Der Scheitel der Parabel $p_{1}p\_\{\backslash tiny\; 1\}\backslash$ist laut Angabe: $S_1(-4\mathrm{\mid }1)\backslash \; S\_\{\backslash tiny\; 1\}(-4|1)\backslash$

Die Scheitelform lautet:

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

Setze die entsprechenden Werte des Scheitels$S_1\backslash \; S\_\{\backslash tiny\; 1\}$ ein.

$p_1:y=(x+4{)}^2+1p\_\{\backslash tiny\; 1\}:y\backslash \; =\backslash \; (x+4)^2+1$

die Normalform ist dann:

$p_1:y=x^2+8x+17p\_\{\backslash tiny\; 1\}:y\backslash \; =\backslash \; x^2+8x+17$

Die allgemeine Form oder auch Hauptform einer quadratischen Funktion lautet:

$f\left(x\right)=a{x}^2+bx+cf\backslash left(x\backslash right)=ax^2+bx+c$

Setze die gegebenen Punktepaare in die Funktionsgleichung ein.

Da es sich um eine nach unten geöffnete Normalparabel handelt ist der Vorfaktor $a=-1a=-1$

$\text{Aus }A(-4\mathrm{\mid }1):(I)1=-1\cdot (-4{)}^2+b\cdot (-4)+c\backslash text\{Aus\; \}A(-4|1):\backslash \; \backslash \; \backslash \; \backslash \; (I)\; \backslash \; \backslash \; 1=\; -1\backslash cdot\; (-4)^2+b\backslash cdot\; (-4)+c$

$\text{Aus }B(0\mathrm{\mid }1):(II)1=-1\cdot (0{)}^2+b\cdot (0)+c\Rightarrow c=1\backslash text\{Aus\; \}\; B\backslash hspace\{1.5mm\}(0|1)\backslash hspace\{1.5mm\}:\backslash \; (II)\backslash \; \backslash \; \backslash \; 1=\; -1\backslash cdot\; (0)^2+b\backslash cdot\; (0)+c\backslash quad\backslash Rightarrow\backslash quad\; c=1$

$\backslash hspace\{74mm\}$setze $cc$ in $(I)(I)$ ein und berechne $bb$

Die Normalform von $p_{2}p\_\{2\}\backslash$lautet:

$p_2:y=-x^2-4x+1p\_\{2\}:y=-x^2-4x+1\backslash$

Umformen der allgemeinen Form von $p_2\backslash \; p\_\{2\}\backslash$in die Scheitelpunktform:

$p_2:y=-x^2-4x+1p\_\{2\}:y=-x^2-4x+1$

$p_2:y=-(x+4x-1)p\_\{2\}:y=-(x+4x-1)$

$p_2:y=-(x+2{)}^2+4+1p\_\{2\}:y=-(x+2)^2+4+1$

$p_2:y=-(x+2{)}^2+5p\_\{2\}:y=-(x+2)^2+5$

$\underset{‾}{\underset{‾}{S_2(-2\mathrm{\mid }5)}}\backslash underline\{\backslash underline\{S\_\{2\}(-2|5)\}\}$

oder, du kannst auch gleich die Scheitelpunktform ermitteln:

Die Funktionswerte von $AA$ und $BB$ sind gleich nämlich: $y_A=y_B=1\backslash \; \backslash \; y\_\{\backslash tiny\{A\}\}=y\_\{\backslash tiny\{B\}\}=1$

d.h. es gilt:$x_S={\displaystyle \frac{x_A+xB}{2}}\Rightarrow x_S={\displaystyle \frac{-4+0}{2}}\Rightarrow \boxed{{\displaystyle x_S=-2}}\backslash \; x\_\{\backslash tiny\{S\}\}=\backslash dfrac\{x\_\{\backslash tiny\{A\}\}+x\{\backslash tiny\{B\}\}\}\{2\}\backslash quad\backslash Rightarrow\backslash quad\; x\_\{\backslash tiny\{S\}\}=\backslash dfrac\{-4+0\}\{2\}\backslash quad\backslash Rightarrow\backslash quad\; \backslash boxed\{x\_\{\backslash tiny\{S\}\}=-2\}$

setze $x_{S}x\_\{\backslash tiny\{S\}\}\backslash$und z.B. $x_{A}x\_\{\backslash tiny\{A\}\}\backslash$in die Scheitelpunktform$(x-x_S{)}^2+y_S\backslash \; (x-x\_\{\backslash tiny\{S\}\})^2+y\_\{\backslash tiny\{S\}\}\backslash$ ein und berechne $y_S\backslash \; y\_\{\backslash tiny\{S\}\}$.

