graphing-functions
I'm only a seventh grader, so this question might seem a little easy, and my vocabulary isn't the best either.
I using Desmos and my graph is here.
Basically, I have two spiral looking shapes across from each other. I want to have the same shape, but instead of the spiral facing the y axis, I want it to face the x axis. Basically, I want to make my creation four-sided.
How's this?
$$\left(x-a-\sqrt{r^2-y^2}\right)\left(x+a+\sqrt{r^2-y^2}\right)(y-r)(y+r)=10^{-n}$$
$a$ and $r$ here are exactly the same parameters as in the diagram you included. $n$ is a "quality factor" of sorts; the larger it gets, the sharper the "corners" of the stadium get (i.e. the points $(\pm\frac{a}{2},\pm a)$). When $n=5$, the graph is nearly indistinguishable from when $n=\infty$ (i.e. when the right side of the equation is $0$).
Here's the Desmos page I used to make the graph: https://www.desmos.com/calculator/1hjfojqisv
When $a=r=4$ and $n=5$, the graph of this equation looks like this:
Funny how this site works sometimes.
I was doing a little research on Desmos for this completely unrelated question when I realized that the method of restricting domain and range I ended up using to answer that question would also work perfectly here. I had to switch to the rectangular versions of the petal equations, but that's a small price to pay.
Just as a quick example, I ended up graphing $$x^2+y^2=1.21$$ $$\begin{align} \frac{(x\cos\phi_1-y\sin\phi_1-h_1)^2}{a_1^2}+\frac{(y\cos\phi_1+x\sin\phi_1-k_1)^2}{b_1^2}&=c_1^2\{x\le-0.5\}\{y\ge0.96465\} \\ \frac{(x\cos\phi_1-y\sin\phi_1-h_1)^2}{a_1^2}+\frac{(y\cos\phi_1+x\sin\phi_1-k_1)^2}{b_1^2}&=c_1^2\{x\ge-0.49999\}\{y\ge1.09571\} \\ \frac{(x\cos\phi_2-y\sin\phi_2-h_2)^2}{a_2^2}+\frac{(y\cos\phi_2+x\sin\phi_2-k_2)^2}{b_2^2}&=c_2^2\{x\le-1.29445\} \\ \frac{(x\cos\phi_2-y\sin\phi_2-h_2)^2}{a_2^2}+\frac{(y\cos\phi_2+x\sin\phi_2-k_2)^2}{b_2^2}&=c_2^2\{x\le-0.99377\}\{y\le0.47522\} \\ \frac{(x\cos\phi_2-y\sin\phi_2-h_2)^2}{a_2^2}+\frac{(y\cos\phi_2+x\sin\phi_2-k_2)^2}{b_2^2}&=c_2^2\{x\ge-0.72153\}\{y\ge0.84065\} \end{align}$$
to obtain the center and the first two petals, and the result looked exactly like what I had hoped for:
I wouldn't even dream of typing out all $40$ equations I ended up graphing, but here's the final result, which looks considerably nicer than what I started with:
And here's the link to the final version, in case anyone's curious.
Best Answer
It seems that A---B might have forgotten about this question, but as someone who used to play with graphing calculators as well around your age, I admire your enthusiasm and would be happy to pick up where they left off!
I don't want to confuse you with exactly why the following is true, but I'll be happy to explain more if you'd like, and as you continue learning more mathematics, you'll understand it soon enough.
Simply put, to rotate any graph in the $xy$-plane by an angle $\phi$, make the following substitutions: $$\begin{align} x &\mapsto x\cos\phi-y\sin\phi \\ y &\mapsto x\sin\phi+y\cos\phi \end{align}$$
As for what this would look like in practice, say you wanted to rotate the graph of $y=x^2$ by $45^{\circ}$. To graph both the original parabola and the rotated parabola, you would enter the following into Desmos: $$\begin{align} y&=x^2 \\ x\sin\left(\frac{\pi}{4}\right)+y\cos\left(\frac{\pi}{4}\right)&=\left(x\cos\left(\frac{\pi}{4}\right)-y\sin\left(\frac{\pi}{4}\right)\right)^2 \end{align}$$
I've converted the angle $45^{\circ}$ into radians here, because that's just the way Desmos wants it; are you familiar with radians?
Anyway, lo and behold, we end up with the following graph:
For even more fun with rotating graphs, you can just leave $\phi$ (phi) in the equation, add a slider for it when prompted, and drag it around to watch what happens!