shobber wrote: Given a function $f(x)=x^{2}+ax+b$ with $a,b \in R$, if there exists a real number $m$ such that $\left| f(m) \right| \leq \frac{1}{4}$ and $\left| f(m+1) \right| \leq \frac{1}{4}$, then find the maximum and minimum of the value of $\Delta=a^{2}-4b$. $x^{2}+ax+b=\frac{(2x+a)^{2}-\Delta }{4}$. Let y=2x+a-1, then $|(y+1)^{2}-\Delta |\le 1, |(y-1)^{2}-\Delta |\le 1$. It give $0\le \Delta \le 2$.