Beispiel Nr: 05
$\text{Aufstellen von Funktionsgleichungen}\\$
$\textbf{Aufgabe:}$ eine ganzrationale Funktion 3.Grades, deren Graph durch Punkte $P(-2/2)$,$Q(0/0)$,$R(1/2)$,$T(2/-1)$.\\
$\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(0\right)=0 \qquad a\cdot 0^3+b\cdot 0^2+c\cdot 0+d=0 \\ f\left(1\right)=2 \qquad a\cdot 1^3+b\cdot 1^2+c\cdot 1+d=2 \\ f\left(2\right)=(-1) \qquad a\cdot 2^3+b\cdot 2^2+c\cdot 2+d=(-1) \\\small \begin{array}{l} -8a+4b-2c+d=2 \\ d=0 \\ a+b+c+d=2 \\ 8a+4b+2c+d=-1 \\ \\ \end{array} \qquad \small \begin{array}{cccc|cc } a & b & c & d & & \\ \hline-8 & 4 & -2 & 1 & 2 \\ 0 & 0 & 0 & 1 & 0 \\ 1 & 1 & 1 & 1 & 2 \\ 8 & 4 & 2 & 1 & -1 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}3=\text{Zeile}3\text{-Zeile}1\cdot \frac{1}{-8}\\z3s1=1-(-8)\cdot \frac{1}{-8}=0 \\ z3s2=1-4\cdot \frac{1}{-8}=1\frac{1}{2} \\ z3s3=1-(-2)\cdot \frac{1}{-8}=\frac{3}{4} \\ z3s4=1-1\cdot \frac{1}{-8}=1\frac{1}{8} \\ z3s5=2-2\cdot \frac{1}{-8}=2\frac{1}{4} \\ \end{array}\qquad \small \begin{array}{cccc|cc } a & b & c & d & & \\ \hline-8 & 4 & -2 & 1 & 2 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 1\frac{1}{2} & \frac{3}{4} & 1\frac{1}{8} & 2\frac{1}{4} \\ 8 & 4 & 2 & 1 & -1 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}4=\text{Zeile}4\text{-Zeile}1\cdot \frac{8}{-8}\\z4s1=8-(-8)\cdot \frac{8}{-8}=0 \\ z4s2=4-4\cdot \frac{8}{-8}=8 \\ z4s3=2-(-2)\cdot \frac{8}{-8}=0 \\ z4s4=1-1\cdot \frac{8}{-8}=2 \\ z4s5=-1-2\cdot \frac{8}{-8}=1 \\ \end{array}\qquad \small \begin{array}{cccc|cc } a & b & c & d & & \\ \hline-8 & 4 & -2 & 1 & 2 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 1\frac{1}{2} & \frac{3}{4} & 1\frac{1}{8} & 2\frac{1}{4} \\ 0 & 8 & 0 & 2 & 1 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeilen vertauschen } \\ \end{array}\qquad \small \begin{array}{cccc|cc } a & b & c & d & & \\ \hline-8 & 4 & -2 & 1 & 2 \\ 0 & 1\frac{1}{2} & \frac{3}{4} & 1\frac{1}{8} & 2\frac{1}{4} \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 8 & 0 & 2 & 1 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}1=\text{Zeile}1\text{-Zeile}2\cdot \frac{4}{1\frac{1}{2}}\\z1s2=4-1\frac{1}{2}\cdot \frac{4}{1\frac{1}{2}}=0 \\ z1s3=-2-\frac{3}{4}\cdot \frac{4}{1\frac{1}{2}}=-4 \\ z1s4=1-1\frac{1}{8}\cdot \frac{4}{1\frac{1}{2}}=-2 \\ z1s5=2-2\frac{1}{4}\cdot \frac{4}{1\frac{1}{2}}=-4 \\ \end{array}\qquad \small \begin{array}{cccc|cc } a & b & c & d & & \\ \hline-8 & 0 & -4 & -2 & -4 \\ 0 & 1\frac{1}{2} & \frac{3}{4} & 1\frac{1}{8} & 2\frac{1}{4} \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 8 & 0 & 2 & 1 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}4=\text{Zeile}4\text{-Zeile}2\cdot \frac{8}{1\frac{1}{2}}\\z4s2=8-1\frac{1}{2}\cdot \frac{8}{1\frac{1}{2}}=0 \\ z4s3=0-\frac{3}{4}\cdot \frac{8}{1\frac{1}{2}}=-4 \\ z4s4=2-1\frac{1}{8}\cdot \frac{8}{1\frac{1}{2}}=-4 \\ z4s5=1-2\frac{1}{4}\cdot \frac{8}{1\frac{1}{2}}=-11 \\ \end{array}\qquad \small \begin{array}{cccc|cc } a & b & c & d & & \\ \hline-8 & 0 & -4 & -2 & -4 \\ 0 & 1\frac{1}{2} & \frac{3}{4} & 1\frac{1}{8} & 2\frac{1}{4} \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & -4 & -4 & -11 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeilen vertauschen } \\ \end{array}\qquad \small \begin{array}{cccc|cc } a & b & c & d & & \\ \hline-8 & 0 & -4 & -2 & -4 \\ 0 & 1\frac{1}{2} & \frac{3}{4} & 1\frac{1}{8} & 2\frac{1}{4} \\ 0 & 0 & -4 & -4 & -11 \\ 0 & 0 & 0 & 1 & 0 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}1=\text{Zeile}1\text{-Zeile}3\cdot \frac{-4}{-4}\\z1s3=-4-(-4)\cdot \frac{-4}{-4}=0 \\ z1s4=-2-(-4)\cdot \frac{-4}{-4}=2 \\ z1s5=-4-(-11)\cdot \frac{-4}{-4}=7 \\ \end{array}\qquad \small \begin{array}{cccc|cc } a & b & c & d & & \\ \hline-8 & 0 & 0 & 2 & 7 \\ 0 & 1\frac{1}{2} & \frac{3}{4} & 1\frac{1}{8} & 2\frac{1}{4} \\ 0 & 0 & -4 & -4 & -11 \\ 0 & 0 & 0 & 1 & 0 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}2=\text{Zeile}2\text{-Zeile}3\cdot \frac{\frac{3}{4}}{-4}\\z2s3=\frac{3}{4}-(-4)\cdot \frac{\frac{3}{4}}{-4}=0 \\ z2s4=1\frac{1}{8}-(-4)\cdot \frac{\frac{3}{4}}{-4}=\frac{3}{8} \\ z2s5=2\frac{1}{4}-(-11)\cdot \frac{\frac{3}{4}}{-4}=\frac{3}{16} \\ \end{array}\qquad \small \begin{array}{cccc|cc } a & b & c & d & & \\ \hline-8 & 0 & 0 & 2 & 7 \\ 0 & 1\frac{1}{2} & 0 & \frac{3}{8} & \frac{3}{16} \\ 0 & 0 & -4 & -4 & -11 \\ 0 & 0 & 0 & 1 & 0 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}1=\text{Zeile}1\text{-Zeile}4\cdot \frac{2}{1}\\z1s4=2-1\cdot \frac{2}{1}=0 \\ z1s5=7-0\cdot \frac{2}{1}=7 \\ \end{array}\qquad \small \begin{array}{cccc|cc } a & b & c & d & & \\ \hline-8 & 0 & 0 & 0 & 7 \\ 0 & 1\frac{1}{2} & 0 & \frac{3}{8} & \frac{3}{16} \\ 0 & 0 & -4 & -4 & -11 \\ 0 & 0 & 0 & 1 & 0 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}2=\text{Zeile}2\text{-Zeile}4\cdot \frac{\frac{3}{8}}{1}\\z2s4=\frac{3}{8}-1\cdot \frac{\frac{3}{8}}{1}=0 \\ z2s5=\frac{3}{16}-0\cdot \frac{\frac{3}{8}}{1}=\frac{3}{16} \\ \end{array}\qquad \small \begin{array}{cccc|cc } a & b & c & d & & \\ \hline-8 & 0 & 0 & 0 & 7 \\ 0 & 1\frac{1}{2} & 0 & 0 & \frac{3}{16} \\ 0 & 0 & -4 & -4 & -11 \\ 0 & 0 & 0 & 1 & 0 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}3=\text{Zeile}3\text{-Zeile}4\cdot \frac{-4}{1}\\z3s4=-4-1\cdot \frac{-4}{1}=0 \\ z3s5=-11-0\cdot \frac{-4}{1}=-11 \\ \end{array}\qquad \small \begin{array}{cccc|cc } a & b & c & d & & \\ \hline-8 & 0 & 0 & 0 & 7 \\ 0 & 1\frac{1}{2} & 0 & 0 & \frac{3}{16} \\ 0 & 0 & -4 & 0 & -11 \\ 0 & 0 & 0 & 1 & 0 \\ \end{array} \\ \\ a=\frac{7}{-8}=-\frac{7}{8}\\b=\frac{\frac{3}{16}}{1\frac{1}{2}}=\frac{1}{8}\\c=\frac{-11}{-4}=2\frac{3}{4}\\d=\frac{0}{1}=0\\L=\{-\frac{7}{8}/\frac{1}{8}/2\frac{3}{4}/0\} \\ \text{Funktion} \\ f\left(x\right)=-\frac{7}{8}x^3+\frac{1}{8}x^2+2\frac{3}{4}x$