Trying to find angle at which projectile was fired. So let's say I have a parabolic function that describes the displacement of some projectile without air resistance.  Let's say it's
$$y=-4.9x^2+V_0x.$$
I want to know at what angle the projectile was fired.  I notice that 
$$\tan \theta_0=f'(x_0)$$ so the angle should be
$$\theta_0 = \arctan(f'(0)).$$ or
$$\theta_0 = \arctan(V_0).$$
Is this correct?  I can't work out why it wouldn't be, but it doesn't look right when I plot curves.

From what you are saying, it looks like you are thinking of $f(x)$ as a real valued function, and in effect you are only considering linear motion. (This would be fine, if the cannon were firing straight up or straight down or straight up.)
However, you are interested in knowing the interesting two-dimensional trajectories, and this means that $V_0$ is a 2-D vector pointing in the direction of the initial trajectory, and that the coefficient of $x^2$ actually accompanies a vector which is pointing straight down (to show the contribution of gravity.)
Once you find what $V_0$ is, you can then compute its angle with respect to horizontal, and have your answer.
