$\begin{array}{c}y=x-\mathrm{1,5}\\ y=x^2-4x+\mathrm{2,5}\end{array}\backslash def\backslash arraystretch\{1.25\}\; \backslash begin\{array\}\{l\}\backslash ;\backslash ;y=x-1\{,\}5\backslash \backslash \backslash ;\backslash ;y=x^2-4x+2\{,\}5\backslash end\{array\}$

Funktionen gleichsetzen.

$x-\mathrm{1,5}=x^2-4x+\mathrm{2,5}x-1\{,\}5=x^2-4x+2\{,\}5$

Rechne: $-x+\mathrm{1,5}\{\}-x+1\{,\}5$

$0=x^2-4x-x+\mathrm{2,5}+\mathrm{1,5}0=x^2-4x-x+2\{,\}5+1\{,\}5$

$0=x^2-5x+40=x^2-5x+4$

Mitternachtsformel anwenden $\to \backslash ;\backslash rightarrow$ Diskriminante $DD$ berechnen.

$D=25-4\cdot 4=9D=25-4\backslash cdot4=9$

$\begin{array}{c}{x}_1=\frac{5-\sqrt{9}}{2}=1\\ x_2=\frac{5+\sqrt{9}}{2}=4\end{array}\backslash def\backslash arraystretch\{1.25\}\; \backslash begin\{array\}\{l\}x\_1=\backslash frac\{5-\backslash sqrt9\}2=1\backslash \backslash x\_2=\backslash frac\{5+\backslash sqrt9\}2=4\backslash end\{array\}$

$yy$-Werte berechnen. $\to \backslash rightarrow$ $x_{1}x\_1$ in Gleichung einsetzen.

$\begin{array}{c}{x}_1=1\\ \Rightarrow y_1=f(x_1)=1-\mathrm{1,5}=-\mathrm{0,5}\end{array}\backslash def\backslash arraystretch\{1.25\}\; \backslash begin\{array\}\{l\}x\_1=1\backslash ;\backslash \backslash \backslash Rightarrow\backslash ;y\_1\backslash ;=f(x\_1)=1-1\{,\}5=-0\{,\}5\backslash end\{array\}$

$\Rightarrow S_1(1\mid -\mathrm{0,5})\backslash Rightarrow\; S\_\{1\backslash ;\}(1\backslash mid-0\{,\}5)$

$x_{2}x\_2$ in Gleichung einsetzen.

$\begin{array}{c}{x}_2=4\\ \Rightarrow y_2=f(x_2)=4-\mathrm{1,5}=\mathrm{2,5}\end{array}\backslash def\backslash arraystretch\{1.25\}\; \backslash begin\{array\}\{l\}x\_2=4\backslash \backslash \backslash Rightarrow\backslash ;\backslash ;y\_2=f(x\_2)=4-1\{,\}5=2\{,\}5\backslash end\{array\}$

$\Rightarrow S_2(4\mid \mathrm{2,5})\backslash Rightarrow\backslash ;S\_\{2\backslash ;\}\backslash left(4\backslash mid2\{,\}5\backslash right)$