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I am currently learning about the asymptotes of rational functions in precalculus. However, 4 things about it confuse me, specifically about the asymptotes of it, and how to calculate them.

  1. Why is it when given a rational function with 2 polynomials of the same degree on their numerators and denominators, the asymptote can be found by dividing the coefficients of the highest degree terms?

  2. Why is it if the polynomial in the numerator is a lower degree than the denominator, the x-axis (y = 0) is the horizontal asymptote?

  3. Why is it if the polynomial in the numerator is a higher degree than the denominator, there is no horizontal asymptote?

  4. Why is it when the polynomial in the numerator is a higher degree than the polynomial in the denominator, there will be a slant asymptote which you can find through polynomial long division?

Can you explain all of this using simple algebra, without complicated techniques such as Euler's division of polynomials? I don't understand any complicated techniques and theorems beyond the quadratic formula. Can you also show and explain your working, so it is easier for me to follow through? I am still a beginner, so that would help very much

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    $\begingroup$ The key idea here is that $\frac{a_nx^n + a_{n-1}x^{n-1} +\cdots+a_0}{b_mx^m + b_{m-1}x^{m-1} +\cdots+b_0} \sim \frac{a_n}{b_m}x^{n-m}$ which follows by dividing the numerator and denominator by $x^m$ and noticing the remaining terms because insignificant for large enough $x$ (which can be formalized through limits) $\endgroup$ – Brevan Ellefsen Jul 17 '18 at 9:34
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Why is it when given a rational function with 2 polynomials of the same degree on their numerators and denominators, the asymptote can be found by dividing the coefficients of the highest degree terms?

Say you have $$\dfrac{Ax^2+Bx+C}{Dx^2+Ex+F}$$

You can rewrite that as

$$\dfrac{\frac{A}D(Dx^2+Ex+F)+(B-\frac{AE}{D})x+(C-\frac{AF}{D})}{Dx^2+Ex+F}=\dfrac{A}{D}+\dfrac{(B-\frac{AE}{D})x+(C-\frac{AF}{D})}{Dx^2+Ex+F}$$

So you have $y=\frac{A}D$ as one of the asymptotes. This works for polynomials of all degrees, but a rigorous proof needs to use a pretty unfriendly notation for precalculus student.

Why is it if the polynomial in the numerator is a lower degree than the denominator, the x-axis (y = 0) is the horizontal asymptote?

You can write that as $$\dfrac{0x^2+Bx+C}{Dx^2+Ex+F}$$ So you have $\frac{A}D=0$. Hence, $y=0$ becomes an asymptote.

Why is it if the polynomial in the numerator is a higher degree than the denominator, there is no horizontal asymptote? Why is it when the polynomial in the numerator is a higher degree than the polynomial in the denominator, there will be a slant asymptote which you can find through polynomial long division?

When you divide the numerator by the denominator, you end up with something of the form $$\text{some polynomial of at least degree 1}+\frac{\text{some polynomial}}{\text{denominator}}$$

You know that a horizontal line only happens with degree 0 polynomial so you get some curve or non-horizontal line instead. If the polynomial in front is of degree 1, i.e. of the form $px+q$, the asymptote is a line. In a way, your horizontal asymptote becomes that slant line/curve instead.


In general, if you have $$\dfrac{P(x)}{Q(x)}$$ where $$P(x)=a_0+a_1x+a_2x^2+\cdots+a_nx^n$$ and $$Q(x)=b_0+b_1x+b_2x^2+\cdots+b_mx^m$$

You can always rewrite it as

$$f(x)+\dfrac{R(x)}{Q(x)}$$

so that $R(x)$ is of degree $(m-1)$, one less than $Q(x)$.

$y=f(x)$ is an asymptote.

If $n<m$, $f(x)=0$, you have $x$-axis as an asymptote.

If $n=m$, $f(x)=\text{some constant}\ne0$, you have a horizontal asymptote that's not the $x$-axis.

If $n=m+1$, $f(x)=ax+b$, you have a slant asymptote.

If $n>m+1$, $f(x)$ will be some polynomial curve of degree $(n-m)$.