$-(-4+2{)}^2+y_S=1\Rightarrow y_S=1+4\Rightarrow \boxed{{\displaystyle y_S=5}}\backslash hspace\{10mm\}-(-4+2)^2+y\_\{\backslash tiny\{S\}\}=1\backslash quad\backslash Rightarrow\backslash quad\; y\_\{\backslash tiny\{S\}\}=1+4\backslash quad\backslash Rightarrow\backslash quad\; \backslash boxed\{y\_\{\backslash tiny\{S\}\}=5\}$

Die Scheitelpunkform lautet dann: $p_2:y=-(x+2{)}^2+5\Rightarrow \underset{‾}{\underset{‾}{S_2(-2\mathrm{\mid }5)}}\backslash \; p\_\{2\}:y=-(x+2)^2+5\backslash quad\backslash Rightarrow\backslash quad\; \backslash underline\{\backslash underline\{S\_\{2\}(-2|5)\}\}$

$\text{Koordinaten der Schnittpunkte: }\boxed{{\displaystyle Q(-4\mathrm{\mid }1)}}\boxed{{\displaystyle R(-2\mathrm{\mid }5)}}\backslash text\{Koordinaten\; der\; Schnittpunkte:\}\backslash \; \backslash hspace\{2mm\}\backslash boxed\{Q\backslash \; (-4|1)\}\backslash hspace\{4mm\}\backslash boxed\{R\backslash \; (-2|5)\}$

Berechne die Schnittpunkte $MM$ und $NN$ der Normalparabel$p_3:\backslash \; \backslash \; p\_3:$ $y=x^2+2x-2\backslash \; y=x^2+2x-2$

mit der Normalparabel $p_4:y=\text{–}{x}^2\text{–}2x+4p\_4:\; y\; =\; –x^2\; –\; 2x\; +\; 4$ , indem du die Funktionsterme gleichsetzt und diese Gleichung anschließend nach $xx$ auflöst.

Löse jetzt die Quadratische Gleichung z.B. mit der Mitternachtsformel

$x_{\mathrm{1,2}}={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}\Rightarrow x_{\mathrm{1,2}}={\displaystyle \frac{-2\pm \sqrt{2^2-4\cdot 1\cdot (-3)}}{2\cdot 1}}x\_\{1\{,\}2\}=\backslash dfrac\{-b\backslash pm\backslash sqrt\{b^2-4ac\}\}\{2a\}\backslash qquad\backslash Rightarrow\backslash \; x\_\{1\{,\}2\}=\backslash dfrac\{-2\backslash pm\backslash sqrt\{2^2-4\backslash cdot1\backslash cdot(-3)\}\}\{2\backslash cdot1\}$

$x_{\mathrm{1,2}}={\displaystyle \frac{-2\pm \sqrt{16}}{2\cdot 1}}\backslash hspace\{4.8cm\}x\_\{1\{,\}2\}=\backslash dfrac\{-2\backslash pm\backslash sqrt\{16\}\}\{2\backslash cdot1\}$

$x_{\mathrm{1,2}}={\displaystyle \frac{-2\pm 4}{2\cdot 1}}=-1\pm 2\backslash hspace\{4.8cm\}x\_\{1\{,\}2\}=\backslash dfrac\{-2\backslash pm\{4\}\}\{2\backslash cdot1\}\backslash \; =\backslash \; -1\backslash pm2$

$x_1=-3x\_1\backslash \; =\backslash \; -3$

$x_2=1x\_2\backslash \; =\backslash \; \backslash \; \backslash \; \backslash \; 1$

Setze $x_{1}x\_1$ und$x_2\backslash \; x\_2$ in z.B. $p_3:y=x^2+2x-2\backslash \; p\_3:\; y\backslash \; =\backslash \; x^2+2x-2\backslash$ ein und bestimme $y_{1}y\_1$ und$y_2\backslash \; y\_2$.

Du kannst $x_{1}x\_1\backslash$und$x_2\backslash \; x\_2\backslash$auch in $p_4:y=-x^2-2x+4p\_4:y=-x^2-2x+4$ einsetzen.

$y_1=1\cdot (-3{)}^2+2\cdot (-3)-2\Rightarrow y_1=9-8\Rightarrow M(-3\mathrm{\mid }1)y\_1\backslash \; =\backslash \; 1\backslash cdot(-3)^2+2\backslash cdot(-3)-2\backslash \; \backslash quad\backslash Rightarrow\backslash quad\; y\_\{1\}=9-8\backslash quad\backslash Rightarrow\backslash quad\; M\backslash \; (-3|1)$

$y_2=1\cdot (1{)}^2+2\cdot 1-2\Rightarrow y_2=1+0\Rightarrow N(1\mathrm{\mid }1)y\_\{2\}\backslash \; =\backslash \; 1\backslash cdot(1)^2+2\backslash cdot1-2\backslash hspace\{12.5mm\}\backslash Rightarrow\backslash quad\; y\_\{2\}=1+0\backslash quad\backslash Rightarrow\backslash quad\; N\backslash \; (1|1)$