Analysis-Aufstellen von Funktionsgleichungen-Ganzrationale Funktion


  • $\text{Funktionsgraph}$
    $\text{Wertetable}$
    $\text{Terme aufstellen}$
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Beispiel Nr: 07
$ \\ \text{ Aufstellen von Funktionsgleichungen}\\ \\ \textbf{Aufgabe:}\\eine ganzrationale Funktion 3.Grades, mit der Wendetangente $y=2x+1=0$ an der Stelle $x_2=1$ und dem Schnittpunkt mit der x-Achse $x=-2$. \\ \\ \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)=0 \qquad a\cdot (-2)^3+b\cdot (-2)^2+c\cdot (-2)+d=0 \\ f\left(1\right)=3 \qquad a\cdot 1^3+b\cdot 1^2+c\cdot 1+d=3 \\ f'\left(1\right)=0 \qquad 3a\cdot 1^2+2b\cdot 1+c=0 \\ f''\left(1\right)=0 \qquad 6a\cdot 1+2b=0 \\\small \begin{array}{l} -8a+4b-2c+d=0 \\ a+b+c+d=3 \\ 3a+2b+c=0 \\ 6a+2b=0 \\ \\ \end{array} \qquad \small \begin{array}{cccc|cc } a & b & c & d & & \\ \hline-8 & 4 & -2 & 1 & 0 \\ 1 & 1 & 1 & 1 & 3 \\ 3 & 2 & 1 & 0 & 0 \\ 6 & 2 & 0 & 0 & 0 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}2=\text{Zeile}2\text{-Zeile}1\cdot \frac{1}{-8}\\z2s1=1-(-8)\cdot \frac{1}{-8}=0 \\ z2s2=1-4\cdot \frac{1}{-8}=1\frac{1}{2} \\ z2s3=1-(-2)\cdot \frac{1}{-8}=\frac{3}{4} \\ z2s4=1-1\cdot \frac{1}{-8}=1\frac{1}{8} \\ z2s5=3-0\cdot \frac{1}{-8}=3 \\ \end{array}\qquad \small \begin{array}{cccc|cc } a & b & c & d & & \\ \hline-8 & 4 & -2 & 1 & 0 \\ 0 & 1\frac{1}{2} & \frac{3}{4} & 1\frac{1}{8} & 3 \\ 3 & 2 & 1 & 0 & 0 \\ 6 & 2 & 0 & 0 & 0 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}3=\text{Zeile}3\text{-Zeile}1\cdot \frac{3}{-8}\\z3s1=3-(-8)\cdot \frac{3}{-8}=0 \\ z3s2=2-4\cdot \frac{3}{-8}=3\frac{1}{2} \\ z3s3=1-(-2)\cdot \frac{3}{-8}=\frac{1}{4} \\ z3s4=0-1\cdot \frac{3}{-8}=\frac{3}{8} \\ z3s5=0-0\cdot \frac{3}{-8}=0 \\ \end{array}\qquad \small \begin{array}{cccc|cc } a & b & c & d & & \\ \hline-8 & 4 & -2 & 1 & 0 \\ 0 & 1\frac{1}{2} & \frac{3}{4} & 1\frac{1}{8} & 3 \\ 0 & 3\frac{1}{2} & \frac{1}{4} & \frac{3}{8} & 0 \\ 6 & 2 & 0 & 0 & 0 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}4=\text{Zeile}4\text{-Zeile}1\cdot \frac{6}{-8}\\z4s1=6-(-8)\cdot \frac{6}{-8}=0 \\ z4s2=2-4\cdot \frac{6}{-8}=5 \\ z4s3=0-(-2)\cdot \frac{6}{-8}=-1\frac{1}{2} \\ z4s4=0-1\cdot \frac{6}{-8}=\frac{3}{4} \\ z4s5=0-0\cdot \frac{6}{-8}=0 \\ \end{array}\qquad \small \begin{array}{cccc|cc } a & b & c & d & & \\ \hline-8 & 4 & -2 & 1 & 0 \\ 0 & 1\frac{1}{2} & \frac{3}{4} & 1\frac{1}{8} & 3 \\ 0 & 3\frac{1}{2} & \frac{1}{4} & \frac{3}{8} & 0 \\ 0 & 5 & -1\frac{1}{2} & \frac{3}{4} & 0 \\ \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=0-3\cdot \frac{4}{1\frac{1}{2}}=-8 \\ \end{array}\qquad \small \begin{array}{cccc|cc } a & b & c & d & & \\ \hline-8 & 0 & -4 & -2 & -8 \\ 0 & 1\frac{1}{2} & \frac{3}{4} & 1\frac{1}{8} & 3 \\ 0 & 3\frac{1}{2} & \frac{1}{4} & \frac{3}{8} & 0 \\ 0 & 5 & -1\frac{1}{2} & \frac{3}{4} & 0 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}3=\text{Zeile}3\text{-Zeile}2\cdot \frac{3\frac{1}{2}}{1\frac{1}{2}}\\z3s2=3\frac{1}{2}-1\frac{1}{2}\cdot \frac{3\frac{1}{2}}{1\frac{1}{2}}=0 \\ z3s3=\frac{1}{4}-\frac{3}{4}\cdot \frac{3\frac{1}{2}}{1\frac{1}{2}}=-1\frac{1}{2} \\ z3s4=\frac{3}{8}-1\frac{1}{8}\cdot \frac{3\frac{1}{2}}{1\frac{1}{2}}=-2\frac{1}{4} \\ z3s5=0-3\cdot \frac{3\frac{1}{2}}{1\frac{1}{2}}=-7 \\ \end{array}\qquad \small \begin{array}{cccc|cc } a & b & c & d & & \\ \hline-8 & 0 & -4 & -2 & -8 \\ 0 & 1\frac{1}{2} & \frac{3}{4} & 1\frac{1}{8} & 3 \\ 0 & 0 & -1\frac{1}{2} & -2\frac{1}{4} & -7 \\ 0 & 5 & -1\frac{1}{2} & \frac{3}{4} & 0 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}4=\text{Zeile}4\text{-Zeile}2\cdot \frac{5}{1\frac{1}{2}}\\z4s2=5-1\frac{1}{2}\cdot \frac{5}{1\frac{1}{2}}=0 \\ z4s3=-1\frac{1}{2}-\frac{3}{4}\cdot \frac{5}{1\frac{1}{2}}=-4 \\ z4s4=\frac{3}{4}-1\frac{1}{8}\cdot \frac{5}{1\frac{1}{2}}=-3 \\ z4s5=0-3\cdot \frac{5}{1\frac{1}{2}}=-10 \\ \end{array}\qquad \small \begin{array}{cccc|cc } a & b & c & d & & \\ \hline-8 & 0 & -4 & -2 & -8 \\ 0 & 1\frac{1}{2} & \frac{3}{4} & 1\frac{1}{8} & 3 \\ 0 & 0 & -1\frac{1}{2} & -2\frac{1}{4} & -7 \\ 0 & 0 & -4 & -3 & -10 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}1=\text{Zeile}1\text{-Zeile}3\cdot \frac{-4}{-1\frac{1}{2}}\\z1s3=-4-(-1\frac{1}{2})\cdot \frac{-4}{-1\frac{1}{2}}=0 \\ z1s4=-2-(-2\frac{1}{4})\cdot \frac{-4}{-1\frac{1}{2}}=4 \\ z1s5=-8-(-7)\cdot \frac{-4}{-1\frac{1}{2}}=10\frac{2}{3} \\ \end{array}\qquad \small \begin{array}{cccc|cc } a & b & c & d & & \\ \hline-8 & 0 & 0 & 4 & 10\frac{2}{3} \\ 0 & 1\frac{1}{2} & \frac{3}{4} & 1\frac{1}{8} & 3 \\ 0 & 0 & -1\frac{1}{2} & -2\frac{1}{4} & -7 \\ 0 & 0 & -4 & -3 & -10 