Sketching a tri signal with horizontal shift
The peak of the tri signal is at the point where the argument of tri (that thingy inside the parentheses following tri) equals $0$. The
base of the tri signal extends from the point where the argument equals
$-\frac 12$ to the point where the argument equals $+\frac 12$ (or $-1$ to $+1$ depending on what you are calling tri).  Thus, 
tri$\left(\frac{t-t_0}{T}\right)$ is an isoceles triangle whose apex
is at the point $(t_0,1)$ and whose base extends from $t_0-\frac{T}{2}$
to $t_0+\frac{T}{2}$ (or from $t_0-T$ to $t_0+T$ for the other
interpretation of tri).