1/3 x^3 - 2*t*x^2
1. schneidet y-Achse bei f(0) = 0
schneidet x-Achse, wenn 1/3 x^3 - 2*t*x^2 = 0
x^2 * ( 1/3 x - 2t ) = 0
x=0 oder x= 6t
2. f ' (1) dazu f ' (x) = x^2 - 4 * t * x
also f ' (1) = 1 - 4t
3. f ' (x) = 1 x^2 - 4 * t * x = 1
x = 2t ±√(4t^2 +1 )
4. Wendepu wenn f ' ' (x) = 0 also 2x-4t = 0 also x = 2t
W ( 2t / -16/3 t^3 ) Ortskurve
x=2t y = -16/3 t^3
t=x/2 also y = -2/3 x^3 Gl. der Ortskurve