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Maria999
@Maria999
April 2019
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Bonsoir! Pouvez-vous m'aider à résoudre ce problème s'il vous plaît? J'ai vraiment besoin d'aide ... :/
Trouver les formes canonique et factorisée
f(x) = x²+3x+2
f(x) = x²-x-20
f(x) = x²+6x+8
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esefiha
D'une manière générale : pour f(x) = ax²+bx+c
la forme canonique est f(x) = a[(x+b/2a)²-(b/2a)² +c]
donc
f(x) = x²+3x+2
a= 1 , b=3 et c= 2
b/2a = 3/2
forme canonique :
f(x) = (x+3/2)²-9/4 +2
f(x) = (x+3/2)²-9/4 +8/4
f(x) = (x+3/2)²-1/4
factorisation
f(x) = (x+3/2)²-1/4
f(x) = (x+3/2)²-(1/2)²
f(x) = (x+3/2-1/2)(x+3/2+1/2)
f(x) = (x+2/2)(x+4/2)
f(x) = (x+1)(x+2)
f(x) = x²-x-20
a=1, b=-1 et c=-20
b/2a = -1/2
forme canonique :
f(x) = (x-1/2)² - 1/4 -20
f(x) = (x-1/2)² - 1/4 - 80/4
f(x) = (x-1/2)² - 81/4
factorisation
f(x) = (x-1/2)² - 81/4
f(x) = (x-1/2)² - (9/2)²
f(x) = (x-1/2-9/2)(x-1/2+9/2)
f(x) = (x-1/2-9/2)(x-1/2+9/2)
f(x) = (x-10/2)(x+8/2)
f(x) = (x-5)(x+4)
f(x) = x²+6x+8
a= 1 ,b = 6 et c = 8
b/2a = 6/2 = 3
forme canonique :
f(x) = (x+3)² - 9 +8
f(x) = (x+3)² - 1
factorisation
f(x) = (x+3)² - 1²
f(x) = (x+3-1)(x+3+1)
f(x) = (x+2)(x+4)
2 votes
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la forme canonique est f(x) = a[(x+b/2a)²-(b/2a)² +c]
donc
f(x) = x²+3x+2
a= 1 , b=3 et c= 2
b/2a = 3/2
forme canonique :
f(x) = (x+3/2)²-9/4 +2
f(x) = (x+3/2)²-9/4 +8/4
f(x) = (x+3/2)²-1/4
factorisation
f(x) = (x+3/2)²-1/4
f(x) = (x+3/2)²-(1/2)²
f(x) = (x+3/2-1/2)(x+3/2+1/2)
f(x) = (x+2/2)(x+4/2)
f(x) = (x+1)(x+2)
f(x) = x²-x-20
a=1, b=-1 et c=-20
b/2a = -1/2
forme canonique :
f(x) = (x-1/2)² - 1/4 -20
f(x) = (x-1/2)² - 1/4 - 80/4
f(x) = (x-1/2)² - 81/4
factorisation
f(x) = (x-1/2)² - 81/4
f(x) = (x-1/2)² - (9/2)²
f(x) = (x-1/2-9/2)(x-1/2+9/2)
f(x) = (x-1/2-9/2)(x-1/2+9/2)
f(x) = (x-10/2)(x+8/2)
f(x) = (x-5)(x+4)
f(x) = x²+6x+8
a= 1 ,b = 6 et c = 8
b/2a = 6/2 = 3
forme canonique :
f(x) = (x+3)² - 9 +8
f(x) = (x+3)² - 1
factorisation
f(x) = (x+3)² - 1²
f(x) = (x+3-1)(x+3+1)
f(x) = (x+2)(x+4)