Question 125672
To graph this function, assume some "nice" values for x.  By this I mean that assume that
x is a number that is a perfect square. Then you can find the corresponding values for y by
taking the square root of both sides.
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Notice one thing ... the value you assign to x cannot be negative because that would require 
that when you solve for y you would be taking the square root of both sides which would mean
that you would be taking the square root of a negative number.  This means that the only
values that x can have are zero and positive numbers. So, your graph can only exist on 
the y-axis or to the right side of the y-axis ... because x is only positive there.
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What happens if you set x = 0?  The equation becomes:
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{{{0 = y^2}}} 
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and when you take the square root of both sides you find that:
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{{{0 = y}}}
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This tells you that the point (0,0) is on the graph.
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Now set x = 4 and the equation becomes:
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{{{4 = y^2}}}
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When you take the square root of both sides the answers are:
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{{{2 = y}}} and {{{-2 = y}}}
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So the points (4, 2) and (4, -2) are on the graph.
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If you set x = 9 and do the same thing you find (9, 3) and (9, -3) are on the graph.
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And setting x = 16 results in finding that (16, 4) and (16, -4) are on the graph.
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When you plot these points (and any others you may try) you should find that your graph 
looks like a combination of the graphs shown in "rust red" and green below:
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{{{graph(400,400,-5,50,-20,20,-sqrt(x),sqrt(x))}}}
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You can tell this is not a function because in a function, there can only be one value for
y for each corresponding value of x. In this case, each value you choose for x has two 
corresponding values for y ... one on the red graph and one on the green graph.
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A simple way to view this is if you are given a graph and you draw a vertical line anywhere
on that graph, and that vertical line intersects more than two points on the graph, the
graph does not represent a function.
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In this case, note that if you draw a vertical line anywhere to the right of the y-axis it
crosses the graph twice ... once above the x-axis and once below the x-axis. Therefore,
this graph does not represent a function ...
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Hope this helps you to understand the problem.
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