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Tangent's equation

  1. Nov 2, 2005 #1
    Decide an equation for the tangent to the curve y=6x-x^2 at the point (4,8)

    Well for this I thought I'd use this



    Then I thought I'd Use


    8-(-16)/8-4 =6

    Then I thought the answer was -6X - 16 But it Wasen't the answer should be
    -2x -16

    What am I doing wrong please help me!?
  2. jcsd
  3. Nov 2, 2005 #2


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    Yes, but you didn't need to do that- you were told that when x= 4, y= 8. That's what "at the point (4, 8)" means.
    NO! (4, 8) are not two different x- values: they are the (x,y) values of a single point.
    Basically, you are doing everything wrong! What you found was the equation of the line between the points (4, 8) and (8, -16).
    You were asked to find the equation of the line through the single point (4, 8) that is tangent to the curve there.

    I assume this is a calculus course, since otherwise there is no way to answer this question- but you didn't use any calculus!

    Do you know how to find the derivative of y with respect to x and evaluate it at x= 4?
    Do you know how "slope of the tangent line" is related to the derivative?
  4. Nov 2, 2005 #3
    That was no Help

    Yeah, I did know that the point was x1 = 4 And y1 = 8
    And I know I decide those things with f(x+h)-F(x) / h and so on but I doesn't help me nada. And your quotation didn't help me either please provide me a solution so maybe I understand?
  5. Nov 2, 2005 #4


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    Homework Helper

    Have you seen derivatives yet?
  6. Nov 2, 2005 #5


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    Izekid, you need to tell us about derivatives.
  7. Nov 2, 2005 #6
    You shall not know the derivate for the equation.... !
    Its a simple question where you shall DECIDE(you shall tell the one who does the question about what the equation shall be) the equation for the curve 6x-x^2
    And you have 2 coordinates (4,8)

    And The Answer Should be Y=-2x-16

    But if you want to derivate this it's gonna be
    X^2 = 2x^2-1

    6x - 2x = 4x And that's no answer

    So please help I have written the things that is written in my math book...
  8. Nov 2, 2005 #7
    Let f(x) = 6x-x^2
    therefore f'(x) = 6-2x. At point (4,8), f'(x) = -2. So, we say that the following is true...
    8-y = -2(4-x) => y = 16-2x (as required, I think)
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