AS & A Level · Practice questions · Asymptotes of rational functions

Lines that steer the curve.

Six original Cambridge-style questions on asymptotes of rational functions: reading vertical and horizontal asymptotes, oblique asymptotes by division, approaches from above and below, and finding unknown constants.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Denominator, degrees, divide, then signs.

Vertical: solve \(Q(x) = 0\); keep each root where the numerator is not also zero.
Compare degrees: lower on top gives \(y = 0\); equal gives the ratio of leading coefficients; one higher means divide.
Divide to \(y = mx + k + \dfrac{r}{Q(x)}\) and check by expanding; \(y = mx + k\) is the oblique asymptote.
Sides: decide the sign of \(\dfrac{r}{Q(x)}\) for \(x \to +\infty\), \(x \to -\infty\) and either side of each vertical asymptote.
No real roots of \(Q(x) = 0\) means no vertical asymptote.

Recall

[2 marks]

State the equations of the asymptotes of the curve \(y = \dfrac{3x - 2}{x + 4}\).

\(x + 4 = 0\) at \(x = -4\), where the numerator is \(-14 \ne 0\): vertical asymptote \(x = -4\). ✓
Equal degrees: horizontal asymptote \(y = \dfrac{3}{1} = 3\). ✓

Calculation

[3 marks]

Express \(\dfrac{x^{2} - 4x + 1}{x - 3}\) in the form \(Ax + B + \dfrac{C}{x - 3}\), and hence state the equations of the asymptotes of \(y = \dfrac{x^{2} - 4x + 1}{x - 3}\).

\((x - 3)(x - 1) = x^{2} - 4x + 3\), so \(x^{2} - 4x + 1 = (x - 3)(x - 1) - 2\). ✓
\(y = x - 1 - \dfrac{2}{x - 3}\), so \(A = 1\), \(B = -1\), \(C = -2\). ✓
Asymptotes: \(x = 3\) and \(y = x - 1\). ✓

Calculation

[4 marks]

The curve \(C\) has equation \(y = \dfrac{2x^{2} - 5x + 3}{x^{2} - 4}\).

(a) Find the equations of the asymptotes of \(C\).

(b) Determine whether \(C\) approaches its horizontal asymptote from above or below as \(x \to +\infty\) and as \(x \to -\infty\).

(a) \(x^{2} - 4 = 0\) at \(x = \pm 2\); the numerator \((2x - 3)(x - 1)\) is not zero there, so \(x = 2\) and \(x = -2\) are vertical asymptotes. ✓
Equal degrees: \(y = \dfrac{2}{1} = 2\). ✓
(b) \(2x^{2} - 5x + 3 = 2(x^{2} - 4) + (-5x + 11)\), so \(y = 2 + \dfrac{11 - 5x}{x^{2} - 4}\). ✓
For large positive \(x\), \(11 - 5x < 0\) and \(x^{2} - 4 > 0\): below \(y = 2\). For large negative \(x\), \(11 - 5x > 0\): above \(y = 2\). ✓

Analysis

[4 marks]

The curve \(C\) has equation \(y = \dfrac{x^{2} + x + 2}{x + 1}\).

(a) Find the equations of the asymptotes of \(C\).

(b) Describe how \(C\) approaches each asymptote.

(a) \(x^{2} + x + 2 = (x + 1)x + 2\), so \(y = x + \dfrac{2}{x + 1}\). Asymptotes \(x = -1\) and \(y = x\). ✓
(b) \(x \to +\infty\): \(\dfrac{2}{x + 1} > 0\), so \(C\) is above \(y = x\). \(x \to -\infty\): \(\dfrac{2}{x + 1} < 0\), so below. ✓
\(x \to -1^{+}\): \(x + 1\) small and positive, \(y \to +\infty\). ✓
\(x \to -1^{-}\): \(x + 1\) small and negative, \(y \to -\infty\). ✓

Analysis

[4 marks]

The curve \(y = \dfrac{ax^{2} + bx + 3}{x + 1}\), where \(a\) and \(b\) are constants, has asymptotes \(x = -1\) and \(y = 2x - 1\).

(a) Find \(a\) and \(b\).

(b) State whether the curve lies above or below \(y = 2x - 1\) for large positive \(x\).

(a) Divide by \(x + 1\): \(ax^{2} + bx + 3 = (x + 1)(ax + (b - a)) + (3 - b + a)\), so the oblique asymptote is \(y = ax + (b - a)\). ✓
Compare with \(y = 2x - 1\): \(a = 2\) and \(b - a = -1\), so \(b = 1\). ✓
(b) Remainder \(3 - 1 + 2 = 4\): \(y = 2x - 1 + \dfrac{4}{x + 1}\) (check: \((x + 1)(2x - 1) + 4 = 2x^{2} + x + 3\)). ✓
For large positive \(x\), \(\dfrac{4}{x + 1} > 0\), so the curve is above \(y = 2x - 1\). ✓

Proof

[3 marks]

Show that the curve \(y = \dfrac{x^{2} + 5}{x^{2} + 4}\) has no vertical asymptote, state its horizontal asymptote, and show that the curve never meets that asymptote.

\(x^{2} + 4 \ge 4 > 0\) for all real \(x\), so the denominator is never zero: no vertical asymptote. ✓
Equal degrees: the horizontal asymptote is \(y = 1\). Write \(y = 1 + \dfrac{1}{x^{2} + 4}\). ✓
\(\dfrac{1}{x^{2} + 4} > 0\) for every \(x\), so \(y > 1\) always: the curve stays above \(y = 1\) and never meets it. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.