Horizontal asymptotes
Far out to the left and to the right, a fraction made of two polynomials keeps changing. Where is it heading, and can you tell without plotting a single point?
Where you are 3 in Asymptotes
Why this exists
In What an asymptote is the graphs settled toward a line because a gap was multiplied by the same number between 0 and 1 at every step. A fraction of two polynomials, such as
has no such step. Both the top and the bottom grow without bound as x moves away from 0, so the fraction is a race between two huge numbers, and it is not obvious what their ratio does. A table of values can suggest an answer but cannot settle it, and plotting a few points far out is slow.
This lesson shows that the answer is decided by one small piece of each polynomial: its leading term. Comparing the degrees of the top and the bottom says at once whether the graph levels off, and if it does, at what height. It also explains something a picture can hide: unlike the graphs in What an asymptote is, these graphs often cross the line they are settling toward.
Definition
A rational function is a function of the form
where p and q are polynomials and q is not the zero polynomial (see Rational functions). Its domain is every input where q(x) \neq 0.
The horizontal line y = L is a horizontal asymptote of a function f if the outputs f(x) get as close to L as you like when x is taken far enough out on at least one side. In symbols,
The definition is about the far ends only. It says nothing about what the graph does in the middle, and it does not forbid the graph from touching or crossing the line there.
The leading term of a polynomial is its term of highest degree, and its number is the leading coefficient. For 6x^2 - x + 2 the leading term is 6x^2 and the leading coefficient is 6. A polynomial written as a product has as its leading term the product of the factors’ leading terms: (2x + 3)(x - 5) has leading term 2x \cdot x = 2x^2.
Intuition
Here is \displaystyle f(x) = \frac{6x^2 - x + 2}{3x^2 + 4} far out on each side, rounded to four decimal places.
| x | f(x) | x | f(x) |
|---|---|---|---|
| 10 | 1.9474 | -10 | 2.0132 |
| 100 | 1.9965 | -100 | 2.0031 |
| 1000 | 1.9997 | -1000 | 2.0003 |
The outputs crowd toward 2 from both sides. The reason shows in the sizes of the terms. At x = 1000 the top is 6\,000\,000 - 1000 + 2 = 5\,999\,002: the leading term 6x^2 is 6\,000\,000 and the other two terms together change it by about one part in six thousand. The bottom is 3\,000\,000 + 4, and the 4 hardly matters at all. So far out
The lower terms do not vanish; they become small compared with the leading term. The further out you go, the smaller their share, and the closer the ratio is to the ratio of the leading terms.
Three things can happen in such a race. If the bottom has the higher degree, it wins and the fraction shrinks toward 0. If the degrees are equal, both grow at the same rate and the fraction settles at the ratio of the leading coefficients. If the top has the higher degree, it wins and the fraction grows without bound, so there is no horizontal line to settle on.
Results
Let \displaystyle f(x) = \frac{p(x)}{q(x)} be a rational function whose numerator has leading term a x^n and whose denominator has leading term b x^m, with a \neq 0 and b \neq 0.
Numerator of lower degree (n < m). The line y = 0 is the horizontal asymptote, on both sides.
Equal degrees (n = m). The line \displaystyle y = \frac{a}{b}, the ratio of the leading coefficients, is the horizontal asymptote, on both sides.
Numerator of higher degree (n > m). There is no horizontal asymptote. Far out, f(x) behaves like \displaystyle \frac{a}{b} x^{\,n - m}, so the outputs grow without bound in size and the end behavior is that of this power. When n = m + 1 the graph follows a slanted line instead, studied in Slant asymptotes.
One line for both ends. A rational function has at most one horizontal asymptote, and when it has one, the graph settles onto it both far to the left and far to the right.
Crossings are allowed and are few. The graph meets its horizontal asymptote y = L exactly where p(x) = L \cdot q(x) with q(x) \neq 0. Unless f(x) = L at every input of its domain, this happens at no more than m - 1 inputs.
