РЕШЕНИЕ:
ax2 + bx + c = y
A(0; 4) a ∙0 + b ∙ 0 + c = 4 ⇒ c = 4
B(1; 11) a ∙ 12+ b ∙ 1 + c = 11
C(–5; –1) a ∙( – 5)2 + b ∙ ( – 5) + c = – 1
подставим с= 4 во 2 и 3 уравнения
a ∙ 12+ b ∙ 1 + 4 = 11
a ∙ ( – 5)2 + b ∙ ( – 5) + 4 = – 1
a + b = 7
25a – 5b = – 5
5a + 5b = 35
25a – 5b = – 5
Сложим уравнения
30a = 30
a = 1
a + b = 7
1 + b = 7
b = 6
x0 = -b/(2a) = – 6/2 = – 3
ax2 + bx + c = y
y0 = 1∙( – 3)2 + 6 ∙ ( – 3) + 4 = 9 – 18 + 4 = – 5
Ответ: { – 3 ; – 5}