" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2025 KUDO.TIPS - All rights reserved.
a=-1
b=2
c=3
p=-b/2a
p=-2/-2=1
Δ=b²-4ac
Δ=4-4*(-1)*3=4+12=16
q=-Δ/4a
q=-16/-4=4
f(x)=-(x-1)²+4 postać kanoniczna
miejsc zerowe:
Δ=16 √Δ=4
x₁=-b-√Δ/4a
x₂=-b+√Δ/4a
x₁=-2-4/-4=-6/-4=1 2/4=1½
x₂=-2+4/-4=2/-4=-½
postać iloczynowa:
f(x)=-(x-1½)(x+½)
a)
a=-1
b=2
c=3
p=-b/2a=-2/-2=1
q=-Δ/4a
Δ=b²-4ac=2²+4×3=16
√Δ=4
q=-16/-4=4
y=a(x-p)²+q
y=-(x-1)²+4
b)x₁=(-b-√Δ):2a=(-2-4):2×(-1)=-6:-2=3
x₂=(-2+4):-2=-1
c)
f(x)=a(x-x₁)(x-x₂)
f(x)=-(x-3)(x+1)