Translation of "Normale im Wendepunkt" Please, can you help me to translate into English and solve this exercise: 

Der Term von $f$ wird variiert: $f_{a}(x)=ax(x-3)^{2}$. Bestimme $a$ so dass die Normale im Wendepunkt durch den Punkt $(0|0)$ geht. 

 A: Translation:
For the function $f_a(x)=ax(x-3)^2$ determine $a$ such that the normal at the point of inflection passes through the point $(0,0)$.
Solution summary:
The point of inflection can be found by finding $x_a$ such that $f_a''(x_a)=0$; the point of inflection is then clearly $(x_a,f_a(x_a))$ . As the tangent has slope $ f_a'(x_a)$, the normal has slope $-\frac1{f_a'(x_a)}$. Combine these results to find $a$.
Detailed solution:
By the product rule, the first derivative is: $$f_a'(x)=a(x-3)^2+2ax(x-3)=3a(x-1)(x-3)$$ the second derivative is: $$f_a''(x)=3a(x-3)+3a(x-1)=6a(x-2)$$ and 
$f_a'''(x)=6\,a$, for later reference.
The case $a=0$ is special: If $a=0$, then $f_a(x)=0$, the function graph has no point of inflection there and so has no solution. (At any rate, there is no "the" point of inflection).
Hence we assume $a\ne 0$: Then the only possibility for a point of inflection is at $x_a=2$ because $f_a''(x_a)=0\iff x_a=2$.
Since $f_a'''(x_a)=6a\ne0$, the point $(x_a,f_a(x_a))=(2,2a)$ is indeed a point of inflection. The slope of the normal through that point is $$-\frac1{fa'(x_a)}=-\frac1{-3a}=\frac1{3a}.$$ On the other hand, the slope of the line through $(0,0)$ and $(2,2a)$ is $a$.
These slopes (and hence the normal and the line through the origin) coincide iff $a=\frac1{3a}$, i.e. $$a=\pm\frac{\sqrt 3}{3}.$$
