The problem I am working on is to find the where the curve intersects itself, using the parametric equations.
These are: $x=t^2-t$ and $y=t^3-3t-1$
For the graph to intersect itself, there must be two distinct t-values, $a$ and $b$, that when plugged into the parametric equations, produce the same output. These two t-values create two ordered-pairs that are the same.
My system of equations: $a^2-a=b^2-b$ and $a^3-3a-1=b^3-3b-1$.
I solved for $a$, but am not sure if I did it correctly: $a(a-1)=b^2-b$ then either $a=b^2-b$ or $a-1=b^2-b$. I would then have two values that I have to test. When I plugged in $a$, I ended up with a 6-degree polynomial, did I do something wrong?