bonjour
pb 1
g(x) = 2x²+4x-1
1) forme canonique ?
g(x) = 2 (x²+2x) - 1
= 2 [(x+1)² - 1²] - 1
= 2 (x+1)² -2-1 = 2(x+1)² - 3
2) g(x) = -1
2x²+4x-1 =-1
2x² +4x=0
2x(x+2) = 0 ; x=0 ou -2
3) g(x)= -3
2(x+1)² - 3 = -3
2(x+1)² = 0 ; x=-1
4) g(-1) = 2*(-1+1)²-3 = -3
g(0) = 2*0²+4*0-1 = -1
pb2
f(x) =(x+2)² - 9
1) f(x) = x² + 4x + 4 - 9 = x²+4x-5
2) f(x) = (x+2)²-3² = (x+2+3) (x+2-3) = (x+5) (x-1)
3) f(x)=0
(x+5) (x-1) ; x=-5 ou 1
f(x) = -5
x²+4x-5 = -5
x²+4x=0
x(x+4)=0 ; x=0 ou -4
f(x) = -9
(x+2)² - 9=-9
(x+2)² = 0 ; x=-2
f(0)=0²+4*0-5 = -5
f(- ?)
