Odpowiedź:
1.
y = - 3(x + 2)² - 4 = - 3(x² + 4x + 4) - 4 = - 3x² - 12x - 12 - 4 =
= - 3x² - 12x - 16
2.
y = - 2x² + 8x + 1
a = - 2 , b = 8 , c = 1
Δ = b² - 4ac = 8² - 4 * (- 2) * 1 = 64 - 8 = 56
W - współrzędne wierzchołka paraboli = (p , q)
p = - b/2a = - 8/(- 4) = 8/4 = 2
q = - Δ/4a = - 56/(- 8) = 56/8 = 7
W = ( 2 , 7 )
3.
y = 2x² - 4x + 7
a = 2 , b = - 4 , c = 7
Δ = b² - 4ac = (- 4)² - 4 * 2 * 7 = 16 - 56 = - 40
W = (p , q)
p = - b/2a = 4/4 = 1
q = - Δ/4a = 40/ 8 = 5
W = (1 , 5 )
Postać kanoniczna
y = a(x - p)² + q = 2(x - 1)² + 5
