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ogólny wzór funkcji kanonicznej ---> y=a(x-p)²+q
Δ=b²-4*a*c=144-4(10)=144-40=104
p=-b/2*a
p=12/4=3
q=-Δ/4*a
q=-104/8=13
podstawiasz pod wzor i wychodzi f(x)=2(x-3)²+13
a=2 b=-12 c=5
y=a(x-p)²+q
p=-b/2a= 12/4=3
q=-Δ/4a= -104/8=-13
Δ=b²-4ac= (-12)²-4*2*5=144-10=104
y=2(x-3)²-13
ZW:y∈<-13,∞)
Δ=144-40=104
p=12/4=3 q=13
f(x)=2(x-3)-13