Question 1062842
Which of the following equations does not define y as a function of x?
A. xy^2 = 8
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Solve for y by dividing both sides by x, (where x is not 0)

      {{{y^2 = 8/x}}}
      {{{y = "" +- sqrt(8/x)}}}

That does not define a function because there is NOT just 
ONE y for each x, since there are TWO y's for each x.

For example when x = 2,

      {{{y = "" +- sqrt(8/2)}}}
      {{{y = "" +- sqrt(4)}}}
      {{{y = "" +- 4}}}  
 
This is not a function because, for example, when x = 2, there are
TWO values for y, +2 and -2.  So that's why it doesn't define a function.

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B. x^2 + y^2 = 0
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There is only one ordered pair that satisfies that.

That is (x,y) = (0,0)

It does define a function whose domain is {0} and whose range is {0}

It defines a function because there is just one y, 0, for each x, 0,
since the only x and y are both 0.

So, it's a function because there is just one y for every x.
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C. 3y = 7
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    y = 7/x

That defines a function because there is just one y for each x.
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D. y = square root of x - 3
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   {{{y=sqrt(x-2)}}}

That defines a function because there is just one y for each x.
Edwin</pre>