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00Marie00
@00Marie00
April 2019
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Bonjour, voici mon exercice : La courbe C représente une fonction f définie sur l'intervalle I=(-1;6)
La courbe est celle de la fonction : x^2-5x-2,75
2) Montrer que f(x)=(x-2,5)^2-9. Factoriser f(x) et résoudre l 'équation f(x)=0
3) Le point de coordonnées (4;-6,89) est-il un point de la courbe f.
Merci :)
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F(x)=x²-5x-2,75
=x²-5x+6,25-6,25-2,75
=(x-2,5)²-9
f(x)=(x-2,5)²-3²
=(x-2,5-3)(x-2,5+3)
=(x-5,5)(x+0,5)
f(x)=0 donne (x-5,5)(x+0,5)=0
donc x=5,5 ou x=-0,5
f(4)=(4-2,5)²-9=-6,75≠-6,89
donc A(4;-6,89)∉Cf
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=x²-5x+6,25-6,25-2,75
=(x-2,5)²-9
f(x)=(x-2,5)²-3²
=(x-2,5-3)(x-2,5+3)
=(x-5,5)(x+0,5)
f(x)=0 donne (x-5,5)(x+0,5)=0
donc x=5,5 ou x=-0,5
f(4)=(4-2,5)²-9=-6,75≠-6,89
donc A(4;-6,89)∉Cf