How do I find where the graph of parametric equations crosses itself? I am trying to find the value of $t$ where the graph of the following parametric equations crosses itself:
\begin{align}
x =&\; t^3 − t + 3\\
y =&\; t^2 − 3
\end{align}
I know that the next step toward solving this problem involves creating two parallel equations with a different kind of variable. 
The result will look like:
\begin{align}
t^3 - t + 3 = s^3 - s + 3 \tag1\\
t^2 - 3 = s^2 - 3 \tag2
\end{align}
After this point, I am totally stumped. I know that I will have to solve for $t$, and it looks like doing so will involve some skill I have not yet acquired. Multiplying (2) by $-t$ and adding (1) to (2) helps me to eliminate the $t^3$, but I am left with 
$$s^3 - s(t^3) - (s^2)t + t = 0 $$
an equation for which I have no idea how to solve for $t$. 
Could someone please show me how to find $t$-values at the point(s) where the graph from these equations crosses itself? Thanks in advance! 
 A: It crosses itself where, for two different values of t, you get equal x and y values. So letting $t_1 = u$ and $t_2=v$:
$$u^3-u+3 = v^3-v+3$$
$$u^2-3 = v^2-3$$
These simplify to:
$$u(u^2-1) = v(v^2-1)$$
$$u^2 = v^2$$
We know we want u and v to be different. From the second equation, we can deduce that $u = -v$. Plug this into the first equation:
$$-v^3+v+3 =  v^3-v+3$$
or 
$$2v^3-2v=0$$
$$v(v^2-1)=0$$
$v=0$ would make $u=0$ which disqualifies it so: 
$$v^2-1 = 0$$
$v\pm 1$ and $u\mp 1$ then are the two values of t where the parametric equation crosses itself.
Plug $t=1$ and $t=-1$ into your equations for x and y and you will see this is true.
A: Solving for $t$ from $y=t^2-3$ we get $t=\pm \sqrt{y+3}$. Substituting this into $x = t^3 − t + 3$ we get
$$x  = (y+3)^{(3/2)}-\sqrt{y+3}+3\quad\text{and}\quad x  = -(y+3)^{(3/2)}+\sqrt{y+3}+3.$$
These two functions (which are only defined for $y\ge-3$) intersect at
$$ y=-3\quad\text{and}\quad y=-2.$$
Substituting these back into the equations for $x$ we get
$$x=3.$$
Since the functions are undefined for $y<-3$, $(3,-3)$ is a connection point, not a crossing. The only crossing is, therefore at $(3,-2)$.
