What is the proper way to study curves defined parametrically? For the sake of convenience, say the curve defined by the set of equations: \begin{equation} \begin{cases} x(t)&=t^4+4t\\ y(t)&=t^4-2t^2 \end{cases} \end{equation} If we can eliminate the parameter $t$ and express $y=f(x)$, one can use one-variable calculus but usually the elimination is a difficult task, if not an impossible one. So, one should use a general method that does not need parameter elimination. Intuitively, I take the first derivatives \begin{equation} \left\lbrace\begin{array}{lll} x'(t)&=4t^3+4&=4(t+1)(t^2-t+1)\\ y'(t)&=4t^3-4t&=4t(t-1)(t+1) \end{array}\right. \end{equation} Setting first derivatives equal to zero we get \begin{equation} \left\lbrace\begin{array}{lll} x'(t)&=0\Rightarrow t&=-1\\ y'(t)&=0\Rightarrow t&=\{-1,0,1\} \end{array}\right. \end{equation} Taking the second derivatives \begin{equation} \left\lbrace\begin{array}{ll} x''(t)&=12t^2\\ y''(t)&=12t^2-4t \end{array}\right. \end{equation} Replacing $t=-1$ we get \begin{equation} \left\lbrace\begin{array}{lll} x''(t)&=12&>0\\ y''(t)&=8&>0 \end{array}\right. \end{equation} So I think there is a minimum at $(x,y)=(-3,-1)$. For $t=0$ we get $y''=-4<0$. There is a maximum at $(x,y)=(0,0)$. For $t=1$ we get $y''=8<0$. There is a minimum at $(x,y)=(5,-1)$. The following figure obtained in Matlab supports my reasoning. I guess that $t=-1$ giving a minimum for both $x$ and $y$ explains the cusp-like look at $(-3,1)$.
What I am missing is a sound theoretical basis for previous analysis. Thanks a lot.
Hint: $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$