Results 1 to 5 of 5

Math Help - Reduction of a relationship to linear form

  1. #1
    Newbie
    Joined
    Jan 2009
    Posts
    11

    Reduction of a relationship to linear form

    Sorry this isnt urgent home work but I couldnt figure out what section to put this under.
    I have been given y = 1/(x-a)(x-b)
    and have to explain how i would verify the relationship graphically and then how to determine a and b from the graph. i know i just have to do a sketch.
    im assuming i have to take logs to get it into the format of y = mx+c
    but i keep going round in circles.
    Then im guessing to work out a and b its just going to be the gradient and intercept. Ive done a few of these but none where there are two x's and its thrown me
    Any pointers in the right direction would be appriciated
    thanks in advance and hope this make sense
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Dec 2008
    From
    Scotland
    Posts
    901
    Quote Originally Posted by p3achy View Post
    Sorry this isnt urgent home work but I couldnt figure out what section to put this under.
    I have been given y = 1/(x-a)(x-b)
    and have to explain how i would verify the relationship graphically and then how to determine a and b from the graph. i know i just have to do a sketch.
    im assuming i have to take logs to get it into the format of y = mx+c
    but i keep going round in circles.
    Then im guessing to work out a and b its just going to be the gradient and intercept. Ive done a few of these but none where there are two x's and its thrown me
    Any pointers in the right direction would be appriciated
    thanks in advance and hope this make sense
    I would draw the graph with a table of sign. Vertical asymptotes are points where, as the graph approaches them, it trails off to plus/minus infinity, due to the fact that at these points, the graph is undefined. Your graph is undefined at two points. x = a, and x = b, because in this situation there is a divide by zero. Hence, you could verify the values of a and b by noting where the graph is undefined and trails off to plus/minus infinity.

    Logarithms indeed linearise such equations.

     \log |y| = \log |\frac{1}{(x-a)(x-b)}|

     \log |y| = \log |1| - \log|(x-a)(x-b)|


     \log |y| =  -\log|(x-a)| - \log|(x-b)|

    It isn't in the form "y=mx+c", but it is still a linear equation.
    Last edited by Mush; January 3rd 2009 at 09:06 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2009
    Posts
    11
    It isn't in the form "y=mx+c", but it is still a linear equation.
    could be where i was going wrong was trying to force it into y=mx+c and getting no where
    your explaination sure did help for the next part of the question.
    but knew i had bumbled my wording thou its the linear graph i have to sketch and i have no idea what i write on the x and y axis
    Last edited by p3achy; January 3rd 2009 at 07:52 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jan 2009
    Posts
    11

    Smile

    Sorry cant figure out how to get the maths type to look right, so hope this isnt to bad to read
    i know that if i have y = ab^x then take logs i get
    logy = loga + xlogb

    then if i plot logy by x on a graph i can work out a and b
    so my problem was that if i have
    y = 1/(x-a)(x-b)
    use logs to get

    \log |y| = -\log|(x-a)| - \log|(x-b)|
    im unclear as to what gets plotted against what
    hope this makes sense and any help much appreciated
    Last edited by p3achy; January 4th 2009 at 03:53 AM. Reason: mathstype issues
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jan 2009
    Posts
    11

    Smile

    just in case anyone is really bored and fancies a peek
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Linear relationship of medians?
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: November 14th 2011, 12:54 AM
  2. Matrix Linear relationship
    Posted in the Advanced Algebra Forum
    Replies: 8
    Last Post: March 15th 2011, 04:10 PM
  3. Replies: 3
    Last Post: September 25th 2010, 03:48 PM
  4. reduction of a relationship to a linear law
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 6th 2009, 01:57 AM
  5. 2 linear regression lines, relationship?
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: April 1st 2009, 02:37 PM

Search Tags


/mathhelpforum @mathhelpforum