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Are the functions the same?

  1. Sep 12, 2011 #1
    Suppose you have these three functions:

    I. y = x-2

    II. y = (x2-4)/(x+2)

    III. (x+2)y = (x2-4)

    It asks whether these functions have the same graph.

    I thought II and III were the same, but it says none of the above functions have the same graph.

    Am I missing something??:cry:
     
  2. jcsd
  3. Sep 12, 2011 #2

    berkeman

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    Please show us your work. Particularly your factoring skills....
     
  4. Sep 12, 2011 #3

    micromass

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    Hi BF! :smile:

    You are missing the behaviour in -2. When you fill in x=-2 in II, then it is undefined; as you divide by 0. But filling in x=-2 in III doesn't yield an error such as division by 0.

    That said, I'm quite troubled by III, as it doesn't specify a function to me.
     
  5. Sep 12, 2011 #4
    I knew of the behavior of some of the functions at -2, but I didn't think it would be a homework question because I saw this in a previous AMC-12.

    http://www.ncssm.edu/courses/math/NCSSM%20Student%20Materials/AMC%20Problems/Sample%20Questions%20from%20past%20AMC.pdf [Broken]

    I. y = x-2
    II. y = x-2 with hole at -2
    III. y = x-2 with hole at -2 (from what I originally thought, but I guess that may not be true...)
     
    Last edited by a moderator: May 5, 2017
  6. Sep 12, 2011 #5

    berkeman

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    It's certainly schoolwork (doesn't matter if it's assigned homework or self-study).

    As to the equations, are we not allowed to factor and cancel to get to the simplest equation y=f(x) before graphing? It sounds from micromass' reply that we are not allowed to do that...
     
    Last edited by a moderator: May 5, 2017
  7. Sep 12, 2011 #6

    dynamicsolo

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    What I believe he means is that you can't just do that: the (x+2) in the denominator of the rational function tells you that the function is not defined at x = -2 . Algebraically speaking, you will get y = x - 2 in all three cases. However, you must beware of any divisor in an algebraic equation that can be zero for some (one or more) value(s) of x . (So, yes, you can divide it out, but don't forget that you did that...)

    So the first function is just the straight line y = x - 2 and is defined for all values of x . The other two functions act like y = x - 2 everywhere except at x = -2 . Thus, they are technically different functions and we must leave a "hole" in the straight line for those graphs at ( -2, -4 ). [Some graphing calculators and computer software will even do this -- when you "zoom in" on that region, you would see a break in the line at that point.]
     
  8. Sep 12, 2011 #7

    micromass

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    But that's the point. The third one IS defined in x=-2. Indeed, if a treat it as an implicit function, then

    [itex](x+2)y = (x^2-4)[/itex]

    is defined in x=-2 for all y-values. So to the x-value -2, there corresponds multiple y-values. This makes it not a function. This is why I'm confused.
     
  9. Sep 12, 2011 #8

    dynamicsolo

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    I wonder if there is some range of interpretation as to how people would read that equation. I would say this: the equation has a well-defined solution for y for every value of x not equal to -2 . As you point out, the value of y is indeterminate at x = -2 (any real value of y works!). So this still behaves as a function except at x = -2 and its graph looks just like that of (II). I believe that effectively (II) and (III) describe the same function, which is distinct from (I).
     
  10. Sep 12, 2011 #9

    micromass

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    Yes, that is a good point-of-view. But I usually interpret the graph of a implicit function as

    [tex]\{(x,y)\in \mathbb{R}^2~\vert~(x+2)y = (x^2-4)\}[/tex]

    So the graph would look like (II) + the line x=-2.

    However, my intepretation is probably not what they mean. But I would like to see what they DO mean... Your interpretation is probably the right one.
     
  11. Sep 12, 2011 #10

    dynamicsolo

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    I honestly hadn't considered that since I don't see equations handled that way in the work I usually do. But given the source of this problem, that could be exactly what they're after for (III), in which case it isn't a function (generally speaking) and all three equations describe different sets of points. (I'll be wary of that sort of thing henceforth...)

    [EDIT: Ah, going back to BloodyFrozen's attachment, the AMC-12 problem doesn't say the three equations all describe functions; it simply asks which of the three equations have the same graphs. In that case, I think you're right about (III) -- the given answer to the problem is in fact (E). ]
     
  12. Sep 13, 2011 #11

    micromass

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    No, only (2,0) satisfies the equation.
     
  13. Sep 13, 2011 #12
    I deleted my post since it was incorrect.
     
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