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nastynka98
@nastynka98
July 2022
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написать уравнение касательной к графику функции y=x^3+x^2-2x+1 в точке x0=-1
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sinusoida2000
Y(x0)=-1+1+2+1=3
y'=3x^2+2x-2
y'(x0)=3-2-2=-1
y=3-1(x+1)=3-x-1=2-x
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Answers & Comments
y'=3x^2+2x-2
y'(x0)=3-2-2=-1
y=3-1(x+1)=3-x-1=2-x