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Precalculus · F.IF.C.7D

Slant asymptotes

When the top of a fraction outgrows the bottom by one power of x, the graph never levels off. Does it still settle down to something simple far out?

You need
Horizontal asymptotesAlgebra 2Polynomial long divisionAlgebra 2
Where you are 3 in Asymptotes
Functions13
Rational functions5
Asymptotes3
Horizontal asymptotes
Slant asymptotes
Vertical asymptotes

Why this exists

Horizontal asymptotes settle the far-out behavior of a rational function when the numerator’s degree is at most the denominator’s. When the numerator has the higher degree, the outputs grow without bound and there is no horizontal line to settle toward. Yet something simple remains. Take

f(x) = \frac{3x^2 - 2x + 4}{x - 1}.

At x = 101 it gives f(101) = 304.05, and at x = 1001 it gives 3004.005. Those are almost exactly 3x + 1, which gives 304 and 3004. The graph keeps climbing, but it climbs along a straight line, and the line can be found exactly. A slanted line is the next simplest thing after a horizontal one, and naming it finishes the picture of the Asymptotes a rational function can have.

Definition

A line y = mx + b with m \neq 0 is a slant asymptote (also called an oblique asymptote) of the graph of f when the vertical gap between the graph and the line shrinks toward 0 far out:

\begin{aligned} &f(x) - (mx + b) \to 0 \\ &\text{as } x \to \infty \text{ or as } x \to -\infty. \end{aligned}

For a rational function \displaystyle f(x) = \frac{P(x)}{Q(x)} this happens exactly when the degree of P is one more than the degree of Q and Q does not divide P exactly. Long division then writes

f(x) = \underbrace{mx + b}_{\text{quotient}} + \frac{R(x)}{Q(x)},

where the remainder R \neq 0 has lower degree than Q, and the quotient line y = mx + b is the slant asymptote.

Intuition

Division splits f into a part that is a straight line and a leftover fraction. The leftover \displaystyle \frac{R(x)}{Q(x)} has a bottom of higher degree than its top, so far out its bottom wins and the fraction is tiny. What is left visible is the line.

The leftover also carries two more facts: its sign says which side of the line the graph is on, and its zeros are where the graph meets the line.

−4 4 −8 8 x y
x = 1
y = 3x + 1
Figure 1Example 1. The graph of \displaystyle f(x) = \frac{3x^2 - 2x + 4}{x - 1} hugs the line y = 3x + 1 far out, above it on the right and below it on the left.

Results

Where the slant line comes from. If \deg P = \deg Q + 1, dividing gives P(x) = Q(x)(mx + b) + R(x) with \deg R < \deg Q, and when R \neq 0 the line y = mx + b is a slant asymptote on both ends. Its slope m is the leading coefficient of P divided by that of Q.

Which side. The graph lies above the slant line wherever \displaystyle \frac{R(x)}{Q(x)} > 0 and below it wherever \displaystyle \frac{R(x)}{Q(x)} < 0. Far out only the leading terms of R and Q matter for this sign.

Crossing. The graph meets its slant line exactly at the inputs where R(x) = 0 and Q(x) \neq 0. When Q is linear, R is a nonzero constant (if R were 0, f would be the line itself with a hole), so the graph never meets the line. When Q has degree 2 or more, R can have zeros and the graph can cross.

At most one. A rational function has at most one horizontal or slant asymptote, and it is the same line on the left and on the right.

Why it is true

Write R(x) = r_k x^k + \dots and Q(x) = q_n x^n + \dots with k < n. Divide the top and bottom of \displaystyle \frac{R(x)}{Q(x)} by x^n. Every term on top becomes a constant over a positive power of x, because k < n, so the whole top shrinks toward 0 as x \to \pm\infty. The bottom becomes q_n plus terms that shrink toward 0, so it settles toward q_n \neq 0. A quantity near 0 divided by a quantity near q_n is near 0. So \displaystyle f(x) - (mx + b) = \frac{R(x)}{Q(x)} \to 0, which is the definition.

The slope is the ratio of leading coefficients because the first step of the division divides the leading term p\,x^{n+1} of P by q_n x^n, giving \displaystyle \frac{p}{q_n}x.

Only one line can work: if f approached two lines on the same end, their difference (m_1 - m_2)x + (b_1 - b_2) would approach 0, which forces m_1 = m_2 and b_1 = b_2. And one division identity holds for every x, so the same quotient serves both ends.

