Vertical asymptotes: where the denominator dies — after factoring.
x − 3 = 0 → x = 3
the function blows up approaching it
f(2.99) ≈ −698, f(3.01) ≈ 702
plunging one side, soaring the other
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Asymptotes of Rational Functions
Vertical asymptotes come from the denominator's uncancelled zeros; the horizontal one comes from comparing degrees. Factor first — a cancelled zero is a hole, not an asymptote.
The problem
f(x) = 2x + 1x − 3 — what are its asymptotes?
Vertical: the denominator dies at x = 3. Horizontal: degrees tie, so y = ratio of leading coefficients = 21 = 2. B solves x + 3 = 0 by sign-slip; C flips the coefficient ratio upside down; D swaps the two answers' roles.
Horizontal asymptote: the degree comparison.
| Degrees | Horizontal asymptote |
|---|---|
| top < bottom | y = 0 |
| top = bottom | y = ratio of leading coefficients |
| top > bottom | none — the function escapes |
This is the limits-at-infinity table wearing its graphing hat: the horizontal asymptote IS the limit as x → ±∞. Here both degrees are 1, so y = 21 = 2.
Backwards: for f(x) = ax − 14x + 3 to have the horizontal asymptote y = 5, the ratio of leading coefficients must be 5: a4 = 5, so a = 20.
The hole trap: cancelled zeros are not asymptotes.
x² − 4x − 2 = x + 2 (x ≠ 2)
the (x − 2) cancels completely
at x = 2: a hole, not an asymptote
the graph is a line with one missing point
Factor numerator and denominator BEFORE reading zeros. Survives the cancelling → asymptote. Cancels away → hole.
So a wall can be made to vanish. f(x) = x + cx + 7 has no vertical asymptote only when the top cancels the bottom: x + c = x + 7, so c = 7. Then f(x) = 1 with a hole at x = −7.
What the curve does near each asymptote.
The graph hugs y = 2 far out on both sides — it may cross a horizontal asymptote in the middle, but never escapes it at the ends. It NEVER touches a vertical asymptote: x = 3 is a wall. Sketching both dashed lines first turns graphing a rational function into connecting the obvious dots.
Variations: x written last, a bottom that is never zero, a top that is too big.
The x terms come second. Find the asymptotes of f(x) = 1 − 6x3x + 2.
3x + 2 = 0 → x = −23
vertical; the top is 5 there, not 0
leading terms: −6x on top, 3x below
leading means highest power, not first written
y = −63 = −2
equal degrees; f(1000) = −59993002 ≈ −1.998 ✓
What is different: the coefficient of x carries its minus sign, and the vertical asymptote is a fraction.
Not y = 12, not y = 2
12 divides the constants, which matter only near x = 0. Dropping the minus on −6x gives 2. Take the coefficients of the highest power, signs included.
A bottom that is never zero. Find the asymptotes of f(x) = x² − 1x² + 4.
x² + 4 = 0 → x² = −4
no real x
no vertical asymptote
x² + 4 is always at least 4
y = 11 = 1
equal degrees
What is different: “set the bottom to zero” finds nothing, and that is the answer — the graph has no walls.
The top has the bigger degree. Find the asymptotes of f(x) = x² + 1x − 1.
x − 1 = 0 → x = 1
top at 1 is 2, so no cancelling: vertical
degree 2 over degree 1
top > bottom
no horizontal asymptote
f(1000) = 1 000 001999 ≈ 1001, still climbing
What is different: the big-number test never settles. Do not divide the leading coefficients (11) here — that rule is for equal degrees only.
Two follow-ups: where the hole sits, and where the curve crosses y = L.
(a) Find the hole of f(x) = x² − 9x² − 2x − 3. (b) Where does g(x) = 3x² + 2xx² + 4 cross its horizontal asymptote?
(a) (x − 3)(x + 3) / [(x − 3)(x + 1)]
factor both
= x + 3x + 1, x ≠ 3
the hole is at x = 3
height: 3 + 33 + 1 = 64 → hole at (3, 32)
put x into the CANCELLED form
(b) equal degrees → y = 31 = 3
the horizontal asymptote
3x² + 2x = 3(x² + 4) → 2x = 12
set g(x) = 3; the x² terms cancel
x = 6: crosses at (6, 3)
check: (108 + 12) ÷ 40 = 3 ✓
The original form gives 00 at the hole
Putting x = 3 into x² − 9x² − 2x − 3 gives 00 — no height at all. Always use the cancelled form. And x = −1, which survives the cancelling, is the vertical asymptote, not a second hole.
On the exam
Items ask you to name the asymptotes, spot the false statement, match a graph, or find a k that sets an asymptote. Factor first, then read: surviving zeros are walls, cancelled zeros are holes, the degree comparison gives y. To match a graph, check the vertical line first — it usually eliminates all but one choice.
Check the horizontal with a big number.
x = 1,000
f = 2001997 ≈ 2.007
snuggling up to y = 2 ✓
x = −1,000
f ≈ 1.993
from the other side ✓
Now you try
Practice problem
The problem
g(x) = x² − 1x² − 4 — what are its asymptotes?
Denominator zeros: x = ±2, and neither cancels with (x − 1)(x + 1) up top. Degrees tie at 2, so y = 11 = 1. A reads the NUMERATOR's zeros — those are x-intercepts, where the graph politely crosses zero, not walls.