Find the equation of the tangent to $ \ y= \frac{1}{x-4} $ at the point L where $\ x=b \ $ This is what I've done so far: 
$$y'  \ = \  -\frac{1}{(x-4)^ 2} \ \ \rightarrow \ \
y' (b) \ = \ -\frac{1}{( b-4)^2 } $$
Not sure what to do after this. Am I doing it right?

Notice, we have $$y=\frac{1}{x-4}$$ substituting $x=b$, we get y-coordinate of the point L as follows  $$y=\frac{1}{b-4}$$
The slope of the tangent  at general point
$$\frac{dy}{dx}=\frac{-1}{(x-4)^2}$$ Hence, the slope at the point $L\left(b, \frac{1}{b-4} \right)$
$$\frac{dy}{dx}=\frac{-1}{(b-4)^2}$$
The equation of a atraight line passing through the point $(x_1, y_1)$ & having a slope $m$ is given as $$\color{blue}{y-y_1=m(x-x_1)}$$
Hence the equation of the tangent passing through the point $L\left(b, \frac{1}{b-4} \right)$ & having slope $\frac{-1}{(b-4)^2}$ $$y-\frac{1}{b-4}=\frac{-1}{(b-4)^2}(x-b)$$ $$(b-4)^2y-(b-4)=-x+b$$ 
$$x+(b-4)^2y+4-2b=0$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{equation of the line:}\ x+(b-4)^2y+4-2b=0}}$$
