Find the slope of the graph of the function y=5x/x-2 at (3,15). then find an equation for the line tangent to the graph at that point.

The first derivative gives you the formula for the slope of the tangent line to the curve at a given point.

Find the first derivative of the function y=5x/x-2:
       
               (x-2)(5) - (5x)(1)
dy/dx =
                     (x-2)^2

simplify the algebra here.
  
                                                      (3-2)(5) - (15)(1)
Then, at the point (3,15), dy/dx = = -10
                                                          (3-2)^2

What is the equation of the tangent line?  In other words, what is the eqn of the tan line to the given curve at when x = 3, if the slope is -10?

     y-15 = -10(x-3)

This can be re-written in other forms if desired.

55. i don’t no. but zhongshan