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Homework Help: At what points on this curve is the tangent line horizontal?

  1. Mar 6, 2013 #1
    1. The problem statement, all variables and given/known data
    At what points on the curve y = (x^2)/(2x+5) is the tangent line horizontal?

    2. Relevant equations
    Quotient rule

    3. The attempt at a solution
    I figured out the derivative which is
    2x(2x+5) - 2x^2

    I also know that for the equation of the tangent to be horizontal, it needs to have a slope of 0, so y' = 0

    Therefore I tried to do 2x(2x+5) - 2x^2 = 0, and I got 2x^2 + 10x = 0 which means that x must be -5. However when I put it into the calculator, it doesn't agree with me, so I went wrong somewhere.

    Thanks in advance!
  2. jcsd
  3. Mar 6, 2013 #2


    User Avatar
    Science Advisor

    I have no idea what you mean by "put it into the calculator". However, "2x^2+ 10x= 0" does NOT mean "that x must be -5". How did you get that?
  4. Mar 6, 2013 #3
    I got a ti84, and i can draw a tangent line and it'll tell me the formula, so when I drew a tangent line through -5, the line was not horizontal completely. If you plug in -5 into 2x^2 + 10x= 0, it works out, but when you plug it back into the first one you don't get 0.
  5. Mar 6, 2013 #4


    Staff: Mentor

    There are two critical numbers; x = -5 is one of them.

    If you substitute x = -5 into y = x2/(2x + 5), you'll get the y value at one of the points where the tangent is horizontal.
  6. Mar 6, 2013 #5
    But the calculator disagrees with the fact that x = -5 is one of them, and also how can I find the other point?
  7. Mar 6, 2013 #6


    Staff: Mentor

    I have no idea what you're entering into the calculator, so I can't say that its objections are valid.

    To find the other critical number, find both solutions of 2x2 + 10x = 0.

    I hope you aren't doing this:

    2x2 = - 10x

    and then dividing both sides by 2x.
  8. Mar 6, 2013 #7
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