the second shifting theorem
When you delay something in time — start a process two seconds late — what happens in the s-domain? The second shifting theorem is the precise answer: a delay in time turns into multiplication by an exponential e^(-as) in the transform. It is the partner of the first shifting theorem (which handled multiplying by e^(at) in time); this one handles sliding in time.
Here is the exact rule. If f(t) has transform F(s), then the delayed-and-switched-on function u(t - a) f(t - a) has transform e^(-as) F(s). Read it carefully: you take the original signal f, shift its argument by a so it starts at t = a instead of t = 0, multiply by the step u(t - a) so it is silent before then, and the payoff is just an extra factor e^(-as) out front. The reverse direction is the one you use most: seeing e^(-as) F(s) in your answer, you invert F to get f(t), then delay it — write u(t - a) f(t - a). The trickiest part in practice is rewriting your forcing so its argument really reads (t - a), not t.
This theorem is the engine that lets the Laplace method handle forcing that switches on partway through — a relay that closes, a switch thrown, a pulse delivered late. Without it, piecewise and delayed inputs would be a mess of cases; with it, every delay is just a clean exponential tag you carry along.
The transform of u(t - 3) (t - 3)^2 is e^(-3s) · 2/s^3, because the transform of t^2 is 2/s^3 and the delay a = 3 simply attaches e^(-3s).
Delay by 3 in time ⟶ multiply by e^(-3s) in the s-domain; the argument must read (t - 3).
The function inside must be expressed in terms of (t - a), not t. The transform of u(t - 2) t^2 is NOT e^(-2s) · 2/s^3; first rewrite t^2 = (t - 2)^2 + 4(t - 2) + 4, then shift each piece.