Question

Pre-Calculus? Parametric Functions?

Answer 1

Notice that if you just plug in $x=t$ into the equation for $y$, you end up with

$\begin{cases}x=t\\y=3t^2-4\end{cases}\implies y=3x^2-4$

You have to do the same thing with the given choices. For example, if

$\begin{cases}x=2t\\y=12t^2-4\end{cases}$

then we see that $t=\dfrac x2$, and plugging this into the second equation gives

$y=12\left(\dfrac x2\right)^2-4=\dfrac{12x^2}4-4=3x^2-4$

which matches the original set of parametric equations.

So the general strategy is to eliminate the parameter $t$ by solving for it in each $x(t)$ equation. Then substitute this result for the $t$ in the corresponding $y(t)$ equation, and see if it reduces to the same equation at the top.