$\text{ If }f\left(x\right)=a{x}^2+bx+c\text{ and }f(-1)\ge -4,f\left(1\right)\le 0\text{ and }f\left(3\right)\ge 5,\text{ then the least value of }a\text{ is }$

$f(-1)\ge -4----\left(1\right)$

$\begin{array}{l}\Rightarrow a-b+c\ge -4\\ f\left(1\right)\le 0\\ \Rightarrow a+b+c\le 0\end{array}$

$\Rightarrow -a-b-c\ge 0---\left(2\right)$

$f\left(3\right)\ge 5$

$\text{ and }9a+3b+c\ge 5----\left(3\right)$

$\text{ From (i) }+\text{ (ii) }\Rightarrow -2\mathrm{b}\ge -4----\left(4\right)$

$\text{ From (ii) }+\left(\mathrm{iii}\right)+\left(\mathrm{iv}\right)\Rightarrow 8a\ge 1\Rightarrow a\ge 1/8$