382. At what point is the tangent to the curve $y=\frac{1}{3} x^{3}-3 x^{2}+$ $+8 x+4$ parallel to the line $2 x+2 y-5=0$?
Solution. The slope of the given line $k_{1}=-1$. The slope of the tangent $k=y^{\prime}=x^{2}-6 x+8$.

From the condition of parallelism, it follows that $k=k_{1}$. Then $x^{2}-6 x+8=-1$; $x^{2}-6 x+9=0 ; \quad(x-3)^{2}=0$. Therefore, $x=3$ is the abscissa of the point of tangency. Substituting this value of $x$ into the equation of the curve, we get the ordinate of the point of tangency: $y=\frac{1}{3} \cdot 3^{3}-3 \cdot 3^{2}+8 \cdot 3+4=10$.

Thus, at the point $(3 ; 10)$, the tangent to the given curve is parallel to the line $2 x+2 y-5=0$.