# Symmetric bezier curve with unsymmetric control points

I'm solving this problem about Cubic Bezier Curve which have $$4$$ control points $$C_0,C_1,C_2,C_3$$. $$C_0 = (0,0),\: C_1 = (x_1,y_1),\: C_2=(x_2,y_2),\: C_3=(1,0)$$ where $$0 < x_1 < x_2 < 1$$ and $$y_1, y_2>0$$.

If $$C_1$$ and $$C_2$$ are symmetric with respect to axis $$x = 0.5$$ (such as $$C_1 = (0.25,1)$$, $$C_2 = (0.75,1)$$), obviously whole bezier curve is symmetric (w.r.t. $$x=0.5$$). But I'm still confusing if converse of this statement is still correct. Is any symmetric (cubic) bezier curve can be existed with two unsymmetrical control points $$C_1,C_2$$?

I tried with reduction, if $$C_1, C_2$$ is not symmetric and the image $$B(C_0,C_1,C_2,C_3)$$ is symmetric, there should be $$C_1^\prime$$ that symmetric with respect to $$C_1$$ but $$C_1^\prime := C2 \cdot B(C_0,C_1,C_1^\prime,C_3)$$ is symmetric and when I move just one control point $$C_1^\prime$$ to $$C2$$, I'm not sure that symmetric property will broke or not.