The curve $y=a{x}^2+bx+c$  passes through the point (1, 2) and its tangent at origin is the line $y=x$ .The area bounded by the curve, the ordinate of the curve at minima and tangent  line is

$$\begin{gathered} (1,2)\Rightarrow a+b+c=2 \\ (0,0)\Rightarrow c=0 \\ a+b=2 \\ {\left(\frac{dy}{dx}\right)}_{(0,0)}=2\left(a\right)\left(0\right)+b=1 \\ \therefore b=1\&a=1 \end{gathered}$$

So the curve  $y=x^2+x$
 $={(x+\frac{1}{2})}^2-\frac{1}{4}$
$x=\frac{-1}{2}$   is point of minima

Required area =$\underset{\frac{-1}{2}}{\overset{0}{\int }}(x^2+x-x)dx$$=\underset{\frac{-1}{2}}{\overset{0}{\int }}{x}^{2}dx={{\left(\frac{x^3}{3}\right)}_{\frac{-1}{2}}}^0=\frac{1}{24}Sq units$