Question

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.

Answer 1

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.