Originally Posted by

**Jason76** I came up with (-2,2) for the horizontal tangent and (0,-3) for the vertical one (but they are wrong on homework) ???

Horizontal

$\displaystyle y = t^{2} - 3$

$\displaystyle \dfrac{dy}{dt} = 2t$

$\displaystyle 2t = 0$

$\displaystyle t = 0$

$\displaystyle y = (0)^{2} - 3 = -3$

$\displaystyle x = (0)^{3} - 3(0) = 0$

$\displaystyle (0,-3)$

Vertical

$\displaystyle x = t^{3} -3t$

$\displaystyle \dfrac{dx}{dt} = 3t^{3} - 3t $

$\displaystyle 3t^{2} - 3 = 0$

$\displaystyle t = 1$

$\displaystyle y = (1)^{2} - 3 = -2$

$\displaystyle x = (1)^{3} - 3(1) = -2$

$\displaystyle (-2,-2)$