Originally Posted by

**drumist** Is this the correct function?

$\displaystyle f(x) = \frac{3x^3-x+1}{x+3}$

To determine the end-behavior (i.e., what happens at the extreme "left" and extreme "right" of the graph), you can take the largest power term from each the numerator and denominator, and calculate the limit like this:

$\displaystyle \lim_{x\to+\infty} \frac{3x^3-x+1}{x+3} = \lim_{x\to+\infty} \frac{3x^3}{x} = \lim_{x\to+\infty} 3x^2 = +\infty$

You should also check the left side if it's not a simple rational function. In this case it's not necessary, but just to be thorough:

$\displaystyle \lim_{x\to-\infty} \frac{3x^3-x+1}{x+3} = \lim_{x\to-\infty} \frac{3x^3}{x} = \lim_{x\to-\infty} 3x^2 = +\infty$

Since the end-behavior goes to infinity on both sides, there aren't any horizontal asymptotes.