How to represent half of equilateral triangle signal?
Try:
$$
x(t) = \left[ u(t-T_{\rm on}) - u(t-T_{\rm off}) \right] \cdot \frac{T_{\rm off} - t}{T_{\rm off} - T_{\rm on} }
$$

So, to explain:  


$T_{\rm on}$ is the time at which the ramp turns on.
$T_{\rm off}$ is the time at which the ramp turns off.


The piece:
$$
 \left[ u(t-T_{\rm on}) - u(t-T_{\rm off}) \right] 
$$
will be zero before $T_{\rm on}$ and one afterwards, until $T_{\rm off}$ at which time it will be zero.  Note that this assumes that $T_{\rm off} &gt; T_{\rm on}$.

That just looks like a pulse of duration $T_{\rm off} - T_{\rm on}$ with amplitude 1.

As you said in the question, the function $-t$ has the right sign of the gradient, but (possibly) the wrong slope and (possibly) the wrong intercept. For the numbers given, $T_{\rm off} = 5$ and $T_{\rm on} = 4$, the gradient is correct but the intercept is wrong.

So, the rest of the equation just fixes that:


We need our function to be one at $T_{\rm on}$.
We need our function to be zero at $T_{\rm off}$.


So 

$$
1 = m T_{\rm on} + c  \tag{1}
$$
$$
0 = m T_{\rm off} + c \tag{2}
$$

So that
$$
(1) - (2) \rightarrow 1 = m \left( T_{\rm on} - T_{\rm off} \right) \rightarrow m = \frac{1}{T_{\rm on} - T_{\rm off} } \\
$$
and substituting that into (2) gives
$$
0 = \frac{1}{T_{\rm on} - T_{\rm off} } T_{\rm off} + c \rightarrow c = \frac{ - T_{\rm off}}{T_{\rm on} - T_{\rm off} }
$$