Why it is true
The lower-degree case first. Divide the top and the bottom by x^m, which changes nothing for x \neq 0. Every term c x^k of the numerator has k \le n < m, so it becomes \displaystyle \frac{c}{x^{m - k}}, a fixed number divided by a positive power of x. Far out that is as small as you like, because the power of x grows without bound in size. So the new numerator shrinks toward 0. In the new denominator the leading term becomes b, and every other term shrinks toward 0 in the same way, so the new denominator approaches b \neq 0. A quotient whose top approaches 0 while its bottom approaches a number other than 0 approaches 0. This is the same whether x is far to the right or far to the left, since only the size of x was used.
Equal degrees, by the gap. Let \displaystyle L = \frac{a}{b} and look at the gap between the graph and the line:
In p(x) - L \cdot q(x) the two leading terms are a x^m and \displaystyle \frac{a}{b} \cdot b x^m = a x^m, so they cancel. What is left, call it r(x), has degree below m (or is 0). By the first case, \displaystyle \frac{r(x)}{q(x)} approaches 0 on both sides, which says exactly that f(x) approaches L. For the function in the intuition the gap is
Higher degree. Divide the top and the bottom by x^m again. Now the denominator approaches b, while the numerator contains the term a x^{\,n - m}, which grows without bound in size and outgrows all the other terms. So the quotient grows without bound in size and cannot settle on any height.
At most one line, and the crossings. Each case gives one answer for both directions, so there is never a second horizontal asymptote. The graph meets y = L exactly where the gap is 0, that is where r(x) = 0. In the lower-degree case L = 0 and r = p, which has degree n \le m - 1; in the equal case r has degree at most m - 1. A polynomial of degree d that is not the zero polynomial has at most d roots, so there are at most m - 1 crossings. The only way out is r being the zero polynomial, which makes f(x) = L at every input of its domain.
Worked examples
The numerator of lower degree
Find the horizontal asymptote of \displaystyle k(x) = \frac{4x - 3}{x^2 + 2}, where the graph crosses it, and on which side of it the graph lies far to the right and far to the left.
SolutionThe numerator has degree 1 and the denominator degree 2. The bottom has the higher degree, so the horizontal asymptote is y = 0, the horizontal axis. Dividing top and bottom by x^2 shows it directly:
and far out the top approaches 0 while the bottom approaches 1.
The asymptote is y = 0, so the graph crosses it where k(x) = 0. The denominator x^2 + 2 is never 0 (a square plus 2 is at least 2), so this happens exactly where 4x - 3 = 0, at \displaystyle x = \frac{3}{4}.
The denominator is always positive, so k(x) has the sign of 4x - 3. For \displaystyle x > \frac{3}{4} it is positive: far to the right the graph lies just above the axis. For \displaystyle x < \frac{3}{4} it is negative: far to the left the graph lies just below. As a check, \displaystyle k(10) = \frac{37}{102} \approx 0.363 and \displaystyle k(-10) = -\frac{43}{102} \approx -0.422.
Equal degrees, with the leading terms hidden
Find the horizontal asymptote of \displaystyle h(x) = \frac{(2x + 3)(x - 5)}{4 - 3x^2} and the input where the graph crosses it.
SolutionNeither polynomial is written with its leading term first, so find the leading terms. The top is a product, and its leading term is the product of the leading terms of the factors, 2x \cdot x = 2x^2. In full, (2x + 3)(x - 5) = 2x^2 - 7x - 15. The bottom 4 - 3x^2 has leading term -3x^2, even though it is written second. Both have degree 2, so the horizontal asymptote is
The graph crosses the line where \displaystyle h(x) = -\frac{2}{3}. Multiplying both sides by 3(4 - 3x^2), allowed wherever the denominator is not 0:
that is, 6x^2 - 21x - 45 = 6x^2 - 8. The 6x^2 terms cancel, as the gap argument promised, leaving -21x = 37, so \displaystyle x = -\frac{37}{21}. The denominator there is \displaystyle 4 - 3 \cdot \frac{1369}{441}, which is not 0 (it is \displaystyle 4 - \frac{4107}{441}, about -5.31), so this is a real crossing: the graph meets its asymptote at exactly one input, \displaystyle x = -\frac{37}{21}.
