Results 1 to 3 of 3

Math Help - Fit Quadratic to 2 points and 1 condition

  1. #1
    Member
    Joined
    Mar 2007
    Posts
    206
    Awards
    1

    Fit Quadratic to 2 points and 1 condition

    Hi,

    I was hoping to get some advice on the best way to handle this problem. When fitting a quadratic with 3 points, I use a matrix approach to evaluate the coefficients which works nicely. In this case, I have only 2 points but an extra condition. I tried to evaluate the coefficients by simultaneous equations but this was painfully slow and I did not get the right equation.

    The conditions are:

     \begin{matrix}<br />
     x=0 & y=0  \\<br />
     x=12 & y=70  \\<br />
     \frac{dy}{dx}=0 & y=-70  \\<br />
  \end{matrix}<br /> <br />

    My answer is:
    <br />
y=2.1012x^2 -19.3805x<br />

    Which satisfies the first 2 conditions but not the third. Any advice on the easiest way to go about this would be greatly appreciated.

    Regards Elbarto
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by elbarto View Post
    Hi,

    I was hoping to get some advice on the best way to handle this problem. When fitting a quadratic with 3 points, I use a matrix approach to evaluate the coefficients which works nicely. In this case, I have only 2 points but an extra condition. I tried to evaluate the coefficients by simultaneous equations but this was painfully slow and I did not get the right equation.

    The conditions are:

     \begin{matrix}<br />
x=0 & y=0 \\<br />
x=12 & y=70 \\<br />
\frac{dy}{dx}=0 & y=-70 \\<br />
\end{matrix}<br /> <br />

    My answer is:
    <br />
y=2.1012x^2 -19.3805x<br />

    Which satisfies the first 2 conditions but not the third. Any advice on the easiest way to go about this would be greatly appreciated.

    Regards Elbarto
    (0, 0): y = ax^2 + bx.

    (12, 70): 70 = 144a + 12b => 25 = 72a + 6b

    dy/dx = 2ax + b = 0 => x = -b/(2a). Substitute into y = ax^2 + bx to get y in terms of a and b. Equate this to -70. This gives a second equation.

    Solve the two equations for a and b.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Mar 2007
    Posts
    206
    Awards
    1
    Thank you for the reply. This is the approach I used, though I suspect that I made a calculation error when evaluating the 2 equations.

    Thanks again for your time, I will attempt to revise my calculations.

    Regards Elbarto
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: August 24th 2011, 11:35 AM
  2. Replies: 3
    Last Post: December 1st 2010, 12:28 PM
  3. Replies: 4
    Last Post: November 29th 2010, 05:05 PM
  4. Replies: 4
    Last Post: September 21st 2009, 03:39 AM
  5. finding quadratic equation, when given 3 points
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: March 21st 2009, 02:51 AM

Search Tags


/mathhelpforum @mathhelpforum