Question

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Answer 1

Thay \(M\left(0;2\right)\Rightarrow d=2\)

Thay \(N\left(2;-2\right)\Rightarrow8a+4b+2c+2=-2\)

\(y'=3ax^2+2bx+c\)

Từ tọa độ 2 cực trị ta có:

\(\left\{{}\begin{matrix}y'\left(0\right)=0\\y'\left(2\right)=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}c=0\\12a+4b=0\end{matrix}\right.\)

Ta được: \(\left\{{}\begin{matrix}8a+4b+2c=-4\\c=0\\12a+4b=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=1\\b=-3\end{matrix}\right.\)

\(\Rightarrow y=x^3-3x^2+2\Rightarrow y\left(-2\right)=-18\)