12.22 On the graph of the function $y=x(x-4)^{3}$, find the points where the tangents are parallel to the x-axis.

Find the points of extremum of the functions (12.23-12.24):
12.22 Let $\left(x_{0} ; y_{0}\right)$ be the point of tangency. Since the tangent is parallel to the x-axis, the slope of the tangent at this point is zero. We have

$$
\begin{aligned}
& y^{\prime}=(x-4)^{3}+x \cdot 3(x-4)^{2}=(x-4)^{2}(x-4+3 x)= \\
& =4(x-4)^{2}(x-1) ; k=y^{\prime}\left(x_{0}\right)=4\left(x_{0}-4\right)^{2}\left(x_{0}-1\right)=0
\end{aligned}
$$

from which we find $\left(x_{0}\right)_{1}=4,\left(x_{0}\right)_{2}=1$ and, therefore, $\left(y_{0}\right)_{1}=0$, $\left(y_{0}\right)_{2}=-27$. Thus, $(4 ; 0)$ and $(1 ;-27)$ are the points of tangency.

Answer: $(4 ; 0)$ and $(1 ;-27)$.