AS & A Level · Practice questions · Sketching rational functions

Every feature, labelled.

Six original Cambridge-style questions on sketching rational functions: asymptotes and intercepts, turning points, the set of values of \(y\) by the discriminant, full sketches and inequalities solved from the graph.

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

Asymptotes, intercepts, turning points, then join.

Asymptotes: denominator zero; then compare degrees or divide for the horizontal or oblique one.
Intercepts: \(x = 0\), and numerator \(= 0\).
Turning points: quotient rule, set the numerator \(v u' - u v'\) to zero.
Range: \(y\,Q(x) = P(x)\) as a quadratic in \(x\), then \(b^{2} - 4ac \ge 0\); check any \(y\) that kills the \(x^{2}\) term.
Inequalities: multiply by the denominator squared, factorise, and check against the sketch.

Recall

[4 marks]

The curve \(C\) has equation \(y = \dfrac{x^{2} - x - 6}{x^{2} - 1}\).

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

(b) Find the coordinates of the points where \(C\) meets the axes.

(a) \(x^{2} - 1 = 0\): \(x = 1\) and \(x = -1\). Equal degrees, leading coefficients 1 and 1: \(y = 1\). ✓
(b) \(x = 0\): \(y = \dfrac{-6}{-1} = 6\), so \((0, 6)\). \(x^{2} - x - 6 = (x - 3)(x + 2) = 0\): \((3, 0)\) and \((-2, 0)\). ✓
(c) \(x^{2} - x - 6 = x^{2} - 1 \Rightarrow -x = 5 \Rightarrow x = -5\). ✓
The curve crosses \(y = 1\) at \((-5, 1)\). ✓

Calculation

[4 marks]

Find the coordinates of the turning points of \(y = \dfrac{x^{2} + 8}{x + 1}\) and determine their nature.

\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2x(x + 1) - (x^{2} + 8)}{(x + 1)^{2}} = \dfrac{x^{2} + 2x - 8}{(x + 1)^{2}} = \dfrac{(x + 4)(x - 2)}{(x + 1)^{2}}\). ✓
\(x = 2\): \(y = \dfrac{12}{3} = 4\). \(x = -4\): \(y = \dfrac{24}{-3} = -8\). ✓
The denominator \((x + 1)^{2}\) is positive, so the sign of the gradient is the sign of \((x + 4)(x - 2)\): \(+\) then \(-\) at \(x = -4\), \(-\) then \(+\) at \(x = 2\). ✓
\((-4, -8)\) is a maximum and \((2, 4)\) is a minimum. ✓

Analysis

[5 marks]

Use a discriminant to find the set of values taken by \(y = \dfrac{x^{2} + x + 1}{x^{2} + 1}\) for real \(x\). Hence find the coordinates of the turning points of the curve.

\(y(x^{2} + 1) = x^{2} + x + 1\), so \((y - 1)x^{2} - x + (y - 1) = 0\). ✓
If \(y = 1\), the equation is \(-x = 0\), so \(x = 0\) gives \(y = 1\): this value is taken. ✓
Otherwise real \(x\) needs \(1 - 4(y - 1)^{2} \ge 0\), so \((y - 1)^{2} \le \frac{1}{4}\), \(-\frac{1}{2} \le y - 1 \le \frac{1}{2}\). ✓
Set of values: \(\frac{1}{2} \le y \le \frac{3}{2}\). ✓
\(y = \frac{1}{2}\): \(-\frac{1}{2}x^{2} - x - \frac{1}{2} = 0 \Rightarrow (x + 1)^{2} = 0\), \(x = -1\). \(y = \frac{3}{2}\): \(\frac{1}{2}x^{2} - x + \frac{1}{2} = 0 \Rightarrow (x - 1)^{2} = 0\), \(x = 1\). ✓
Turning points: minimum \(\left(-1, \frac{1}{2}\right)\) and maximum \(\left(1, \frac{3}{2}\right)\). ✓

Analysis

[6 marks]

The curve \(C\) has equation \(y = \dfrac{x^{2} + 3}{x - 1}\).

