Answer
Vertical asymptote at $t=-2.$
Work Step by Step
$h(t)=\dfrac{t^2-2t}{t^4-16}=\dfrac{t(t-2)}{(t^2+4)(t^2-4)}=\dfrac{t(t-2)}{(t^2+4)(t-2)(t+2)}$
$=\dfrac{t}{(t^2+4)(t+2)}.$
Vertical asymptotes occur when the denominator alone is $0\to$
$(t+2)(t^2+4)=0\to t=-2.$
