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Math Help - Manipulating cubic functions: x-intercepts

  1. #1
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    Manipulating cubic functions: x-intercepts

    Hi,

    I'm trying to create a cubic function that has only one x-intercept of -5 and a y-intercept of around 10.
    The y-intercept doesn't have to be exact.
    I can't seem to get this as I dont know what to do with the roots.
    I get to:
    (x+5)(x+-a)(x+-b)
    and i dont know what to do from there.
    Any help is greatly appreciated.

    Thanks,
    John
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  2. #2
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    Hello, John!

    I'm trying to create a cubic function that has only one x-intercept of -5
    and a y-intercept of around 10.
    The y-intercept doesn't have to be exact.

    Since x=-5 is a zero of the cubic, (x+5) is a factor.

    Then the cubic is the product of (x+5) and a quadratic.
    . . f(x) \;=\;(x+5)(ax^2+bx + c)

    The y-intercept will be 10 if c = 2.
    . . f(x) \;=\;(x+5)(ax^2 + bx + 2)


    Since x = -5 is the only x-intercept, the quadratic has no real roots.

    . . That is: . b^2 - 4ac \:<\:0

    We have: . b^2 - 8a \:<\:0 \quad\Rightarrow\quad a \:>\:\frac{b^2}{8}


    Let b = 2 \quad\Rightarrow\quad a = 1

    And we have: . f(x) \:=\:(x+5)(x^2 + 2x + 20 \:=\:x^3 + 7x^2 + 12x + 10

    . . . . . one of a brizillion possible cubics.

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  3. #3
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    Oh, I understand now!
    Thanks heaps.
    although, where does b^2 - 4ac \:<\:0 come from?
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