a) parralel zur x-Achse und in Q (0/2) ihren Wendepunkt b) parallel zur 1. Winkelhalbierenden (was ist dass?) und in Q (0/2) eine Tangente parralel zur x-Achse. a) in O(0/0) die x-Achse und in A(2/2) die 1. Winkelhalbierende als Tangente b) in O(0/0) die 1. Winkelhalbierende und in B(2/0) die x-Achse als Tangente. Wir wissen also b=0; d=2 und a+b+c+d=4 => a+0+c+2=4 => a+c=2 => c=2-a 3a+2b+c=0 => 3a+c=0 c=2-a eingesetzt in 3a+c=0, ergibt 3a+2-a=0 2a+2=0 a=-1 => c=2-(-1)=2+1=3 Also lautet die Funktionsgleichung f(x)=-x³+3x+2. Fassen wir zusammen: c=0, d=2, 3a+2b+c=1 => 3a+2b+0=1 => 3a+2b=1 => 2b=1-3a => b=(1/2)(1-3a) a+b+c+d=4 => a+(1/2)(1-3a)+0+2=4 => a+(1/2)-(3/2)a+2=4 => -(1/2)a=3/2 => a=-3 b=(1/2)(1-3a)=(1/2)(1-3*(-3))=(1/2)(1+9)=5.
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