Grade by Grade
B. Derivatives and Their Applications · Curve Sketching: Connecting f, f′ and f″

Asymptotes of Rational Functions: Vertical, Horizontal and Oblique

B1.3About 70 minutes
Learning goals
  • I can find vertical asymptotes and use one-sided limits to describe the behaviour of a rational function near them.
  • I can evaluate limits at infinity and find horizontal asymptotes.
  • I can find oblique asymptotes by division and tell whether the graph is above or below an asymptote.
  • I can tell holes from vertical asymptotes.

Words to know

vertical asymptote
A line x = a near which f(x) → ∞ or −∞ as x → a from one or both sides.
horizontal asymptote
A line y = L with f(x) → L as x → ∞ or x → −∞.
oblique (slant) asymptote
A non-horizontal line y = mx + b that the graph approaches as x → ±∞; occurs when the numerator's degree is one more than the denominator's.
limit at infinity
lim (x→∞) f(x): the value f(x) approaches as x grows without bound.
hole
A single missing point of a graph, caused by a factor that cancels from the numerator and denominator.

Graphs that run off to infinity

Polynomials are well behaved: no breaks, no asymptotes. Rational functions are different. They can shoot up or down near a vertical asymptote, and level off or follow a slanted line far from the origin. Before you can sketch a rational function with calculus, you need to describe this behaviour precisely using limits.

**Remember from MHF4U (Rational Functions and Equations):** you found vertical asymptotes from zeros of the denominator, horizontal asymptotes of functions like y = (ax + b)/(cx + d), and holes from common factors. In Limits and Rates of Change you learned limit notation. Now you will combine the two, and add a new kind of asymptote: the oblique asymptote.

Vertical asymptotes and one-sided limits

If x = a makes the denominator 0 but not the numerator (after cancelling common factors), the line x = a is a vertical asymptote. Near it, f(x) grows without bound. We write, for example,

lim (x→a⁺) f(x) = ∞ (as x approaches a from the right, f(x) increases without bound),

lim (x→a⁻) f(x) = −∞ (from the left, f(x) decreases without bound).

To decide the sign: substitute a value just to the right (or left) of a, such as a + 0.01, and look at the signs of the numerator and denominator. You do not need the exact value, only its sign.

Horizontal asymptotes and limits at infinity

lim (x→∞) f(x) = L means f(x) gets as close to L as we like when x is large. If this happens (as x → ∞ or x → −∞), y = L is a horizontal asymptote.

Technique: divide every term of the numerator and denominator by the highest power of x in the denominator. Every term like 3/x or 5/x² approaches 0.

For f(x) = p(x)/q(x):

  • degree of p < degree of q → HA y = 0;
  • degree of p = degree of q → HA y = (leading coefficient of p)/(leading coefficient of q);
  • degree of p = degree of q + 1 → no HA, but an oblique asymptote (found by division);
  • degree of p ≥ degree of q + 2 → no linear asymptote; the ends behave like a polynomial.
Worked example

Example 1: Vertical and horizontal asymptotes

Find all asymptotes of f(x) = (2x + 1)/(x − 3) and describe the behaviour near each.

  1. VA: x − 3 = 0 at x = 3, and the numerator there is 7 ≠ 0. So x = 3 is a vertical asymptote.
  2. Right of 3 (x = 3.01): numerator ≈ 7 > 0, denominator 0.01 > 0, so f → +∞: lim (x→3⁺) f(x) = ∞.
  3. Left of 3 (x = 2.99): denominator < 0, so lim (x→3⁻) f(x) = −∞.
  4. HA: divide by x: f(x) = (2 + 1/x)/(1 − 3/x) → (2 + 0)/(1 − 0) = 2 as x → ±∞. So y = 2.
  5. Above or below? f(x) − 2 = (2x + 1 − 2x + 6)/(x − 3) = 7/(x − 3). This is positive for x > 3 (graph above y = 2) and negative for x < 3 (below).
  6. Intercepts: y-int f(0) = −1/3; x-int where 2x + 1 = 0, x = −1/2.

VA x = 3 (−∞ on the left, +∞ on the right); HA y = 2 (approached from above as x → ∞, from below as x → −∞).

y = (2x + 1)/(x − 3)xy−4−2246810−8−448x = 3y = 2
Near x = 3 the graph shoots up on the right and down on the left. The intercepts are (−1/2, 0) and (0, −1/3). Far from the origin it hugs the horizontal asymptote y = 2: above it for x > 3, below it for x < 3.
Worked example

Example 2: Equal degrees

Find the asymptotes of f(x) = 3x²/(x² − 4) and decide whether the graph ever crosses its horizontal asymptote.

  1. VAs: x² − 4 = (x − 2)(x + 2) = 0 at x = ±2; the numerator 3x² is 12 ≠ 0 there. So x = 2 and x = −2.
  2. Near x = 2: at x = 2.01, denominator (0.01)(4.01) > 0, so f → +∞; at x = 1.99, f → −∞. By symmetry (f is even), near x = −2: f → −∞ from the right and +∞ from the left.
  3. HA: divide by x²: 3/(1 − 4/x²) → 3. So y = 3.
  4. Crossing? Solve 3x²/(x² − 4) = 3: 3x² = 3x² − 12, which gives 0 = −12, impossible. No crossing.

VAs x = ±2, HA y = 3, never crossed. (Some graphs do cross their horizontal asymptote, such as y = x/(x² + 1), which crosses y = 0 at the origin. An HA describes the ends only.)

Oblique asymptotes

When the degree of the numerator is exactly one more than the degree of the denominator, use long division (or synthetic division) to write

f(x) = (mx + b) + r(x)/q(x), with the remainder r(x) of lower degree than q(x).

As x → ±∞, r(x)/q(x) → 0, so the graph approaches the line y = mx + b, the oblique (slant) asymptote. The sign of r(x)/q(x) tells you whether the graph is above or below the line.

Worked example

Example 3: An oblique asymptote

Find all asymptotes of f(x) = (x² − x − 2)/(x − 3).

  1. VA: x = 3 (numerator 9 − 3 − 2 = 4 ≠ 0).
  2. Divide: x² − x − 2 = (x − 3)(x + 2) + 4. So f(x) = x + 2 + 4/(x − 3).
  3. As x → ±∞, 4/(x − 3) → 0. Oblique asymptote y = x + 2.
  4. 4/(x − 3) > 0 for x > 3: the graph is above the line on the right; below it on the left.
  5. Intercepts: zeros from x² − x − 2 = (x − 2)(x + 1): x = 2, −1. y-int f(0) = −2/(−3) = 2/3.

VA x = 3; OA y = x + 2; zeros −1 and 2; y-intercept 2/3.

y = (x² − x − 2)/(x − 3)xy−4−22468−6−24812x = 3y = x + 2(0, 2/3)
An oblique (slant) asymptote. Dividing gives f(x) = x + 2 + 4/(x − 3). For large |x| the fraction 4/(x − 3) is tiny, so the graph approaches the line y = x + 2: from above on the right and from below on the left.
Worked example

Example 4: A hole, not an asymptote

Find the asymptotes and holes of f(x) = (x² − 1)/(x² − x − 2).

  1. Factor: f(x) = (x − 1)(x + 1)/[(x − 2)(x + 1)].
  2. The factor x + 1 cancels: f(x) = (x − 1)/(x − 2) for x ≠ −1. Hole at x = −1, with y = (−2)/(−3) = 2/3.
  3. VA: x = 2 (it does not cancel).
  4. HA: equal degrees, leading coefficients 1 and 1: y = 1.

Hole (−1, 2/3); VA x = 2; HA y = 1. Always factor and cancel before naming vertical asymptotes.

y = (x² − 1)/(x² − x − 2)xy−3−2−112345−4−3−2−112345x = 2y = 1hole (−1, 2/3)
The factor x + 1 cancels, leaving a hole at x = −1 (open circle) instead of an asymptote. The remaining factor x − 2 gives the vertical asymptote x = 2.

