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.

For a fraction $\frac{a}{b}$ to be equal to 0, we must have $a=0$. In this case we have $$x^2+4x+1=0$$ and the result follows.
EDIT:
Note that $$\frac{x-1}{(x+1)^2}+2-\frac{x}{x+1}=\frac{x^2+4x+1}{(x+1)^2}$$
You might wanna check your steps.
