Given the function $f(x) = \dfrac{(x-1)}x$, find the equation of the tangent line to the graph of $f$ that pass though the point $(4,1)$. NOTE: The point (4,1) is NOT on the graph of f.

Okay so first I found the slope of the tangent through the following:

$m=f'(x)=(f'(x)g(x)-f(x)g'(x))/(g(x))^2$

$= [(1)(x)-(x-1)(1)]/ x^2$

$= (x-(x-1))/x^2$

$= 1/x^2$
So now what do I do?