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Question
- -/1 points details my notes ask your teacher practice another find the vertical asymptotes (if any) of the graph of the function. (enter your answers as a comma - separated list. if an answer does not exist, enter dne.) \\( h(t) = \frac{t^2 - 3t}{t^4 - 81} \\) \\( t = \\) resources read it watch it tutorial
Step1: Set denominator to 0
$t^4 - 81 = 0$
Step2: Factor denominator
$(t^2 - 9)(t^2 + 9) = 0$
$(t-3)(t+3)(t^2+9)=0$
Step3: Solve for t
$t-3=0 \implies t=3$; $t+3=0 \implies t=-3$; $t^2+9=0$ has no real solutions.
Step4: Check numerator at t-values
For $t=3$: $3^2 - 3(3)=9-9=0$; simplify function: $\frac{t(t-3)}{(t-3)(t+3)(t^2+9)}=\frac{t}{(t+3)(t^2+9)}$ (t≠3), so t=3 is a hole, not asymptote.
For $t=-3$: $(-3)^2 - 3(-3)=9+9=18≠0$, so t=-3 is a vertical asymptote.
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$t=-3$