Составьте уравнение касательной к графику функции y = – x2

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June 2022 1 7 Report

Составьте уравнение касательной к графику функции y = – x2 – 4x + 2 , проходящей через точку М(-3;5)

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$y=-x^2-4x+2\ \ ,\ \ \ M(-3\ ;\ 5\ )\\\\y(M)=y(-3)=-(-3)^2-4\cdot (-3)+2=5\\\\y'(x)=-2x-4\ \ ,\ \ y(-3)=6-4=2\\\\kasatelnaya:\ \ y=5+2(x+3)\\\\{}\qquad \qquad \quuad \qquad \ \ \boxed {\ y=2x+11\ }$

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