How do you write an equation for a parametric equation given a graph and a domain for T? I'm taking a Precalculus summer course and this question came up in a practice quiz. This is the graph we are given and the domain for T is 0 less than equal to T less than equal to 2.

I know how to write a parametric equation for a graph that's infinite and I can do the same thing here. The slope is 1/3 so you replace x with t and end up with x = t and y = (T/3) + (2/3). However I don't know how to account for the domain.
 A: If we write $$\array x=t\\ y=\frac{t}{3}+\frac{2}{3} \tag{i}$$ where $t$ is a parameter such that $t \in [0,2]$,
the equation will definitely not cover the full line segment shown in the graph, for that you need $t\in [-2,4]$ but since you've already been given domain which is $[0,2]$ so you need to change the parameter $t$ with $T$ such that $T= 3t-2$ (this is a one-one onto function $T(t)$ which will give all real numbers lying in $[-2,4] $ when $t\in[0,2]$) so your new parametric equation will be $$\array x=3t-2\\ y=t$$ where $t$ is a parameter such that $t \in [0,2]$
But here at $t =0$, $(x,y)=(-2,0)$, you can see that the direction is reversed.
So, for the direction (shown in the graph) you need
$(x,y)=(4,2)$ at $t=0$ and $(x,y)=(-2,0)$ at $t=2$.
Therefore the parameter $t$  in parameter equation (i) must be replaced  with $T'$ such that $T'= -3t+4$ (this is another one-one onto function T'(t) which will give all real numbers lying in $[-2,4]$ when $t\in[0,2]$) so your new parametric equation will be $$\array x=4-3t\\ y=-t+2$$ where $t$ is a parameter such that $t \in [0,2]$
which is the required parametric equation of the directed line segment shown in the graph.
$\mathbf{Note:}$
The answer is written in the manner you can understand, otherwise it's very simple and can be easily done by parameterisation which you might not have read considering you are taking a precalculus course. (You may refer to the precise answer written by another user using parameterisation)
A: I interpret the direction of the arrow to mean the increment direction of $g$.
When $t=0$, we are at $(4,2)=2(2,1)$.
When $t=2$, we are at $(-2,0)=2(-1,0)$.
$(x,y)=(2-t)(2,1) + t(-1,0)=(4-3t, 2-t)$.
That is
$$x(t) = 4-3t$$
$$y(t) = 2-t$$
where $0 \le t \le 2$.
