# The curve has a parametric equation of x=(t^3)-12t and y=3(t^2), how would you find the two points the parameter t=2 crosses the curve?

##### 1 Answer
Oct 7, 2017

We have parametric equations:

$x = {t}^{3} - 12 t$
$y = 3 {t}^{2}$

When $t = 2$ we have:

$x = 8 - 24 = - 16$
$y = 3 \cdot 4 \setminus \setminus \setminus \setminus = 12$

So, there is only one point associated with $t$ which has cartesian coordinates $\left(- 16 , 12\right)$

The graph of the curve is as follows: