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Wattzef
@Wattzef
May 2019
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J'ai un dm de math pour demain et je men sort vraiment pas , quelqu'un pourais t'il m'aider svp ?
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loulakar
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Forme développée :
f(x) = (-x + 1)(x - 1)
f(x) = -x^2 + x + x - 1
f(x) = -x^2 + 2x - 1
f(x) = ax^2 + bx - 1
f(x) = 0
(-x + 1)(x - 1) = 0
-x + 1 = 0 ou x - 1 = 0
Même solution pour les 2
x = 1
Coordonnées du sommet :
xS = (-b) / (2a)
xS = (-2) / (2 * (-1))
xS = (-2) / (-2)
xS = 1
yS = f(xS)
yS = f(1)
yS = (-1 + 1)(1 - 1)
yS = 0
S = (1 ; 0)
Forme développée :
g(x) = (2x + 2)(x - 3)
g(x) = 2x^2 - 6x + 2x - 6
g(x) = 2x^2 - 4x - 6
g(x) = 2(x^2 - 2x - 3)
g(x) = 0
(2x + 2)(x - 3) = 0
2x + 2 = 0 ou x - 3 = 0
2x = -2 ou x = 3
x = -2 / 2 ou x = 3
x = (-1) ou x = 3
Coordonnées du sommet :
xS = (-b) / (2a)
xS = (2) / (2 * 1)
xS = 2 / 2
xS = 1
yS = g(xS)
yS = g(1)
yS = (2 * 1 + 2)(1 - 3)
yS = (2 + 2)(-2)
yS = 4 * (-2)
yS = (-8)
Je te laisse faire la suite
1 votes
Thanks 1
wattzef
Merci beaucoup <3 !
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Lista de comentários
Verified answer
Forme développée :f(x) = (-x + 1)(x - 1)
f(x) = -x^2 + x + x - 1
f(x) = -x^2 + 2x - 1
f(x) = ax^2 + bx - 1
f(x) = 0
(-x + 1)(x - 1) = 0
-x + 1 = 0 ou x - 1 = 0
Même solution pour les 2
x = 1
Coordonnées du sommet :
xS = (-b) / (2a)
xS = (-2) / (2 * (-1))
xS = (-2) / (-2)
xS = 1
yS = f(xS)
yS = f(1)
yS = (-1 + 1)(1 - 1)
yS = 0
S = (1 ; 0)
Forme développée :
g(x) = (2x + 2)(x - 3)
g(x) = 2x^2 - 6x + 2x - 6
g(x) = 2x^2 - 4x - 6
g(x) = 2(x^2 - 2x - 3)
g(x) = 0
(2x + 2)(x - 3) = 0
2x + 2 = 0 ou x - 3 = 0
2x = -2 ou x = 3
x = -2 / 2 ou x = 3
x = (-1) ou x = 3
Coordonnées du sommet :
xS = (-b) / (2a)
xS = (2) / (2 * 1)
xS = 2 / 2
xS = 1
yS = g(xS)
yS = g(1)
yS = (2 * 1 + 2)(1 - 3)
yS = (2 + 2)(-2)
yS = 4 * (-2)
yS = (-8)
Je te laisse faire la suite