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}2=\text{Zeile}2\text{-Zeile}3\cdot \frac{\frac{3}{4}}{-1\frac{1}{2}}\\z2s3=\frac{3}{4}-(-1\frac{1}{2})\cdot \frac{\frac{3}{4}}{-1\frac{1}{2}}=0 \\ z2s4=1\frac{1}{8}-(-2\frac{1}{4})\cdot \frac{\frac{3}{4}}{-1\frac{1}{2}}=0 \\ z2s5=3-(-7)\cdot \frac{\frac{3}{4}}{-1\frac{1}{2}}=-\frac{1}{2} \\ \end{array}\qquad \small \begin{array}{cccc|cc } a & b & c & d & & \\ \hline-8 & 0 & 0 & 4 & 10\frac{2}{3} \\ 0 & 1\frac{1}{2} & 0 & 0 & -\frac{1}{2} \\ 0 & 0 & -1\frac{1}{2} & -2\frac{1}{4} & -7 \\ 0 & 0 & -4 & -3 & -10 \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}4=\text{Zeile}4\text{-Zeile}3\cdot \frac{-4}{-1\frac{1}{2}}\\z4s3=-4-(-1\frac{1}{2})\cdot \frac{-4}{-1\frac{1}{2}}=0 \\ z4s4=-3-(-2\frac{1}{4})\cdot \frac{-4}{-1\frac{1}{2}}=3 \\ z4s5=-10-(-7)\cdot \frac{-4}{-1\frac{1}{2}}=8\frac{2}{3} \\ \end{array}\qquad \small \begin{array}{cccc|cc } a & b & c & d & & \\ \hline-8 & 0 & 0 & 4 & 10\frac{2}{3} \\ 0 & 1\frac{1}{2} & 0 & 0 & -\frac{1}{2} \\ 0 & 0 & -1\frac{1}{2} & -2\frac{1}{4} & -7 \\ 0 & 0 & 0 & 3 & 8\frac{2}{3} \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}1=\text{Zeile}1\text{-Zeile}4\cdot \frac{4}{3}\\z1s4=4-3\cdot \frac{4}{3}=0 \\ z1s5=10\frac{2}{3}-8\frac{2}{3}\cdot \frac{4}{3}=-\frac{8}{9} \\ \end{array}\qquad \small \begin{array}{cccc|cc } a & b & c & d & & \\ \hline-8 & 0 & 0 & 0 & -\frac{8}{9} \\ 0 & 1\frac{1}{2} & 0 & 0 & -\frac{1}{2} \\ 0 & 0 & -1\frac{1}{2} & -2\frac{1}{4} & -7 \\ 0 & 0 & 0 & 3 & 8\frac{2}{3} \\ \end{array} \\ \\ \small \begin{array}{l}\text{Zeile}3=\text{Zeile}3\text{-Zeile}4\cdot \frac{-2\frac{1}{4}}{3}\\z3s4=-2\frac{1}{4}-3\cdot \frac{-2\frac{1}{4}}{3}=0 \\ z3s5=-7-8\frac{2}{3}\cdot \frac{-2\frac{1}{4}}{3}=-\frac{1}{2} \\ \end{array}\qquad \small \begin{array}{cccc|cc } a & b & c & d & & \\ \hline-8 & 0 & 0 & 0 & -\frac{8}{9} \\ 0 & 1\frac{1}{2} & 0 & 0 & -\frac{1}{2} \\ 0 & 0 & -1\frac{1}{2} & 0 & -\frac{1}{2} \\ 0 & 0 & 0 & 3 & 8\frac{2}{3} \\ \end{array} \\ \\ a=\frac{-\frac{8}{9}}{-8}=\frac{1}{9}\\b=\frac{-\frac{1}{2}}{1\frac{1}{2}}=-\frac{1}{3}\\c=\frac{-\frac{1}{2}}{-1\frac{1}{2}}=\frac{1}{3}\\d=\frac{8\frac{2}{3}}{3}=2\frac{8}{9}\\L=\{\frac{1}{9}/-\frac{1}{3}/\frac{1}{3}/2\frac{8}{9}\} \\ \text{Funktion} \\ f\left(x\right)= \frac{1}{9}x^3-\frac{1}{3}x^2+\frac{1}{3}x+2\frac{8}{9} $