Over a polynomial problem Assume that $P$ is a quadratic polynomial such that $P(1+x) = P(3-x)$, and $P(x)\geq 1$. If $P(0) = 13$, how could we find $P(1)$?
If $P(0) = 13$, then $P$ is of the form $P(x) = ax^2+bx+13$. Since $P(1+x) = P(3-x)$, $P(4) = P(0) =13$ and therefore, $16a+4b = 0$, $b = -4a$.  And if $P(-\frac{b}{2a}) = 1$, then $4a-8a + 13 = 1$, leading us to $a = 3$ and $P(x) = 3x^2-12x+13$. From which we could conclude that $P(1) = 4$.

Hint…consider the minimum point$$P(-\frac{b}{2a})=1$$
