1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Equation of tangent line to curve

  1. Dec 13, 2011 #1
    1. The problem statement, all variables and given/known data

    Find all x coordinates of points (x, y) on the curve y = (x − 2)5 /(x − 4)3 where the tangent line is horizontal

    2. Relevant equations

    Quotient Rule - Differentiation

    3. The attempt at a solution

    So I could guess, that if tangent line is horizontal, the equation of tangent line to curve is y = c, where gradient is 0.

    So, first derivative of the curve is suppose equal to zero.

    dy/dx;
    0 = ((5)(x - 2)4(1)(x - 4)3 - (x - 2)5(3)(x - 4)2(1)) / ((x - 4)3)2
    0 = (5(x - 2)4(x - 4)3 - 3(x - 2)5(x - 4)2) / (x - 4)6

    Okay now I'm stuck.
     
  2. jcsd
  3. Dec 13, 2011 #2

    Mark44

    Staff: Mentor

    That's a good start.

    What you need to do is to write the numerator in a factored form. The best way to do that is to find the common factors in 5(x - 2)4(x - 4)3 - 3(x - 2)5(x - 4)2.

    The result will look like this: (x - 2)m(x - 4)n(<what's left>)
     
  4. Dec 13, 2011 #3
    So it would be like this,
    ((x - 2)4(x - 4)2(5(x - 4) - 3(x - 2))) / (x - 4)6
    ( (x - 2)4(5(x - 4) - 3(x - 2)) ) / (x - 4)4

    Now I stuck again.
     
    Last edited: Dec 13, 2011
  5. Dec 13, 2011 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You can simplify (5(x - 4) - 3(x - 2)) a bit more. Then if the derivative is going to zero, then the numerator has to be zero. Which values of x make that happen?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook