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## 'Quadratic Rotation' printed from http://nrich.maths.org/

$x^{2} - 6x + 11 = (x - 3)^{2} + 2$.

When the curve is rotated $180^\circ$ about the origin, the equation of the new curve will be

$y = -(x + 3)^{2} - 2$

$= -x^{2} - 6x - 9 - 2$

$= -x^{2} - 6x - 11$.

Note: the image of point $(a,b)$ under a $180^\circ$ rotation about the origin is the point $(-a,-b)$. An alternative method, therefore, is to replace $x$ and $y$ in the original equation by $-x$ and $-y$ respectively.

*This problem is taken from the UKMT Mathematical Challenges.*

*View the archive of all weekly problems grouped by curriculum topic*