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 $y=20 g(x)$. Find all values of $A$ such that the graph of the function $y=(g(x))^{2}$ touches the graph of the function $y=\frac{f(x)}{A}$.
Answer: $-0.05$.

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 $y=20 g(x)$ is equivalent to the equation $(f(x))^{2}=20 g(x)$ having exactly one solution. We get $(a x+b)^{2}=20(a x+c), a^{2} x^{2}+2 a(b-10) x+b^{2}-20 c=0$. One quarter of the discriminant of this equation is $a^{2}(b-10)^{2}-a^{2}\left(b^{2}-20 c\right)=20 a^{2}(5-b+c)$, from which $b-c=5$.

Similarly, the tangency of the graphs $y=(g(x))^{2}$ and $y=\frac{f(x)}{A}$ means that the equation $(g(x))^{2}=\frac{f(x)}{A}$ has a unique solution. This equation (under the condition $A \neq 0$) is equivalent to the following: $A(a x+c)^{2}=a x+b, A a^{2} x^{2}+a(2 A c-1) x+A c^{2}-b=0$. The discriminant is $D=a^{2}(2 A c-1)^{2}-$ $4 A a^{2}\left(A c^{2}-b\right)=a^{2}(4 A b-4 A c+1)$. It becomes zero when $A=\frac{1}{4(c-b)}=-\frac{1}{20}$.