## Calculus 10th Edition

$g(t)=\dfrac{t-1}{t^{2}+1}$ 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.