Answer
$y = \frac{1}{2}x + \frac{7}{2}$
$y = \frac{1}{2}x - \frac{1}{2}$
Work Step by Step
Solve for $y$:
$x - 2y = 2$
$-2y = 2 -x$
$ y = \frac{1}{2}x -1$
The slope of the line is $m = \frac{1}{2}$
Then we find the derivative to the curve:
$ y = \frac{x-1}{x+1}$
$y' = \frac{\frac{d(x-1)}{dx}(x+1) - (x-1)\frac{d(x+1)}{dx}}{(x+1)^{2}}$
$y' = \frac{(1)(x+1) - (x-1)(1)}{(x+1)^{2}}$
$y' = \frac{(x+1) - (x-1)}{(x+1)^{2}}$
$y' = \frac{x+1 - x+1}{(x+1)^{2}}$
$y' = \frac{2}{(x+1)^{2}}$
Now we use $y'$ which is the slope of a tangent line to find where $y' = \frac{1}{2}$
$ \frac{1}{2} = \frac{2}{(x+1)^{2}}$
Now multiply:
$4 = (x + 1)^{2}$
Now rationalize:
$\sqrt 4 = \sqrt (x + 1)^{2}$
$ \pm{2} = x+1$
$x = 2 - 1 = 1$
$x = -2 -1 = -3$
Now we calculate the coordinates using the two points we found:
$ y = \frac{x-1}{x+1}$
$x = -3$
$y = \frac{-3-1}{-3+1}$
$y = \frac{-4}{-2}$
$y = 2$; this becomes $(-3,2)$
$x = 1$
$y = \frac{1-1}{1+1}$
$y = \frac{0}{2}$
$y = 0$; this becomes $(1,0)$
Now we use the line equation to find the tangent line:
$y = m(x - x_{1})+y_{1}$
$(-3,2)$:
$y = \frac{1}{2}(x -(-3))+2$
$y = \frac{1}{2}x + \frac{3}{2}+2$
$y = \frac{1}{2}x + \frac{7}{2}$
$(1,0)$:
$y = \frac{1}{2}(x -1)+0$
$y = \frac{1}{2}x - \frac{1}{2}+0$
$y = \frac{1}{2}x - \frac{1}{2}$
