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anonim7917
@anonim7917
December 2022
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Напишите уравнение касательной к графику функции =y=5-2x/x-2, проведенной через точку с абсциссой x0=1
Срочно!!
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fenix6810
Verified answer
y=4x+3
y(x0)=y(1)=5-2/(1-2)=5+2=7
y'(x)=(-2(x-2)+2x)/(x-2)^2=4/(x-2)^2
y'(1)=4
y=y(x0)+y'(x0)(x-x0)
y=7+4*(x-1)=7+4x-4=4x+3
6 votes
Thanks 23
bogdankasymov684
а есть остальные ответы на задания в этом же соре?
anatoltolik4
тоже нужны
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Answers & Comments
Verified answer
y=4x+3y(x0)=y(1)=5-2/(1-2)=5+2=7
y'(x)=(-2(x-2)+2x)/(x-2)^2=4/(x-2)^2
y'(1)=4
y=y(x0)+y'(x0)(x-x0)
y=7+4*(x-1)=7+4x-4=4x+3