# Convert segment of parabola to quadratic bezier curve

How do I convert a segment of parabola to a cubic Bezier curve?

The parabola segment is given as a polynomial with two x values for the edges.

My target is to convert a quadratic piecewise polynomial to a Bezier path (a set of concatenated Bezier curves).

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The title does not reflect the question. –  lhf Mar 20 '13 at 19:36
Fixed that. Thanks. –  Ben-Uri Mar 20 '13 at 20:17
–  lhf Mar 20 '13 at 20:58
–  lhf Mar 20 '13 at 22:13
Just minor correction - I had submitted an edit but it was rejected for not being "substantive," and I don't have the rep to comment. To calculate the first control point the formula should be: $C=(\frac{x_1+x_2}{2},f(x_1)+f'(x_1)\cdot \frac{x_2-x_1}{2})$ Note the minor difference in computing the point's Y component, without which the formula results in an incorrect control point for segments of the parabola where $x1 \neq 0$. The conversion to a cubic Bezier works fine. –  Roland Nov 19 '13 at 19:03

Let $f(x)=Ax^2+Bx+C$ be the parabola and let $x_1$ and $x_2$ be the edges of the segment on which the parabola is defined.
Then $P_1=(x_1,f(x_1))$ and $P_2=(x_2,f(x_2))$ are the Bezier curve start and end points and $C=(\frac{x_1+x_2}{2},f(x_1)+f'(x_1)\cdot \frac{x_1+x_2}{2})$ is the control point for the quadratic Bezier curve.
Now you can convert this quadratic Bezier curve to a cubic Bezier curve by define the two control points as: $C_1=\frac{2}{3}C+\frac{1}{3}P_1$ and $C_2=\frac{2}{3}C+\frac{1}{3}P_2$.