Sketching Cubics and Cubic Inequalities

Sketch a cubic given as three linear factors, and its modulus, with intercepts labelled, find a cubic from its graph, and solve cubic inequalities from the graph.

In the exam for IGCSE Additional Maths 0606 · 8 real past-paper questions below

اقرأ هذا الدرس مشروحاً بالعربي ←

Brackets give the shape

Cut a square of side x from each corner of a 20 cm by 30 cm card, and fold up the sides to make an open box. Its volume is $V = x (20 - 2 x) (30 - 2 x)$ : three brackets multiplied together. That is a cubic.

When a cubic comes as a product of three brackets, the brackets tell you everything you need to sketch it. You never need to expand.

Formula

y = k (x - p) (x - q) (x - r)

$p, q, r$ — the roots: where the curve meets the x-axis. Each bracket is zero at one of them.
$k$ — the number in front of $x^{3}$ once you multiply out. Its sign decides the shape.

Rule

Nothing here is on the formula list. Learn the shape rule: if the $x^{3}$ term is positive, the curve starts bottom left and ends top right. If it is negative, it starts top left and ends bottom right.

Worked example

1
Is the $x^{3}$ term of $y = (3 - x) (x + 1) (2 x - 1)$ positive or negative?
2
$(- x) (x) (2 x)$ — Multiply only the x part of each bracket. In 3 − x, the x part is −x.
3
$= - 2 x^{3}$ — Negative. So this curve starts top left and ends bottom right.

Common mistake

Reading $(3 - x)$ as if it were $(x - 3)$ . The x part is $- x$ , which flips the whole curve.

Key idea

Multiply the x parts of the brackets. The sign of the answer tells you which way up the curve goes.

Sketching a cubic

A sketch is not an accurate drawing. It shows the shape, and the points where the curve meets the axes, labelled with their values.

Steps

1
1. x-intercepts: set each bracket equal to zero.
2. y-intercept: put $x = 0$ , which means multiplying the three number parts.
3. Shape: use the sign of the $x^{3}$ term to fix the ends.
4. Draw one smooth curve through the intercepts, with one peak and one dip, and ends that carry on past the axis.

Worked example

1
Sketch $y = (x + 2) (x - 1) (x - 4)$ .
2
$x = - 2, x = 1, x = 4$ — Each bracket equal to zero gives an x-intercept.
3
$y = (2) (- 1) (- 4) = 8$ — Put x = 0. The y-intercept is 8.
4
$(x) (x) (x) = x^{3}$ — Positive x³: start bottom left, end top right.
5
$up through - 2, down through 1, up through 4$ — From the bottom left, cross at −2, rise to a peak above 8, fall through 1, dip, then rise through 4.

Worked example

1
Sketch $y = (3 - x) (x + 1) (2 x - 1)$ .
2
$x = 3, x = - 1, x = \frac{1}{2}$ — Each bracket equal to zero.
3
$y = (3) (1) (- 1) = - 3$ — Put x = 0. The y-intercept is −3.
4
$start top left$ — The x³ term is −2x³, negative. Come down through −1, dip below the axis through −3, rise through one half, peak, then fall through 3.

Mark schemes give marks for the correct shape and for the correct intercepts, so every intercept must be labelled. You do not need the peak or dip coordinates unless the question asks.

Common mistake

Stopping the curve at the outer roots. The ends must carry on past the x-axis, heading off to infinity.

Key idea

Three crossings, one y-intercept, the right way up. That is a full-mark sketch.

Repeated roots

Sometimes a bracket appears twice, like $(x - 2)^{2}$ . That root is called a repeated root. The curve does not cross the axis there: it touches the axis and turns back.

Worked example

1
Sketch $y = (x - 2)^{2} (x + 1)$ .
2
$x = - 1, x = 2$ — Two different roots. x = 2 comes from a squared bracket.
3
$y = (- 2)^{2} (1) = 4$ — Put x = 0. The y-intercept is 4.
4
$(x)^{2} (x) = x^{3}$ — Positive x³: start bottom left, end top right.
5
$cross at - 1, touch at 2$ — Rise through −1 to a peak at (0, 4), fall to just touch the axis at 2, then rise again.

Common mistake

Crossing the axis at a repeated root. At $x = 2$ the curve sits on the axis for a moment and turns, like a ball bouncing on the floor.

Key idea

Single bracket: cross. Squared bracket: touch and turn back.

Equation from a graph

Exams also run the sketch backwards: you see the graph and must write its equation. Each x-intercept gives a bracket. The y-intercept gives the number in front.

Worked example

1
A cubic crosses the x-axis at $- 3$ , 1 and 2, and meets the y-axis at $- 12$ . Find its equation.
2
$y = k (x + 3) (x - 1) (x - 2)$ — One bracket for each root. k is still unknown.
3
$- 12 = k (3) (- 1) (- 2) = 6 k$ — The point (0, −12) is on the curve, so put x = 0 and y = −12.
4
$k = - 2$ — Divide by 6.
5
$y = - 2 (x + 3) (x - 1) (x - 2)$ — Check the shape: k is negative, so the graph must start top left. It should match the picture.

