Đáp án:
\(\left[ \begin{array}{l}
y = 3x + 3 - 4\sqrt 2 \\
y = 3x - 3 + 4\sqrt 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
y' = 3{x^2} - 6x\\
y'\left( {{x_0}} \right) = k = 3{x_0}^2 - 6{x_0} = 3\\
\to \left[ \begin{array}{l}
{x_0} = 1 + \sqrt 2 \\
{x_0} = 1 - \sqrt 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
{y_0} = - \sqrt 2 \\
{y_0} = \sqrt 2
\end{array} \right.\\
\to PTTT:\left[ \begin{array}{l}
y = 3\left( {x - 1 - \sqrt 2 } \right) - \sqrt 2 \\
y = 3\left( {x - 1 + \sqrt 2 } \right) + \sqrt 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
y = 3x + 3 - 4\sqrt 2 \\
y = 3x - 3 + 4\sqrt 2
\end{array} \right.
\end{array}\)
