Đáp án:
\(y=- 2x + \frac{{10}}{3}\) và \(y=- 2x + 2\)
Giải thích các bước giải:
$\begin{array}{l}
y' = {x^2} - 4x + 1\\
Goi\,M\left( {{x_0};{y_0}} \right)\,la\,tiep\,diem\\
Tiep\,tuyen\,tai\,M\,song\,song\,duong\,thang\,y = - 2x + 5\,nen\,hsg\,k = - 2\\
\Rightarrow x_0^2 - 4{x_0} + 1 = - 2 \Leftrightarrow x_0^2 - 4{x_0} + 3 = 0 \Leftrightarrow \left[ \begin{array}{l}
{x_0} = 1 \Rightarrow {y_0} = \frac{4}{3}\\
{x_0} = 3 \Rightarrow {y_0} = - 4
\end{array} \right.\\
Voi\,M\left( {1;\frac{4}{3}} \right)\,thi\,{d_1}:y = - 2\left( {x - 1} \right) + \frac{4}{3} = - 2x + \frac{{10}}{3}\\
Voi\,M\left( {3; - 4} \right)\,thi\,{d_2}:y = - 2\left( {x - 3} \right) - 4 = - 2x + 2
\end{array}$
