1. Determine the curve on which the vertices of the parabolas

$$
y=-x^{2}+b x+c
$$

lie, which are tangent to the parabola $y=x^{2}$.
Solution. Let the parabolas $y=-x^{2}+b x+c$ and $y=x^{2}$ touch each other. Then the equation $-x^{2}+b x+c=x^{2}$ has one solution. The equation $2 x^{2}-b x-c=0$ has one solution, i.e., it has a double root if and only if $b^{2}+8 c=0$. According to this, the parabolas touch if and only if $c=-\frac{b^{2}}{8}$, i.e., if and only if the parabola $y=-x^{2}+b x+c$ has the form $y=-x^{2}+b x-\frac{b^{2}}{8}$. Its vertex is $T\left(\frac{b}{2}, \frac{b^{2}}{8}\right)$. Since $\frac{b^{2}}{8}=\frac{1}{2}\left(\frac{b}{2}\right)^{2}$, we get that the vertices of the parabolas lie on the parabola $\frac{b}{2} \rightarrow \frac{1}{2}\left(\frac{b}{2}\right)^{2}$, i.e., $y=\frac{1}{2} x^{2}$.