Answer
$y=\displaystyle \frac{2-x}{2x-1},\ \ x\displaystyle \lt \frac{1}{2}.$
Work Step by Step
From the parametric equation for x, x varies from $\displaystyle \frac{-1}{-1-1}=\frac{1}{2}$ to $-\infty$ (as t nears 1).
So, the restriction on x is $x\displaystyle \lt \frac{1}{2}.$
Express t in terms of x:
$\displaystyle \frac{t}{t-1}=\frac{t-1+1}{t-1}=1+\frac{1}{t-1}$
$x-1=\displaystyle \frac{1}{t-1}$
$t-1=\displaystyle \frac{1}{x-1}$
$t=\displaystyle \frac{1}{x-1}+1=\frac{1+x-1}{1-x}$
$t=\displaystyle \frac{x}{x-1},\quad x\lt \frac{1}{2}.$
Substituting into the parametric equation for y,
$y=\displaystyle \frac{\frac{x}{x-1}-2}{\frac{x}{x-1}+1}=\frac{\frac{x-2(x-1)}{x-1}}{\frac{x+x-1}{x-1}}=\frac{2-x}{2x-1}$
Include the restriction
$ y=\displaystyle \frac{2-x}{2x-1},\ \ x\displaystyle \lt \frac{1}{2}.$
To graph, create a function value table using values for t, in ascending order of t, calculating the x- and y-coordinates of points on the graph.
Plot and join the points obtained with a smooth curve, noting the direction in which t increases.
