Odpowiedź:
zad 1
y = - x² - 4x + 3
a = - 1 , b = - 4 , c = 3
Δ = b² - 4ac = (- 4)² - 4 * ( - 1) * 3 = 16 - 12 = 4
Postać kanoniczna
y = a(x - p)² + q
p = - b/2a = 4/(- 2) = - 4/2 = - 2
q = - Δ/4a = - 4/(- 4) = 4/4 = 1
y = - (x + 2)² + 1
zad 2
y = - 1/2x² + x + 4
a = - 1/2 , b = 1 , c = 4
Δ = b² - 4ac = 1² - 4 * (- 1/2) * 4 = 1 + 2 * 4 = 1 + 8 = 9
√Δ = √9 = 3
x₁ = (- b - √Δ)/2a = ( - 1 - 3)/(- 1/2 * 2) = - 4/(- 1) =4/1 = 4
x₂ =(- b + √Δ)/2a = (- 1 + 3)/(- 1/2 * 2) = 2/(- 1) = - 2/1 = - 2
Postać iloczynowa
y = a(x - x₁)(x - x₂) = - 1/2(x - 4)(x + 2)
