Page 1 of 2 12 LastLast
Results 1 to 15 of 16
Like Tree5Thanks

Thread: Vertical Asymptotes

  1. #1
    Senior Member
    Joined
    Mar 2012
    Posts
    465
    Thanks
    5

    Vertical Asymptotes

    f(x) = (1 - x) / [(1 - x)(5 + x)]

    My professor argues with us that the value 1 is considered a vertical asymptote.

    Is he correct?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,155
    Thanks
    3672

    Re: Vertical Asymptotes

    no vertical asymptote ... there is a point discontinuity (a hole) at x = 1

    see attached TI-84 graph ... tick mark on the x-axis is x = 1
    Attached Thumbnails Attached Thumbnails Vertical Asymptotes-hole.png  
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Mar 2012
    Posts
    465
    Thanks
    5

    Re: Vertical Asymptotes

    i know that there is a hole at x = 1, but my professor consider this point of discontinuity as a vertical asymptote.

    How to convince him that it is not?

    Arguing with professors is dangerous cuz they are always right!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,155
    Thanks
    3672

    Re: Vertical Asymptotes

    Sounds like it's time to stand up, show your evidence, and argue your point ... do you have face-to-face contact with this "professor", or is he/she some online hack? Don't be intimidated. What can he/she do to you? ... take away your birthday?
    Thanks from topsquark and joshuaa
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Mar 2012
    Posts
    465
    Thanks
    5

    Re: Vertical Asymptotes

    He is with me face to face in class. I feel that he made me wrong cuz he does not want to show students that he got a mistake, but if x = 1, will never be a vertical asymptote, i can discuss this issue with him again in his office. I just want to make sure he is wrong cuz if he is right, i am confused about this answer.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,071
    Thanks
    708
    Awards
    1

    Re: Vertical Asymptotes

    Quote Originally Posted by joshuaa View Post
    He is with me face to face in class. I feel that he made me wrong cuz he does not want to show students that he got a mistake, but if x = 1, will never be a vertical asymptote, i can discuss this issue with him again in his office. I just want to make sure he is wrong cuz if he is right, i am confused about this answer.
    Just for validation, skeeter is 100% correct about this. Feel free to ask him to come online and discuss it with us.

    -Dan
    Thanks from joshuaa
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Mar 2012
    Posts
    465
    Thanks
    5

    Re: Vertical Asymptotes

    Thanks alot guys. now, i can show him my muscles in this issue.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member
    Joined
    Mar 2012
    Posts
    465
    Thanks
    5

    Re: Vertical Asymptotes

    This is the problem we discussed in class.

    Determine all horizontal and vertical asymptotes. For each vertical asymptote, determine whether f(x) goes to infinity or f(x) goes to negative infinity on either side of the asymptote.

    f(x) = (1 - x) / (x^2 + x - 2)
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,155
    Thanks
    3672

    Re: Vertical Asymptotes

    fyi, this is not the same function you originally posted $\rightarrow$
    f(x) = (1 - x) / [(1 - x)(5 + x)]

    f(x) = (1 - x) / (x^2 + x - 2)
    $f(x) = \dfrac{1-x}{(x+2)(x-1)} = - \dfrac{x-1}{(x+2)(x-1)}$

    this function still has a point discontinuity at x = 1
    Thanks from joshuaa
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,420
    Thanks
    2889

    Re: Vertical Asymptotes

    What definition does your teacher or text give for "vertical asymptote"?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Senior Member
    Joined
    Mar 2012
    Posts
    465
    Thanks
    5

    Re: Vertical Asymptotes

    But it has the same concept.

    I know that x = -2 is a vertical asymptote.

    But would you consider x = 1 as a vertical asymptote?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,155
    Thanks
    3672

    Re: Vertical Asymptotes

    Quote Originally Posted by joshuaa View Post
    But it has the same concept.

    I know that x = -2 is a vertical asymptote.

    But would you consider x = 1 as a vertical asymptote?
    a vertical asymptote is an infinite discontinuity, not a point discontinuity

    for $f(x) = -\dfrac{x-1}{(x+2)(x-1)}$ ...

    x = -2 is a vertical asymptote , point discontinuity at x = 1, horizontal asymptote is y = 0
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Senior Member
    Joined
    Mar 2012
    Posts
    465
    Thanks
    5

    Re: Vertical Asymptotes

    May be this is the definition.

    Examine lim 1/x as x goes to 0

    Of course, we can draw a graph and compute a table of function values easily, by hand. While we say that the limits lim 1/x as x goes to 0+ and lim 1/x as x goes to 0- do not exist, they do so for different reasons. Specifically, as x goes 0+, 1/x increases without bound, while as x goes 0-, 1/x decreases without bound. To indicate this, we write

    (1) lim 1/x as x goes 0+ = Infinity

    and

    (2) lim 1/x as x goes 0- = -Infinity

    Graphically, this says that the graph of y = 1/x approaches the vertical line x = 0, as x goes to 0. When this occurs, we say that the line x = 0 is a vertical asymptote. It is important to note that while the limits (1) and (2) do not exist, we say that they equal Infinity and -Infinity, respectively, only to be specific as to why they do not exist. Finally, in view of the one-sided limits (1) and (2), we say as before that

    lim 1/x as x goes 0 does not exist
    Last edited by joshuaa; Oct 18th 2016 at 11:10 AM.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Senior Member
    Joined
    Mar 2012
    Posts
    465
    Thanks
    5

    Re: Vertical Asymptotes

    Quote Originally Posted by skeeter View Post
    a vertical asymptote is an infinite discontinuity, not a point discontinuity

    for $f(x) = -\dfrac{x-1}{(x+2)(x-1)}$ ...

    x = -2 is a vertical asymptote , point discontinuity at x = 1, horizontal asymptote is y = 0
    Thanks a Lot skeeter. This is exactly what I want to reach to.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,459
    Thanks
    2728
    Awards
    1

    Re: Vertical Asymptotes

    Quote Originally Posted by joshuaa View Post
    I know that x = -2 is a vertical asymptote.
    But would you consider x = 1 as a vertical asymptote?
    f(x) = \frac{{(1 - x)}}{{(1 - x)(5 + x)}} = \frac{1}{{(5 + x)}},\text{ if }x \ne 1 then \displaystyle{\lim _{x \to 1}}f(x) = \frac{1}{6}

    If your instructor sees this and then insists he/she is correct, then take the matter to the head of school.
    Thanks from joshuaa
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Vertical Asymptotes
    Posted in the Algebra Forum
    Replies: 5
    Last Post: Oct 15th 2016, 08:49 AM
  2. vertical asymptotes
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: Oct 20th 2009, 05:22 PM
  3. Vertical Asymptotes
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Oct 4th 2009, 06:53 PM
  4. vertical asymptotes
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: Oct 2nd 2008, 07:17 PM
  5. vertical asymptotes
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Jul 7th 2006, 07:56 AM

Search Tags


/mathhelpforum @mathhelpforum