IMPORTANT Simplification!!!

Feynman

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Feb 5, 2012
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So I am trying to find the Asymptotes of this function, but I would like it simplified.

I dont know how to do this thought.


Please help!

Here it is:

y = 3(x-1)/(x+1)(x-2) + 2(x-1)/(x-2)(x-4)

I need to know how to simplify this

thanks
 
So I am trying to find the Asymptotes of this function, but I would like it simplified.

I dont know how to do this thought.


Please help!

Here it is:

y = 3(x-1)/(x+1)(x-2) + 2(x-1)/(x-2)(x-4)

You might want to combine the two terms on the right side of the equation into a single fraction. To do this, you need a common denominator, which would be (x + 1)(x - 2)(x - 4). Multiply the first term by (x - 4)/(x - 4) and the second term by (x + 1)/(x + 1). Then combine the two terms. Can you continue from there?
 
You might want to combine the two terms on the right side of the equation into a single fraction. To do this, you need a common denominator, which would be (x + 1)(x - 2)(x - 4). Multiply the first term by (x - 4)/(x - 4) and the second term by (x + 1)/(x + 1). Then combine the two terms. Can you continue from there?

First off, thank you so much for genuinely helping me! :)


So is the final answer:

10/(x+1)(x-2)(x-4)


?
 
Originally: 3(x-1)/(x+1)(x-2) + 2(x-1)/(x-2)(x-4)

I'm assuming you meant this:
3(x-1) / [(x+1)(x-2)] + 2(x-1) / [(x-2)(x-4)]
and
10 / [(x+1)(x-2)(x-4)]
(the extra brackets are VERY important...see why?)

As an aside, x cannot equal -1, 2 and 4 : see why?

Your solution is not correct: you can check these yourself by assigning a
value to x (as example, 5) and substituting in original and in your solution;
if both have same results, then you've hit pay dirt!

I got down to this:
5(x-1) / [(x+1)(x-4)] : it does check out ok...
So you were not too far out ;)

OH i see how you got the 5(x-1) / [(x+1)(x-4)]... i was over-thinking way to hard (as I usually do) >.<


so are -1, 2, and 4 the only asymptotes of the graph? I have not checked using and formulas but just by looking at the graph it seems that way. Also, for future use, can I find all asymptotes simply by finding values that make x = 0 for the denominator? just curious
 
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