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Graphs equal to zero

  1. Jul 14, 2007 #1
    I'm have a hard time with graphs equal to zero like these:

    A. [tex] x^{2}+y^{2}-4x+2y+5=0[/tex]

    B. [tex]xy=0[/tex]

    C. [tex]2x^{2}-3y^{2}+8x+6y+5=0[/tex]


    I know how to simplify them:

    A. [tex] (x-2)^{2}+(y+1)^{2}=0[/tex]

    C. [tex] 2(x+2)^{2}-3(y-1)^{2}=0[/tex]

    However, I just can't seem to figure out there graphs. They seem so strange with the zero. How come graph B. are two intersecting lines (x-axis and y-axis)? Please tell me how I can determine the graphs of these equations easily w/o plotting or plugging in points. Thanks.

    I guess some of them are just points but how do I find out the exact point?
     
    Last edited: Jul 14, 2007
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  3. Jul 14, 2007 #2

    CompuChip

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    Actually, the right hand being the constant zero instead of [itex]\pi[/itex] or [itex]\sqrt{2}[/itex] or any other number, is just a special case of a level curve.

    For B, you have just one term on the right, which is a product. When does a product vanish? Alternatively, you could try dividing out a variable. For example, what do you get when you divide both sides by x (and when is this allowed?)
     
    Last edited: Jul 14, 2007
  4. Jul 14, 2007 #3

    HallsofIvy

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    First of all, "graphs" aren't equal to 0. There are, typically, points on graphs where one or the other of the variables is equal to 0. Exacly what are you trying to do- solve the equations for x or y?


    Oh, so you are trying to figure out what the graphs look like! Use a little common sense.

    You know that real numbers squared cannot be negative and so that the sum of two squares is 0 only if each number is 0: the only way (x-2)2+ (y+1)2 can equal 0 is if x-2= 0 and y+1= 0 separatly. In other words, the graph is just the single point (2, -1). This is sometimes thought of as a "degenerate" ellipse.

    You can write 2(x+2)2- 3(y-1)2= 0 as 3(y-1)2= 2(x+2)2 and take square roots of both sides:
    [tex]\sqrt{3}(y-1)= \pm\sqrt{2}(x+2)[/tex]
    Since those are linear they are straight lines- the two different signs give you two lines. This can be thought of as a "degenerate" hypebola.
     
  5. Jul 15, 2007 #4
    Wow, I can't believe I've never come across a graph like this before! Anyways, I think I get how you solve it now. I'm reading a review book and it does talk about the "degenerate case" but this is my first time seeing it before.

    Now, I still don't get fully what the "degenerate case" is. It says: The degenerate case occurs when one of the variables drops out, or the terms with the variables equal zero, or a negative number after the square has been completed. I think for the problems that I've posted, the terms with the variables equal zero,right? Can someone explain the other two?

    Also, how do you determine the graph for [tex]xy=0[/tex] The book says that the graph is 2 intersecting lines (the x- and y- axis) but how is this so? Can't the graph also be (x-axis and y=6)? The equal would still work out!
     
  6. Jul 15, 2007 #5
    Ah, I guess x-axis and y=6 would be a point of the two intersecting lines. I think I meant x=0 and y=6. But still, how do you know that the graph is 2 intersecting x- and y-axis lines?
     
  7. Jul 15, 2007 #6

    HallsofIvy

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    You are forgetting what graphs are: "x-axis and y= 6" is NOT the point (0,6). For one thing, the x-axis is the set of all points of the form (x,0): y=0 there, not x. Also you are not talking about the intersection of the two sets of points, you are talking about their union,. If the graph were the two lines "x-axis and y= 6" then it would include all points on the x-axis (of the form (x, 0)) AND all points on y= 6 (of the form (x, 6)). That second line includes, for example, (1, 6). (1)(6)= 6, not 0.

    If xy= 0, then x=0 or y= 0. The first, x= 0, is all points of the form (0,y), the y-axis. The second, y=0, is all points of the form (x, 0), the x-axis.
     
  8. Jul 15, 2007 #7

    CompuChip

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    Hmm, is my post missing? I'd really say I had already given you a couple of hints...
     
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