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Find the equation for a parabola given 3 points

  1. Jun 1, 2008 #1
    1. The problem statement, all variables and given/known data

    Find an equation for a parabola that passes through the following points. To solve you must use triple elimination, you may check you answer any way you wish

    2. Relevant equations



    3. The attempt at a solution
    okay so the points given were (1,-2) (-3,10) and (4,31)
    I got to the point of making equations to olve with but then my brain hit a wall. What do i need to do next?
    -2=a+b+c
    10=9a -3b +c
    and
    31= 16a +4b +c
     
  2. jcsd
  3. Jun 1, 2008 #2
    i really am not good at quadratics
     
  4. Jun 1, 2008 #3
    This is a system of linear equations. There are several ways to solve for a,b, and c. One way is to use linear algebra and set up a matrix equation "Ax=b". Where A is the matrix with coefficients, x is vector [a b c] and b is vector [-2 10 31]
    [tex]\[ \left( \begin{array}{ccc}
    1 & 1 & 1 \\
    9 & -3 & 1 \\
    16 & 4 & 1 \end{array} \right)\] [/tex]
    Now, solve for x
     
  5. Jun 1, 2008 #4

    dynamicsolo

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    Homework Helper

    What you found is now a system of three simultaneous equations in three unknowns (which you've set up correctly), those unknowns being the coefficients of the equation for your parabola. Are you comfortable with solving such systems?
     
  6. Jun 1, 2008 #5
    im not very comforable with them
     
  7. Jun 1, 2008 #6
    okay i got down to b=2a-3
    but i dont know where to go from here
     
  8. Jun 1, 2008 #7
    I have to use triple elimination
     
  9. Jun 1, 2008 #8
    okay i got the right answer with the matrices:
    y=2x^2+x-5
    but how do i find this through triple elimination???
     
  10. Jun 1, 2008 #9
    Could you explain what a triple elimination is? I've never heard this term before. Is this another name for Gaussian elimination?
     
  11. Jun 2, 2008 #10
    no it's the classic way of doing things, we do it in high school.

    Ok, so you have 3 equations:

    [1]. -2=a + b +c
    [2]. 10=9a -3b +c
    [3]. 31= 16a +4b +c

    Eliminate a by using 9 * [1] - [2] and then 16 * [1] - [3]

    (-18 = 9a + 9b + 9c) - (10 = 9a - 3b + c) ---> -28 = 12b + 8c [4]
    (-32 = 16a + 16b + 16c) - (31= 16a +4b +c) ---> -63 = 12b + 15c [5]

    eq [4] - eq [5], should be left with c now.

    Now solve for c

    Plug c back into [4]

    Solve for b

    Now use eq [1], plug back b and c

    Solve for a
     
  12. Jun 2, 2008 #11
    yeah ok thx
    im actually in highschool and feel like a total idiot but thx for helping me
    i much appreciate it
     
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