Problem 5. (Option 2).

Find the form of all quadratic trinomials $f(x)=a x^{2}+b x+c$, where $a, b, c$ are given constants, $\quad a \neq 0$, such that for all values of $x$ the condition $f(3.8 x-1)=f(-3.8 x)$ is satisfied.
# Solution:

$f(3.8 x-1)=f(-3.8 x) \prec=\succ f(-1)=f(0) \prec=\succ a-b+c=c \prec=\succ a-b=0 \prec=\succ b=a$ $\prec=\succ(x)=a x^{2}+a x+c$.

Verification: $\quad f(3.8 x-1)=f(-3.8 x) \prec=\succ$

$a(3.8 x-1)^{2}+a(3.8 x-1)+c=a(-3.8 x)^{2}+a(-3.8 x)+c$

$\prec=14.44 a x^{2}-7.6 a x+a+3.8 a x-a=14.44 a x^{2}-3.8 a x \prec=\succ 0=0$.

Answer. $f(x)=a x^{2}+a x+c$.