The intuition behind slant asymptotes

Question

I'm asking this question to get a better understanding of oblique asymptotes.

As regards vertical asymptotes, I know that they represent the numbers that make a function undefined. An example of this is a $0$ in the denominator, or a negative argument to a square root since that would return a complex number.

For horizontal asymptotes, I know that if the degree of the numerator is greater than the degree of the denominator of a rational function, there are no horizontal asymptotes because as x tends to infinity, the result of the function keeps getting larger and larger. Also, if the degree of the denominator is greater than the degree of the numerator, there is a horizontal asymptote at $y = 0$ because as $x$ gets larger and larger, the value of the denominator greatly exceeds the value of the denominator and thus the value gets closer and closer to $0$. And finally, if the degrees are equal, the horizontal asymptote will be $y = \frac{n}{d}$ where $n$ are the leading coefficients on the numerator and denominator respectively.

But for slant asymptotes, my book just says that they will exist only when the degree of the numerator is exactly one more than that of the denominator (and that you need to divide the polynomials to find the equation of the line of the slant asymptote).

My question is, why is it that slant asymptotes only occur when the degree of the numerator is exactly one more than that of the denominator?

Answer 1

Just to elaborate on Michael Hardy's answer a little bit...

Suppose we have a rational function written like $f(x)=\frac{p(x)}{q(x)}$. Performing the long division $p(x)\div q(x)$ gives you a different way of writing the same function, $$f(x)=a(x)+\frac{r(x)}{q(x)},$$ as a polynomial plus a rational function whose numerator's degree is smaller than its denominator's degree. (To be more specific, $a(x)$ has degree equal to the degree of the original numerator $p(x)$ minus the degree of the denominator $q(x)$. The degree of $r(x)$ is smaller than that of $q(x)$ because $r(x)$ is the remainder of the long division.)

As you already understand, $\frac{r(x)}{q(x)}$ tends to zero as $x$ tends to infinity, because the degree of the numerator is smaller than the degree of its denominator. That means that as $x\rightarrow\infty$, the function $f(x)$ tends towards the polynomial $a(x)$.

If $p(x)$ has degree exactly one more than that of $q(x)$, then $a(x)$ has degree one: it's a line. We call this line a slant or oblique asymptote of $f(x)$. If the degree of $p(x)$ is more than one higher than that of $q(x)$, then $a(x)$ is a quadratic, or a cubic, or whatever, but it's not a line. So $f(x)$ doesn't have a slant asymptote in this direction. (We still might say something like $f(x)$ asymptotically approaches a(x).)

Hopefully this explains why asymptotes only occur when the degree of the numerator is exactly one more than that of the denominator. It also might give you a hint for how you can find slant asymptotes of functions that aren't rational: if you can rewrite your function as a line plus something that goes to zero, you've got yourself an asymptote!

Answer 2

It is not true that a rational function always has a vertical asymptote when there's a $0$ in the denominator. When there are $0$s in both the numerator and denominator, then it often---in fact usually---happens that there is no vertical asymptote, but rather the function approaches a finite limit. When there is a $0$ in the denominator and something else in the numerator, then there's a vertical asymptote.

As for slant asymptotes, do long division. For example suppose you have $$ f(x) = \frac{18x^5 + 2x^4-91x^3 + \cdots}{3x^4 + 11x^3 -10x^2 + \cdots} $$ Then do long division: $$ \begin{array}{ccccccccccccccccccc} & & 6x & {} + \tfrac{-64}{3} \\ \\ 3x^4 + 11x^3 -10x^2 + \cdots & {\Big)} & 18x^5 & {}+ 2x^4 & {} - 91x^3 & {} +\; \cdots \\ & & 18x^5 & {} + 66x^4 & {} - 60 x^3 & {} +\; \cdots \\ \\ \\ & & & {} - 64x^4 & {} -31x^3 & {} + \;\cdots \\ & & & {} -64x^4 & {} -\tfrac{704}{3}x^3 & {} + \;\cdots \\ \\ \\ & & & & \tfrac{611}{3} x^3 & {} +\;\cdots \end{array} $$ Then $$ f(x) = \underbrace{6x - \frac{64}{3}} + \underbrace{\frac{\frac{611}{3}x^3+\cdots}{3x^4 + 11x^3 -10x^2 + \cdots}} $$

The graph of $$ y = 6x - \frac{64}{3} $$ is a slanted line.

The remaining expression $$ \frac{\frac{611}{3}x^3+\cdots}{3x^4 + 11x^3 -10x^2 + \cdots} $$ approaches $0$ as $x$ approaches $\pm\infty$.

Answer 3

If you try performing the division of the two polynomials, you'll note that you can always separate your rational function into a "linear" portion and a "rational" portion, where the numerator's degree is less than the denominator. This is akin to decomposing improper fractions as mixed numbers (integer + proper fraction). As $x$ increases in magnitude, the rational portion goes to zero (why?), and in that range, the linear part of your function dominates. Hence the oblique asymptotes you observe.

Answer 4

For $\frac{f}{g}$, with degree of $f = m$ > degree of $g=n$, dividing will give you a polynomial of degree $m - n$, and a remainder. If $m-n =1$, then you have a linear polynomial (plus remainder).