Sketching a tri signal with horizontal shift I'm unsure how to sketch a tri signal with horizontal shift.

Eg , let's say we want to sketch tri((t-4)/4). Can I reduce it to tri( (t/4) - 1 )? This would mean sketching a tri(t/4) signal and shifting it to the right by 1. Or should it be a tri(t/4) signal shifted to the right by 4?

I've found out on wikipedia that the rect signal is defined as rect((X-Y)/T) where you shift the center of the signal by varying Y. But I'm not sure if sketching tri applies the same concept.
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).