SOLUTION: Show that the equation of the line which passes through (0,6) and is a tangent to the graph of y=1/x is y=-9x+6, and find the point of intersection.

How you would approach this problem depends on whether you
are taking college algebra or calculus.

If you are taking algebra and have not studied calculus,
then

The equation of a line through (0,6) which is the y-intercept
and b=6, is

y = mx+b or

y = mx+6

We solve that system of equations

y = mx+6
y = 1/x

mx+6 = 1/x

Multiply through by x

  mx²+6x = 1
mx²+6x-1 = 0

A line is tangent to a curve when there is 
just one point of intersection.  That is,
when the quadratic equation has discriminant = 0

discriminant = B²-4AC = 6²-4(m)(-1) = 0

6²-4(m)(-1) = 0
      36+4m = 0
         4m = -36
          m = -9

So  y = mx+b becomes:
    y = -9x+6

which shows that that is a tangent line.

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Regardless of which you are taking, you will have to find the point of intersection:

Solve the system:

           y = 1/x
           y = -9x+6

Set the right sides equal:

         1/x = -9x+6

Multiply through by x

           1 = -9x²+6x
    9x²-6x+1 = 0
(3x-1)(3x-1) = 0
        3x-1 = 0
          3x = 1
           x = 

Substitute in

    y = -9x+6
    y = -9(+6
    y = -3+6
    y = 3

So the point of intersection, which is the point of tangency, is:

      (,3)

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If you are studying calculus, find the derivative of

y = 1/x
Rewrite as
y = x-1
 = -1x-2
 = 

Substitute  for x in  

 =  =  = -9

which is the same as the slope of the line.

Therefore the line is tangent to the curve at their
point of intersection. 

Edwin