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    mattimeo_boyd's Avatar
    mattimeo_boyd Posts: 31, Reputation: 1
    Junior Member
     
    #1

    Jun 5, 2010, 11:23 AM
    Solving Polynomials and Descarte's Rule?
    Analyze the function: x^2 + 3 / X + 1
    X Intercepts:
    Y Intercepts:
    Asymptotes:

    Descartes Rule: f(x) = -X^4 + 3X^2 + X + 7
    How do I find the positive and negative zeros?
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
    Uber Member
     
    #2

    Jun 5, 2010, 12:02 PM

    1.



    Put x = 0 to find the y intercepts. (y = 3)
    Put y = 0 to find the x intercepts. (x = no real solution)

    At the asymptote, you'll have the denominator either as infinity, negative infinity or zero.

    This way, you get the line y = 1 as an asymptote (when x tends to infinity, y tends to 1), the line y = -1 as an asymptote (when x tends to negative infinity, y tends to -1) and when x = -1, you get a zero denominator, making this time a vertical asymptote at x = -1.

    Sorry if I made a mistake in my calculations, it's really late here... shouldn't be here that late...
    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
    Ultra Member
     
    #3

    Jun 5, 2010, 03:39 PM
    Descartes Rule:
    How do I find the positive and negative zeros?
    Descartes rule of signs says that the number of positive real roots is either equal to the number of changes in sign of f(x) or less than that by an even number.

    How many times does it change sign? 1 time.

    So, there must only be 1 positive real root because we can not be less than 1 by an even integer.

    The rule also says that the number of negative real roots is equal to the number of changes in sign of f(-x) or less than that by an even number.

    How many sign changes are in f(-x)? Sub in -x for x and we get



    How many sign changes now? 3.

    So, there are either 3 or 1 negative roots. See?

    Solving, f(x) has 1 positive real root and 1 negative real root, as DRS said. The other 2 are complex.

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