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Vert and horiz asymptotes, axis same transformations and other things.

  1. Aug 31, 2011 #1
    oblique and horiz asymptotes, axis same transformations and other things.

    i do not understand these at all and the text makes no sense (see attachment).

    i can find the asymptotes of all types, but i do not understand how the methods i use work. please explain the methods and the reasoning behind them in detail.

    the horizontal and oblique ones are what confuse me.


    also, for some reason my book says that one cannot have bot horizontal and oblique asymptotes yet i am pretty sure i can draw a graph that does. am i somehow wrong?


    and also in regards to flipping things about axises

    -1(1/x) = 1/-x = -1/x so i could flip such a graph around either axis and get the same result, right?

    but

    -1(1/x^2) = 1/-x^2 = -1/x^2 yet this only allows a rotation about the x axis? why?
     
    Last edited: Aug 31, 2011
  2. jcsd
  3. Aug 31, 2011 #2

    Mark44

    Staff: Mentor

    Yes. Horizontal and oblique asymptotes have to do with the behavior of the graph of a function for large values of the independent variable and for very negative values of the variable.

    A function has an oblique asymptote if, for large |x|, the graph approaches the graph of a line whose equation is y = mx + b, for some constants m and b. A function has a horizontal asymptote if, for large |x|, the graph approaches the graph of a line whose equation is y = b, for some constant b. For a horizontal asymptote, the slope of this line is zero. For oblique asymptotes, the slope is necessarily nonzero.
    The graph of y = 1/x has what is called symmetry about the origin. This means that for any point (x, y) on the graph, the point (-x, -y) is also on the graph. If you reflect the graph across the x-axis or across the y-axis, you get the graph of y = -1/x.
    The transformation you're talking about is a reflection, not a rotation. The graph of y = 1/x2 is symmetric about the y-axis. This means that if (x, y) is on the graph, the point (-x, y) is also on the graph. Reflecting the graph of y = 1/x2 across the y-axis gives you a graph with identical appearance. Reflecting the graph across the x-axis, however, gives you a different graph.
     
  4. Oct 6, 2011 #3
    So, how can a graph have both horizontal and oblique asymptotes?

    Also, what I really want to know is why the operations for finding these things work.

    why does dividing the quotient out give me the asymptote?

    and why... well, see the attached image for this one as I have no way to type it.

    why is it that when we get y= c + a/b is it that c is the oblique asymptote? or should c be a constant the horizontal asymptote?

    and i still don't get why when the equation is proper 0 is the horizontal asymptote.

    please explain this
     

    Attached Files:

  5. Oct 6, 2011 #4

    SammyS

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    Re: oblique and horiz asymptotes, axis same transformations and other things.

    You apparently are confused about several concepts.

    The functions whose graphs you refer to regarding asymptotes are likely rational functions.
    I see no such attachment.
    How canyou find them if you can't explain the method?
    Draw us such a graph and attach it.
    Mark44 has explained the "flipping".
     
  6. Oct 6, 2011 #5
    i can follow the steps needed to complete the task. that does not mean that i can understand why the steps work. I am asking for an explanation of why they work. not a list of steps.

    and you are right, i do not understand the concepts behind them. i am asking for explanations of those concepts. i think someone else would be better suited to answering my question.

    there is an image attached.

    such a graph need not be drawn. just stick a horizontal asymptote and a vertical one or two or even and oblique in there and draw some lines that adhere to them. as for finding an equation of one, i will provide that later. i will have to look them up as it's been a while.

    this would probably be a good example of an equation with both.

    3/((x-1)(x^2-4))

    an image of it is attached. it is not an exact image.
     

    Attached Files:

    Last edited: Oct 6, 2011
  7. Oct 6, 2011 #6

    symbolipoint

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    Curd,
    A rational function which has a numerator of degree higher than the denominator's degree, may have an oblique asymtote. When stated in its strictly numerator & denominator form, that exact asymtote is difficult to see. You perform polynomial division and THEN examine the limiting behavior as x approaches extreme values. The limiting behavior is much easier to investigate after tranformed into this long-division-processed form.
     
  8. Oct 6, 2011 #7
    I know that is the process you follow and I know that according to the step that you have to do that to get it into a form that allows you to define the asymptote.

    I am not asking what you do. although I've noticed that this is very hard to do with "proper" equations.

    I am asking why does it work.
     
