Solution

Three useful geometric properties, with their concrete consequences here, are:

Endpoint interpolation. $F (0) = A = (- 2, 0)$ and $F (1) = C = (0, 0)$ . The middle control point $B$ is a handle, not an interpolated point: $F_{y} (t) \leq 1/2$ , so the curve never reaches $B = (0, 1)$ .
Endpoint tangents. $F^{'} (0) = 2 (B - A) = (4, 2)$ and $F^{'} (1) = 2 (C - B) = (0, - 2)$ . Thus the tangent lines are parallel to the first and last sides of the control polygon, respectively; both endpoint derivatives are nonzero.
Convex-hull containment. The three Bernstein basis weights are nonnegative and sum to one on $[0, 1]$ . Every $F (t)$ is therefore a convex combination of $A, B, C$ , so the entire Bézier curve lies in their filled triangle.

These give the compact answer

endpoint interpolation; control-edge endpoint tangents; convex-hull containment .

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