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. It is known that the graph of the function $y=(f(x))^{2}$ touches the graph of the function $y=11 g(x)$. Find all values of $A$ such that the graph of the function $y=(g(x))^{2}$ touches the graph of the function $y=A f(x)$.
Answer: $A=-11, A=0$.

Solution. Let $f(x)=a x+b, g(x)=a x+c$, where $a \neq 0$. The tangency of the graphs $y=(f(x))^{2}$ and $y=11 g(x)$ is equivalent to the equation $(f(x))^{2}=11 g(x)$ having exactly one solution. We get $(a x+b)^{2}=11(a x+c), a^{2} x^{2}+a(2 b-11) x+b^{2}-11 c=0$. The discriminant of this equation is $a^{2}(2 b-11)^{2}-4 a^{2}\left(b^{2}-11 c\right)=11 a^{2}(11-4 b+4 c)$, from which $4 b-4 c=11$.

Similarly, the tangency of the graphs $y=(g(x))^{2}$ and $y=A f(x)$ means that the equation $(g(x))^{2}=A f(x)$ has a unique solution. This equation is equivalent to the following: $(a x+c)^{2}=A(a x+b)$, $a^{2} x^{2}+a(2 c-A) x+c^{2}-A b=0$. The discriminant is $D=a^{2}(2 c-A)^{2}-4 a^{2}\left(c^{2}-A b\right)=a^{2} A(A-$ $4 c+4 b)$. It becomes zero when $A=0$ and $A=4 c-4 b=-11$.