Beispiel Nr: 04
$\text{Aufstellen von Funktionsgleichungen}\\$
$\textbf{Aufgabe:}$ eine ganzrationale Funktion 3.Grades, deren Graph durch den Ursprung geht und den Terrassenpunkt $T(2/2)$ hat.\\
$\textbf{Rechnung:}\\ \\ \text{Funktion} \\ f\left(x\right)=a\cdot x^3+b\cdot x^2+c\cdot x+d\\ f'\left(x\right)=3a\cdot x^2+2b\cdot x+c\\ f''\left(x\right)=6a\cdot x+2b\\ \text{Gegeben:}\\ f\left(2\right)=2 \qquad a\cdot 2^3+b\cdot 2^2+c\cdot 2+d=2 \\ f'\left(2\right)=0 \qquad 3a\cdot 2^2+2b\cdot 2+c=0 \\ f''\left(2\right)=0 \qquad 6a\cdot 2+2b=0 \\ f\left(0\right)=0 \qquad a\cdot 0^3+b\cdot 0^2+c\cdot 0+d=0 \\\small \begin{array}{l} 8a+4b+2c+d=2 \\ 12a+4b+c=0 \\ 12a+2b=0 \\ d=0 \\ \\ \end{array} \qquad \small \begin{array}{cccc|cc } a & b & c & d & & \\ \hline8 & 4 & 2 & 1 & 2 \\ 12 & 4 & 1 & 0 & 0 \\ 12 & 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}2=\text{Zeile}2\text{-Zeile}1\cdot \frac{12}{8}\\z2s1=12-8\cdot \frac{12}{8}=0 \\ z2s2=4-4\cdot \frac{12}{8}=-2 \\ z2s3=1-2\cdot \frac{12}{8}=-2 \\ z2s4=0-1\cdot \frac{12}{8}=-1\frac{1}{2} \\ z2s5=0-2\cdot \frac{12}{8}=-3 \\ \end{array}\qquad \small \begin{array}{cccc|cc } a & b & c & d & & \\ \hline8 & 4 & 2 & 1 & 2 \\ 0 & -2 & -2 & -1\frac{1}{2} & -3 \\ 12 & 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}3=\text{Zeile}3\text{-Zeile}1\cdot \frac{12}{8}\\z3s1=12-8\cdot \frac{12}{8}=0 \\ z3s2=2-4\cdot \frac{12}{8}=-4 \\ z3s3=0-2\cdot \frac{12}{8}=-3 \\ z3s4=0-1\cdot \frac{12}{8}=-1\frac{1}{2} \\ z3s5=0-2\cdot \frac{12}{8}=-3 \\ \end{array}\qquad \small \begin{array}{cccc|cc } a & b & c & d & & \\ \hline8 & 4 & 2 & 1 & 2 \\ 0 & -2 & -2 & -1\frac{1}{2} & -3 \\ 0 & -4 & -3 & -1\frac{1}{2} & -3 \\ 0 & 0 & 0 & 1 & 0 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}1=\text{Zeile}1\text{-Zeile}2\cdot \frac{4}{-2}\\z1s2=4-(-2)\cdot \frac{4}{-2}=0 \\ z1s3=2-(-2)\cdot \frac{4}{-2}=-2 \\ z1s4=1-(-1\frac{1}{2})\cdot \frac{4}{-2}=-2 \\ z1s5=2-(-3)\cdot \frac{4}{-2}=-4 \\ \end{array}\qquad \small \begin{array}{cccc|cc } a & b & c & d & & \\ \hline8 & 0 & -2 & -2 & -4 \\ 0 & -2 & -2 & -1\frac{1}{2} & -3 \\ 0 & -4 & -3 & -1\frac{1}{2} & -3 \\ 0 & 0 & 0 & 1 & 0 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}3=\text{Zeile}3\text{-Zeile}2\cdot \frac{-4}{-2}\\z3s2=-4-(-2)\cdot \frac{-4}{-2}=0 \\ z3s3=-3-(-2)\cdot \frac{-4}{-2}=1 \\ z3s4=-1\frac{1}{2}-(-1\frac{1}{2})\cdot \frac{-4}{-2}=1\frac{1}{2} \\ z3s5=-3-(-3)\cdot \frac{-4}{-2}=3 \\ \end{array}\qquad \small \begin{array}{cccc|cc } a & b & c & d & & \\ \hline8 & 0 & -2 & -2 & -4 \\ 0 & -2 & -2 & -1\frac{1}{2} & -3 \\ 0 & 0 & 1 & 1\frac{1}{2} & 3 \\ 0 & 0 & 0 & 1 & 0 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}1=\text{Zeile}1\text{-Zeile}3\cdot \frac{-2}{1}\\z1s3=-2-1\cdot \frac{-2}{1}=0 \\ z1s4=-2-1\frac{1}{2}\cdot \frac{-2}{1}=1 \\ z1s5=-4-3\cdot \frac{-2}{1}=2 \\ \end{array}\qquad \small \begin{array}{cccc|cc } a & b & c & d & & \\ \hline8 & 0 & 0 & 1 & 2 \\ 0 & -2 & -2 & -1\frac{1}{2} & -3 \\ 0 & 0 & 1 & 1\frac{1}{2} & 3 \\ 0 & 0 & 0 & 1 & 0 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}2=\text{Zeile}2\text{-Zeile}3\cdot \frac{-2}{1}\\z2s3=-2-1\cdot \frac{-2}{1}=0 \\ z2s4=-1\frac{1}{2}-1\frac{1}{2}\cdot \frac{-2}{1}=1\frac{1}{2} \\ z2s5=-3-3\cdot \frac{-2}{1}=3 \\ \end{array}\qquad \small \begin{array}{cccc|cc } a & b & c & d & & \\ \hline8 & 0 & 0 & 1 & 2 \\ 0 & -2 & 0 & 1\frac{1}{2} & 3 \\ 0 & 0 & 1 & 1\frac{1}{2} & 3 \\ 0 & 0 & 0 & 1 & 0 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}1=\text{Zeile}1\text{-Zeile}4\cdot \frac{1}{1}\\z1s4=1-1\cdot \frac{1}{1}=0 \\ z1s5=2-0\cdot \frac{1}{1}=2 \\ \end{array}\qquad \small \begin{array}{cccc|cc } a & b & c & d & & \\ \hline8 & 0 & 0 & 0 & 2 \\ 0 & -2 & 0 & 1\frac{1}{2} & 3 \\ 0 & 0 & 1 & 1\frac{1}{2} & 3 \\ 0 & 0 & 0 & 1 & 0 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}2=\text{Zeile}2\text{-Zeile}4\cdot \frac{1\frac{1}{2}}{1}\\z2s4=1\frac{1}{2}-1\cdot \frac{1\frac{1}{2}}{1}=0 \\ z2s5=3-0\cdot \frac{1\frac{1}{2}}{1}=3 \\ \end{array}\qquad \small \begin{array}{cccc|cc } a & b & c & d & & \\ \hline8 & 0 & 0 & 0 & 2 \\ 0 & -2 & 0 & 0 & 3 \\ 0 & 0 & 1 & 1\frac{1}{2} & 3 \\ 0 & 0 & 0 & 1 & 0 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}3=\text{Zeile}3\text{-Zeile}4\cdot \frac{1\frac{1}{2}}{1}\\z3s4=1\frac{1}{2}-1\cdot \frac{1\frac{1}{2}}{1}=0 \\ z3s5=3-0\cdot \frac{1\frac{1}{2}}{1}=3 \\ \end{array}\qquad \small \begin{array}{cccc|cc } a & b & c & d & & \\ \hline8 & 0 & 0 & 0 & 2 \\ 0 & -2 & 0 & 0 & 3 \\ 0 & 0 & 1 & 0 & 3 \\ 0 & 0 & 0 & 1 & 0 \\ \end{array} \\ \\ a=\frac{2}{8}=\frac{1}{4}\\b=\frac{3}{-2}=-1\frac{1}{2}\\c=\frac{3}{1}=3\\d=\frac{0}{1}=0\\L=\{\frac{1}{4}/-1\frac{1}{2}/3/0\} \\ \text{Funktion} \\ f\left(x\right)= \frac{1}{4}x^3-1\frac{1}{2}x^2+3x$