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

(b) Find the coordinates of the turning points of \(C\).

(a) \(x^{2} + 3 = (x - 1)(x + 1) + 4\), so \(y = x + 1 + \dfrac{4}{x - 1}\): asymptotes \(x = 1\) and \(y = x + 1\). ✓
(b) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 1 - \dfrac{4}{(x - 1)^{2}} = 0 \Rightarrow (x - 1)^{2} = 4 \Rightarrow x = 3\) or \(x = -1\). ✓
\(x = 3\): \(y = 6\); \(x = -1\): \(y = -2\). Maximum \((-1, -2)\), minimum \((3, 6)\). ✓
(c) \(y\)-intercept \((0, -3)\); no \(x\)-intercepts since \(x^{2} + 3 > 0\), so the left branch (for \(x < 1\)) lies entirely below the \(x\)-axis. ✓
As \(x \to 1^{+}\), \(y \to +\infty\); as \(x \to 1^{-}\), \(y \to -\infty\). As \(x \to \pm\infty\) the curve is above \(y = x + 1\) on the right and below it on the left. ✓
Sketch: left branch rising from below \(y = x + 1\) to the maximum \((-1, -2)\), through \((0, -3)\) and down the asymptote; right branch coming down the asymptote to the minimum \((3, 6)\), then up towards \(y = x + 1\). ✓

Analysis

[5 marks]

The curve \(y = \dfrac{x^{2} + 3}{x - 1}\) has a maximum at \((-1, -2)\) and a minimum at \((3, 6)\). Solve the inequality \(\dfrac{x^{2} + 3}{x - 1} \le 7\).

Multiply by \((x - 1)^{2} > 0\) (\(x \ne 1\)): \((x^{2} + 3)(x - 1) - 7(x - 1)^{2} \le 0\). ✓
\((x - 1)(x^{2} + 3 - 7x + 7) \le 0\), so \((x - 1)(x^{2} - 7x + 10) \le 0\). ✓
\((x - 1)(x - 2)(x - 5) \le 0\), with \(x \ne 1\). ✓
The cubic is negative for \(x < 1\), positive for \(1 < x < 2\), negative for \(2 < x < 5\) and positive for \(x > 5\). ✓
Solution: \(x < 1\) or \(2 \le x \le 5\). The sketch agrees: the whole left branch is below \(-2\), and the right branch is at most 7 only round its minimum \((3, 6)\). ✓

Proof

[6 marks]

The curve \(C\) has equation \(y = \dfrac{x^{2} + k}{x - 2}\), where \(k\) is a constant and \(k \ne -4\).

(a) Show that \(C\) has no turning points when \(k < -4\).

(b) For \(k > -4\), show that \(y\) cannot take any value between \(4 - 2\sqrt{4 + k}\) and \(4 + 2\sqrt{4 + k}\).

(a) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2x(x - 2) - (x^{2} + k)}{(x - 2)^{2}} = \dfrac{x^{2} - 4x - k}{(x - 2)^{2}}\). ✓
\(x^{2} - 4x - k = 0\) has discriminant \(16 + 4k\), which is negative when \(k < -4\): no stationary points. ✓
(b) \(y(x - 2) = x^{2} + k \Rightarrow x^{2} - yx + (2y + k) = 0\). Real \(x\): \(y^{2} - 8y - 4k \ge 0\). ✓
\(y^{2} - 8y - 4k = 0\) has roots \(y = \dfrac{8 \pm \sqrt{64 + 16k}}{2} = 4 \pm 2\sqrt{4 + k}\), real since \(k > -4\). ✓
So \(y \le 4 - 2\sqrt{4 + k}\) or \(y \ge 4 + 2\sqrt{4 + k}\), and no value strictly between them is taken. ✓
(c) The minimum is the lowest point of the upper branch: \(4 + 2\sqrt{4 + k} = 10 \Rightarrow \sqrt{4 + k} = 3 \Rightarrow k = 5\). ✓

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.