bjr
f(x) = 2x² + x - 3
1) factoriser..
Δ = 1² - 4*2*(-3) = 1+24=25 = 5²
x1 = (-1-5)/4 = -6/4 = -3/2
x2 = (-1+5)/4 = 1
=> f(x) = 2 (x+3/2) (x-1) = (3x+2) (x-1)
2)
forme canonique
f(x) = 2 (x² + 1/2x) - 3 = 2 [(x + 1/4)² - (1/4)²] - 3
= 2 (x+1/4)² - 2/16 - 3 = 2(x+1/4)² - 2/16-48/16 = 2(x+1/4)² - 50/16
= 2(x+1/4)² - 25/8 ouf
3a)
image de -3/2 => calcul de f(-3/2)
prenons f(x) = (3x+2)(x-1)
f(-3/2) = (3*(-3/2) + 2) (3/2-1) = ....
b)
f(-1/4) - tu prends la forme canonique..
c) antécédents de -3
donc résoudre f(x) = -3
d) antécédents de 0
donc résoudre f(x) = 0 => prendre la forme factorisée
e) f(x) = -49/8
