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Math Help - Ordered Pairs (x,y)

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    Ordered Pairs (x,y)

    For each question below, tell whether the set of ordered pairs (x,y) defined by each equation is a function.

    (a) x + 2y^2 = 1

    (b) y = (2/x)
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    Quote Originally Posted by blueridge View Post
    For each question below, tell whether the set of ordered pairs (x,y) defined by each equation is a function.

    (a) x + 2y^2 = 1

    (b) y = (2/x)
    hint: we have a function if each x value maps to one and only 1 y value. is this true for both these formulas?
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    (a) x + 2y^2 = 1
    No. Because (x,y) = \left( 0, \frac{1}{\sqrt{2}}\right) = \left(0, -\frac{1}{\sqrt{2}}\right)
    Both work. For the same x we get more than one y.
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    tell me...

    What do you mean that both work?

    Are these one-to-one functions?

    Where did you get those values in terms of (x,y)?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by blueridge View Post
    What do you mean that both work?
    he found a single value for x that maps to two values of y, so it is not a function, since both y's work for 1 x


    Are these one-to-one functions?
    the first one is not a function at all, the second is a 1-1 function


    Where did you get those values in terms of (x,y)?
    through inspection. he realized that the y is squared, so for any positive y-value we plug in, we will get the same answer as when we plug in the negative of that value
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