Написать уравнение касательной к графику функции y=f(x) в точке с абсциссой x0: f(x)= x-3x^2, x0=2
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Уравнение касательной: ![Y(x)=y(x_{0})+y'(x_{0})*(x-x_{0}) Y(x)=y(x_{0})+y'(x_{0})*(x-x_{0})](https://tex.z-dn.net/?f=Y%28x%29%3Dy%28x_%7B0%7D%29%2By%27%28x_%7B0%7D%29%2A%28x-x_%7B0%7D%29)
![x_{0}=2 x_{0}=2](https://tex.z-dn.net/?f=x_%7B0%7D%3D2)
![y(x)=x-3x^{2} y(x)=x-3x^{2}](https://tex.z-dn.net/?f=y%28x%29%3Dx-3x%5E%7B2%7D)
![y(x_{0})=x_{0}-3*(x_{0})^{2} y(x_{0})=x_{0}-3*(x_{0})^{2}](https://tex.z-dn.net/?f=y%28x_%7B0%7D%29%3Dx_%7B0%7D-3%2A%28x_%7B0%7D%29%5E%7B2%7D)
![y'(x)=1-6x y'(x)=1-6x](https://tex.z-dn.net/?f=y%27%28x%29%3D1-6x)
![y'(x_{0})=1-6x_{0} y'(x_{0})=1-6x_{0}](https://tex.z-dn.net/?f=y%27%28x_%7B0%7D%29%3D1-6x_%7B0%7D)
![Y(x)=x_{0}-3*(x_{0})^{2}+(1-6x_{0})*(x-x_{0})=x_{0}-3*(x_{0})^{2}+x*(1-6x_{0})-x_{0}+6*(x_{0})^{2}=x*(1-6x_{0})+3*(x_{0})^{2} Y(x)=x_{0}-3*(x_{0})^{2}+(1-6x_{0})*(x-x_{0})=x_{0}-3*(x_{0})^{2}+x*(1-6x_{0})-x_{0}+6*(x_{0})^{2}=x*(1-6x_{0})+3*(x_{0})^{2}](https://tex.z-dn.net/?f=Y%28x%29%3Dx_%7B0%7D-3%2A%28x_%7B0%7D%29%5E%7B2%7D%2B%281-6x_%7B0%7D%29%2A%28x-x_%7B0%7D%29%3Dx_%7B0%7D-3%2A%28x_%7B0%7D%29%5E%7B2%7D%2Bx%2A%281-6x_%7B0%7D%29-x_%7B0%7D%2B6%2A%28x_%7B0%7D%29%5E%7B2%7D%3Dx%2A%281-6x_%7B0%7D%29%2B3%2A%28x_%7B0%7D%29%5E%7B2%7D)
- уравнение касательной
Ответ: -5х+3
Ответ: -5х+3
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