Creating a Parametric Equation The problem is as follows:
"In each of the following cases, you will be asked to write down a family of parametric curves that have the property that at $$t = 1$$ we have $$x'(t) = y'(t) = 0$$ but the slope of the curve has the property listed here:
a. Horizontal tangent.
b. Vertical tangent.
c. A slope of $4.2$.
d. Slope of the curve is undefined."
So far I have figured out: $x(t)$ and $y(t)$ need to be a cyclical function with some kind of polynomial (I think) whose derivative will subtract to $0$ when evaluated at $1$. So for problem a, Once a function for $y(t)$ and $x(t)$ is found that satisfies the conditions given. $$\frac{d^2y}{d^2t}$$ Will need to be $0$ and $$\frac{d^2x}{d^2t}$$ will need to be equal to any non $0$ number. As the derivative $dy/dx$ will equal $0$ and you would need to use l'hopital's rule. With this information I am having trouble finding parametrics that satisfy al of these conditions.
 A: Because $x'(1)=y'(1)=0$, we can assume that
$$x(t)=a(t-1)^p,\quad y(t)=b(t-1)^q\quad(a,b\in\mathbb R,\quad 2\le p,q\in\mathbb N)$$
There can be other functions that have $x'(1)=y'(1)=0$ but let's just simply assume polynomials as they can also do the job.
I also claim that $p$ and $q$ should be odd numbers (so that $x(t)$ and $y(t)$ would be odd functions with respect to $t=1$) to make the slope defined, because if one is odd and the other is even then the curve's slope will change its sign between $t=1^-$ and $t=1^-$ (like $y=|x|$) unless the even side has zero slope, so the slope won't be defined (will be useful for the Q#4 though).
If both $p$ and $q$ are even, then the curve will bounce back at $t=1$ and will look like it just ended at $(x(1),y(1))$.
So I came to the answers #1 through #3. ($p,q\ge 2$ and should be all odd numbers).
\begin{align}
&\text{1. }(x(t),y(t))=((t-1)^3,(t-1)^5)\\
&\text{2. }(x(t),y(t))=((t-1)^5,(t-1)^3)\\
&\text{3. }(x(t),y(t))=((t-1)^3,4.2(t-1)^3)\\
&\text{4. }(x(t),y(t))=((t-1)^3,(t-1)^2)\\
\end{align}
Note that to make the slope ($=\frac{dy/dt}{dx/dt}$) either zero or infinite, I assigned bigger odd number for $x(t)$ or $y(t)$ than the other side.
Finally, regarding the #4, you can choose one of $p$ or $q$ as an even number and (I really suggest you to draw the graph) it will look like the graph of $y=\sqrt{|x|}$.
For a good graphing website, I recommend here https://www.desmos.com/calculator.
Try to type in "((t-1)^2,(t-1)^3)" to an empty panel and be sure to set the range of $t$ wider (it's set to $[0,1]$ by default so you won't see the whole curve unless you change it.)
