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Math Help - Maths - Parabola Equations

  1. #1
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    Maths - Parabola Equations

    Hey guys! I'm new here & browse more thanb I post
    I have an emergency, and a lot of guys who are here are pretty knowledgable.
    If you could help me out with my question below I would be so happy! Thanks in advance,
    Damien.
    Question:
    I need the EQUATION of a PARABOLA in the form of y = ax^2 + bx + c
    I have three points; A (17, 3), B (51,3) and C (34, z)
    Z is unkown, however, point C is a local minimum.
    So y' = 2ax + b
    I need the EQUATION of a PARABOLA in the form of y = ax^2 + bx + c
    I have three points; A (17, 3), B (51,3) and C (34, z)
    Z is unkown, however, point C is a local minimum.
    So y' = 2ax + b. Can anyone help me find the equation of the parabola and the value of z?
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  2. #2
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    Hello & welcome

    Quote Originally Posted by damien275x View Post
    Hey guys! I'm new here & browse more thanb I post
    I have an emergency, and a lot of guys who are here are pretty knowledgable.
    If you could help me out with my question below I would be so happy! Thanks in advance,
    Damien.
    Question:
    I need the EQUATION of a PARABOLA in the form of y = ax^2 + bx + c
    I have three points; A (17, 3), B (51,3) and C (34, z)
    Z is unkown, however, point C is a local minimum.
    So y' = 2ax + b
    I need the EQUATION of a PARABOLA in the form of y = ax^2 + bx + c
    I have three points; A (17, 3), B (51,3) and C (34, z)
    Z is unkown, however, point C is a local minimum.
    So y' = 2ax + b. Can anyone help me find the equation of the parabola and the value of z?
    f(x)=ax^2+bx+c

    The points A, B, C are on the curve.

    Therefore, f(17)=3, because at the point A, when the absciss is 17, the ordinate is 3.

    This goes the same way for B & C : f(51)=3 and f(34)=z

    Substitute in the expression of f(x)

    These three equations will give you relations between a,b,c, and z.



    Now, there is one more information : there is a local minimum at the point C.
    This means that at the point C, the derivative is equal to 0.
    Therefore f'(34)=0.

    This will give you a relation between a and b.




    So now it's OK, you have 4 equations with 4 unknown variable... Try to get out from this
    If you don't understand, just ask...
    Last edited by Moo; June 22nd 2008 at 06:34 AM.
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  3. #3
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    When I did these as simultaneous equations, it kept equating to 0. I was using b = -68a, then trying to find c, with no luck whatsoever. I don't know the relations part.
    Could you please go into more detail. I would really appreciate it! Thanks for your help.
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  4. #4
    Moo
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    Quote Originally Posted by damien275x View Post
    When I did these as simultaneous equations, it kept equating to 0. I was using b = -68a, then trying to find c, with no luck whatsoever. I don't know the relations part.
    Could you please go into more detail. I would really appreciate it! Thanks for your help.
    I would prefer you show what you've done ^^
    It's easier to explain & point out your mistakes
    And I don't see why it would "keep equating to 0"
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  5. #5
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    Hi again. I hope you are still there. Not a problem, I will show you what I have come up with so far.
    Here it is:

    For the form of y = ax^2 + bx +c

    For point (17, 3)
    3 = a(17^2) + b(17) + c
    3 = 289a + 17b + c ... labelling this equation (1)

    For point (51, 3)
    3 = a(51^2) + b(51) + c
    3 = 2601a + 51b + c ... labelling this equation (2)

    Stationary points (such as a local minimum) occur when y' = 0
    y' = 2ax + b
    0 = 2ax + b
    0 = 2a(34) + b
    0 = 68a + b
    b = -68a

    Now we have b = -68a
    Substitute b = -68a into equation (1)
    Therefore, 3 = 289a + 17(-68a) + c
    3 = -867a + c
    c = 867a + 3


    .. and i'm basically stuck after there. Help ! :P
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  6. #6
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    you have 4 unknown variables, a,b,c,z.. thus you should have 4 equations at first..

    use that f(34) = z
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  7. #7
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    more:

    3 = 289a + 17b + c (1)
    3 = 2601a + 51b + c (2)

    .... (1) x 3 (to eliminate b)
    Therefore, 9 = 867a + 51b + 3c (1')

    3 = 2601a + 51b + c (2)
    9 = 867a + 51b + 3c (1')
    (2) - (1')
    -6 = 1734a - 2c

    f(34) = z
    z = a(34^2) + 34b + c
    z = 1156a + 34b + c
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  8. #8
    MHF Contributor kalagota's Avatar
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    did you substitute that  b = -68a to your (1) and (2)?
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  9. #9
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    For b = -68a,

    For equation (2)
    3 = 2601a + 51(-68a) + c
    3 = 2601a - 3468a + c
    3 = -867a + c


    For equation (1)
    3 = 289a + 17(-68a) + c
    3 = 289a - 1156a + c
    3 = -867a + c

    Using b = -68a and c = 3 + 867a,
    (Substituting these values into equation (1))
    3 = 289a + 17(-68a) + 867a + 3
    3 = 289a - 1156a + 867a + 3
    Which then just gives.. 3 = 0 + 3
    3 = 3
    Last edited by damien275x; June 22nd 2008 at 07:14 AM.
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  10. #10
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    Okay I have worked everything out:

    I have three equations;
    289a + 17b + c = 3
    2601a + 51b + c = 3
    1088a + 33b + c = 0
    `
    What do I do now ?
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  11. #11
    Moo
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    Quote Originally Posted by damien275x View Post
    Okay I have worked everything out:

    I have three equations;
    289a + 17b + c = 3
    2601a + 51b + c = 3
    1088a + 33b + c = 0
    `
    What do I do now ?
    You're still forgetting this one :

    z=a(34)^2+b(34)+c

    -----------------------------
    To sum up :
    f(17)=3 \implies 289a+17b+c=3

    f(51)=3 \implies 2601a+51b+c=3

    f(34)=z \implies 1156a+34b+c=z

    f'(34)=0 \implies 68a+b=0

    From the last one, b=-68a.

    Substituting in the previous equations, we get :

    \left\{\begin{array}{lll} 289a-1156a+c=3 \\ 2601a-3468a+c=3 \\ 1156a-2312a+c=z \end{array} \right.

    \left\{\begin{array}{lll} -867a+c=3 \ (1) \\ -867a+c=3 \\ -1156a+c=z \end{array} \right.

    From (1), we get :

    c=3+867a

    Substituting in the last one, we get :

    -1156a+(3+867a)=z
    \implies z=3-289a

    ....

    If there is no mistake, there is an infinity of solutions for z !

    Why ? because we see that A and B have same ordinate. Plus, a parabola is always symmetric in regard to a line. Here, it would be symmetric in regard to the line x=34, because it's half way from A and B.
    If it's symmetric in regard to this line, it means that the minimum is on x=34, which is what we want. But there are many solutions !
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