# Problem 5. (Option 1).

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:

$y(0.01 x+1)=y(-0.01 x) \prec=\succ y(1)=y(0) \prec=\succ a+b+k=k \prec=\succ a+b=0 \prec=\succ$ $b=-a \prec=\succ y(x)=a x^{2}-a x+k$.

Verification: $y(0.01 x+1)=y(-0.01 x) \prec=\succ$

$a(0.01 x+1)^{2}-a(0.01 x+1)+k=a(-0.01 x)^{2}-a(-0.01 x)+k$

$\prec \Rightarrow 0.0001 a x^{2}+0.02 a x+a-0.01 a x-a=0.0001 a x^{2}+0.01 a x \prec=\succ 0=0$.

Answer: $y(x)=a x^{2}-a x+k$.