2. Given two linear functions $f(x)$ and $g(x)$ such that the graphs $y=f(x)$ and $y=g(x)$ are parallel lines, not parallel to the coordinate axes. Find the minimum value of the function $(g(x))^{2}-$ $3 f(x)$, if the minimum value of the function $(f(x))^{2}-3 g(x)$ is $\frac{11}{2}$.
Answer: -10.

Solution. Let $f(x)=a x+b, g(x)=a x+c$, where $a \neq 0$. Consider $h(x)=(f(x))^{2}-3 g(x)$. Expanding the brackets, we get $h(x)=(a x+b)^{2}-3(a x+c)=a^{2} x^{2}+a(2 b-3) x+b^{2}-3 c$. The graph of $y=$ $h(x)$ is a parabola opening upwards, and the minimum value is attained at the vertex. The x-coordinate of the vertex is $x_{\mathrm{B}}=-\frac{2 b-3}{2 a}$; the y-coordinate of the vertex is $h\left(x_{\mathrm{B}}\right)=3 b-\frac{9}{4}-3 c$.

Similarly, we find that the minimum value of the expression $(g(x))^{2}-3 f(x)$ is $3 c-\frac{9}{4}-3 b$. Note that the sum of these two minimum values is $-\frac{9}{2}$, therefore, if one of these minimum values is $\frac{11}{2}$, then the other is $-\frac{9}{2}-\frac{11}{2}=-10$.