Answer
$$\langle 4-4t,16-32t \rangle $$
Work Step by Step
Since $ r (t)=\langle t^2,t^4 \rangle, $ then $ r' (t)=\langle 2 t,4t^3 \rangle $
then, the parametrization of the tangent line at $ t=-2$ is given by
$$ L(t)=r (-2)+tr'(-2)\\
=\langle 4,16 \rangle +t \langle-4, -32 \rangle \\
=\langle 4-4t,16-32t \rangle \\
$$
