Find a trajectory such that the action is a minimum

A particle is subjected to the potential V (x) = −F x, where F is a constant. The particle travels from x = 0 to x = a in a time interval t0 . Assume the motion of the particle can be expressed in the form $x(t)=A+Bt+C{t}^2$ . Find the values of A, B, and C such that the action is a minimum.

I was thinking it can solved using Lagrangian rather than Hamilton. There's no frictional force.

$$L=\frac{1}{2}m{\dot{x}}^2+Fx$$

$$\frac{d}{dt}(\frac{\partial L}{\partial \dot{x}})-\frac{\partial L}{\partial x}=0$$

$$m\ddot{x}=F$$

$$\ddot{x}=\frac{F}{m}$$

Differentiate $x(t)$ twice. $$2C=\frac{F}{m}=>C=\frac{F}{2m}$$

For finding B I was thinking to integrate $\ddot{x}$ once. $$\dot{x}=\int \ddot{x}\mathrm{d}t$$

$$=\ddot{x}t$$

initial position is 0 so, not writing constant.

$$\dot{x}=\frac{F}{m}$$

Differentiate $x(t)$ once.

$$B+2Ct=\frac{F}{m}$$

$$B=\frac{F}{m}-\frac{2Ft}{2m}$$

$$=-\frac{Ft}{2m}$$

Again, going to integrate $\ddot{x}$ twice.

$$x=\int \int \ddot{x}dtdt$$

$$=\frac{\ddot{x}{t}^2}{2}$$

initial velocity and initial position is 0.

$$x=\frac{F{t}^2}{2m}$$

$$A+Bt+C{t}^2=\frac{F{t}^2}{2m}$$

$$A=\frac{F{t}^2+Ft-F}{2m}$$

According to my, I think that C is the minimum (I think B is cause, B is negative; negative is less than positive). And, A is maximum.

A person were saying that It asked you to minimise the action; it told you the particle moved from $0$ to $a$ in time $t_0$; it gave you the equation of the trajectory.

In my work where should I put the interval?