d) Für welches a hat fa einen Wendepunkt mit der Ordinate y=-2 ?
\( f_{a}(x)=x^{3}-a x^{2} \)
\( f_{a}^{\prime}(x)=3 x^{2}-2 a x \)
\( f''_{a}(x)=6 x-2 a \)
\( 6 x-2 a=0 \)
\( 6 x=2 a \)
\( x=\frac{1}{3} a \)
\( \begin{aligned} f\left(\frac{1}{3} a\right) &=\left(\frac{1}{3} a\right)^{3}-a\left(\frac{1}{3} a\right)^{2} \\ &=\frac{1}{27} a^{3}-\frac{1}{9} a^{3} \\ &=-\frac{2}{27} a^{3} \end{aligned} \)
\( \begin{aligned}-\frac{2}{27} a^{3} &=-2 \\ a^{3} &=27 \\ a &=3 \end{aligned} \)
e) Welcher Graph der Schar hat die Wendetangente y(x)=-12x+8?
\( \quad t(x)=-12 x+8 \)
\(WP\left(\frac{1}{3} a \mid-\frac{2}{27} a^{3}\right) \)
\( \begin{aligned} f^{\prime}\left(\frac{1}{3 }a\right)=& 3 \cdot\left(\frac{1}{3} a\right)^{2}-2 a \cdot \frac{1}{3} a \\ =& \frac{1}{3} a^{2}-\frac{2}{3} a^{2} \\ =&-\frac{1}{3} a^{2} \\ -\frac{1}{3 }a^{2}=-12 \\ a^{2}=36 \\ a=\pm 6 \\[10pt] a=6\Rightarrow & f(x)=x^{3}-6 x^{2} \\[10pt] WP(2|-16) \\ -16=&-12 \cdot 2+n \\ -16=&-24+n \\ 8=& n \end{aligned} \)