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f(x)=-2(x-2)(x+3)=-2(x^2+3x-2x-6)=
= -2x^2-2x+12
kanoniczna:
x1=2 x2=-3
p=(2+(-3))/2=-1/2
q=f(p)=f(-1/2)=-2(-1/2-2)(-1/2+3)=
=-2*(-5/2)*5/2=12 1/2
więc:
f(x)=-2(x+1/2)^2 + 12 1/2
ogolna: y=ax^2+bx+c
kanonicza: y=a(x-p)^2+q