Odpowiedź:
f(x) = x² - 1
a = 1 , b = 0 , c = - 1
Δ = b² - 4ac = 0² - 4 * 1 * (- 1) = 4
Postać kanoniczna
f(x) = a(x - p)² + q
p = - b/2a = 0/2 = 0
q = - Δ/4a = - 4/4 = - 1
f(x) = 1 * (x - 0)² - 1 = x² - 1
Postać iloczynowa
f(x) = x² - 1
obliczamy miejsca zerowe
x² - 1 = 0
(x - 1)(x + 1) = 0
x - 1 = 0 ∨ x + 1 = 0
x₁ = 1 ∨ x₂ = - 1
f(x) = a(x - x₁)(x - x₂) = (x - 1)(x + 1)