A graph that crosses its asymptote
For \displaystyle g(x) = \frac{2x^2 + 3x}{x^2 + 5}, find the horizontal asymptote, the crossing, and how the graph approaches the line far to the right.
SolutionEqual degrees, leading coefficients 2 and 1, so the horizontal asymptote is y = 2. The gap is
The denominator is always positive, so the gap is 0 only at \displaystyle x = \frac{10}{3}: the graph meets the line at the point \displaystyle \left(\frac{10}{3}, 2\right). For \displaystyle x > \frac{10}{3} the gap is positive, so just past the crossing the graph is above the line, and it stays above for every larger input. Yet the gap must shrink toward 0, so the graph rises a little above 2 and then comes back down toward the line. Values bear this out:
| x | 5 | 10 | 100 | 1000 |
|---|---|---|---|---|
| g(x) | \displaystyle \frac{13}{6} \approx 2.167 | \displaystyle \frac{46}{21} \approx 2.190 | 2.029 | 2.003 |
The highest value, about 2.20, comes near x \approx 7.35. For \displaystyle x < \frac{10}{3} the gap is negative, so far to the left the graph approaches y = 2 from below.
Crossing does not break the definition. The asymptote is a statement about the far ends, and far to the right the graph does get as close to 2 as you like: it simply arrives from above after one crossing in the middle.
When the numerator wins
Describe the end behavior of (a) \displaystyle u(x) = \frac{x^4 - 1}{2x^2 + 3} and (b) \displaystyle v(x) = \frac{-2x^3 + x}{x^2 + 7}.
Solution(a) The numerator has degree 4 and the denominator degree 2, so there is no horizontal asymptote. Far out,
which grows without bound on both sides. So u(x) \to \infty as x \to \infty and as x \to -\infty: both ends of the graph rise. At x = 100, \displaystyle u(100) = \frac{99\,999\,999}{20\,003} \approx 4999.25, close to \displaystyle \frac{1}{2} \cdot 100^2 = 5000.
(b) Degrees 3 and 2, so again no horizontal asymptote. Far out, \displaystyle v(x) \approx \frac{-2x^3}{x^2} = -2x. So v(x) \to -\infty as x \to \infty and v(x) \to \infty as x \to -\infty: the graph comes down from the upper left and falls away to the lower right. Because the degrees differ by exactly one, the graph follows a slanted line far out, which Slant asymptotes finds.
Two different asymptotes, from a function that is not rational
Find the horizontal asymptotes of \displaystyle w(x) = \frac{6x}{\sqrt{4x^2 + 9}}.
SolutionThe bottom is not a polynomial, so the results above do not apply, and the work has to be done by hand. Far out, \displaystyle \sqrt{4x^2 + 9} is close to \displaystyle \sqrt{4x^2} = 2|x|, and the absolute value is what makes the two sides differ.
For x > 0, take x out of the root: \displaystyle \sqrt{4x^2 + 9} = \sqrt{x^2\left(4 + \frac{9}{x^2}\right)} = x\sqrt{4 + \frac{9}{x^2}}, since \displaystyle \sqrt{x^2} = x when x > 0. Then
and as x \to \infty the bottom approaches \displaystyle \sqrt{4} = 2, so w(x) \to 3.
For x < 0, \displaystyle \sqrt{x^2} = -x, so the same step gives \displaystyle w(x) = \frac{-6}{\sqrt{4 + \frac{9}{x^2}}}, and w(x) \to -3 as x \to -\infty.
So w has two horizontal asymptotes: y = 3 on the right and y = -3 on the left. As a check, \displaystyle w(10) = \frac{60}{\sqrt{409}} \approx 2.967 and w(-10) \approx -2.967. The graph never reaches either line, since \displaystyle \sqrt{4x^2 + 9} > 2|x| gives |w(x)| < 3 for every x.
