How Do You Find an Oblique (Slant) Asymptote?

Divide the numerator by the denominator using polynomial division. If the quotient is linear and the remainder term approaches zero as $x \to \pm\infty$, that quotient is the oblique asymptote.

The Mechanism

For a rational function

$$f(x)=\frac{P(x)}{D(x)},$$

first simplify any common factors. An oblique asymptote normally occurs when the degree of $P(x)$ is exactly one greater than the degree of $D(x)$.

This check matters because cancellation can create a removable discontinuity rather than a vertical asymptote. However, cancellation does not alter the end-behaviour asymptote: perform the degree comparison using the simplified expression.

Polynomial division gives

$$P(x)=D(x)Q(x)+R(x),$$

and therefore

$$f(x)=Q(x)+\frac{R(x)}{D(x)}.$$

The remainder must satisfy $\deg R<\deg D$. Consequently, $\frac{R(x)}{D(x)}\to 0$ as , so the vertical difference between the function and approaches zero. If , the asymptote is .

Worked Example

Find the oblique asymptote of

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

To divide, first calculate $2x^2\div x=2x$. Multiplying gives $2x^2-4x$; subtraction leaves . Next, ; multiplying and subtracting leaves the remainder . These lines make the quotient and remainder clear without relying on a calculator.

Polynomial division gives

$$2x^2+3x+1=(x-2)(2x+7)+15.$$

Hence,

$$f(x)=2x+7+\frac{15}{x-2}.$$

Since $\frac{15}{x-2}\to 0$ as $x\to\pm\infty$,

$$\boxed{y=2x+7}.$$

This describes end behaviour only. The graph may cross an oblique asymptote, unlike a common but incorrect assumption that an asymptote can never be crossed.

Another common misconception is that the remainder belongs in the asymptote. It does not: $\frac{15}{x-2}$ measures the function's decreasing vertical difference from the line.

IB Exam Technique

This is IB Mathematics AA HL, AHL 2.13: Rational functions and can appear on Paper 1 or Paper 2. For the command term Find, show the polynomial division, rewrite the function as quotient plus remainder term, explain why that term approaches zero, and state the asymptote as $y=mx+c$. A GDC graph alone is not sufficient working.