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Math Help - Find Polynomial from properties

  1. #1
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    Question Find Polynomial from properties

    Hello,

    I'm preparing for my final exams and I got stuck in a question concerning differentiation:

    Let p(x) be a polynomial of degree four whose graph touches the x-axis at the origin and has a horizontal point of inflection at T= (2,2).
    Determine p(x).

    Well, I understand that the function must be of the form:

    p(x) = ax^4 + bx^3 + cx^2 + dx + e

    and that

    p(0)=0
    p'(0)=0
    p(2)=2

    but how do you know proceed?

    I would be very gladful if someone could help me
    Last edited by CaptainBlack; July 13th 2010 at 02:16 AM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by zimtaepfelchen View Post
    Hello,

    I'm preparing for my final exams and I got stuck in a question concerning differentiation:

    Let p(x) be a polynomial of degree four whose graph touches the x-axis at the origin and has a horizontal point of inflection at T= (2,2).
    Determine p(x).

    Well, I understand that the function must be of the form:

    p(x) = ax^4 + bx^3 + cx^2 + dx + e

    and that

    p(0)=0
    p'(0)=0
    p(2)=2

    but how do you know proceed?

    I would be very gladful if someone could help me
    Touches the x-axis at the origin means that the polynomial is of the form:

    p(x)=x^2 (ax^2+bx+c)

    and that $$ x=2, $$ y=2 is a point of inflection means that p(2)=2 and p''(2)=0 (is a root of odd multiplicity)

    CB
    Last edited by CaptainBlack; July 13th 2010 at 01:31 PM. Reason: missing order of derivative
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  3. #3
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    Quote Originally Posted by zimtaepfelchen View Post
    Hello,

    I'm preparing for my final exams and I got stuck in a question concerning differentiation:

    Let p(x) be a polynomial of degree four whose graph touches the x-axis at the origin and has a horizontal point of inflection at T= (2,2).
    Determine p(x).

    Well, I understand that the function must be of the form:

    p(x) = ax^4 + bx^3 + cx^2 + dx + e

    and that

    p(0)=0
    p'(0)=0
    p(2)=2

    but how do you know proceed?

    I would be very gladful if someone could help me
    see that x=0, that means the equation has a multiplicity of 2 roots.

    use what you know about multiplicity, and you shall find the answer to those unknown variables.
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  4. #4
    A Plied Mathematician
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    You could also use the fact that since \langle 2,2\rangle is a point of inflection, you must have p''(2)=0.
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  5. #5
    A Plied Mathematician
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    Indeed, I would probably just bootstrap my way into the solution by using differentiation all the way.

    From p(0)=0, you can immediately conclude that your e=0. Thus, your polynomial becomes

    p(x)=ax^{4}+bx^{3}+cx^{2}+dx.

    Next, you set p'(0)=0, which implies that

    (4ax^{3}+3bx^{2}+2cx+d)|_{x\to 0}=0.

    From this equation, you can see that d=0.

    Rewrite your new polynomial, and continue plugging in the equations you have.

    [EDIT]: This I post merely as an alternative to CB's perfectly valid approach.
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