The strength of a mixing tank
Thaddea’s tank starts with 48 liters of pure water. A salt solution with 42 grams of salt per liter runs in at 6 liters per minute, the tank is kept stirred, and nothing drains out. Write the concentration of salt in the tank after t minutes, find its horizontal asymptote and say what it means, and find when the concentration reaches 35 grams per liter.
SolutionAfter t minutes, 6t liters of solution have run in, carrying 42 \cdot 6t = 252t grams of salt, and the tank holds 48 + 6t liters. Concentration is salt divided by volume, so
Top and bottom both have degree 1, with leading coefficients 252 and 6, so the horizontal asymptote is \displaystyle C = \frac{252}{6} = 42. Only the right end matters, since time does not run back before the start. The meaning: as more and more solution pours in, the original water becomes a smaller and smaller part of the tank, so the mixture gets ever closer to the strength of what is pouring in, 42 grams per liter, without ever reaching it. The gap shows this exactly:
which is positive and shrinks toward 0.
For the concentration to reach 35, solve \displaystyle \frac{252t}{48 + 6t} = 35. Multiplying by 48 + 6t, which is positive, gives 252t = 1680 + 210t, so 42t = 1680 and t = 40 minutes. Check: the gap then is \displaystyle \frac{2016}{288} = 7, and 42 - 7 = 35.
Boundaries
The degree comparison is for rational functions. It needs a polynomial on top and a polynomial on the bottom. Other functions follow other rules. Example 5 has different asymptotes on its two sides, and y = 2^x has the asymptote y = 0 on the left only, with no asymptote on the right (see Graphs of exponential functions). For a rational function, “same line on both sides” is a result, not a rule for every graph.
Find the true leading terms. The comparison uses the term of highest degree, wherever it is written and whether or not the polynomial is multiplied out, as in Example 2. Canceling a factor common to top and bottom does not change the answer, since it lowers both degrees by the same amount and leaves the leading coefficients’ ratio alone.
Far ends only. A horizontal asymptote says nothing about the middle of the graph. The graph may cross the line (Examples 1 to 3), and a rational function may also have vertical asymptotes where its denominator is 0, which are a different matter (Vertical asymptotes).
When the graph lies on the line. If p(x) = L \cdot q(x) for every x, as for \displaystyle \frac{2x^2 + 2}{x^2 + 1} = 2, the gap is 0 at every input. The outputs still approach 2, so by the definition y = 2 is a horizontal asymptote, but the graph lies on it rather than settling toward it, and the count of at most m - 1 crossings does not apply.
A table or a window is evidence, not proof. The gap in Example 3 is only about 0.2 at its largest, and the gap for the function in the intuition is about 0.014 at its largest on the left, too small to see in the figure. Only the degrees, or the gap written as a fraction, settle where a graph is heading.
Where this goes next
The horizontal asymptote is one of the first lines drawn when graphing a rational function, alongside its vertical asymptotes (Vertical asymptotes) and its zeros (Domain and zeros of a rational function). When the numerator wins by exactly one degree, the graph follows a slanted line instead, found by division in Slant asymptotes. In applications, such as rational functions in context, the asymptote is the long-run value: the strength a mixture tends to, the average cost a factory tends to. The S-shaped curves of logistic growth have two horizontal asymptotes, one on each side: the carrying capacity on the right and 0 on the left. Later, the phrase “as close as you like” is made exact, and the argument in this lesson becomes a computation of limits at infinity.
Exercises
Finding the asymptote
For each function, give the horizontal asymptote, or say that there is none and describe the end behavior.
