<s>[INST] Determine the value of $k$ such that the line determined by the points $(3,2)$ and $(1,-2)$ is tangent to the graph of $y=\frac{k}{x+1}$. I have done the following work but I am stuck on solving for k. Am I doing this right? If so what do I do next?

 [/INST] We have 
$1.$ $k/(x+1)=2x-4$
$2.$ $-k/(x+1)^2=2$
First equation
Multiplying it by $(x+1)$
$k=(x+1)\cdot (2x-4)=2x^2-4x+2x-4=2x^2-2x-4 \quad (1)$
Second equation
Multiply it by $(x+1)^2$
$-k=2(x+1)^2=2x^2+4x+2 \Rightarrow k=-2x^2-4x-2 \quad (2)$
Set $(1)$ equal to $(2)$
$2x^2-2x-4=-2x^2-4x-2$
$4x^2+2x-2=0$
$2x^2+1x-1=0$
Solve for $x$. This gives $x_1=-1$ and $x_2=0.5$ 
$x_1$ is not in the domain of $y=k/(x+1)$.
Now insert $x=0.5$ in  $k/(x+1)=2x-4$ to obtain the value of $k$

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