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Finding a tangent in a Cubic Function

  1. Jun 5, 2010 #1
    1. The problem statement, all variables and given/known data
    My cubic function is y=(x-6)(x-1)(x-9) or y=x^3-16x^2+69x-54
    I need to find the tangent at the point x=2.5


    2. Relevant equations



    3. The attempt at a solution
    All that I have managed to do is work out the y value for x=2.5, that is y=34.125
    Please help someone!
     
    Last edited: Jun 5, 2010
  2. jcsd
  3. Jun 5, 2010 #2

    lanedance

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    Homework Helper

    the gradient is the derivative of your function, so differentiate, ie. find [itex]\frac{dy}{dx}[/itex]
    same as other post
     
  4. Jun 5, 2010 #3
    Derive and plug in x = 2.5

    In other words, solve for [tex]\frac{dy}{dx}[/tex], and plug in 2.5 for x:

    y = x3-16x2+69x-54

    [tex]\frac{dy}{dx}[/tex] = 3x2-32x+69

    At x = 2.5, [tex]\frac{dy}{dx}[/tex], which is the tangent, equals

    3(2.5)2-32(2.5)+69 = ?

    I am too lazy to do the calculation, but here is the basic setup.

    If you are looking for a tangent line, then use y'(2.5) as the slope for the slope

    equation: y-y0 = f'(2.5)(x-x0)
     
    Last edited: Jun 5, 2010
  5. Jun 5, 2010 #4
    thankyou ever so much, if I have any more questions I will ask!!
     
  6. Jun 5, 2010 #5
    Ok, so I now have the Gradient and the X and Y values of the tangent. How do I get the equation of the tangent.

    ps. Realy sorry for all the trouble, you are a big help!
     
  7. Jun 5, 2010 #6
    Well, if you have the slope of a line and a point on the line, how do you find the equation of the line? Remember back to algebra!
     
  8. Jun 5, 2010 #7
    i cant think back that far, Please explain
     
  9. Jun 5, 2010 #8
    ahh I know now, Point slope formula!
     
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