a=1 b=-3 c=-10
Δ=b²-4ac = 9+40=49
postac kanoniczna:
y=a(x-p)²+q
p=-b/2a = -(-3)/(2*1)=3/2
q=-Δ/4a=-49/4
odp:
y=(x-3/2)² -49/4
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a=1 b=-3 c=-10
Δ=b²-4ac = 9+40=49
postac kanoniczna:
y=a(x-p)²+q
p=-b/2a = -(-3)/(2*1)=3/2
q=-Δ/4a=-49/4
odp:
y=(x-3/2)² -49/4