me: damn, how do i find a cubic that crosses the x-axis at these three points
me:
me: oh lmao
troublesome fox girl
- she/her
- eev.ee/
hello i like to make video games and stuff and also have a good time on the computer. look @ my pinned for some of the video games and things. sometimes i am horny on @squishfox
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eevee.itch.io/bluesky
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in reply to @lexyeevee's post:
i assume the joke is that you just wanted the line y = 0? x)
but ofc there's an infinite number of cubic curves that satisfy that; like, any cubic bezier A,B,C,D for which A=X₁, D=X₃, and B & C are on the y=x₃ line, on on opposite sides of and equidistant from the x-axis
...im sure you already know that i just really like bezier curves
edit: i don't think they even need to be equidistant, i just woke up and am sleepy
most (nearly all?) cubic beziers on a cartesian plane will not actually be functions though
aside from a scaling factor, there is exactly one cubic fitting this condition, and it is... y = (x - a)(x - b)(x - c)
oh i didn't realize you were looking for a function x) i usually hang out in parametric land
i guess actually this did end up being a parametric thing but i did the orthogonal and perpendicular directions completely separately haha
I thought the joke is that it’s just y = D(x-A)(x-B)(x-C)?
Where D is any nonzero value that scales the amplitude of the curve (and could be omitted entirely), and A,B,C are the X-axis intercepts, since the right-hand side will evaluate to 0 when X is equal to any of A,B,C?
edit: ah, eevee beat me to it
thanks! i sketched the basic design myself, but all credit for making it actually pretty goes to my tattoo artist 
