f (X) = ax^4 + bx^2+ c
f'(x)=4ax^3+ 2bx
f"(x) = 12ax^2+2b
f(2) = 0 = 16a + 4b + c (1)
f'(2)= 2 = 32a + 4b (2)
f" (-1) = 0 = 12a + 2b (3)
(3) → b = - 6a (4)
(2) → 2 = 32a - 24a = 8a → a=0,25
(4) --> b = -6a = -6*0,25 = -1,5
(1) → 0 = 16a + 4b + c = 16*0,25 +4*(-1,5) + c = 4 - 6 + c --> c = 2
c = 2 ist der y-Achsenabschnitt.