Finding an equation of a parabola given three points.

  1. Sep 15, 2007 #1
    1. The problem statement, all variables and given/known data

    Find an equation of a parabola given three points without a vertex point.

    2. Relevant equations

    y=a(x-h)+k

    3. The attempt at a solution

    The parabola is upside down to I know a is negative.
     
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  3. Sep 15, 2007 #2

    radou

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    Any thoughts about the problem? What does the general form of the equation of a parabola look like?
     
  4. Sep 15, 2007 #3
    y=ax^2+bx+c
     
  5. Sep 15, 2007 #4
    I don't know how to find the variables given the three points.
     
  6. Sep 15, 2007 #5

    symbolipoint

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    The general form will give you your method. If you have three points, then you can form three equations in general form. The unknowns will be the coefficients. You can simply solve the system of linear equations. This would be introductory level algebra.
     
  7. Sep 15, 2007 #6
    if you have 3 points and you know the general equation of a parabola what could you make?
     
  8. Sep 15, 2007 #7
    the 3 points are: (-2,2), (0,1), (1,-2.5)

    I substituted the numbers, giving three equations.

    2=4a-2b+c

    1=0x^2+0b+C

    -5/2=a+b+c

    so c=1

    -2b=-4a+1

    b=(4a+1)/2


    I can't get the numbers to work.


    This is actually in my calculus book. chapter 1
     
  9. Sep 15, 2007 #8
    I got the numbers to work. The equation in general form is 2x^2-11/2x+1. How would I convert that to standard form. What is the trick to solving these? I feel like I got lucky to get the answer.
     
  10. Sep 15, 2007 #9
    To solve these you will just need to know algebra to solve a system of equations.

    if instead of (0,1) you were given some other point you would go ahead and combine 2 equations into 1 hopefully eliminating an x^2 or x or c term, and then working with that one and the remaining equation try to solve for a,b,or c, and using that to solve for the other co-efficients.

    to get that equation to the standard form you need to complete the square, but you first need to remove the 2 in front of x^2 first.
     
  11. Sep 15, 2007 #10

    HallsofIvy

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    You said you had 3 equations- you've only written 2 there!
    The three equations you get are:
    (-2,2) 4a- 2b+ c= 2
    (0, 1) c= 1
    (1, -2.5) a+ b+ c= -2.5

    Yes, from the second equation, you get c= 1. Putting that into the other two equations, 4a- 2b= 1 and a+ b= -3.5. Can you solve those two equations for a and b?

    Sorry, but it's too early to feel lucky! That's not at all right. Putting a= 2, b= 11/2, c= 1 into the first equation, you get 4(2)- 2(11/2)+ 1= 2- 11+ 1= -8, not 2.

    Once you do have the correct equation, if by "standard form" you mean y= a(x- h)2+ k, you get that form by completing the square.
     
  12. Sep 15, 2007 #11
    I had 3 equations.

    -2=4a+2b+c
    -5/2=a+b+c
    1=0+0+c
     
  13. Sep 15, 2007 #12
    the solutions I got were 2, -11/2, and 1. They worked in all 3 equations.
     
  14. Sep 15, 2007 #13
    Every time I do this problem, I get different answers. I have always had trouble with these. I can handle trig easily compared to these problems. I am so frustrated.
     
  15. Sep 15, 2007 #14
    I think I switched the signs around. Now I get
    2=4a-2b+c
    1=0+0+c
    -5/2=a+b+c

    and got

    a=-1
    b=-5/2
    c=1

    Does that sound better?
     
  16. Sep 15, 2007 #15

    learningphysics

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    looks good to me.
     
  17. Aug 12, 2011 #16
    If I am given the points (-6; -2), (4; -2) and (1; -8) with the equation y = 2 x^2 + bx + c, CAN I use any of the two points two simultaneously solve for b and c?

    By the way I tried and get different values with different points. The only reasonable solution is if I solve the problem using the fact that x = -1 is an axis of symmetry and then use the general equation y = 2(x + 1)^2 + q. Then use the point (1; -8) to find q. Rewriting the equation gives b = 4 and c = -14.

    Why don't I get the same when I solve simultaneously?
     
    Last edited: Aug 12, 2011
  18. Aug 12, 2011 #17

    SammyS

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    The three points are not on the same parabola -- at least they're not consistent with being on the same parabola if it's of the form: y = 2 x2 + b x + c .
     
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