Bonjour,
1) f(x) = 0 ⇔ -2x² + 4x + 6 = 0
Δ = b² - 4ac = 4² - 4 × (-2) × 6 = 16 + 48 = 64
⇒ Δ > 0 donc deux solutions dans ℝ
X₁ = (-b - √Δ)/2a = (-4 -8)/(-4) = (-12)/(-4) = 3
X₂ = (-b + √Δ)/2a = (-4 + 8)/(-4) = 4/(-4) = - 1
Tableau de signe de f sur ℝ :
x | -∞ - 1 3 +∞
f(x)| - 0 + 0 -
2) f(x) = -2(x-α)² + β
α = -b/(2a) = -4/(-4) = 1
β = -(b² - 4ac)/4a = - (Δ/4a) = - 64/ (-8) = 8
ainsi f(x) = -2(x - 1)² + 8
Vérification : f(x) = -2(x - 1)² + 8 = -2(x² - 2x + 1) + 8 = -2x² + 4x - 2 + 8 = -2x² + 4x + 6
