= Solution
Three useful geometric properties, with their concrete consequences here, are:
* \b[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>: $F_y(t)\le1/2$, so the <curve> never reaches $B=(0,1)$.
* \b[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.
* \b[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
$$
\boxed{\text{endpoint interpolation; control-edge endpoint tangents; convex-hull containment}.}
$$
The same sum-to-one calculation also explains <affine equivariance of a geometric basis>: applying an <affine map> to the handles applies that map to the <curve>.
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