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Postac kanoniczna:
f(x)=a(x-p)²+q
f(x)=a(x+3)²+2
f(x)=a(x²+6x+9)+2=ax²+6ax+9a+2
9a+2=-4
9a=-4-2
9a=-6 /:9
a=-2/3
b=6*(-2/3)=-4
f(x)=-2/3x²-4x-4
-2/3(x²+6x+6)=0
Δ=6²-4*6=36-24=12
√Δ=2√3
x₁=1/2*(-6-2√3)=-3-√3, x₂=-3+√3
Postac iloczynowa:
f(x)=-2/3*(x+3+√3)(x+3-√3)