The Laplace transform works by multiplying a function by the decaying exponential \( e^{-st} \) and adding up the result over all time, so it is really testing how the function behaves when weighted by exponentials of different decay rates \( s \). If the function itself already carries an extra decaying factor \( e^{-at} \), then multiplying it by \( e^{-st} \) is exactly the same as multiplying the original, unmodified function by a faster-decaying exponential \( e^{-(s+a)t} \), since the two exponential factors simply combine into one. In other words, tacking on \( e^{-at} \) in the time domain has the same overall effect as replacing every occurrence of \( s \) with \( s + a \) inside the transform formula for \( f(t) \) alone. This means the new transform is just the old transform function evaluated at a shifted argument, giving \( F(s+a) \) rather than \( F(s-a) \) or any sign-flipped version. So the correct choice is option (D).