Zapisz trójmian kwadratowy w postaci iloczynowej i kanonicznej:
a) y=4x2-4x+1
b) y=x2 -6x
POMOCY :)
Przydatny wzór:
delta=b2-4 a c
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
a) y=4x2-4x+1
Δ = b² +4ac
Δ = 16 - 16 = 0
x₀ = -b/2a
x₀ = 4/8 = ½
postać iloczynowa: y= 4(x-½)²
p = -b/2a
p=½
q= -Δ/4a
q = 0
postać kanoniczna: y= 4(x-½)²
b) y=x2 -6x
Δ = 36 √Δ = 6
x₁ = -b-Δ/2a
x₁ = 6-6/2 =0
x₂ = -b+Δ/2a
x² = 6+6/2 = 6
postać iloczynowa : y= (x-0)(x-6)
p = -b/2a
p = 6/2 =3
q= -Δ/4a
q= -36/4 = -9
postać kanoniczna: y=(x -3) -9