5. Напишите уравнение касательной к графику функции
f(x) = x2+2x+1 в точке с
абсциссой х0 = - 2.
f(x) = x2+2x+1 в точке с
абсциссой х0 = - 2.
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y = f '(x0) · (x − x0) + f(x0)f '(x) = 2x + 2
f ' (-2) = 2*(-2) + 2 = - 4 + 2 = - 2
f (-2) = (-2)^2 + 2(-2) + 1 = 1
y = - 2(x + 2) + 1 = - 2x - 3