## Calculus: Early Transcendentals 8th Edition

Horizontal tangent at $(0,-3)$ Vertical tangents at $(2,-2)$ and $(-2,-2)$ Graphed with desmos:
$\displaystyle \frac{dy}{dx}=\frac{dy/dt}{dx/dt}$ The tangent is horizontal for $\displaystyle \frac{dy}{dt}=0$ $y=t^{2}-3$ $\displaystyle \quad \frac{dy}{dt}=2t,$ $2t=0\Rightarrow t=0$ When $t=0, (x,y)=(0,-3)$ The tangent is vertical for $\displaystyle \frac{dx}{dt}=0$ $x=t^{3}-3t$ $\displaystyle \quad \frac{dx}{dt}=3t^{2}-3=3(t+1)(t-1),$ $t=-1$ or $1$ When t=1, $(x,y)=(2,-2)$, and when t=$-1, (x,y)=(-2,-2).$