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orlovaorlovaorlova
@orlovaorlovaorlova
July 2022
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Помогите с заданием по алгебре, пожалуйста! Запишите уравнение касательной к графику функции f ' (x) = x^2 - 3x, в точке x(нулевое) = 2
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sedinalana
Verified answer
F(2)=4-3*2=4-6=-2
f`(x)=2x-3
f`(2)=2*2-3=1
Y=-2+1(x-2)=-2+x-2=x-4
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Answers & Comments
Verified answer
F(2)=4-3*2=4-6=-2f`(x)=2x-3
f`(2)=2*2-3=1
Y=-2+1(x-2)=-2+x-2=x-4