# Why is this an asymptote?

#### m3dicat3d

##### New member
Why 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) Thanks anyways though!

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#### Prove It

##### Well-known member
MHB Math Helper
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