Let $ f_m: \mathbb R \to \mathbb R$, $ f_m(x)=(m^2+m+1)x^2-2(m^2+1)x+m^2-m+1,$ where $ m \in \mathbb R$.
1) Find the fixed common point of all this parabolas.
2) Find $ m$ such that the distance from that fixed point to $ Oy$ is minimal.
Let $r$ be a solution to the equation. The, by the quadratic formula:
$r=\frac{2(m^2+1) \pm \sqrt{4(m^2+1)^2-4(m^2+m+1)(m^2-m+1)}}{2m^2+2m+2} $
$r=\frac{2(m^2+1) \pm \sqrt{4(m^2+1)^2-4(m^4+m^2+1)}}{2m^2+2m+2} $
$r=\frac{2(m^2+1) \pm \sqrt{4m^2}}{2m^2+2m+2} $
$r=\frac{2m^2 \pm 2m + 2}{2m^2+2m+2} $
It follows that $r=1$ will be a root regardless of $m$, and therefore $(1,0)$ is a fixed point.