Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding the equation of the line of a cubic function

  1. May 30, 2009 #1

    I am trying to find the equation in the form ax^3+bx^2+cx+d for the curve passing through the origin and (40 sq root 6, -20).
    How do I find the a, b, c, and d values?
  2. jcsd
  3. May 30, 2009 #2


    User Avatar
    Science Advisor
    Homework Helper

    If the curve passes through the origin, what does that tell you about 'd'?

    Now if the curve passes through (40 sqrt6, -20) then you have one equation and three unknowns (a,b,c). Evidently, there will not be a unique solution, and you'll have some flexibility in choosing a,b, and c.
  4. May 31, 2009 #3
    Ok so I take it d shifts the curve off the origin, so if it is going through the origin, there is no d value?
    And how many points would I need to come up with a, b, and c values?
  5. May 31, 2009 #4


    User Avatar
    Science Advisor

    Yes, setting x= 0 gives y= a03+ b02+ c0+ d= d. "Going through the origin" means x= 0 gives y= 0. You need 3 equations to solve for three variables. Each point gives an x and y value to put into the equation so you need three points to solve for the three variables a, b, and c.
  6. May 31, 2009 #5
    Ok thanks so much. I will work on getting a third point.
  7. May 31, 2009 #6


    User Avatar
    Science Advisor

    No, you need two more points. In your original form, [itex]y= ax^3+ bx^2+ cx+ d= 0[/itex], you have 4 numbers to determine, a, b, c, and d. You used the origin, (0,0) to determine d. Now you need 3 other points to determine a, b, and c.

    You probably learned in geometry that "two points determine a line". Taking a= b= 0 you get a line, with equation y= cx+ d passing through those two points. Three points will determine a quadratic and it requires four points to determine a cubic.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook