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annakot121212
@annakot121212
July 2022
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Напишите уравнение касательной к графику функции f(X)=X^2+2X-8 В ТОЧКЕ С АБСЦИССОЙ Х0=3
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m11m
Verified answer
F(x)=x²+2x-8
f(3)=3²+2*3-8=9+6-8=7
f ' (x)=2x+2
f ' (3)=2*3+2=8
y=7+8(x-3)=7+8x-24=8x-17
y=8x-17 - уравнение касательной.
1 votes
Thanks 2
m11m
Была ошибка в решении, исправлено, обновить страницу.
Lesben
F(x)=x²+2x-8, T(3,?)
f(3)=3²+2.3-8=9+6-8=7, T(3,7)
f´(x)=2x+2, f´(3)=2.3+2=6+2=8
f´(x)=k, k=8
y-7=8(x-3)
y-7=8x-24
y=8x-24+7
y=8x-17
=======
3 votes
Thanks 1
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Answers & Comments
Verified answer
F(x)=x²+2x-8f(3)=3²+2*3-8=9+6-8=7
f ' (x)=2x+2
f ' (3)=2*3+2=8
y=7+8(x-3)=7+8x-24=8x-17
y=8x-17 - уравнение касательной.
f(3)=3²+2.3-8=9+6-8=7, T(3,7)
f´(x)=2x+2, f´(3)=2.3+2=6+2=8
f´(x)=k, k=8
y-7=8(x-3)
y-7=8x-24
y=8x-24+7
y=8x-17
=======