Proof clamped nurbs curve is tangent to first and last legs of its control points

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I want to prove that $\mathbf{C}$ a quadratic NURBS curve is tangent to the first and last legs of its control polygon.

Given three control points $\mathbf{P}_0$, $\mathbf{P}_1$, $\mathbf{P}_2$ with positive real weights $w_0$, $w_1$, $w_2$

and

$$\mathbf{C}(t) = \dfrac{\sum_{i=0}^{2} N_{i,2}(t) \cdot w_i \cdot \mathbf{P}_i}{\sum_{i=0}^{2} N_{i,2}(t) \cdot w_i}$$

With the knot vector

$$ U = (0,0,0,1,1,1) $$

I know it's clamped and that's Nurbs are basically sharing b-splines properties so if we have clamped b-spline it will be tangent to it's first and last legs but I don't know how to give a proper proof for this.