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a=1, p=1, q=3
Postac kanoniczna:
f(x)=(x-1)²+3
x²-2x+1+3=x²-2x+4
Odp. a=1, b=-2, c=4.
W=(1,3) tzn. p=1 q=3
postać kanoniczna y=a(x-p)^2+q
y=1(x-1)^2+3=(x-1)^2+3=x^2-2x+1+3=x^2-2x+4
zatem a=1, b=-2, c=4