How exactly do you do a Laplace Transform of a time delayed signal?

Question

We know that the Laplace Transform of a time-delayed signal is $$\mathscr{L} \left[ f(t-a) \cdot u(t-a) \right] = e^{-as} F(s)$$ by definition, but how exactly does that work in practice, say for the general function below? $$f(t)=c_1 t + u(t-a) (c_2(t-a))$$

Answer 1

$$ \begin{array}{rcl} \mathscr{L}[f(t)](s) &=& \displaystyle \mathscr{L}[c_1t + c_2(t-a)u(t-a)](s) \\ &=& \displaystyle c_1\mathscr{L}[t](s) + c_2e^{-sa}\mathscr{L}[t-a](s) \\ &=& \displaystyle \frac{c_1}{s^2} + c_2e^{-sa}\left(\frac{1}{s^2}-\frac{a}{s}\right) \\ \end{array} $$