R – Convert a quadratic curve points to polynomial representation

Using your formula for B'(t), evaluated at t=1/2, we get

B'(1/2) = -P0 + P2

From the look of your graph, P0 = (0,0) and P2 = (400,0). So

B'(1/2) = (400,0).

This is the "velocity" of a point traveling along B(t) at t=1/2.

(400,0) is a horizontal vector, with magnitude 400.

So all is as it should be. Since B'(t) is horizontal, it does have "slope" 0.

Use the quadratic Bézier formula, found, for instance, on the Wikipedia page for Bézier Curves:

In pseudo-code, that's

t = 0.5; // given example value
x = (1 - t) * (1 - t) * p[0].x + 2 * (1 - t) * t * p[1].x + t * t * p[2].x;
y = (1 - t) * (1 - t) * p[0].y + 2 * (1 - t) * t * p[1].y + t * t * p[2].y;

p[0] is the start point, p[1] is the control point, and p[2] is the end point. t is the parameter, which goes from 0 to 1.