Answer
This function has no vertical asymptote.
Work Step by Step
$g(t)=\dfrac{t-1}{t^{2}+1}$
To find the vertical asymptote of this function, set its denominator equal to $0$ and solve for $t$:
$t^{2}+1=0$
Take $1$ to the right side:
$t^{2}=-1$
Take the square root of both sides:
$\sqrt{t^{2}}=\sqrt{-1}$
$t=\sqrt{-1}$
Since $\sqrt{-1}$ is not a real number, this function has no vertical asymptote.
