Results 1 to 5 of 5

Math Help - Plot the function and find any asymptotes?

  1. #1
    Newbie scott_237's Avatar
    Joined
    Feb 2008
    Posts
    13

    Plot the function and find any asymptotes?

    Plot the following function and find the equation of any asymptotes [10 marks]

    could somebody please explain the answer and method using repeated factorisation used here in Q7) (pdf file)

    THANKS
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Matt Westwood's Avatar
    Joined
    Jul 2008
    From
    Reading, UK
    Posts
    824
    Thanks
    33
    It's a less obvious method of doing long division.

    Effectively you work out how many times the highest power of the divisor (the bit on the bottom) goes into the highest power of the dividend (the bit on the top).

    In this case it's "x" times.

    Then you multiply the bottom by that value, and subtract it from the top.

    Thus you have "x + ..." where what you've got left after you've subtracted that "x" is a polynomial of a reduced index (because you've extracted as many "x^2 - x - 2" factors out as you can.

    Then you do the same thing again, each time you're reducing the order of the polynomial at the top till it has a smaller order than the one at the bottom.


    Notice how "x^3" becomes "x(x^2 - x - 2) + x^2 + 2" in the first line, multiply it out and you see it comes to x^3.


    Oh yeah, I notice the "polynomial long division" at the bottom as well, that's doing in symbols what I did with words above.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie scott_237's Avatar
    Joined
    Feb 2008
    Posts
    13
    thanks matt, I eventually ended up doing the polynomial long divison method as it seemed easier. One more question about this, my answer looks identical to the polynomial method answer in the link in the first post but now that I worked out the final answer - how do I use (-3x+11) and (x/4 + 3/4) to draw the graph? and how do I conclude that there are vertical asymtotes at x = -1 and 2 from the answer that I got, I just factorise the divisor right? Thanks a lot for your help btw
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Matt Westwood's Avatar
    Joined
    Jul 2008
    From
    Reading, UK
    Posts
    824
    Thanks
    33
    We were given the "Sidamo" procedure when plotting graphs:

    S - Sign of y for given x. Work out where the zeroes are and work out where it changes sign.

    I - Intercepts. Work out where it crosses the axes.

    D - Derivatives (although you haven't done calculus yet so you won't have this in your toolbox).

    A - Asymptotes - that's the lines the function gets near but doesn't reach when the function gets very big in whatever direction.

    M and O I can't remember, maybe someone else can.

    To get a rough idea of the asymptotes, see what happens when x tends towards plus and minus infinity - in this example the constants 11 and 3/4 can effectively be ignored (as they'll reduce in importance relative to x as x gets bigger). So see what you have left and that may give you an idea as to where the asymptotes are. This is of course a horrribly inexact way of looking at things but it gives you a rough idea.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,551
    Thanks
    12
    Awards
    1
    Quote Originally Posted by scott_237 View Post
    thanks matt, I eventually ended up doing the polynomial long divison method as it seemed easier. One more question about this, my answer looks identical to the polynomial method answer in the link in the first post but now that I worked out the final answer - how do I use (-3x+11) and (x/4 + 3/4) to draw the graph? and how do I conclude that there are vertical asymtotes at x = -1 and 2 from the answer that I got, I just factorise the divisor right? Thanks a lot for your help btw
    From your original rational function, the function is undefined at x=-1 and x=2. These values cause the denominator to go to zero. The rational function cannot be reduced further, so these values make up vertical asymptotes.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: June 2nd 2009, 07:55 PM
  2. How to Plot a 'sum' function
    Posted in the Math Software Forum
    Replies: 4
    Last Post: May 25th 2009, 12:02 PM
  3. Plot a function
    Posted in the Math Software Forum
    Replies: 2
    Last Post: May 23rd 2009, 12:06 AM
  4. Find Asymptotes of Rational Function
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: July 26th 2008, 03:44 PM
  5. how do you find the asymptotes of a function?
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: March 23rd 2007, 09:46 PM

Search Tags


/mathhelpforum @mathhelpforum