Odpowiedź:
f(x) = - 3x² + 5x + 2
a = - 3 , b = 5 , c = 2
- 3x² + 5x + 2 = 0
Δ = b² - 4ac = 5² - 4 * (- 3) * 2 = 25 + 24 = 49
√Δ = √49 = 7
x₁ = ( - b - √Δ)/2a = (- 5 - 7)/(- 6) = - 12/(- 6) = 12/6 = 2
x₂ = (- b + √Δ)/2a = (- 5 + 7)/(- 6) = 2/(- 6) = - 2/6 = - 1/3
Postać kanoniczna
f(x) = a(x - p)² + q
p = - b/2a = - 5/(- 6) = 5/6
q = - Δ/4a = - 49/(- 12) = 49/12 = 4 1/12
f(x) = - 3(x - 5/6)² + 4 1/12
Postać iloczynowa
f(x) = a(x - x₁)(x - x₂) = - 3(x - 2)(x + 1/3)