- \displaystyle \frac{7}{x^2 + 3}
- \displaystyle \frac{9x + 2}{3x - 5}
- \displaystyle \frac{4x^2 - 1}{x^3 + x}
- \displaystyle \frac{5 - 2x^2}{4x^2 + x}
- \displaystyle \frac{x^3 + 2x}{6 - x}
- \displaystyle \frac{(3x + 2)(x - 4)}{(2x + 1)^2}
- \displaystyle \frac{8x^5 - 3}{2x^5 + x^4 + 1}
- \displaystyle \frac{x^2 + 1}{x - 3}
Crossings and sides
For each function, find the horizontal asymptote, every input where the graph crosses it, and on which side of it the graph lies far to the right and far to the left.
- \displaystyle \frac{3x^2 - 2x}{x^2 + 1}
- \displaystyle \frac{2x - 6}{x^2 + 4}
- \displaystyle \frac{x^2 + 4x}{x^2 + 1}
- \displaystyle \frac{5x^2 + 1}{x^2 + 2}
- For \displaystyle f(x) = \frac{8x + 3}{4x}, write the gap f(x) - 2 as a single fraction, and find the positive inputs at which f(x) is less than 0.001 away from its horizontal asymptote.
In situations
- Oriel starts a business printing mugs. Setting up costs \$1200 and each mug then costs \$3 to print, so the average cost per mug for n mugs is \displaystyle A(n) = \frac{1200 + 3n}{n} dollars. Find the horizontal asymptote and say what it means. How many mugs bring the average cost below \$3.50?
- The concentration of a medicine in the blood t hours after a dose is \displaystyle C(t) = \frac{12t}{t^2 + 4} milligrams per liter. Find the horizontal asymptote and say what it means. Find C(2), and show that C(t) \le 3 for every t \ge 0.
- A pond is stocked with fish, and the population after t years is modeled by \displaystyle P(t) = \frac{900t + 200}{3t + 2}. Find the starting population, the long-run population, and when the population reaches 250.
Building and explaining
- Write a rational function whose horizontal asymptote is y = -4 and whose graph has no vertical asymptote. Explain why yours works.
- Find the number a for which \displaystyle \frac{ax^2 + 1}{2x^2 - 7} has the horizontal asymptote y = 5.
- Explain why a rational function cannot settle toward one height far to the right and a different height far to the left.
- For which numbers c does the graph of \displaystyle \frac{2x^2 + cx + 1}{x^2 + 3} never meet its horizontal asymptote?
- A rational function has numerator and denominator both of degree m. Prove that its graph crosses its horizontal asymptote at no more than m - 1 inputs, unless it equals the asymptote’s height at every input of its domain.
- Find the horizontal asymptotes of \displaystyle \frac{5x}{|x| + 1} on each side, and explain why this does not contradict the result that a rational function has one line for both ends.
Answers
- y = 0. Degree 0 on top, 2 below. The top is never 0, so the graph never crosses; it lies above the axis everywhere.
- y = 3. Equal degrees, leading coefficients 9 and 3, and \displaystyle \frac{9}{3} = 3.
- y = 0. Degree 2 on top, 3 below.
- \displaystyle y = -\frac{1}{2}. Equal degrees; the leading terms are -2x^2 and 4x^2, and \displaystyle \frac{-2}{4} = -\frac{1}{2}.
- No horizontal asymptote. Far out it behaves like \displaystyle \frac{x^3}{-x} = -x^2, so both ends fall: the outputs go to -\infty as x \to \infty and as x \to -\infty.
- \displaystyle y = \frac{3}{4}. The top’s leading term is 3x \cdot x = 3x^2, the bottom’s is (2x)^2 = 4x^2.
- y = 4. Both of degree 5, and \displaystyle \frac{8}{2} = 4; the x^4 term does not matter.
- No horizontal asymptote. It behaves like \displaystyle \frac{x^2}{x} = x: the outputs go to -\infty on the left and to \infty on the right. The degrees differ by one, so there is a slant asymptote.
