$\begin{align}  & y=3-2x+{{x}^{2}} \\ & \frac{dy}{dx}=-2+2x \\ & \text{At turning point }\frac{dy}{dx}=0 \\ & -2+2x=0 \\ & x=1 \\ & \text{To test for min (max) point} \\ & \frac{{{d}^{2}}y}{d{{x}^{2}}}=2>0 \\ & \therefore x=1\text{ is minimium is a point of the function} \\\end{align}$