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Slant Asymptote Calculator

Find the oblique asymptote of a rational function by long division, step by step
slant asymptote of (x^2 + 3x + 2)/(x - 1)
oblique asymptote of (2x^2 - 5x + 3)/(x + 2)
end behaviour of (x^3 + 2)/(x - 1)
does (x^2 + 1)/(x^3 - x) have a slant asymptote?

When a Slant Asymptote Exists

A slant (or oblique) asymptote is a non-horizontal line y=mx+by = mx + b that the graph approaches as x→±∞x \to \pm\infty.

For a rational function f(x)=P(x)Q(x)f(x) = \dfrac{P(x)}{Q(x)} the condition is exact:

deg⁡P=deg⁡Q+1\deg P = \deg Q + 1

One degree more — no more, no less. If the degrees are equal you get a horizontal asymptote; if the numerator is smaller you get y=0y = 0; if it exceeds the denominator by two or more, the end behaviour follows a curve (parabolic or higher), not a line.

A function never has a slant asymptote and a horizontal one, since the two describe the same limit. It can, however, have a slant asymptote together with several vertical asymptotes. Reduce the fraction first: cancelling a common factor can change the degrees and therefore the answer.

How to Find It by Long Division

The method

  1. Check the degree condition deg⁡P=deg⁡Q+1\deg P = \deg Q + 1. If it fails, stop — there is no slant asymptote.
  2. Divide P(x)P(x) by Q(x)Q(x) with polynomial long division (synthetic division works when the divisor is linear).
  3. Write the result as

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

  1. Discard the remainder term. Because deg⁡R<deg⁡Q\deg R < \deg Q, the fraction R(x)/Q(x)→0R(x)/Q(x) \to 0 as x→±∞x \to \pm\infty.
  2. The slant asymptote is the quotient: y=mx+by = mx + b.

Why the remainder can be dropped

That vanishing remainder is the whole justification, and it also tells you which side the curve approaches from. If R(x)/Q(x)>0R(x)/Q(x) > 0 for large positive xx, the graph sits above the line there; if negative, below. Evaluating the remainder term at one large value is enough to sketch it.

Limit version

Equivalently m=lim⁡x→∞f(x)xm = \lim\limits_{x \to \infty} \dfrac{f(x)}{x} and b=lim⁡x→∞[f(x)−mx]b = \lim\limits_{x \to \infty} \left[f(x) - mx\right], which works for non-rational functions too.

Common Mistakes to Avoid

  • Skipping the degree check: dividing a function whose degrees differ by two produces a parabola, not a line. That is a curvilinear asymptote, and calling it slant is wrong.
  • Keeping the remainder: the asymptote is the quotient only. Writing y=x+4+6x−1y = x + 4 + \frac{6}{x-1} describes the function, not its asymptote.
  • Forgetting missing terms in the division: write x3+2x^3 + 2 as x3+0x2+0x+2x^3 + 0x^2 + 0x + 2 before dividing, or every column shifts.
  • Cancelling too late: simplify the fraction first, since a common factor can drop the numerator's degree and remove the slant asymptote entirely.
  • Assuming the graph stays on one side: it can cross the slant asymptote wherever R(x)=0R(x) = 0.

示例题目

Step 1: Degrees: numerator 22, denominator 11, difference exactly 11 — a slant asymptote exists
Step 2: Divide: x2+3x+2x^2 + 3x + 2 by x−1x - 1. First term xx, since x⋅(x−1)=x2−xx \cdot (x-1) = x^2 - x; subtracting leaves 4x+24x + 2
Step 3: Next term 44, since 4(x−1)=4x−44(x-1) = 4x - 4; subtracting leaves remainder 66
Step 4: So f(x)=x+4+6x−1f(x) = x + 4 + \dfrac{6}{x-1}, and 6x−1→0\dfrac{6}{x-1} \to 0 as x→±∞x \to \pm\infty
Step 5: The remainder term is positive for large x>1x > 1, so the graph approaches from above on the right
Answer: y=x+4y = x + 4

Step 1: Degree difference is 11, so proceed
Step 2: 2x⋅(x+2)=2x2+4x2x \cdot (x+2) = 2x^2 + 4x; subtracting from 2x2−5x+32x^2 - 5x + 3 leaves −9x+3-9x + 3
Step 3: −9(x+2)=−9x−18-9(x+2) = -9x - 18; subtracting leaves remainder 2121
Step 4: f(x)=2x−9+21x+2f(x) = 2x - 9 + \dfrac{21}{x+2}
Step 5: Check: (x+2)(2x−9)+21=2x2−5x−18+21=2x2−5x+3(x+2)(2x-9) + 21 = 2x^2 - 5x - 18 + 21 = 2x^2 - 5x + 3 ✓
Answer: y=2x−9y = 2x - 9 (with a vertical asymptote at x=−2x = -2)

Step 1: Degrees: numerator 33, denominator 11, difference 22 — the condition deg⁡P=deg⁡Q+1\deg P = \deg Q + 1 fails
Step 2: Divide anyway to describe the end behaviour: x3+0x2+0x+2x^3 + 0x^2 + 0x + 2 divided by x−1x - 1 gives quotient x2+x+1x^2 + x + 1 with remainder 33
Step 3: Check: (x−1)(x2+x+1)+3=x3−1+3=x3+2(x-1)(x^2+x+1) + 3 = x^3 - 1 + 3 = x^3 + 2 ✓
Step 4: The graph approaches the parabola y=x2+x+1y = x^2 + x + 1, which is a curvilinear asymptote, not a line
Answer: No slant asymptote; the end behaviour follows y=x2+x+1y = x^2 + x + 1

常见问题

The numerator's degree must be exactly one greater than the denominator's, after the fraction has been fully reduced. Equal degrees give a horizontal asymptote, a smaller numerator gives y = 0, and a gap of two or more gives a curved asymptote instead of a line.

Yes, the two names are completely interchangeable, and textbooks use both. Each means a straight-line asymptote with a non-zero, finite slope, found as the quotient of the polynomial long division. Some sources reserve oblique for the general non-horizontal case, but the computation is identical.

Not in the same direction, since each end of the graph has only one limiting behaviour. A rational function has one or the other. Non-rational functions can differ between the two directions — for example, one end horizontal and the other slanted.

Yes, whenever the denominator is linear of the form x - c. It is faster and gives the same quotient and remainder. For a quadratic or higher denominator you need full polynomial long division.

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