# Reduction of a relationship to linear form

• Jan 3rd 2009, 07:44 AM
p3achy
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(Worried)
Any pointers in the right direction would be appriciated
thanks in advance and hope this make sense(Happy)(Happy)
• Jan 3rd 2009, 08:10 AM
Mush
Quote:

Originally Posted by p3achy
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(Worried)
Any pointers in the right direction would be appriciated
thanks in advance and hope this make sense(Happy)(Happy)

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.
• Jan 3rd 2009, 08:36 AM
p3achy
Quote:

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
• Jan 4th 2009, 04:48 AM
p3achy
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
• Jan 4th 2009, 02:18 PM
p3achy
just in case anyone is really bored and fancies a peek(Wink)