РЕШЕНИЕ:
ax2 + bx + c = y
A(0; – 4) a ∙0 + b ∙ 0 + c = – 4 ⇒ c = – 4
B(– 1; – 11) a ∙( – 1)2+ b ∙ ( – 1) + c = – 11
C(4; 4) a ∙42 + b ∙ 4 + c = 4
подставим с= – 4 во 2 и 3 уравнения
a ∙( – 1)2+ b ∙ ( – 1) – 4 = – 11
a ∙42 + b ∙ 4 – 4 = 4
a – b = – 7
16 a + 4b = 8
4a – 4b = – 28
16a + 4b = 8
Сложим уравнения
20a = – 20
a = – 1
a – b = – 7
– 1 – b = – 7
b = 6
x0 = -b/(2a) = – 6/( – 2) =3
ax2 + bx + c = y
y0 = – 1∙32 + 6∙3 – 4 = – 9 + 18 – 4 = 5
Ответ: { 3 ; 5}
