AS & A Level · Practice questions · Related graphs: y squared, reciprocal and modulus

Sketch once, use it four times.

Six original Cambridge-style questions on the graphs of \(y^{2} = f(x)\), \(y = \frac{1}{f(x)}\), \(y = |f(x)|\) and \(y = f(|x|)\) for rational functions, from naming features to solving an inequality and counting roots.

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

Know \(y = f(x)\) first, then transform it.

List the features of \(y = f(x)\): zeros, asymptotes, turning points, intercepts, sign.
\(y^{2} = f(x)\): keep \(f \ge 0\), take \(\pm\sqrt{\ }\), vertical tangents at zeros, symmetric in the \(x\)-axis.
\(y = \frac{1}{f(x)}\): swap zeros and vertical asymptotes, reciprocate \(y\)-values, keep signs, flip max and min.
\(y = |f(x)|\): reflect negative parts up. \(y = f(|x|)\): reflect the right half to the left.
Count or solve with a horizontal line; find exact boundary values algebraically.

Recall

[3 marks]

Given \(f(x) = \dfrac{x + 2}{x - 3}\), state the equations of the asymptotes of \(y = \dfrac{1}{f(x)}\) and the coordinates of the points where it meets the axes.

\(\dfrac{1}{f(x)} = \dfrac{x - 3}{x + 2}\). ✓
Asymptotes: \(x = -2\) (from the zero of \(f\)) and \(y = 1\) (since \(f \to 1\) and \(\frac{1}{1} = 1\)). ✓
Meets the axes at \((3, 0)\) (from the asymptote \(x = 3\) of \(f\)) and \(\left(0, -\dfrac{3}{2}\right)\). ✓

Calculation

[4 marks]

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

(a) State the set of values of \(x\) for which \(C\) exists.

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

(a) \(x^{2} + 1 > 0\), so need \(x^{2} - 1 \ge 0\): \(x \le -1\) or \(x \ge 1\). ✓
(b) As \(x \to \pm\infty\), \(\dfrac{x^{2} - 1}{x^{2} + 1} \to 1\), so \(y \to \pm 1\): asymptotes \(y = 1\) and \(y = -1\). There is no vertical asymptote since the denominator is never zero. ✓
(c) \(f(x) = 0\) at \(x = \pm 1\): \(C\) meets the \(x\)-axis at \((-1, 0)\) and \((1, 0)\) with vertical tangents, the two halves symmetric in the \(x\)-axis. ✓
Since \(0 \le f(x) < 1\), \(C\) lies between the lines \(y = -1\) and \(y = 1\) and further from the \(x\)-axis than \(y = f(x)\). ✓

Analysis

[5 marks]

The curve \(y = \dfrac{x^{2} + 3}{x - 1}\) has a maximum point at \((-1, -2)\), a minimum point at \((3, 6)\), the vertical asymptote \(x = 1\) and the oblique asymptote \(y = x + 1\).

(a) Verify that \((3, 6)\) is a stationary point.

(b) State the turning points of \(y = \dfrac{x - 1}{x^{2} + 3}\), with their nature, and its asymptote. Give the point where it meets the \(x\)-axis.

(a) \(\dfrac{dy}{dx} = \dfrac{2x(x - 1) - (x^{2} + 3)}{(x - 1)^{2}} = \dfrac{x^{2} - 2x - 3}{(x - 1)^{2}}\), which is \(0\) at \(x = 3\); and \(y = \dfrac{12}{2} = 6\). ✓
(b) \(\dfrac{x - 1}{x^{2} + 3}\) is the reciprocal of \(f(x) = \dfrac{x^{2} + 3}{x - 1}\). ✓
The minimum \((3, 6)\) becomes a maximum \(\left(3, \dfrac{1}{6}\right)\); the maximum \((-1, -2)\) becomes a minimum \(\left(-1, -\dfrac{1}{2}\right)\). ✓
\(f(x) \to \pm\infty\) as \(x \to \pm\infty\), so the reciprocal has asymptote \(y = 0\); \(f\) has no zeros (\(x^{2} + 3 > 0\)), so there is no vertical asymptote. ✓
The asymptote \(x = 1\) of \(f\) becomes the point \((1, 0)\). ✓

Calculation

[4 marks]

Find the set of values of \(x\) for which \(\left|\dfrac{2x + 1}{x - 1}\right| < 1\).

