Finding tangent lines to a curve 
Find the equations of all tangent lines to the curve $y=x/(x+1)$ that intersect the point $(1,2)$. Note that the point $(1,2)$ does not lie on the curve. Please simplify your final answer as much as possible.

-I am so lost in trying to solve this question. I found the derivative of $f(x)$ which is $1/(x+1)^2$ which is equal to the slope. Then I wrote the equation in the $y=mx +c$ format to solve for an equation.
I got the equation $y=[1/(x+1)^2]x +1.75$
I have a feeling this is incorrect and do not know what to do next. Am I approaching this question right? 
 A: The derivative is correct.
As stated in the execise, the point $(1,2)$ is not on the curve; so you need to find such a point $(x_0,y_0)$ which a) lies on the curve; b) the tangent line to the curve in this point passes through $(1,2)$. Are these hints sufficient?
Edit
Our line is described by the equation $y=kx+c$. It is tangent to the curve int he point $(x_0,y_0)$, thus $y_0=kx_0+c$; in addition, the derivatives of the line and of the curve must coincide in this point, i.e. $k=\frac{1}{(x_0+1)^2}$. Finally, the line passes though $(1,2)$, hence $2= k+c$.
Thus, we obtain the equations:
$$y_0=kx_0+c\\y_0=x_0/(x_0+1)\\k=\frac{1}{(x_0+1)^2}\\2= k+c$$
Can you solve these equations to obtain all possible pairs $(k,c)$?
A: The general form of a line which intersects the point $\left(1,2\right)$ is given by $y = 2 + a\left(x - 1\right)$ where $a$ is the line slope. The line and curve intersection is determined by
$$
2 + a\left(x - 1\right) = x/\left(x + 1\right)
$$
Since the line slope $a$ must be equal to the curve slope at the intersection point, we have
$$
a
=
{{\rm d} \over {\rm d}x}\left(x \over x + 1\right)
=
{1 \over \left(x + 1\right)^{2}}
$$
In solving those equations we find two values of
$a\ \left(7 \mp 4 \sqrt{3\,}\right)$ which yield two lines:
$$
\begin{array}{|lcr|}\hline\\
\quad\color{#ff0000}{\large y}
& \color{#000000}{\large\ =\ } &
\color{#ff0000}{\large%
2 + \left(7 - 4\sqrt{3\,}\,\right)\left(x - 1\right)\quad} 
\\[2mm]
\quad\color{#ff0000}{\large y}
& \color{#000000}{\large\ =\ } &
\color{#ff0000}{\large%
2 + \left(7 + 4\sqrt{3\,}\,\right)\left(x - 1\right)\quad}
\\ \\ \hline
\end{array}
$$
A: A slightly different way of attacking the problem:
Let $(X,Y)$ be a point on the given curve, and also on a straight line through  $(1,2)$
Since $(X,Y)$ is on the curve, we have Equation #$1$ $$Y=\frac{X}{X+1}$$
Using your derivative, the slope of the curve at $(X,Y)$ is:$$\text{Curve Slope} =\frac{1}{(X+1)^2} $$The slope of a straight line through $(X,Y)$ and $(1,2)$, $\text{m}$, is given by:$$\text{m}=\frac{Y-2}{X-1}$$Setting the curve slope equal to the slope of the straight line. we get Equation #$2$:$$\frac{1}{(X+1)^2}=\frac{Y-2}{X-1}$$If we use Equation#$1$ to eliminate $Y$ from Equation #$2$ we get an equation in $X$ that can be re-arranged into a quadratic.
EDIT:$$\frac{1}{(X+1)^2}=\frac{\frac{X}{X+1}-2}{X-1}=\frac{X-2X-2}{(X+1)(X-1)}$$ $$\frac{1}{(X+1)}=\frac{-X-2}{(X-1)}$$
Solve the quadratic, sub back into #$1$ to get $Y$.  Each $(X,Y)$ combined with $(1,2)$ gives two points on a straight line, and thus the equation of the two tangent lines.  
