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DrPro
@DrPro
June 2022
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Прямая y=-4x+1 является касательной к графику функции [tex]y=x^3+7x^2+7x+6[/tex]. Найдите абсциссу точки касания.
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sangers1959
Verified answer
Y=-4x+1 y=x³+7x²+7x+6
-4x+1=x³+7x²+7x+6
x³+7x²+11x+5=0
x₁=-5
x³+7x²+11x+5 I_x+5_
x³+5x² I x²+2x+1
----------
2x²+11x
2x²+10x
------------
x+5
x+5
-----
0
x²+2x+1=0
(x+1)²=0
x₂=-1
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Answers & Comments
Verified answer
Y=-4x+1 y=x³+7x²+7x+6-4x+1=x³+7x²+7x+6
x³+7x²+11x+5=0
x₁=-5
x³+7x²+11x+5 I_x+5_
x³+5x² I x²+2x+1
----------
2x²+11x
2x²+10x
------------
x+5
x+5
-----
0
x²+2x+1=0
(x+1)²=0
x₂=-1