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Math Help - Quadratic curve

  1. #1
    Senior Member Mukilab's Avatar
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    Quadratic curve

    How to find the minimum point on a quadratic curve?

    Or highest point for that matter?

    E.X: I have the equation y=x^2-8x+23
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by Mukilab View Post
    How to find the minimum point on a quadratic curve?

    Or highest point for that matter?

    E.X: I have the equation y=x^2-8x+23

    Hi Mukilab,

    This curve will have a minimum because the coefficient of the quadratic term is positive.

    The minimum point is the vertex.

    You can find the x-coordinate of the vertex by using this little formula:

    x=\frac{-b}{2a}

    Once you have the x-coordinate of the vertex, plug it into the quadratic function to find the corresponding y-coordinate.

    You could also complete the square to find the vertex, but who wants to do that, right?
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  3. #3
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by Mukilab View Post
    How to find the minimum point on a quadratic curve?

    Or highest point for that matter?

    E.X: I have the equation y=x^2-8x+23
    Take the derivative and equate to 0.

     y^{ \prime } = 2x - 8 = 0

    Min happens at

     x = 4

    If we don't use derivatives (which I suspect you are not supposed to) we complete the square.

     y = x^2-8x+23 = (x^2 - 8x + 16 - 16) + 23 = (x-4)^2 + 7

    From here we can find the max/min. Since this parabola is facing upwards, we know there in a minimum.

    This is known as "vertex form" and the min/max points happen at the vertex. In this case,

    the vertex is  (4,7) which comes from the negative of the number inside the bracket with the x, and the number that is added.

    In general

     y = (x-p)^2 + k

    has the vertex

     (p,k)
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  4. #4
    Senior Member Mukilab's Avatar
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    Quote Originally Posted by AllanCuz View Post
    Take the derivative and equate to 0.

     y^{ \prime } = 2x - 8 = 0

    Min happens at

     x = 4

    If we don't use derivatives (which I suspect you are not supposed to) we complete the square.

     y = x^2-8x+23 = (x^2 - 8x + 16 - 16) + 23 = (x-4)^2 + 7

    From here we can find the max/min. Since this parabola is facing upwards, we know there in a minimum.

    This is known as "vertex form" and the min/max points happen at the vertex. In this case,

    the vertex is  (4,7) which comes from the negative of the number inside the bracket with the x, and the number that is added.

    In general

     y = (x-p)^2 + k

    has the vertex

     (p,k)
    Surely its far easier to use the method master stated? Or does the one you have stated work well under different circumstances?
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  5. #5
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by Mukilab View Post
    Surely its far easier to use the method master stated? Or does the one you have stated work well under different circumstances?
    Completing the square takes 15 seconds and is a requisite skill for higher level math (i.e. derivatives and integrals). I would learn to master it as it will help you on your current exams (for other problems not of this nature and when Master's equation does not apply) and in the future.
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  6. #6
    Senior Member Mukilab's Avatar
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    Do not know what the derivative is. Do no know what 'equate' means, is this change it to zero and change others as well in ratio?

    what is y^{/prime}??????

    Lastly, why did you add in +16, -16? Somehow you changed -8x, +16, -16 all into -4 and changed 23 into 7?!
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  7. #7
    A riddle wrapped in an enigma
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    Quote Originally Posted by Mukilab View Post
    Do not know what the derivative is. Do no know what 'equate' means, is this change it to zero and change others as well in ratio?

    what is y^{/prime}??????

    Lastly, why did you add in +16, -16? Somehow you changed -8x, +16, -16 all into -4 and changed 23 into 7?!
    This is all done in the process of 'completing the square'.
    If you haven't studied how to do this, then go back to
    the way I showed you.
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  8. #8
    Senior Member Mukilab's Avatar
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    Quote Originally Posted by masters View Post
    This is all done in the process of 'completing the square'.
    If you haven't studied how to do this, then go back to
    the way I showed you.
    I need this for another question where someone told me to 'complete the square' and redirected me here
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  9. #9
    A riddle wrapped in an enigma
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    Quote Originally Posted by Mukilab View Post
    I need this for another question where someone told me to 'complete the square' and redirected me here
    Maybe this will help: How to Complete the Square
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