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du benötigst hierfür also 4. Bedingungen: bei der 1a zum beispiel: Zitat: in 0(0/0) die x-achse und in A(2/2) die 1. Winkelhalbierende als Tangente. Erstmal Danke schon mal für die vielen Antworten.Bin ganz überrascht , dass das so schnell gehen kann. Die normalen Bedingungen kenne ich ja , ich kann sie nur nicht auf solchen Aufgaben beziehen. 1a)wurde ja schonvorgefertigt f(x) = ax³ + bx² + cx + d 1.Bdg.: Punkt P(1|4)--> f(1)=4 2.Bdg.:Punkt Q(0|2)--->f(0)=2 3.Bdg.: Tangente parallel zur x-AChse---> f'(x)=0 4.Bdg.: Wendepunkt in Q(0|2)-->f''(0)=2. 8a + 4b 12a + 4b = 1. Und dann probierst du das Ganze nochmal aufzustellen! Nach gut. Bet365 dk.

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