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Sagot :

AYUDA

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

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