Worked examples

Example 1

The slant line, the gap, and the side

Find the slant asymptote of \displaystyle f(x) = \frac{3x^2 - 2x + 4}{x - 1}, say which side of it the graph is on at each end, and decide whether the graph ever meets it.

Solution

The numerator has degree 2 and the denominator degree 1, so there is a slant asymptote. Divide: 3x^2 \div x = 3x, and 3x(x - 1) = 3x^2 - 3x leaves x + 4; then x \div x = 1, and 1 \cdot (x - 1) leaves 5. So

f(x) = 3x + 1 + \frac{5}{x - 1}.

Check: (x - 1)(3x + 1) is 3x^2 - 2x - 1, and adding 5 gives the numerator. The slant asymptote is y = 3x + 1.

The gap \displaystyle \frac{5}{x - 1} shrinks as x grows: it is 0.5 at x = 11, 0.05 at x = 101 and 0.005 at x = 1001. For x > 1 it is positive, so on the right the graph is above the line. For x < 1 the denominator is negative, so the gap is negative and on the left the graph is below the line (at x = -9 the gap is -0.5).

The graph meets the line only where \displaystyle \frac{5}{x - 1} = 0, and a fraction with numerator 5 is never 0. So it never meets the line. The denominator is 0 at x = 1 while the numerator is 3 - 2 + 4 = 5 there, so x = 1 is a vertical asymptote. The two branches sit on opposite sides of the slant line, as the figure shows.

Example 2

A denominator with a leading coefficient

Find the slant asymptote of \displaystyle g(x) = \frac{6x^2 + x - 5}{2x + 1} and the side of it the graph takes on each end.

Solution

Degrees 2 and 1, so there is a slant asymptote. Divide: 6x^2 \div 2x = 3x, and 3x(2x + 1) = 6x^2 + 3x leaves -2x - 5; then -2x \div 2x = -1, and -1(2x + 1) = -2x - 1 leaves -4. So

g(x) = 3x - 1 - \frac{4}{2x + 1}.

The slant asymptote is y = 3x - 1. Its slope is \displaystyle \frac{6}{2} = 3, the ratio of the leading coefficients, not the numerator’s 6.

For large positive x, 2x + 1 > 0, so \displaystyle -\frac{4}{2x + 1} < 0 and the graph is below the line on the right (at x = 10 the gap is \displaystyle -\frac{4}{21}). For large negative x, 2x + 1 < 0, so the gap is positive and the graph is above the line on the left. The vertical asymptote is \displaystyle x = -\frac{1}{2}, where the numerator is \displaystyle \frac{6}{4} - \frac{1}{2} - 5 = -4 \neq 0.

Example 3

Crossing the slant asymptote

Find the slant asymptote of \displaystyle k(x) = \frac{x^3 - 2x^2 + 3x + 1}{x^2 + 1} and every point where the graph meets it.

Solution

Degrees 3 and 2, one apart. Divide: x^3 \div x^2 = x, and x(x^2 + 1) = x^3 + x leaves -2x^2 + 2x + 1; then -2x^2 \div x^2 = -2, and -2(x^2 + 1) = -2x^2 - 2 leaves 2x + 3. So

k(x) = x - 2 + \frac{2x + 3}{x^2 + 1}.

The slant asymptote is y = x - 2. This time the remainder is linear, so it can be 0: 2x + 3 = 0 at \displaystyle x = -\frac{3}{2}, and x^2 + 1 is never 0. The graph meets the line at \displaystyle x = -\frac{3}{2}, \displaystyle y = -\frac{3}{2} - 2 = -\frac{7}{2}, the point \displaystyle \left(-\frac{3}{2}, -\frac{7}{2}\right). Check in the original: the top is \displaystyle -\frac{91}{8}, the bottom is \displaystyle \frac{13}{4}, and \displaystyle -\frac{91}{8} \div \frac{13}{4} = -\frac{7}{2}.

The denominator is always positive, so the side depends on 2x + 3 alone: below the line for \displaystyle x < -\frac{3}{2}, above it for \displaystyle x > -\frac{3}{2}. There is no vertical asymptote. Far out the gap still shrinks (at x = 10 it is \displaystyle \frac{23}{101}), so the line is an asymptote even though the graph crosses it.

