# Finding the standard form of a rational function

• Mar 30th 2011, 04:08 PM
brumby_3
Finding the standard form of a rational function
Find the standard form for (20x^2+11x-1)/(4x+3)?

Not sure what 'standard' form means, help would be appreciated (Happy)
• Mar 30th 2011, 04:14 PM
pickslides
I'm not sure what is meant either, but maybe the requirement is to perform polynomial long division here?

I.e make $\displaystyle \frac{20x^2+11x-1}{4x+3} = a+\frac{b}{4x+3}$
• Mar 30th 2011, 04:19 PM
brumby_3
well after finding the standard form, I'm asked to find all straight line asymptotes (vertical, horizontal or oblique), maybe that might give a clue?
• Mar 30th 2011, 04:23 PM
pickslides
Well now knowing this, my advice in post #2 is the way forward.
• Mar 30th 2011, 04:34 PM
brumby_3
How do I find a and b? Would you be able to show me an example (I've got quite a few of these to do and seeing the steps would help a lot!)
• Mar 30th 2011, 04:37 PM
pickslides
Polynomial long division is the road I would take.

Polynomial Long Division

Spoiler:
$\displaystyle \frac{20x^2+11x-1}{4x+3} = 5x-1+\frac{2}{4x+3}$
• Mar 30th 2011, 04:41 PM
brumby_3
Thanks a lot, I really appreciate it :)
• Mar 30th 2011, 04:47 PM
pickslides
Are you aware of what this means in context of the following question about asymptotes?
• Mar 30th 2011, 04:59 PM
brumby_3
I just did all the working out for the question using the example, thanks a lot! For the asymptotes, I'm not sure how to figure them out in this form. Don't I have to find where (4x+3) is 0 so that it is infinity or something along those lines?

Is the slant/oblique asymptote just 5x-1? and I think there is no vertical or horizontal asymptotes
• Mar 30th 2011, 05:51 PM
pickslides
Quote:

Originally Posted by brumby_3
Don't I have to find where (4x+3) is 0

Yep, solve 4x+3=0, that is the vertical asymptote.

Quote:

Originally Posted by brumby_3

Is the slant/oblique asymptote just 5x-1?

Yep!