rotating an equation about a point

Ys*

New member
I'm trying to rotate the equation below about the point (p, q). The equation below is similar to y = x^2 however it's been altered to allow me to slightly alter it. Any help?

f(x) = (r(x+p)^2.8+q)
OR
y = f(x)

ksdhart2

Senior Member
What have you tried? What are your thoughts? Please comply with the rules as laid out in the Read Before Postinghttps://www.freemathhelp.com/forum/threads/54004-Read-Before-Posting thread that's stickied at the top of every sub-forum, and share with us any and all work you've done on this problem, even including the parts you know for sure are wrong. Thank you.

HallsofIvy

Elite Member
To rotate (x, y) through angle $$\displaystyle \theta$$ about the origin, do the matrix multiplication $$\displaystyle \begin{bmatrix}cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta)\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}$$.

To rotate (x, y) through angle $$\displaystyle \theta$$ about a point (a, b) that is not the origin
1) translate (a, b) to the origin using (x, y)-> (x- a, y- b)
2) do the matrix multiplication above to get the new (x, y)
3( translate back using (x, y)-> (x+ a, y+ b).

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Ys*

New member
I figure this out on my own. I was learning this for a personal program I'm coding. Let's just say this is quite a bit more complex than I'm letting on. Thanks for the resources anyways.