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Thread: Sketching a curve. Not sure where does this transformation comes from

  1. #1
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    Sketching a curve. Not sure where does this transformation comes from

    Not sure where last two equasions comes from. Question is rewritten from a book preparing to A-levels. Thanks for help in advance.

    Sketch the curve given by y=(3x+3)/(x(3-x)) ***(1)***

    "Range of values of y":

    from (1): 3xy - (x^2)y = 3x +3 ***(2)***

    from (2): (x^2)y + 3x(1-y)+3=0 ***(3)***

    so for y=0 => 3x+3=0 => x=-1

    Now for y!=-1 this is a quadratic in x.

    Thus, for real x,

    [3(1-y)]^2 - 4y(3)>= 0
    i.e. (3y-1)(y-3) >= 0

    ...question continues with sketching the curve and is easy apart from last two lines.

    Kind regards,

    aski
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  2. #2
    Junior Member Cervesa's Avatar
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    Re: Sketching a curve. Not sure where does this transformation comes from

    Not sure where last two equasions comes from.
    The last two equations come from the discriminant of the quadratic formula, $b^2-4ac$

    from (2): (x^2)y + 3x(1-y)+3=0
    $a = y$

    $b = 3(1-y)$

    $c = 3$

    $b^2 - 4ac = [3(1-y)]^2 - 4(y)(3)$
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  3. #3
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    Re: Sketching a curve. Not sure where does this transformation comes from

    A few things.
    The restriction on the domain of x is x != 0,3 so when you say "Thus, for real x" this isn't true.

    Second, when you say y != -1 this is a quadratic in x, it's also a quadratic in x when y = -1, namely -x^2 + 6x + 3 = 0 which has no real roots.
    It is a quadratic for all values y except for y=0, if the definition of quadratic is that the highest degree with a non-zero coefficient is 2.

    The last two lines refer to the determinant of quadratic (3). ie. D = b^2 - 4ac, where a is the coefficient of x^2, b is the coefficient of x, and c is the constant. If D >= 0 this means the quadratic has real roots. (If 0, repeated real roots)
    a = y, the coefficient of x^2
    b = 3(1-y), the coefficient of x
    c = 3

    edited because tex tags didn't format.
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    Re: Sketching a curve. Not sure where does this transformation comes from

    Quote Originally Posted by MacstersUndead View Post
    A few things.
    The restriction on the domain of x is x != 0,3 so when you say "Thus, for real x" this isn't true.
    Thanks, that is true, just copied that line from the book and did not think about that.

    Quote Originally Posted by MacstersUndead View Post
    A few things.

    Second, when you say y != -1 this is a quadratic in x, it's also a quadratic in x when y = -1, namely -x^2 + 6x + 3 = 0 which has no real roots.
    It is a quadratic for all values y except for y=0, if the definition of quadratic is that the highest degree with a non-zero coefficient is 2.
    Sorry, made a mistake while rewriting from the book. Edited in the question.

    Edited note: sorry,no chance to edit now.
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