How do you determine the end behavior of a rational function?
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.
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$\begingroup$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.
$\endgroup$ $\begingroup$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.
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