Asymptote checklist

  1. Factor numerator and denominator; cancel common factors (each gives a hole).
  2. Remaining zeros of the denominator → vertical asymptotes. Test each side for +∞ or −∞.
  3. Compare degrees → horizontal, oblique or none.
  4. Find f(x) minus the asymptote to see whether the graph is above or below it at each end.
Just for fun

A horizontal asymptote and a rational function have been dating for years. Friends keep asking when they will finally meet. "At infinity," they say, "we're very patient." (Though some rational functions do cross their HA once or twice on the way.)

Hands-on activity

Sharing a pizza with infinity

You need: a calculator, paper and a pencil.

  1. A $24 pizza is shared equally by x people, plus a $2 delivery charge per person. The cost per person is C(x) = 24/x + 2. Make a table for x = 1, 2, 4, 8, 16, 100, 1000.
  2. What value does C(x) approach? Which asymptote is that?
  3. Now evaluate C(x) for x = 0.5, 0.1 and 0.01 (as if only a fraction of a person paid). Which asymptote does that illustrate?
  4. Sketch C(x) for x > 0.

What to look for: C(x) gets closer and closer to $2 (the horizontal asymptote y = 2) but never reaches it, because the pizza cost is always split. As x → 0⁺, C(x) grows without bound (the vertical asymptote x = 0): 24/0.01 + 2 = $2402. (In real life x is a whole number, so this second part is only about the formula.)

Your turn

1

Find the vertical and horizontal asymptotes of f(x) = (4x − 1)/(2x + 6).

2

Evaluate lim (x→∞) (5x² − 3x)/(2x² + 7).

3

Evaluate lim (x→∞) (3x + 1)/(x² + 4).

4

For f(x) = 1/(x − 4), find lim (x→4⁺) f(x) and lim (x→4⁻) f(x).

5

Find the oblique asymptote of f(x) = (x² + 3x + 5)/(x + 1).

6

Find the holes and asymptotes of f(x) = (x² − 9)/(x − 3).

7

Is the graph of f(x) = (2x + 1)/(x − 3) above or below y = 2 when x = −10?

More practice: warm-up

1

HA of y = 7/(x + 2)?

2

HA of y = (6x + 1)/(3x − 5)?

3

VA of y = (x + 1)/(x − 8)?

4

Does y = (x³ + 1)/(x − 1) have a horizontal asymptote?

5

lim (x→∞) 10/x = ?

More practice: on your own

1

Find all asymptotes of f(x) = (x² + 1)/(x² − 9).

2

Find all asymptotes of f(x) = (2x² − 3x)/(x + 1).

3

Describe the behaviour of f(x) = x/(x + 2)² near x = −2.

4

Find the holes, VAs and HA of f(x) = (x² + 2x)/(x² − 4).

5

Find where f(x) = (x² + x)/(x² + 1) crosses its horizontal asymptote.

Challenge

1

Find the oblique asymptote of f(x) = (x³ − 2x² + 1)/(x² + 1) and state where the graph crosses it.

2

Find a rational function with VA x = 1, HA y = −2 and x-intercept 4.

3

Use limits to show that y = √(x² + 1) − x approaches 0 as x → ∞ (a horizontal asymptote of a non-rational function).

Check your understanding

Quiz

1.

What is the horizontal asymptote of f(x) = (6x² + 1)/(2x² − x)?

2.

Which function has an oblique asymptote?

3.

For f(x) = (x + 5)/(x − 2), what is lim (x→2⁻) f(x)?

4.

f(x) = (x² − 4)/(x² + x − 6). What happens at x = 2?

5.

Can the graph of a rational function cross its horizontal asymptote?

6.

What is the oblique asymptote of f(x) = (x² + 3x)/(x + 1)?

For parents and teachers

Big idea (B1.3): rational functions can have vertical asymptotes (where they blow up), horizontal asymptotes (where they level off), oblique asymptotes (where they follow a slanted line) and holes. Limits describe each one precisely.

How to help: the pizza activity shows a horizontal asymptote with real numbers. Ask your teen to explain why the cost per person never quite reaches $2.

Common difficulties:

  • Declaring a vertical asymptote before cancelling common factors (it may be a hole).
  • Mixing up the degree rules for horizontal asymptotes.
  • Getting the sign wrong near a vertical asymptote; testing a value such as a ± 0.01 fixes this.
  • Thinking a graph can never cross a horizontal asymptote.