  9. Oct 6, 2011 #8

    SammyS

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    By "proper" equation, I take it that you are referring to a rational expression in which the degree of the numerator is less than the degree of the denominator -- somewhat like the case of a proper fraction.

    If, on the other hand, the degree of the numerator is equal to or greater than that of the denominator, the rational expression is somewhat like an improper fraction. Of course, rational expressions are more complicated than fractions, so the analogy isn't perfect.

    In my opinion, the place to start on this is to consider the behavior of "proper" rational functions, those in which the degree of the numerator is less than the degree of the denominator, and then move on to the long-division thing.

    Why are there vertical asymptotes? And why is there a horizontal asymptotes along the x-axis? Yes, for a "proper" rational function, the horizontal asymptote is always the x-axis. That's the horizontal line, y=0.
     
  10. Oct 6, 2011 #9

    SammyS

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    No rational function has both horizontal and oblique asymptotes.

    The function y = 3/((x-1)(x^2-4)), for which you sketched a graph, has 3 vertical asymptotes and one horizontal asymptote.
     
  11. Oct 6, 2011 #10
    i think i ran across one that did have that, but it may not have been rational.

    i'll look it up later.

    what about the other questions?
     
  12. Oct 6, 2011 #11
    again, i know this.

    I want to know WHY these operations tell us what the asymptotes are.
     
  13. Oct 6, 2011 #12

    SammyS

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    So, you do know why the graph of a "proper" rational function has a horizontal asymptote of y=0 ?

    I thought that's one of the things you need to have explained.
     
  14. Oct 7, 2011 #13
    that's right to some extent. although i think i have a better grasp of it now. what I need explained is why these operations are what we do to find these asymptotes. why do these operations work?

    I know that as x approaches infinity that the value of the function approaches zero.

    my current theory is that they (when dividing for non proper ones anyway) put the equation into a workable form in which one has an equation in the form of a quotient that shows that as x approaches infinity in either direction that the value of the overall quotient section goes toward zero, which is then modified with a transformation symbolized by C in C + A/B so that the slope of that equation is changed as it approaches zero.


    i'm not sure if that's an accurate view though.

    so far you've told me that these operations work. I do not think you are telling me how and why they work.


    also, why does one, when dividing these polynomials, add the result that's on top of the long division sign to the result in front of it divided by the result below it? (a picture useful for this question is in post #3)
     
    Last edited: Oct 7, 2011
  15. Oct 7, 2011 #14

    Mark44

    Staff: Mentor

    A proper rational function is the quotient of two polynomials p(x)/q(x), where the degree of the numerator is less than the degree of the denominator.

    When the degree of p(x) is less than the degree of q(x) (e.g., f(x) = x/(x^2 + 1)) the horizontal asymptote is the x-axis. The reason for this is that as x gets very large or very negative, the denominator grows large (or very negative) more quickly than the numerator, so the whole fraction gets closer to zero.

    When the degree of p(x) equals the degree of q(x), you can perform polynomial long division to write the function as a polynomial plus a proper rational function. The division that is done is similar to what you show in your attachment.

    Let's look at a simple example with just numbers: 7 divided by 3. Back when you were first learning division you probably would have done something like this:
    Code (Text):

    ...___2 R 1
    3) 7
    IOW, 3 goes into 7 two times, with a remainder of 1

    So 7 = 2 * 3 + 1
    If you divide both sides of this equation by 3, you get
    7/3 = 2 + 1/3

    You get an integer (2) plus a rational number (1/3)
    Polynomial long division works a lot like this.

    If your (improper) rational function is g(x) = (x + 3)/(x + 1), you can do poly. long division to get

    (x + 3)/(x + 1) = 1 + 2/(x + 1), a polynomial (1) plus a proper rational function (2/(x + 1))

    You can check that this is correct by multiplying both sides of the equation above by x + 1, which gives you
    x + 3 = 1*(x + 1) + 2 = x + 1 + 2 = x + 3, which is identically true.

    This particular rational function, g(x) = 1 + 2/(x + 1) has a horizontal asymptote: the line y = 1. As x gets very large or very negative, the 2/(x + 1) part approaches 0, so that the graph gets closer and closer to the line y = 1.

    If the degree of the numerator in an improper rational function is one more than the degree of the denominator, division will produce a polynomial of the form kx plus a proper rational function. In this case, there is an oblique asymptote, the line y = kx.

    Hopefully, this covers the how and why you're asking about.
     
  16. Oct 7, 2011 #15
    it's a lot better. i think the rest i'm going to have to come to on my own.
     
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