Sagot :
bjr
1)
a)
f(x) = 2x(x+4)
f(x) = 2(x² + 4x)
= 2[x² + 4x + 4 - 4]
= 2[(x + 2)² - 4]
= 2(x + 2)² - 8
b)
f(x) = a(x - α)² + β avec α = -b/2a et β = - (b² - 4ac)/4a
f(x) = 2x(x+4)
= 2x² + 8x
ici : a = 2 ; b = 8 et c = 0
α = -8/4 α = -2
β = - (8² -4*2*0)/4*2 = - 8²/8 = - 8
on retrouve
f(x) = 2(x+2)² - 8
2)
a)
f(x) = 3x²- 6x+5
= 3( x² - 2x) + 5
= 3(x² - 2x + 1 - 1) + 5
= 3[(x - 1)² -1) + 5
= 3(x - 1)² - 3 + 5
= 3(x - 1)² + 2
b)
a = 3 ; b = -6 ; c = 5
α = -(-6)/3*2 = 1
β = - [(-6)² - 4*3*5)/3*4)
= -[(36 - 60)/12]
= - (-24/12)
= 2
d'où la réponse
Réponse :
Mettre sous forme canonique
1ère façon
f(x) = 2 x(x + 4) = 2 x² + 8 x = 2(x² + 4 x) = 2(x² + 4 x + 4 - 4)
= 2(x² + 4 x + 4 - 4) = 2((x + 2)² - 4) = 2(x + 2)² - 8
f(x) = 2(x + 2)² - 8
f(x) = 3 x² - 6 x + 5 = 3(x² - 2 x + 5/3) =3(x² - 2 x + 5/3 + 1 - 1)
= 3(x² - 2 x + 1 + 2/3) = 3((x - 1)² + 2/3) = 3(x - 1)² + 2
f(x) = 3(x - 1)² + 2
2ème façon
f(x) = 2 x² + 8 x
on veut mettre f(x) = a(x - α)²+ β
a = 2
α = - b/2a = - 8/4 = - 2
β = f(α) = f(- 2) = 2*(- 2)² + 8*(-2) = 8 - 16 = - 8
donc f(x) = 2(x + 2)² - 8
f(x) = 3 x² - 6 x + 5
a = 3
α = - b/2a = 6/6 = 1
β = f(1) = 3 - 6 +5 = 2
donc f(x) = a(x -α)²+ β est f(x) = 3(x - 1)² + 2
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