Calculus-1

I assumed, $f(x)$ to be $f(x) = ax^2 + bx + d$ and substituted in the given three conditions to get the values of $a,b,d$

$\begin{aligned} f(0) = & d = 2 \\ f'(0) = & b = -2 \\ f(1) = & a - 2 + 2 = 1 \implies a = 1 \\ f(c) \cdot f'(c) + f''(c) & = 0 \\ (c^2 - 2c + 2)(2c - 2) + 2c & = 0 \\ c^3 - 3c^2 + 5c - 2 & = 0 \end{aligned}$

The only real solution to the above equation in $c$ is $0.5467$ and the other two are imaginary roots. As $c = 0.54 \in (1,0)$ it is the only possible solution.

My only doubt here is that, can we assume $f(x) = ax^2 + bx + d$ ?