Answer
There are two tangent lines.
The tangent lines touch the curve at these points:
$(-2-\sqrt{3},\frac{1+\sqrt{3}}{2})$
$(-2+\sqrt{3},\frac{1-\sqrt{3}}{2})$
Work Step by Step
$y = \frac{x}{x+1}$
$\frac{dy}{dx} = \frac{(x+1)-x}{(x+1)^2} = \frac{1}{(x+1)^2}$
Note that $\frac{dy}{dx}$ is the slope of the graph at each point $x$
If a line through the point $(1,2)$ is tangent to the curve at $x$, then the slope of the line is $~~m = \frac{1}{(x+1)^2}$
The equation of the line is:
$y-2 = \frac{1}{(x+1)^2}(x-1)$
$y = \frac{1}{(x+1)^2}(x-1)+2$
We can find the points of intersection:
$\frac{x}{x+1} = \frac{1}{(x+1)^2}(x-1)+2$
$\frac{x}{x+1}-2 = \frac{x-1}{(x+1)^2}$
$\frac{x}{x+1}-\frac{2x+2}{x+1} = \frac{x-1}{(x+1)^2}$
$\frac{-x-2}{x+1} = \frac{x-1}{(x+1)^2}$
$(-x-2)(x+1) = x-1$
$-x^2-3x-2 = x-1$
$x^2+4x+1 = 0$
Using the quadratic formula:
$x = \frac{-4\pm\sqrt{4^2-4(1)(1)}}{2(1)}$
$x = \frac{-4\pm\sqrt{12}}{2}$
$x = -2\pm\sqrt{3}$
There are two solutions.
We can find $y$ when $x = -2-\sqrt{3}$:
$y = \frac{x}{x+1}$
$y = \frac{-2-\sqrt{3}}{-2-\sqrt{3}+1}$
$y = \frac{-2-\sqrt{3}}{-1-\sqrt{3}}\cdot \frac{-1+\sqrt{3}}{-1+\sqrt{3}}$
$y = \frac{-1-\sqrt{3}}{-2}$
$y = \frac{1+\sqrt{3}}{2}$
We can find $y$ when $x = -2+\sqrt{3}$:
$y = \frac{x}{x+1}$
$y = \frac{-2+\sqrt{3}}{-2+\sqrt{3}+1}$
$y = \frac{-2+\sqrt{3}}{-1+\sqrt{3}}\cdot \frac{-1-\sqrt{3}}{-1-\sqrt{3}}$
$y = \frac{-1+\sqrt{3}}{-2}$
$y = \frac{1-\sqrt{3}}{2}$
The tangent lines touch the curve at these points:
$(-2-\sqrt{3},\frac{1+\sqrt{3}}{2})$
$(-2+\sqrt{3},\frac{1-\sqrt{3}}{2})$