−4 4 −4 4 x y
y = x - 2
\left(-\tfrac{3}{2}, -\tfrac{7}{2}\right)
Figure 2Example 3. With a quadratic denominator the remainder can be zero, so the graph of k crosses its slant asymptote once.
Example 4

Degree higher by two: no line at all

Show that \displaystyle p(x) = \frac{x^3 + 1}{x - 2} has no slant asymptote, and find the curve it follows far out.

Solution

The degrees are 3 and 1, two apart. Division gives x^3 + 1 = (x - 2)(x^2 + 2x + 4) + 9, since (x - 2)(x^2 + 2x + 4) = x^3 - 8. So

p(x) = x^2 + 2x + 4 + \frac{9}{x - 2}.

Suppose some line y = mx + b were a slant asymptote. The gap would be

\begin{aligned} p(x) - (mx + b) ={} &x^2 + (2 - m)x + (4 - b) \\ &+ \frac{9}{x - 2}. \end{aligned}

The fraction shrinks, but the quadratic x^2 + (2 - m)x + (4 - b) grows without bound on both ends whatever m and b are. So the gap does not shrink, and no line works.

The graph follows the parabola y = x^2 + 2x + 4 instead: the gap to it is \displaystyle \frac{9}{x - 2}, which shrinks, positive on the right and negative on the left. In general the quotient gives the far-out shape: degree 0 a horizontal line, degree 1 a slant line, degree 2 or more a curve.

−4 4 −10 20 x y
x = 2
y = x^2 + 2x + 4
Figure 3Example 4. The degree is higher by two, so far out the graph follows the parabola y = x^2 + 2x + 4 and no straight line at all.
Example 5

Building a function from its asymptotes

Write a rational function whose graph has vertical asymptote x = 3, slant asymptote y = 2x - 1, and passes through (4, 12).

Solution

The form \displaystyle f(x) = 2x - 1 + \frac{c}{x - 3} with c \neq 0 has both asymptotes: the fraction blows up at x = 3 and shrinks far out. The point fixes c: \displaystyle f(4) = 7 + \frac{c}{1} = 12, so c = 5. Over one denominator,

f(x) = \frac{(2x - 1)(x - 3) + 5}{x - 3} = \frac{2x^2 - 7x + 8}{x - 3}.

Check: \displaystyle f(4) = \frac{32 - 28 + 8}{1} = 12, and at x = 3 the numerator is 18 - 21 + 8 = 5 \neq 0, so x = 3 really is a vertical asymptote and not a hole.

Example 6

Average cost at a bike-rental company

Ayelet runs a bike-rental company. Keeping n bikes on the road costs C(n) = 0.02n^2 + 6n + 400 dollars a day: 400 dollars of fixed costs, 6 dollars a bike, and a term that grows as a crowded fleet needs more repairs. Find the slant asymptote of the average cost per bike \displaystyle A(n) = \frac{C(n)}{n}, compare the two at n = 200, and find how large the fleet must be for the average to be within 1 dollar of the line.

Solution

Dividing each term by n is the whole division here:

A(n) = 0.02n + 6 + \frac{400}{n}.

The slant asymptote is y = 0.02n + 6. Only n > 0 makes sense, so only the right end matters, and there the gap \displaystyle \frac{400}{n} is positive: the average sits above the line. At n = 200 the average is 4 + 6 + 2 = 12 dollars and the line gives 10, a gap of 2 dollars, the fixed cost spread over 200 bikes.

Within 1 dollar means \displaystyle \frac{400}{n} \le 1, so n \ge 400 bikes. Past that, the average rises almost exactly along the line, 0.02 dollars for each extra bike.

Boundaries

  • The degree must be higher by exactly one. Higher by two or more, the quotient is not a line and no slant asymptote exists (Example 4). Equal or lower degree, the asymptote is horizontal and Horizontal asymptotes applies.
  • A zero remainder gives no asymptote. If Q divides P exactly, f is the quotient line itself, with holes at the zeros of Q: \displaystyle \frac{x^2 - 9}{x + 3} = x - 3 for x \neq -3. The graph is the line, not something approaching it.
  • Not only rational functions. x + 2^{-x} has the slant asymptote y = x on the right only. Division and the “both ends” result belong to rational functions.
  • An asymptote does not forbid crossing. It describes the ends only (Example 3).

Where this goes next

Graphing rational functions puts all three kinds of asymptote together with intercepts and signs into a full sketch, and Rational functions in context reads them as long-run behavior. For a divisor x - a, Synthetic division finds the quotient line faster. In calculus the shrinking gap becomes a limit at infinity, as in Limits at infinity and horizontal asymptotes.