Both sides are non-negative, so square (with \(x \ne 1\)): \((2x + 1)^{2} < (x - 1)^{2}\). ✓
\(4x^{2} + 4x + 1 < x^{2} - 2x + 1 \Rightarrow 3x^{2} + 6x < 0\). ✓
\(3x(x + 2) < 0 \Rightarrow -2 < x < 0\). ✓
Check: at \(x = -1\), \(\left|\frac{-1}{-2}\right| = \frac{1}{2} < 1\). \(x = 1\) is outside the interval, so nothing to remove. ✓

Analysis

[4 marks]

Given \(f(x) = \dfrac{x - 3}{x + 1}\).

(a) Explain why \(y = f(|x|)\) has no vertical asymptote, and state its lowest point.

(b) Find the set of values of \(k\) for which \(f(|x|) = k\) has exactly two real solutions.

(a) \(f(|x|) = \dfrac{|x| - 3}{|x| + 1}\) and \(|x| + 1 \ge 1\), so the denominator is never zero. (The asymptote \(x = -1\) of \(f\) is in the part that is discarded.) ✓
For \(x \ge 0\), \(f(x) = 1 - \dfrac{4}{x + 1}\) increases from \(f(0) = -3\) towards \(1\). Reflecting in the \(y\)-axis, the lowest point is \((0, -3)\). ✓
(b) A line \(y = k\) meets each half once when \(-3 < k < 1\). ✓
\(k = -3\) gives only \(x = 0\); \(k \ge 1\) or \(k < -3\) gives none. So \(-3 < k < 1\). ✓

Analysis

[6 marks]

The function \(f\) is defined by \(f(x) = \dfrac{4x}{x^{2} + 4}\) for all real \(x\).

(a) Show that the turning points of \(y = f(x)\) are \((2, 1)\) and \((-2, -1)\).

(b) State the number of real solutions of \(|f(x)| = \frac{1}{2}\).

(c) For the curve \(y = \dfrac{1}{f(x)}\), state the turning points with their nature and the equations of the asymptotes.

(a) \(f'(x) = \dfrac{4(x^{2} + 4) - 4x \cdot 2x}{(x^{2} + 4)^{2}} = \dfrac{4(4 - x^{2})}{(x^{2} + 4)^{2}} = 0 \Rightarrow x = \pm 2\). ✓
\(f(2) = \dfrac{8}{8} = 1\) and \(f(-2) = -1\): maximum \((2, 1)\), minimum \((-2, -1)\), with \(f(x) \to 0\) as \(x \to \pm\infty\). ✓
(b) \(y = |f(x)|\) has maxima \((\pm 2, 1)\), touches \(0\) at the origin and tends to \(0\). The line \(y = \frac{1}{2}\) meets each of the two humps twice: \(4\) solutions (\(x = \pm 4 \pm 2\sqrt{3}\)). ✓
(c) \(f(\pm 2) = \pm 1\), so these points are fixed: the maximum \((2, 1)\) becomes a minimum \((2, 1)\) and the minimum \((-2, -1)\) becomes a maximum \((-2, -1)\). ✓
The zero at \(x = 0\) gives the vertical asymptote \(x = 0\). Since \(\dfrac{1}{f(x)} = \dfrac{x^{2} + 4}{4x} = \dfrac{x}{4} + \dfrac{1}{x}\), the oblique asymptote is \(y = \dfrac{x}{4}\). ✓

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.