381. Given the curve $y=-x^{2}+4$. Draw a tangent to it at the point where the abscissa is $x=-1$.
Solution. Let's find the ordinate of the point of tangency: $y_{x=-1}=\left(-x^{2}+\right.$ $+4)_{x=-1}=-(-1)^{2}+4=3$; hence, $A(-1 ; 3)$ is the point of tangency.

The equation of any line passing through point $A$ is $y-3=k(x+1)$. For the line to be a tangent, it is necessary and sufficient that $k=y_{x=-1}^{\prime}$. Let's find the slope $k$:

$$
y^{\prime}=\left(-x^{2}+4\right)^{\prime}=-2 x ; k=y_{x=-1}^{\prime}=(-2)(-1)=2
$$

Thus, the equation of the tangent line is $y-3=2(x+1)$, i.e., $y=$ $=2 x+5$.