At what point on the graph of y=1+(1/x) will the tangent line pass through the point (3,0)?

The tagent to the curve y = 1+1/x is given by
.

y-y1 = (dy/dx) (x-x1).

So
the slope of the tangent dy/dx is gotten by differentiating y =
1+1/x.

dy/dx = (1+1/x)' =
-1/x^2.

Therefore  at x1 dy/dx = -1/x1^2. At x1,  y1 =
1+1/x1.

So the tangent (at x1,y1) is given
by:

y- (1+1/x1) = (-1/x2^2 (x-x1).....(1) Since this line
passes through (3,0) , it should satisfy eq (1).

0-1-1/x1 =
(-1/x1^2) (3-x1)

(-1-1/x1)(-x1^2)  =
3-x1.

x1^2+x1+x1-3 = 0.

x^2
+2x-3 = 0.

 (x1+3)(x1-1).

So
x1 = -3 , or x1 = 1.

Therefore y1 =  1+1/x1 . Or y1 =
1-1/3 = 2/3 when x= -3

 y1 = 1+1/1 = 2 , when x1 =
1.

Therefore the tangents at (-3,2/3) and (1, 2) on the
curve y = 1+1/x passes the point (3,0).