Answer
Horizontal : $(0,-3)$
Vertical : $(2,-2)$ and $(-2,-2)$
Work Step by Step
1. Calculate $\frac{dy}{dt}$ and $\frac{dx}{dt}$:
$\frac{dy}{dt} = \frac{d(t^2 - 3)}{dt} = 2t$
$\frac{dx}{dt} = \frac{d(t^3 - 3t)}{dt} = 3t^2 - 3$
2. The tangent is horizontal when $\frac{dy}{dt} = 0$
$2t = 0$
$\frac {2t} 2 = \frac 0 2$
$t = 0$
Cartesian coordinates: $x = (0)^3-3(0) = 0;y = (0)^2-3 = -3$
3. The tangent is vertical when $\frac{dx}{dt} \approx 0$:
$3t^2 -3 =0$
$3(t^2 - 1) = 0$
$\frac {3(t^2-1)}{3} = \frac 0 3$
$t^2 - 1 = 0$
$t^2 = 1$
Therefore, $t \approx 1$ or $t \approx -1$.
Cartesian coordinates: $x = (1)^3-3(1) = -2;y = (1)^2-3 = -2$
Cartesian coordinates: $x = (-1)^3-3(-1) = 2;y = (-1)^2-3 = -2$
