Laplace Transform of a Heaviside function Find the Laplace transform.
$$g(t)= (t-1) u_1(t) - 2(t-2) u_2(t) + (t-3) u_3(t)$$
I understand that the $\mathcal{L}\{u_c(t) f(t-c)\} = e^{-cs}*F(s)$
Finding $F(s)$ is the hard part for me. My professor has used, for example, 
$$f(t-2)=t^2$$
let $$s = t-2$$
$$t= s+2$$
$$f(s) = (s+2)^2$$
therefore $f(t) = (t+2)^2$
But then he said that $f(t-2) = 1$ therefore $f(t) = 1$. But why/how?
By the previous logic if you let $s = t-2$ then $t= s+2$, and $f(s) = s+2$, so $f(t) = t+2$ not $1$.
I'm having a tough time figuring this out.

Your function is $f(t)=t$ but not $f(t)=1$.
You need to use the formula 
$$\mathcal{L}\{u_c(t) f(t-c)\} = e^{-cs}*F(s)$$
3 times with $c=1,2, 3$ to get the answer.
The answer derived by Leucippus looks correct. 
