Welcome to our community

Be a part of something great, join today!

Why is this an asymptote?

m3dicat3d

New member
Jan 11, 2013
19
slant asymtope.JPGWhy is the slant asymptote pictured here correct for this function? I was under the impression an asymptote was never crossed by the function. I get that the dividend gives the equation for the asymptote for a non zero remainder, but seeing this graphically is a bit confusing. Thanks! (EDIT: NVM, I just answered my own question:p) Thanks anyways though!
 
Last edited:

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
If you long divide, you'll see that \(\displaystyle \displaystyle \frac{-6x^3 + 4x^2 - 1}{2x^2 + 1} \equiv -3x + 2 + \frac{x}{2x^2 + 1} = -3x + 2 + \frac{1}{2x + \frac{1}{x}} \).

You can see that the denominator easily overpowers the numerator, and so gets closer to 0, which means the entire function eventually works like \(\displaystyle \displaystyle -3x + 2\). However, since there is always that tiny bit added, it never actually will be \(\displaystyle \displaystyle -3x + 2\), and so \(\displaystyle \displaystyle y = -3x + 2\) must be an asymptote :)