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  • $\begingroup$ @Karn_Watcharasupat Thank you so much. Last 2 questions. First, why is it if the polynomial in the numerator is a higher degree than the denominator, there is no horizontal asymptote? Second, what is the implication of the fact that you can always rewrite it as You can always rewrite it as f(x)+R(x)/Q(x)? $\endgroup$ – Ethan Chan Jul 17 '18 at 9:52
  • $\begingroup$ Wow. That initial equality is so much harder than simply writing $\frac{Ax^2+Bx+C}{Dx^2+Ex+F} = \frac{A+\mathcal{O}(1/x)}{D+\mathcal{O}(1/x)}$ and letting $x$ be large. Limits are a wonderful thing $\endgroup$ – Brevan Ellefsen Jul 17 '18 at 9:55
  • $\begingroup$ @BrevanEllefsen I'm trying to follow how Ethan thinks. He's just starting out on precalculus so I don't want to do the potentially-devastating-for-high-school-student 'divide by $x$' trick yet... $\endgroup$ – Karn Watcharasupat Jul 17 '18 at 9:57
  • $\begingroup$ @EthanChan the horizontal asymptote and the slant asymptote is just a different manifestation of $f(x)$ when it has different degrees. That's why it's an either or issue. For the second question, try writing $P=fQ+R$ with a low degree polynomial first. You will see for yourself :) $\endgroup$ – Karn Watcharasupat Jul 17 '18 at 9:59
  • $\begingroup$ @KarnWatcharasupat I should have asked this much earlier. But I just thought of it. But you wrote that y=A/D will be the asymptote because the rational function can be rewritten as A/D+(B−AED)x+(C−AFD)/Dx2+Ex+F. But what if the bit that A/D is added to becomes 0, or negative when inputed with a certain value of x? If so, wouldn't that invalidate the claim that the asymptote is y=A/D, because there will be values that touch it, if the bit added to becomes 0, or negative when inputed with a certain value of x?Can you please explain to me why the bit will never have a value of 0 or negative? $\endgroup$ – Ethan Chan Jul 17 '18 at 11:35
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  1. Why is it when given a rational function with 2 polynomials of the same degree on their numerators and denominators, the asymptote can be found by dividing the coefficients of the highest degree terms?

You will understand with this simple example :

$$y(x)=\frac{a_2x^2+a_1x+a_0}{b_2x^2+b_1x+b_0}$$ $$y(x)=\frac{a_2+a_1\frac{1}{x}+a_0\frac{1}{x^2}}{b_2+b_1\frac{1}{x}+b_0\frac{1}{x^2}}$$ When $x\to\infty$ all terms with $\frac{1}{x}$ and $\frac{1}{x^2}$ tend to $0$. $$y(x)\to\frac{a_2}{b_2}\quad\text{thus it is the asymptote.}$$ That's the same for other polynomials of higher degree.

  1. Why is it if the polynomial in the numerator is a lower degree than the denominator, the x-axis (y = 0) is the horizontal asymptote?

$$y(x)=\frac{a_2x^2+a_1x+a_0}{b_3x^3+b_2x^2+b_1x+b_0}$$ $$y(x)=\frac{a_2\frac{1}{x}+a_1\frac{1}{x^2}+a_0\frac{1}{x^3}}{b_3+b_2\frac{1}{x}+b_1\frac{1}{x^2}+b_0\frac{1}{x^3}}$$ When $x\to\infty$
$$y(x)\to\frac{0}{b_3}\quad\text{thus } y=0 \text{ is the asymptote.}$$

  1. Why is it if the polynomial in the numerator is a higher degree than the denominator, there is no horizontal asymptote?

$$y(x)=\frac{a_3x^3+a_2x^2+a_1x+a_0}{b_2x^2+b_1x+b_0}$$ $$y(x)=x\frac{a_3+a_2\frac{1}{x}+a_0\frac{1}{x^2}+a_0\frac{1}{x^3}}{b_2+b_1\frac{1}{x}+b_0\frac{1}{x^2}}$$ When $x\to\infty$ $$y(x)\to\frac{a_3}{b_2}x\to\infty\quad\text{thus there is no horizontal asymptote.}$$ In fact, in this case there is a slant asymptote. The equation of the asymptote can be found in expending into power series of $\frac{1}{x}$. Or equivalently by long division.

$y(x)=\frac{a_3}{b_2}x+\frac{a_1b0-a_0b1}{b_0^2}+c_1\frac{1}{x}+c_2\frac{1}{x^2}+...$

So, the equation of the asymptote is : $$y(x)=\frac{a_3}{b_2}x+\frac{a_1b_0-a_0b_1}{b_0^2}$$ No need to express explicitly $c_1$ , $c_2$, … which should be boring and without use.

  1. Why is it when the polynomial in the numerator is a higher degree than the polynomial in the denominator, there will be a slant asymptote which you can find through polynomial long division?

There is a slant linear asymptote only if the degree of the polynomial at numerator is equal to the degree of the polynomial at denominator plus $1$ . This was the case of the preceding example.

If the degree at numerator is higher, this is not a straight asymptote, but a curve asymptote. You can imagine how to find the equation of the curve asymptote with the above method of series expansion of powers of $\frac{1}{x}$.

The long division leads to a first part which is a polynomial (that is the straight or the curve asymptote), plus the remaining part which is a rational fraction of two polynomials. As done in the above examples, the rational fraction is a function of $\frac{1}{x}$ which terms are eliminated when $x\to\infty$. The result is the same as the with the series expansion method.

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