How many tangent lines to the curve $y = \frac{x}{x+1}$ pass through the point $(1,2)$? The question:
How many tangent lines to the curve $y = \frac{x}{x+1}$ pass through the point $(1,2)$? At which points do these tangent lines touch the curve?
My attempt:
To find the slope of the tangent line to the curve, we need to get the first derivative of $y = \frac{x}{x+1}$, which is $y' = \frac{1}{(x+1)^2}$.
So, the line equation of the tangent line is $y = m(x-1) + 2$, where $m = \frac{1}{(x+1)^2}$ for the given $x$.
We also know that the point where the tangent line intersects the curve must be both on the tangent line and the curve, so using that,
$$\frac{x}{x+1} = \frac{x-1}{(x+1)^2} + 2$$
$$\frac{x-1}{(x+1)^2} + 2 - \frac{x}{x+1} = 0$$
$$\frac{x - 1 + 2x^2 + 4x + 2 - x^2}{(x+1)^2} = 0$$
$$\frac{x^2 + 5x + 1}{(x+1)^2} = 0$$
And I'm stuck. I'd appreciate if you can point out where I'm wrong and how to better approach this problem.

hint
The equation of the tangent line to the curve at the point $ (a,\frac{a}{a+1}) $ is
$$y=\frac{1}{(a+1)^2}(x-a)+\frac{a}{a+1}$$
it passes through the point $ (1,2) $ if
$$2=\frac{1}{(a+1)^2}(1-a)+\frac{a}{a+1}$$
So,
$$2(a+1)^2=(1-a)+a(a+1)$$
or
$$a^2+4a+1=0$$
thus
$$a=-2\pm \sqrt{3}$$
