\}\}\{2\backslash cdot\backslash left(-\backslash displaystyle\backslash frac23\backslash right)\}\backslash approx3\{,\}61$

$\to \backslash ;\backslash rightarrow\backslash ;$ ${\mathrm{P}}_{\mathrm{x}2}\left(\mathrm{3,61}\mathrm{\mid }0\right)\{\backslash mathrm\; P\}\_\{\backslash mathrm\; x2\}\backslash left(3\{,\}61\backslash vert0\backslash right)$

Schnittpunkt mit der y-Achse

$f\left(x\right)=-\frac{2}{3}{x}^2+\frac{3}{4}x+6f\backslash left(x\backslash right)=-\backslash frac23x^2+\backslash frac34x+6$

x gleich 0 setzen.

$f\left(0\right)=0+0+6=6f\backslash left(0\backslash right)=0+0+6=6$

$\to \backslash ;\backslash rightarrow\backslash ;$ ${\mathrm{P}}_{\mathrm{y}}\left(0\mathrm{\mid }6\right)\{\backslash mathrm\; P\}\_\backslash mathrm\; y\backslash left(0\backslash vert6\backslash right)$

Normalparabel: $f\left(x\right)=x^{2}f\backslash left(x\backslash right)=x^2$

Die verschiedenen Verschiebungen lassen sich an der Scheitelform ablesen.

$=-\frac{2}{3}{(x-\frac{9}{16})}^2+\frac{795}{128}=-\backslash frac23\backslash left(x-\backslash frac9\{16\}\backslash right)^2+\backslash frac\{795\}\{128\}$

$\to \backslash ;\backslash rightarrow\backslash ;$ Nach unten geöffnet.

$\to \backslash ;\backslash rightarrow\backslash ;$ Gestaucht durch den Faktor $\frac{2}{3}\backslash frac23$.

$\to \backslash ;\backslash rightarrow\backslash ;$ Verschiebung um  $\frac{9}{16}\backslash frac9\{16\}$ LE nach rechts.

$\to \backslash ;\backslash rightarrow\backslash ;$ Verschiebung um $\frac{795}{128}\backslash frac\{795\}\{128\}$ LE nach oben.