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Just a question about vertical and horizontal line tests

  1. Nov 24, 2012 #1
    There's no real question per say. I've been looking in books and on the internet i can't find anything.

    I don't know when to use the vertical and horizontal line test to test for injective functions and surjective.

    Let's say if I was to sketch the graph of x2+4x - 5

    I would get a parabola, it's a function cause a vertical line cuts it once. Now a horizontal line cuts it twice... so what does that mean? It's surjective?
     
  2. jcsd
  3. Nov 24, 2012 #2

    Dick

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    If a horizontal line cuts it twice that means it's not injective. But not all horizontal lines cut it twice. Some don't cut it at all. What does that mean? Think back to what 'injective' and 'surjective' really mean. That will explain why the line test works.
     
  4. Nov 24, 2012 #3
    Hm... if it cuts it twice like 1 value of y can be mapped to x so it's not 1 to 1?
     
  5. Nov 24, 2012 #4

    Dick

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    If you mean if it cuts twice then two values of x correspond to one value of y, then yes, not injective.
     
  6. Nov 24, 2012 #5
    When do the line tests fail?
     
  7. Nov 24, 2012 #6

    Dick

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    Why would you ask that? What makes you think they do? What exactly are the 'line tests' you've been taught?
     
  8. Nov 24, 2012 #7
    Like I once heard my teacher say when it fails umm show it's one to one or w/e algebraically.
     
  9. Nov 24, 2012 #8

    Dick

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    That's kind of vague. Like, sort of.
     
  10. Nov 25, 2012 #9
    The horizontal and vertical line test is an intuitive way to see if a graph belongs to an injective function. To be clear, if the graph belongs to a function, and a horizontal line drawn anywhere on the graph means that the line will only intersect the graph at most once, then the function is injective, that is, the function evaluated at any two points means that the evaluation will be different for these two points. Continuing, a graph belongs to a function if a vertical line can be drawn anywhere on the graph and it will happen that the line intersections the graph at maximum one time. The idea here is that a graph is only a function if an evaluation at one point means you are only going to get one evaluation, and not, say, two.

    This should make everything clear enough so that you can do that example yourself. If not, say so, :).
     
  11. Nov 25, 2012 #10
    Dick I mean I heard my teacher say the tests aren't always accurate so it's better to show it algebraically. Also I I believe I have a problem showing a function is surjective algebraically
     
  12. Nov 25, 2012 #11

    Dick

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    If your teacher means that you might not be able to draw graphs accurately enough to be sure, then that makes sense. Which function are you having problems with?
     
  13. Nov 25, 2012 #12
    Determine whether or not g is onto f(x)=x^2+4x-5, x>-2
     
  14. Nov 25, 2012 #13
    Oh and thank u 5hassay for that explanation
     
  15. Nov 25, 2012 #14

    Dick

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    Did you look at the graph? What did you decide from that? You said you got a parabola. Where was the vertex?
     
    Last edited: Nov 25, 2012
  16. Nov 25, 2012 #15
    0,-5. It is not one to one
     
  17. Nov 25, 2012 #16

    Dick

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    (0,-5) is not the vertex, but it IS not 1-1. Give me a better reason. If you complete the square you can show f(x)=(x+2)^2-9. Does that help?
     
  18. Nov 25, 2012 #17
    The vertex is -2,9
     
  19. Nov 25, 2012 #18

    Dick

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    No, it's not. It's (-2,-9). But I'm not really even interested in that. I am interested in what you think about surjectivity.
     
  20. Nov 25, 2012 #19
    Oh I meant to type-9 I'm on a mobile device but um it's surjective since mOre two things in x map to one y.
     
  21. Nov 25, 2012 #20

    Dick

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    Oh, come on. Surjective doesn't have anything to do with two things mapping onto one. Look up surjective on the mobile device.
     
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