- y = 3. Gap \displaystyle \frac{3x^2 - 2x - 3(x^2 + 1)}{x^2 + 1} = \frac{-2x - 3}{x^2 + 1}, zero only at \displaystyle x = -\frac{3}{2}. For \displaystyle x > -\frac{3}{2} the gap is negative, so far to the right the graph is below the line; far to the left it is above.
- y = 0. It crosses where 2x - 6 = 0, at x = 3. The bottom is always positive, so far to the right (where x > 3) the graph is above the axis and far to the left below it.
- y = 1. Gap \displaystyle \frac{x^2 + 4x - (x^2 + 1)}{x^2 + 1} = \frac{4x - 1}{x^2 + 1}, zero only at \displaystyle x = \frac{1}{4}. Far to the right the graph is above the line, far to the left below it.
- y = 5. Gap \displaystyle \frac{5x^2 + 1 - 5(x^2 + 2)}{x^2 + 2} = \frac{-9}{x^2 + 2}, which is never 0 and always negative: no crossing, and the graph lies below the line everywhere.
- \displaystyle f(x) - 2 = \frac{8x + 3 - 8x}{4x} = \frac{3}{4x}. For x > 0 this is positive and below 0.001 exactly when 4x > 3000, that is for x > 750.
- A = 3: as the number of mugs grows, the setup cost is shared over more and more mugs, so the average cost approaches the \$3 printing cost of one mug. \displaystyle \frac{1200 + 3n}{n} < 3.5 means 1200 < 0.5n, so n > 2400: at least 2401 mugs.
- C = 0 (degree 1 on top, 2 below): in the long run the medicine leaves the blood. \displaystyle C(2) = \frac{24}{8} = 3. And 3(t^2 + 4) - 12t = 3(t - 2)^2 \ge 0, so 12t \le 3(t^2 + 4), which divided by the positive t^2 + 4 gives C(t) \le 3.
- \displaystyle P(0) = \frac{200}{2} = 100 fish. Long run: equal degrees, so \displaystyle P \to \frac{900}{3} = 300 fish. 900t + 200 = 250(3t + 2) gives 150t = 300, so t = 2 years.
- One answer: \displaystyle \frac{-4x^2}{x^2 + 1}. Equal degrees with leading coefficients -4 and 1 give y = -4, and x^2 + 1 is never 0, so there is no vertical asymptote. Any top of degree m with leading coefficient -4b over a bottom of degree m with leading coefficient b and no real zeros works.
- a = 10. The degrees are equal, so the asymptote is \displaystyle y = \frac{a}{2}, and \displaystyle \frac{a}{2} = 5.
- Each case of the degree comparison uses only the size of x, not its sign: the terms \displaystyle \frac{c}{x^{m - k}} shrink toward 0 whether x is far to the right or far to the left. So the same limit, or the same unbounded growth, comes out in both directions, and there is never a second height.
- Only c = 0. The asymptote is y = 2 and the gap is \displaystyle \frac{2x^2 + cx + 1 - 2(x^2 + 3)}{x^2 + 3} = \frac{cx - 5}{x^2 + 3}. For c \neq 0 it is 0 at \displaystyle x = \frac{5}{c}; for c = 0 it is \displaystyle \frac{-5}{x^2 + 3}, never 0.
- Let the leading terms be a x^m and b x^m and \displaystyle L = \frac{a}{b}. The graph meets y = L where p(x) - L q(x) = 0. In p - Lq the x^m terms are \displaystyle a x^m - \frac{a}{b} b x^m = 0, so p - Lq has degree at most m - 1. If it is not the zero polynomial it has at most m - 1 roots, so at most m - 1 crossings. If it is the zero polynomial, f(x) = L at every input of the domain.
- y = 5 on the right and y = -5 on the left. For x > 0 it is \displaystyle \frac{5x}{x + 1}, which approaches 5; for x < 0 it is \displaystyle \frac{5x}{-x + 1}, which approaches -5. There is no contradiction: |x| is not a polynomial, so \displaystyle \frac{5x}{|x| + 1} is not a rational function. It is two different rational functions glued together at 0, one for each side.