Witam was mam 3 zadania do zrobienia z matmy na jutro:
1.oblicz postac kanoniczna i iloczynowa funkcji:
f(x)=-2x∧2+3x-1
2.Sprowadz do postaci ogolnej wzor:
f(x)=(x+2)∧2-3
3.Rozwiąż rownanie:
a)-x∧2+9=0
b)x∧2-7x+12=0
c)(2x+3)(2-x)=0
Wielkie dzieki za rozwiazanie
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1)
postac kanoniczna:
y=a(x-p)²+q
Δ= 9-8 = 1
√Δ=1
p=-b/2a =-3/-4 = 3/4
q=-1/-8 = 1/8
więc:
y=-2(x-3/4)²+1/8
postac iloczynowa:
y=a(x-x1)(x-x2)
x1=-3-1/-4 = -4/-4 = 1
x2 = -3+1/-4 = -2/-4 = 1/2
więc:
y=-2(x-1)(x-1/2)
2)
f(x) = x²+4x+4 - 3 = x²+4x+1
3)
a)
-x²+9 = 0
x² = 9
x=±3
b)
x²-7x+12=0
Δ=49-48 = 1
√Δ=1
x1=7-1 / 2 = 6/2 = 3
x2 = 7+1 / 2 = 8/2 = 4
c)
(2x+3)(2-x)=0
2x+3 =0 U 2-x=0
x=-3/2 x=2