$x^2+y^2=x^{(\sqrt\pi)}$ I was messing around on desmos, and graphed this and cannot figure out how to flip it over the x axis. Appreciate the help
-
1What do you mean with "flip it over the x-axis"? – Kevin Dietrich Oct 24 '22 at 21:30
-
Instead of the two main points being (0,0) and (1,0) I want it to be (0,0) and (-1,0) to make it look like an infinite symbol – Kateiscursed Oct 24 '22 at 21:32
-
What is the point of this question? Just some random equation and just some random transformation? – David G. Stork Oct 24 '22 at 21:34
-
2Yes? What is wrong with that? I'm just looking for help, I don't really know that much about math in this field – Kateiscursed Oct 24 '22 at 21:36
-
Replace $x$ with $-x$ in the equation. Incidentally, I’d describe this as reflecting in the $y$-axis, not the $x$-axis. – Theo Bendit Oct 24 '22 at 21:48
-
I appreciate the help, but that doesn't work. It completely changes the graph – Kateiscursed Oct 24 '22 at 21:50
1 Answers
Instead of the two main points being (0,0) and (1,0) I want it to be (0,0) and (0,-1).
Ah that's simple.
how to do that
If you exchange the x with the - you get the graph rotated $90^{\circ}$ counter-clockwise (this is always the case). This is because $+y$ is a $90^{\circ}$ counter-clockwise rotated $+x$.:
$ y^{2} + x^{2} = y^{\sqrt{\pi}} $ is $ y^{2} + x^{2} = x^{\sqrt{\pi}} $ but rotated $90^{\circ}$ counter-clockwise.
If you now multiply all $y$ by $-1$, the graph that you rotated $90^{\circ}$ counterclockwise now rotates $180^{\circ}$ counterclockwise (this is always the case too). This is because we use it to mirror the therm about the x-axis.
$ (-y)^{2} + x^{2} = y^{2} + x^{2} = (-y)^{\sqrt{\pi}} $ is $ y^{2} + x^{2} = x^{\sqrt{\pi}} $ but rotated $90^{\circ}$ clockwise.
soluton
Aka you're soluton is $ y^{2} + x^{2} = (-y)^{\sqrt{\pi}} $
- 1,862