Exercises

Finding the slant asymptote

For each function, find the slant asymptote and any vertical asymptotes.

  1. \displaystyle \frac{x^2 + 5x + 2}{x + 3}
  2. \displaystyle \frac{2x^2 - 3x - 1}{x - 2}
  3. \displaystyle \frac{-x^2 + 4x + 3}{x + 1}
  4. \displaystyle \frac{4x^2 + 2x + 3}{2x - 1}
  5. \displaystyle \frac{x^3 - x + 4}{x^2 + 2}, and the point where the graph meets the slant line.
  6. \displaystyle \frac{3x^3 + x^2 - 2}{x^2 - 4}, and the point where the graph meets the slant line.

Sides and crossings

  1. For \displaystyle \frac{x^2 - 3x + 5}{x - 2}, say which side of the slant line the graph is on at each end, and how far from the line it is at x = 32.
  2. For \displaystyle \frac{x^2 + x - 9}{x - 2}, say which side of the slant line the graph is on at each end, and find the input on the right where the graph is exactly 0.01 below the line.
  3. Find where the graph of \displaystyle \frac{2x^3 + x^2 + 5x}{x^2 + 1} meets its slant asymptote, and which side it is on at each end.
  4. Find where the graph of \displaystyle \frac{x^3 + x^2 + 4}{x^2 + 4} meets its slant asymptote, and which side it is on at each end.
  5. Does the graph of \displaystyle \frac{5x^2 - 3x + 2}{x + 1} ever meet its slant asymptote? Find the asymptote and say why or why not.

Degree higher by more than one

  1. Does \displaystyle \frac{x^3 - 2x + 1}{x + 1} have a slant asymptote? Find the curve its graph follows far out.
  2. Find the curve the graph of \displaystyle \frac{x^4 + 3}{x^2 + 1} follows far out, and say which side of it the graph is on.

Building and reading

  1. Write a rational function with vertical asymptote x = -2 and slant asymptote y = x + 4 whose graph passes through (0, 3).
  2. Find a and b so that \displaystyle \frac{ax^2 + bx + 3}{x - 1} has slant asymptote y = 2x - 5.
  3. For which k does \displaystyle \frac{x^2 + kx - 6}{x - 2} have no vertical asymptote? What is its graph then?

The bike-rental company

Temuulen’s bike-rental company has daily cost C(n) = 0.05n^2 + 4n + 180 dollars for n bikes.

  1. Find the slant asymptote of the average cost \displaystyle A(n) = \frac{C(n)}{n}, the average cost at n = 60, and how far above the line it is there.
  2. How many bikes are needed for the average cost to be within 0.50 dollars of the line?

