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Math Help - Parabola help

  1. #1
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    Parabola help

    Consider the parabola defined by the equation y = -x^2 + 6.

    Find: axis of symmetry: i got x = 0

    Find: vertex: i got (0,6)

    Express the distance as a function in x between the origin and point P=(x,y) which is on the parabola on the first quadrant: okay i have no idea how to do this one.

    thanks for any help
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  2. #2
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    Quote Originally Posted by NeedHelp18 View Post
    Consider the parabola defined by the equation y = -x^2 + 6.

    Find: axis of symmetry: i got x = 0

    Find: vertex: i got (0,6)

    Express the distance as a function in x between the origin and point P=(x,y) which is on the parabola on the first quadrant: okay i have no idea how to do this one.

    thanks for any help
    A general point on the parabola has coordinates (x, y), that is, (x, -x^2 + 6). The origin has coordinate (0, 0). Now apply the usual formula for the distance between two points.
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  3. #3
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    f(x) = √(x+x^4 + 36) for 0≤x≤√6

    is this the answer?
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  4. #4
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    Dear NeedHelp18,

    I hope you don't belive so: (a-b)^2 = a^2 + b^2

    It would be pity!
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  5. #5
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    wats the answer then? i still keep gettin the same thing.
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  6. #6
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    the rigth formula:

    (a-b)^2 = a^2 - 2*a*b + b^2
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  7. #7
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    Quote Originally Posted by NeedHelp18 View Post
    f(x) = √(x+x^4 + 36) for 0≤x≤√6

    is this the answer?
    No. d = \sqrt{x^2 + (- x^2 + 6)^2} = \, ....
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