From the given condition, we should have $p'(2)=p''(2)$ because their graph are tangent at $(2, 1), and $p(2) = p'(2) = 1$ because $(2, 1)$ are on both graphs. Therefore,

$$g'(x) = 2ax + b,\quad g''(x)= 2a \implies 4a = 2a\implies a=\frac{1}{2}$$

and

$$4a + 2b + c = 4a + b = 1\implies b=-1, c = 1$$

Hence, we conclude there is only one qualified polynomial

$$p(x)=\boxed{\frac{1}{2} x^2 - x + 1 }$$