\(y=x^3-3x^2+2\)
\(\Rightarrow y^`=3x^2-6x\)
\(y^`\left(\dfrac{23}{9}\right)=k=3\cdot\left(\dfrac{23}{9}\right)^2-6\cdot\dfrac{23}{9}=\dfrac{115}{27}\)
\(y=k\left(x-x_0\right)+y_0=\dfrac{115}{27}\cdot\left(x-\dfrac{23}{9}\right)-2\)
\(y=k\left(x-x_A\right)+y_A=k\left(x-\dfrac{23}{9}\right)-2\)
\(\left\{{}\begin{matrix}k\left(x-\dfrac{23}{9}\right)-2=x^3-3x^2+2\left(1\right)\\k=3x^2-6x\left(2\right)\end{matrix}\right.\)
The (2) vo (1)\(\Rightarrow\left(3x^2-6x\right)\left(x-\dfrac{23}{9}\right)-2=x^3-3x^2+2\)
\(\Leftrightarrow3x^3-\dfrac{23}{3}x^2-6x^2+\dfrac{46}{3}x-2=x^3-3x^2+2\)
\(\Leftrightarrow2x^3-\dfrac{32}{3}x^2+\dfrac{46}{3}x-4=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=3\\x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=\dfrac{46}{27}\\y=2\\y=-2\end{matrix}\right.\)
\(y'=3x^2-6x\Rightarrow\left[{}\begin{matrix}y'\left(x\right)=-\dfrac{5}{3}\\y'\left(x\right)=9\\y'\left(x\right)=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}y=-\dfrac{5}{3}\left(x-\dfrac{1}{3}\right)+\dfrac{46}{27}\\y=9\left(x-3\right)+2\\y=-2\end{matrix}\right.\)