Reasoning

  1. Rangi says a rational function might follow a slant line on the right and a horizontal line on the left. Explain why that cannot happen.
  2. Explain why the slope of the slant asymptote of \displaystyle \frac{P(x)}{Q(x)} is the leading coefficient of P divided by the leading coefficient of Q.
  3. Ayelet says a graph can never cross its slant asymptote, because an asymptote is a line the graph approaches but never reaches. Give a counterexample, and say what the word asymptote does promise.
Answers
  1. Slant y = x + 2, vertical x = -3. Since (x + 3)(x + 2) = x^2 + 5x + 6, the function is \displaystyle x + 2 - \frac{4}{x + 3}.
  2. Slant y = 2x + 1, vertical x = 2. Since (x - 2)(2x + 1) = 2x^2 - 3x - 2, the function is \displaystyle 2x + 1 + \frac{1}{x - 2}.
  3. Slant y = -x + 5, vertical x = -1. Since (x + 1)(-x + 5) = -x^2 + 4x + 5, the function is \displaystyle -x + 5 - \frac{2}{x + 1}.
  4. Slant y = 2x + 2, vertical \displaystyle x = \frac{1}{2}. Since (2x - 1)(2x + 2) = 4x^2 + 2x - 2, the function is \displaystyle 2x + 2 + \frac{5}{2x - 1}.
  5. Slant y = x, no vertical asymptote (x^2 + 2 > 0). The function is \displaystyle x + \frac{-3x + 4}{x^2 + 2}, and -3x + 4 = 0 at \displaystyle x = \frac{4}{3}, so it meets the line at \displaystyle \left(\frac{4}{3}, \frac{4}{3}\right).
  6. Slant y = 3x + 1, vertical x = 2 and x = -2 (the numerator is 26 and -22 there). Since (x^2 - 4)(3x + 1) = 3x^3 + x^2 - 12x - 4, the remainder is 12x + 2, zero at \displaystyle x = -\frac{1}{6}; the meeting point is \displaystyle \left(-\frac{1}{6}, \frac{1}{2}\right).
  7. The function is \displaystyle x - 1 + \frac{3}{x - 2}: above the line y = x - 1 on the right, below it on the left. At x = 32 the gap is \displaystyle \frac{3}{30} = 0.1 above.
  8. The function is \displaystyle x + 3 - \frac{3}{x - 2}: below y = x + 3 on the right, above it on the left. \displaystyle \frac{3}{x - 2} = 0.01 gives x - 2 = 300, so x = 302.
  9. The function is \displaystyle 2x + 1 + \frac{3x - 1}{x^2 + 1}. It meets y = 2x + 1 where 3x - 1 = 0, at \displaystyle \left(\frac{1}{3}, \frac{5}{3}\right); below the line on the left, above on the right.
  10. The function is \displaystyle x + 1 - \frac{4x}{x^2 + 4}. It meets y = x + 1 where 4x = 0, at (0, 1); above the line on the left, below on the right.
  11. No. Since (x + 1)(5x - 8) = 5x^2 - 3x - 8, the function is \displaystyle 5x - 8 + \frac{10}{x + 1}, with slant y = 5x - 8. The gap \displaystyle \frac{10}{x + 1} has numerator 10, so it is never 0.
  12. No: the degrees differ by two. Since (x + 1)(x^2 - x - 1) = x^3 - 2x - 1, the function is \displaystyle x^2 - x - 1 + \frac{2}{x + 1}, and the graph follows the parabola y = x^2 - x - 1.
  13. Since (x^2 + 1)(x^2 - 1) = x^4 - 1, the function is \displaystyle x^2 - 1 + \frac{4}{x^2 + 1}. It follows y = x^2 - 1, always above it because \displaystyle \frac{4}{x^2 + 1} > 0.
  14. \displaystyle \frac{x^2 + 6x + 6}{x + 2}. Take \displaystyle x + 4 + \frac{c}{x + 2}; at x = 0 it is \displaystyle 4 + \frac{c}{2} = 3, so c = -2, and (x + 4)(x + 2) - 2 = x^2 + 6x + 6.
  15. a = 2, b = -7. Dividing by x - 1 gives the quotient ax + (a + b) and remainder a + b + 3, and the quotient must be 2x - 5. Check: \displaystyle \frac{2x^2 - 7x + 3}{x - 1} = 2x - 5 - \frac{2}{x - 1}.
  16. k = 1. No vertical asymptote needs the numerator to be 0 at x = 2: 4 + 2k - 6 = 0. Then \displaystyle \frac{x^2 + x - 6}{x - 2} = x + 3 for x \neq 2, so the graph is the line y = x + 3 with a hole at (2, 5), and it has no slant asymptote.
  17. \displaystyle A(n) = 0.05n + 4 + \frac{180}{n}, slant y = 0.05n + 4. At n = 60, A = 3 + 4 + 3 = 10 dollars, which is 3 dollars above the line’s 7.
  18. At least 360 bikes: \displaystyle \frac{180}{n} \le 0.5 gives n \ge 360.
  19. Division gives one identity \displaystyle f(x) = q(x) + \frac{R(x)}{Q(x)} that holds for every x, and the fraction shrinks toward 0 at both ends. So on both ends f follows the same quotient q. A quotient is either constant (horizontal) or of degree one (slant), never both.
  20. The quotient’s first term comes from dividing the leading term of P by the leading term of Q: \displaystyle \frac{p\,x^{n+1}}{q\,x^n} = \frac{p}{q}x. The rest of the division only lowers degrees, so no other x term enters the quotient, and the slope is \displaystyle \frac{p}{q}.
  21. Example 3: \displaystyle k(x) = \frac{x^3 - 2x^2 + 3x + 1}{x^2 + 1} meets y = x - 2 at \displaystyle \left(-\frac{3}{2}, -\frac{7}{2}\right). Exercise 9 is another. An asymptote promises only that the gap shrinks toward 0 far out; it says nothing about the middle of the graph.
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