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Math Help - Sqrt Graph Help

  1. #1
    Junior Member Godfather's Avatar
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    Sqrt Graph Help

    How would i graph the sqrt of (a^2)-(x^2)??
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  2. #2
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    Hello, Godfather!

    How would i graph: . y \;=\;\sqrt{a^2 -x^2} ?

    Square both sides: . y^2 \:=\:a^2-x^2 \quad\Rightarrow\quad x^2+y^2 \:=\:a^2

    This is a circle: center at the origin, with radius a.


    Then: . y \;=\;^+ \sqrt{a^2-x^2} .is the upper semicircle.

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  3. #3
    Junior Member Godfather's Avatar
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    how would i graph it without know a ?
    what about the domain,range,zero,even/odd,periodic and one to one ?
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  4. #4
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    The domain consists of all legal values for x given that the function must be a real number. So, a^2 - x^2 must be nonnegative. For what values of x is this true?

    The range consists of all the values the function can take on its domain. To start, consider what the largest possible value for \sqrt{a^2 - x^2} would be. Can the function give a negative result?

    The zeroes of the function will be the same as the zeroes of a^2 - x^2. For what values of x is the function equal to zero?

    Whether or not the function is even or odd depends on whether or not f(x) = f(-x) or f(x) = -f(-x), respectively.
    Given that f(x) = \sqrt{a^2 - x^2}, is either of these statements true?

    If the function is periodic with period p, then f(x) = f(x + np) for every integer n. Is a semicircle periodic?

    If the function is one-to-one, then every y value must be matched by exactly one x value. Is this the case for a semicircle?
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