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.
This is what I have done so far. I don't know if it is correct.
1/(x+1)= -x-2/(x-1) [cross-multiplied] to get, x-1=-x^2 -3x -2
I made it equal to zero to get a quadratic equation of x^2 + 4x + 1 = 0 to solve for x
I got x = -2 +- root 3
I dont know what to do next.
Best Answer
Pick some point $(a,b)$ which lies on the curve $y=\frac{x}{x+1}$. This means that $b=\frac{a}{a+1}$. Taking the derivative using the quotient rule, we find:
$$y'(x)=\frac{1}{(1+x)^2}$$
so the slope of the tangent line at our point $(a,b)$ is
$$y'(a)=\frac{1}{(1+a)^2}$$
The point slope formula then tells us that the equation of the tangent line to the curve at the point $(a,b)$ is
$$y-b=\frac{1}{(1+a)^2}(x-a)$$
We want the point $(1,2)$ to satisfy this equation, so remembering that $b=\frac{a}{a+1}$, we have:
$$2-b=2-\frac{a}{a+1}=\frac{1}{(1+a)^2}(1-a)$$
This equation reduces to the quadratic $a^2+4a+1=0$ (which is the quadratic you found). The solutions $a=-2\pm\sqrt{3}$ can be plugged back into the equation
$$y=\frac{1}{(1+a)^2}(x-a)+\frac{a}{1+a}=\frac{1}{(1+a)^2}x+\frac{a^2}{(1+a)^2}$$
to obtain the equations of the desired tangent lines.
Here is a helpful picture plotted with Wolfram|Alpha:
Notice where the tangent lines intersect!