rotating an equation about a point

Ys*

New member
Joined
Feb 9, 2017
Messages
9
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
Joined
Mar 25, 2016
Messages
1,075
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.

As a hint, you'll probably want to review your class notes about rotations, specifically seeing if there's a section about rotating about some arbitrary point. You may also find this Youtube videohttps://www.youtube.com/watch?v=FqiGuTtjmMg and/or this KhanAcademy videohttps://www.khanacademy.org/math/geometry-home/transformations/geo-rotations/v/rotating-about-arbitrary-point to be helpful.
 

HallsofIvy

Elite Member
Joined
Jan 27, 2012
Messages
4,949
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).
 
Last edited by a moderator:

Ys*

New member
Joined
Feb 9, 2017
Messages
9
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.
 
Top