$\begin{array}{l} \\ \text{ Aufstellen von Funktionsgleichungen}\\ \\ \textbf{Aufgabe:}\\eine quadratische Funktion mit dem Scheitel S(0/3) und einer Nullstelle bei x=2.\\\\ \textbf{Rechnung:}\\ \\ \text{Funktion} \\ f\left(x\right)=a\cdot x^2+b\cdot x+c\\ f'\left(x\right)=2a\cdot x+b\\ f''\left(x\right)=2a\\ \text{Gegeben:}\\ f\left(2\right)=0 \qquad a\cdot 2^2+b\cdot 2+c=0 \\ f'\left(1\right)=0 \qquad 2a\cdot 1+b=0 \\ f\left(1\right)=3 \qquad a\cdot 1^2+b\cdot 1+c=3 \\\small \begin{array}{l} 4a+2b+c=0 \\ 2a+b=0 \\ a+b+c=3 \\ \\ \end{array} \qquad \small \begin{array}{ccc|cc } a & b & c & & \\ \hline4 & 2 & 1 & 0 \\ 2 & 1 & 0 & 0 \\ 1 & 1 & 1 & 3 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}2=\text{Zeile}2\text{-Zeile}1\cdot \frac{2}{4}\\z2s1=2-4\cdot \frac{2}{4}=0 \\ z2s2=1-2\cdot \frac{2}{4}=0 \\ z2s3=0-1\cdot \frac{2}{4}=-\frac{1}{2} \\ z2s4=0-0\cdot \frac{2}{4}=0 \\ \end{array}\qquad \small \begin{array}{ccc|cc } a & b & c & & \\ \hline4 & 2 & 1 & 0 \\ 0 & 0 & -\frac{1}{2} & 0 \\ 1 & 1 & 1 & 3 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}3=\text{Zeile}3\text{-Zeile}1\cdot \frac{1}{4}\\z3s1=1-4\cdot \frac{1}{4}=0 \\ z3s2=1-2\cdot \frac{1}{4}=\frac{1}{2} \\ z3s3=1-1\cdot \frac{1}{4}=\frac{3}{4} \\ z3s4=3-0\cdot \frac{1}{4}=3 \\ \end{array}\qquad \small \begin{array}{ccc|cc } a & b & c & & \\ \hline4 & 2 & 1 & 0 \\ 0 & 0 & -\frac{1}{2} & 0 \\ 0 & \frac{1}{2} & \frac{3}{4} & 3 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeilen vertauschen } \\ \end{array}\qquad \small \begin{array}{ccc|cc } a & b & c & & \\ \hline4 & 2 & 1 & 0 \\ 0 & \frac{1}{2} & \frac{3}{4} & 3 \\ 0 & 0 & -\frac{1}{2} & 0 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}1=\text{Zeile}1\text{-Zeile}2\cdot \frac{2}{\frac{1}{2}}\\z1s2=2-\frac{1}{2}\cdot \frac{2}{\frac{1}{2}}=0 \\ z1s3=1-\frac{3}{4}\cdot \frac{2}{\frac{1}{2}}=-2 \\ z1s4=0-3\cdot \frac{2}{\frac{1}{2}}=-12 \\ \end{array}\qquad \small \begin{array}{ccc|cc } a & b & c & & \\ \hline4 & 0 & -2 & -12 \\ 0 & \frac{1}{2} & \frac{3}{4} & 3 \\ 0 & 0 & -\frac{1}{2} & 0 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}1=\text{Zeile}1\text{-Zeile}3\cdot \frac{-2}{-\frac{1}{2}}\\z1s3=-2-(-\frac{1}{2})\cdot \frac{-2}{-\frac{1}{2}}=0 \\ z1s4=-12-0\cdot \frac{-2}{-\frac{1}{2}}=-12 \\ \end{array}\qquad \small \begin{array}{ccc|cc } a & b & c & & \\ \hline4 & 0 & 0 & -12 \\ 0 & \frac{1}{2} & \frac{3}{4} & 3 \\ 0 & 0 & -\frac{1}{2} & 0 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}2=\text{Zeile}2\text{-Zeile}3\cdot \frac{\frac{3}{4}}{-\frac{1}{2}}\\z2s3=\frac{3}{4}-(-\frac{1}{2})\cdot \frac{\frac{3}{4}}{-\frac{1}{2}}=0 \\ z2s4=3-0\cdot \frac{\frac{3}{4}}{-\frac{1}{2}}=3 \\ \end{array}\qquad \small \begin{array}{ccc|cc } a & b & c & & \\ \hline4 & 0 & 0 & -12 \\ 0 & \frac{1}{2} & 0 & 3 \\ 0 & 0 & -\frac{1}{2} & 0 \\ \end{array} \\ \\ a=\frac{-12}{4}=-3\\b=\frac{3}{\frac{1}{2}}=6\\c=\frac{0}{-\frac{1}{2}}=0\\L=\{-3/6/0\} \\ \text{Funktion} \\ f\left(x\right)=-3x^2+6x \end{array}$