How do you determine the end behavior of a rational function?

Question

Example

$$\frac{6x + 2}{x^2 - 9} = \frac{6x + 2}{(x + 3)(x - 3)}$$

I know how to find the vertical and horizontal asypmtotes and everything, I just don't know how to find end behavior for a RATIONAL function without plugging in a bunch of numbers.

Answer 1

Horizontal asymptotes (if they exist) are the end behavior. However horizontal asymptotes are really just a special case of slant asymptotes (slope$\;=0$).

The slant asymptote is found by using polynomial division to write a rational function $\frac{F(x)}{G(x)}$ in the form

$$\frac{F(x)}{G(x)} = Q(x) + \frac{R(x)}{G(x)}$$

Where Q(x) (the quotient) is the line of your slant asymptote, and the degree of $R(x)$ (the remainder) is strictly less than the degree of $G(x)$.

In your example, the degree of the numerator is already strictly less than the degree of the denominator, so what is your slant asymptote? This will tell you the end behavior.

Answer 2

If you are concerned by the behavior of the function when $x$ starts to be large, just perform the long division of polynomials.

For $$f(x)=\frac{6x+2}{x^2-9}$$ this will give $$f(x)\approx \frac{6}{x}+\frac{2}{x^2}$$ and then the asymptote would be function $\frac{6}{x}$.

Changing to $$g(x)=\frac{6x^2+2}{x^2-9}$$ this will give $$g(x)\approx 6+\frac{56}{x^2}$$ and then the asymptote would be function $6$, an horizontal asymptote.

Changing to $$h(x)=\frac{6x^3+2}{x^2-9}$$ this will give $$h(x)\approx 6 x+\frac{54}{x}+\frac{2}{x^2}$$ and then the asymptote would be function $6 x$, an oblique asymptote.

You could notice that this simple division gives you the asymptote as well as the manner the function appoaches it.