<s>[INST] 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.
 [/INST] 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
$$a=-2\pm \sqrt{3}$$
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