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Find Equation of Tangent

  1. Mar 9, 2013 #1
    1. The problem statement, all variables and given/known data
    The equation of the tangent to the curve f(x)=ax^3+bx at the point of contact (-1;3) is
    y-x-4=0. Calculate the values of a and b


    2. Relevant equations

    y-y1=m(x-x1)

    3. The attempt at a solution

    I am totally stuck, here is what I could derive:
    Equation of tangent y=x+4
    f'(x)=3ax^2+b

    Dont know what to do, can someone please help?
     
  2. jcsd
  3. Mar 9, 2013 #2
    Hint: Equation of tangent is given by :

    y-y1=f'(x)*(x-x1)

    You are already given x1 and y1. Compare this with the already given equation of tangent.

    Note: f'(x)=(dy/dx) at x=x1 and y=y1
     
  4. Mar 9, 2013 #3

    SteamKing

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    You have one point common to the curve and the tangent line.
    You know the value of the slope at the tangent point.

    You have two equations and two unknowns.
     
  5. Mar 9, 2013 #4
    Ok, set up two equations:
    a-b=3
    3a+b=0
    Make a subject of formula:
    a=3+b

    plug into equation 2 and solved, final answer:
    b=-9/4
    a=3/4

    All good?
     
  6. Mar 9, 2013 #5

    SteamKing

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    What is the slope of the tangent line at point (-1,3)?

    Your second equation is incorrect.

    Always check your solutions by substituting back into original equations.
     
  7. Mar 10, 2013 #6
    The slope would be 1, because y=mx+c and equation is y=x+4
     
  8. Mar 10, 2013 #7

    HallsofIvy

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    So why did you, previously, set 3a+b equal to 0?
     
  9. Mar 10, 2013 #8

    SteamKing

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    How does slope = 1 affect your second equation for a and b?
     
  10. Mar 10, 2013 #9
    Sorry guys, just slipped my mind that derivative is same as gradient, so my final answers should be:
    a=1
    b=-2
     
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