IGCSE · Practice questions · Cubic graphs and their moduli

Ends, roots, intercept, reflect.

Six original Cambridge-style questions on cubic graphs given as three linear factors: end behaviour, roots and intercepts, repeated roots, finding an equation from a sketch, sketching \(y = |f(x)|\) and counting solutions with a horizontal line.

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

The method

Ends, roots, intercept, then reflect.

Ends: multiply the \(x\)-coefficients of the brackets; positive starts bottom left, negative starts top left.
Roots: set each bracket to zero; a squared bracket touches the axis, a single one crosses.
Intercept: put \(x = 0\) and multiply; label every axis point with coordinates.
Modulus: reflect only the parts below the \(x\)-axis; sharp V-points at crossing roots; \(y\)-intercept \(|f(0)|\).
Count solutions of \(|f(x)| = k\) with the line \(y = k\): each arm once, each hump twice if \(k\) is below its top.

Recall

[3 marks]

Sketch the curve \(y = (x + 2)(x - 1)(x - 5)\), stating the coordinates of the points where the curve meets the axes.

The \(x^{3}\) coefficient is \(1 > 0\), so the curve starts bottom left and ends top right. ✓
It crosses the \(x\)-axis at \((-2, 0)\), \((1, 0)\) and \((5, 0)\). ✓
\(x = 0\): \((2)(-1)(-5) = 10\), so it meets the \(y\)-axis at \((0, 10)\). ✓
Shape: up through \((-2, 0)\), a maximum above the axis (through \((0, 10)\)), down through \((1, 0)\), a minimum below the axis, up through \((5, 0)\). ✓

Calculation

[3 marks]

Sketch the curve \(y = (4 - x)(2x + 1)(x + 2)\), stating the coordinates of the points where the curve meets the axes.

The \(x^{3}\) coefficient is \((-1)(2)(1) = -2 < 0\), so the curve starts top left and ends bottom right. ✓
Roots: \(x = 4\), \(x = -\dfrac{1}{2}\) and \(x = -2\), so it crosses at \((-2, 0)\), \((-\frac{1}{2}, 0)\) and \((4, 0)\). ✓
\(x = 0\): \((4)(1)(2) = 8\), so \((0, 8)\). ✓
Shape: down through \((-2, 0)\), a minimum below the axis, up through \((-\frac{1}{2}, 0)\) and \((0, 8)\), a maximum, then down through \((4, 0)\) to bottom right. ✓

Analysis

[4 marks]

A cubic curve touches the \(x\)-axis at \((2, 0)\), crosses it at \((-3, 0)\) and meets the \(y\)-axis at \((0, -12)\).

(a) Find the equation of the curve in the form \(y = a(x - 2)^{2}(x + 3)\).

(b) Write the equation in the form \(y = px^{3} + qx^{2} + rx + s\) and state which way the curve starts and ends.

(a) \((0, -12)\): \(-12 = a(-2)^{2}(3) = 12a\), so \(a = -1\) and \(y = -(x - 2)^{2}(x + 3)\). ✓
(b) \((x - 2)^{2}(x + 3) = (x^{2} - 4x + 4)(x + 3) = x^{3} - x^{2} - 8x + 12\). ✓
So \(y = -x^{3} + x^{2} + 8x - 12\). ✓
The \(x^{3}\) coefficient is \(-1 < 0\): the curve starts top left and ends bottom right. ✓

Calculation

[4 marks]

Sketch the graph of \(y = |(x + 3)(x - 1)(x - 2)|\), stating the coordinates of the points where the graph meets the axes.

First \(y = (x + 3)(x - 1)(x - 2)\): \(x^{3}\) coefficient \(1 > 0\), roots \(-3, 1, 2\), \(y\)-intercept \((3)(-1)(-2) = 6\). ✓
\(f(x) < 0\) for \(x < -3\) (three negative factors) and for \(1 < x < 2\) (one negative factor). ✓
Reflect those two parts in the \(x\)-axis: the left arm now rises to top left, and the small dip between \(1\) and \(2\) becomes a small hump. ✓
The graph meets the axes at \((-3, 0)\), \((1, 0)\), \((2, 0)\) (sharp V-points) and \((0, 6)\); it is never below the \(x\)-axis. ✓

Analysis

[5 marks]

The curve \(y = x^{2}(3 - x)\) has a minimum point at \((0, 0)\) and a maximum point at \((2, 4)\).

(a) Sketch the graph of \(y = |x^{2}(3 - x)|\).

(b) Find the number of solutions of the equation \(|x^{2}(3 - x)| = k\) when (i) \(k = 2\), (ii) \(k = 4\), (iii) \(k = 5\).

(a) \(y = x^{2}(3 - x)\): \(x^{3}\) coefficient \(-1\), so top left to bottom right; it touches the axis at \((0, 0)\) and crosses at \((3, 0)\). ✓
\(f(x) < 0\) only for \(x > 3\), so reflect that arm: it now rises from a V-point at \((3, 0)\). The touch at the origin stays smooth. ✓
The graph: a left arm coming down to \((0, 0)\), a hump of height 4 at \(x = 2\), a V-point at \((3, 0)\) and a right arm rising for ever. ✓
(b)(i) \(k = 2\): left arm once, hump twice, right arm once: 4 solutions. ✓
(ii) \(k = 4\): the line just touches the hump at \(x = 2\): \(1 + 1 + 1 = 3\) solutions. ✓
(iii) \(k = 5\): above the hump, so only the two arms: 2 solutions. ✓

Analysis

[6 marks]

\(f(x) = (1 - x)(x + 2)(x - 4)\). The turning points of \(y = f(x)\) are at approximately \((-0.7, -10.4)\) and \((2.7, 10.4)\).

(a) Sketch \(y = f(x)\), stating the coordinates of the points where it meets the axes.

(b) Sketch \(y = |f(x)|\) on a separate diagram.

(a) \(x^{3}\) coefficient \((-1)(1)(1) = -1\): top left to bottom right. Roots \(1, -2, 4\); \(y\)-intercept \((1)(2)(-4) = -8\). ✓
It crosses at \((-2, 0)\), \((1, 0)\) and \((4, 0)\), with a minimum below the axis between \(-2\) and \(1\) (through \((0, -8)\)) and a maximum above it between \(1\) and \(4\). ✓
(b) \(f(x) < 0\) for \(-2 < x < 1\) and for \(x > 4\); reflect those parts. The graph has a left arm, a hump of height about \(10.4\) on \(-2 < x < 1\) through \((0, 8)\), a hump of height about \(10.4\) on \(1 < x < 4\), and a right arm rising from \((4, 0)\). ✓
(c)(i) \(8 < 10.4\): the line meets each arm once and each hump twice: 6 solutions (one of them is \(x = 0\), since \(|f(0)| = 8\)). ✓
(ii) \(12 > 10.4\): above both humps, so only the two arms: 2 solutions. ✓

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.