Question

ПОМОГИТЕ 80Б! Рівняння дотичної до графіка функції в точці х=2 у = х 2 +5x ................................................................ Уравнение касательной к графику функции в точке х = 2 у = х 2 + 5x

Answer 1

Ответ:

$y=x^2+5x\ \ ,\ \ x_0=2\qquad \boxed{y=y(x_0)+y'(x_0)(x-x_0)}\\\\y(2)=4+10=14\\\\y'(x)=2x+5\ \ ,\ \ \ y'(2)=4+5=9\\\\y=14+9(x-2)\\\\\boxed{\ y=9x-4\ }$

Answer 2

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