Ответы на вопрос:
Нули функции 2(x + 1)^2 - 8 = 0 2(x + 1)^2 = 8 (x + 1)^2 = 4 (x + 1)^2 - 2^2 = 0 (x + 1 - 2)(x + 1 + 2) = 0 (x + 3)(x - 1) = 0 x + 3 = 0 x = - 3 x - 1 = 0 x = 1 ( - 3; 0) ; ( 1; 0) ======================================== y = 2(x + 1)^2 - 8 = 2(x^2 + 2x + 1) - 8 = 2x^2 + 4x - 6 x0 = - b/2a = - 4/4 = - 1 y0 = - 8 ( - 1; - 8) - вершина
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