2B. Find the equation of the common tangent to the parabolas

$$
y=x^{2}+2 x+2 \text{ and } y=x^{2}+x+2
$$
Solution. Let the common tangent have the equation $y=k x+b$. Then each of the equations $x^{2}+2 x+2=k x+n$ and $x^{2}+x+2=k x+n$ has a unique solution. The discriminants of the quadratic equations obtained, $x^{2}+(2-k) x+2-n=0$ and $x^{2}+(1-k) x+2-n=0$, must be equal to zero, i.e., $k^{2}-4 k+4 n-4=0$ and $k^{2}-2 k+4 n-7=0$. If we subtract the second equation from the first, we get $-2 k+3=0$, i.e., $k=\frac{3}{2}$, and by substituting into one of the equations we find $n=\frac{31}{16}$. Therefore, the common tangent to the given parabolas is $y=\frac{3}{2} x+\frac{31}{16}$.