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 DN \[ g(t)=\frac{t-9}{t^{2}+81} \] \( t=\square \)

To find the vertical asymptotes of the function \( g(t) = \frac{t-9}{t^2 + 81} \), we need to determine where the denominator equals zero, as vertical asymptotes occur when the function approaches infinity. Setting the denominator equal to zero: \[ t^2 + 81 = 0 \] However, this equation has no real solutions since \( t^2 = -81 \) leads to imaginary numbers. Therefore, there are no values of \( t \) that cause the denominator to equal zero, meaning there are no vertical asymptotes for this function. So the final answer is: DN.