# Quadratic Rotation

The curve $y=x^2âˆ’6x+11$ is rotated through $180^\circ$ about the origin. What is the equation of the new curve?

## Problem

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The curve $y=x^{2}-6x+11$ is rotated through $180^\circ$ about the origin.

What is the equation of the new curve?

If you liked this problem, here is an NRICH task which challenges you to use similar mathematical ideas.

## Student Solutions

**Answer**: $y=-x^{2} - 6x - 11 $, which is the same as $y= -(x + 3)^{2} - 2$.

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**Completing the square**

New graph is a negative quadratic with maximum $(-3,-2)$

The curve hasn't changed shape so $y=-x^2+...$ (not $y=-2x^2+...$ etc)

$\therefore$ equation is $y=-(x+3)^2-2$

**Comparing coefficients**

New graph has equation $y=ax^2+bx+c$

The curve hasn't changed shape but the quadratic is negative so $a=-1$

Intercept is $-11\Rightarrow c=-11$

$y=-x^2+bx-11$

And $(-3,-2)$ is on the graph so

$-2=-(-3)^2+b(-3)-11\\

\Rightarrow -2=-9-3b-11\\

\Rightarrow 18=-3b\\

\Rightarrow b=-6$

So the equation is $y=-x^2-6x-11$

**Using $-x$ and $-y$**

Rotation of $180^\circ$ about $(0,0)$ is the same as swapping $x$ for $-x$ and $y$ for $-y$

Swap $x$ for $-x$ and $y$ for $-y$ in the equation $y=x^2-6x+11:$

$$\begin{align} -y&=(-x)^2-6(-x)+11\\

\Rightarrow -y&=x^2+6x+11\\

\Rightarrow y&=-x^2-6x-11\end{align}$$