How do I rotate a graph of a function around a point, and show it in the related equation?
An example could be $f(x)=\lvert x\rvert$ (absolute Value) and $f(x)=x^2$
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Graham Kemp
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2The problem is that the resulting equation may no longer be a function. However, in some cases - as in $f(x)=x^2$ - you can take the inverse and write it as $y_2=\pm \sqrt{x}$. – HDE 226868 Jun 08 '15 at 23:09
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3It is very unclear what you are asking. Do you mean reflecting the graph across a horizontal or vertical line? – TravisJ Jun 08 '15 at 23:11
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No, I mean rotating the graph f(x)=x^2 25 degrees around 0,0 – Danish Exchangestudent Jun 08 '15 at 23:21
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1@DanishExchangestudent: Then it won't be the graph of a function anymore. You can certainly rotate the solution set of the equation $y=x^2$, but calling it a function in that case is a bit confusing. – hmakholm left over Monica Jun 08 '15 at 23:24
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1Okay, I get that now. – Danish Exchangestudent Jun 08 '15 at 23:44
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But how do I rotate a graph? – Danish Exchangestudent Jun 08 '15 at 23:44
3 Answers
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In case you are literally after a rotation of the graph of a function as a set, this works:
The graph of the function $f(x)=x^2$ is given by: $$ \{(x,x^2):x\in\mathbb{R}\} $$
Rotation by a degree $\alpha$ is given by the matrix: $$ \left( \begin{array}{cc} \cos\alpha&-\sin\alpha\\ \sin\alpha&\cos\alpha \end{array} \right) $$
Applying this matrix to the above set, you get $$ \{(x\cos\alpha-x^2\sin\alpha,x\sin\alpha+x^2\cos\alpha):x\in\mathbb{R}\} $$
But as mentioned already, this is not the graph of a function anymore.
Fabian
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If you don't care about whether the result is a graph of a function, then you can certainly rotate the set of points that satisfy $y=x^2$.
The general plan is to start by writing down the correspondence between new and old coordinates. If you're rotating by 25° around $(0,0)$, for example, it would be something like
$$ \begin{align} x &= \cos(25^\circ)x' - \sin(25^\circ)y' \\ y &= \sin(25^\circ)x' + \cos(25^\circ)y' \end{align} $$ were $x'$ and $y'$ are new coordinates. You can then insert into your original equation and get $$ \sin(25^\circ)x' + \cos(25^\circ)y' = (\cos(25^\circ)x' - \sin(25^\circ)y')^2 $$
hmakholm left over Monica
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To rotate the curve $y=f(x)$ around point $\langle a, b\rangle$ by angle $\theta$.
First express the curve as a parameterised vector equation. $$\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix} t \\ f(t)\end{pmatrix}$$ Second, translate and rotate by matrix multiplication:
$$\begin{pmatrix}x_2 \\ y_2\end{pmatrix} =\begin{bmatrix}\cos \theta & -\sin \theta\\\sin\theta & \cos\theta\end{bmatrix}\times \begin{pmatrix} t-a \\ f(t)-b\end{pmatrix}+\begin{pmatrix} a \\ b\end{pmatrix}$$
Graham Kemp
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