Any equal form is accepted, for example $y = (x + 3) (1 - x) (2 x - 4)$ , because $(1 - x) (2 x - 4) = - 2 (x - 1) (x - 2)$ .

Common mistake

Leaving out k. Without it, the curve has the right roots but the wrong y-intercept, and possibly the wrong shape.

Key idea

Roots give the brackets. The y-intercept fixes the number in front.

Cubic inequalities

A cubic inequality asks where the curve is above or below the x-axis. Your sketch answers it.

Rule

$f (x) > 0$ : where the curve is above the x-axis. $f (x) < 0$ : where it is below. With $\geq$ or $\leq$ , the roots are included.

Worked example

1
Solve $(x + 2) (x - 1) (x - 4) > 0$ .
2
$x = - 2, 1, 4$ — The roots, from the sketch in this lesson.
3
$- 2 < x < 1$ — The curve is above the axis between −2 and 1.
4
$x > 4$ — It is above the axis again to the right of 4.
5
$- 2 < x < 1 or x > 4$ — Write both parts. Strict signs, because > 0 does not include the roots.

Worked example

1
Solve $(3 - x) (x + 1) (2 x - 1) \geq 0$ .
2
$x \leq - 1$ — This curve starts top left, so it is above the axis up to −1.
3
$\frac{1}{2} \leq x \leq 3$ — It is above the axis again between one half and 3.
4
$x \leq - 1 or \frac{1}{2} \leq x \leq 3$ — Include the roots because of ≥.

Each correct part usually earns its own mark, so write the two parts separately, joined by “or”.

Common mistake

Joining the parts into one, like $- 2 < x > 4$ . Two separate regions need two separate inequalities.

Key idea

Sketch first, then read: above the axis is positive, below is negative.

Comparing with a line

Sometimes you are given an accurate graph and asked for $f (x) \geq d$ , where d is a number. Now compare the curve with the horizontal line $y = d$ , not with the x-axis.

Steps

1
1. Draw the straight line $y = d$ across the grid.
2. Read the x-values where it meets the curve.
3. Find where the curve is above the line (for > or ≥) or below it (for < or ≤).
4. Write the regions as inequalities.

Worked example

1
An accurate graph of $y = (x + 2) (x - 1) (x - 4)$ is drawn. Solve $(x + 2) (x - 1) (x - 4) \geq 8$ .
2
$y = 8$ — Draw this horizontal line on the grid.
3
$x \approx - 1.4, x = 0, x \approx 4.4$ — Read where the line meets the curve. Note x = 0 is the y-intercept, 8.
4
$- 1.4 \leq x \leq 0$ — Between the first two meeting points the curve is above the line.
5
$x \geq 4.4$ — Past the third meeting point it is above the line again.
6
$- 1.4 \leq x \leq 0 or x \geq 4.4$ — The full answer, read to graph accuracy.

The mark scheme gives a mark for the line drawn, meeting the curve where it should, and then marks for the answer. So draw the line on the graph.

Common mistake

Reading where the curve crosses the x-axis. For $f (x) \geq 8$ , the boundary is the line $y = 8$ .

Key idea

The number of times the line meets the curve is the number of solutions of $f (x) = d$ .

The modulus of a cubic

You may also be asked to sketch $y = ∣ f (x) ∣$ . The bars mean the size of the value, so it is never negative: $∣ - 5∣ = 5$ and $∣5∣ = 5$ .

Rule

To sketch $y = ∣ f (x) ∣$ : sketch $y = f (x)$ , then reflect every part below the x-axis up into the positive side. Parts above the axis stay as they are.

Worked example

1
Sketch $y = ∣ (x + 2) (x - 1) (x - 4) ∣$ .
2
$x < - 2 and 1 < x < 4$ — These are the parts of y = (x + 2)(x − 1)(x − 4) below the axis.
3
$reflect in the x-axis$ — The left end now rises upwards, and the dip between 1 and 4 becomes a hump.
4
$(- 2, 0), (1, 0), (4, 0)$ — At each root the curve now meets the axis in a sharp point.
5
$(0, 8)$ — The y-intercept was already positive, so it stays at 8.

Common mistake

Drawing smooth rounded bottoms at the roots. The modulus graph has sharp points there, called cusps, and examiners look for them.

Key idea

Modulus: anything below the axis bounces up. Nothing ever goes below the x-axis.

Which method? Read the question

Read the question first. The wording tells you what to draw and what to read.

You see	You do
“Sketch” a product of three brackets	Roots, y-intercept, shape from the sign of the x³ term
A squared bracket	The curve touches the axis there and turns
A graph, “find the equation”	Brackets from the roots, then k from the y-intercept
$f (x) > 0$ or $f (x) \leq 0$	Read where the curve is above or below the x-axis
$f (x) \geq d$ , accurate graph	Draw $y = d$ ; read where the curve is above it
$y = ∣ f (x) ∣$	Reflect the parts below the axis; sharp points at the roots

Common mistake

Leaving intercepts unlabelled. A sketch earns its marks only if the values are written on the axes.

Key idea

Every question in this lesson starts with the same sketch. Get the roots, the y-intercept and the shape right, and the rest is reading.

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Real past-paper questions on sketching cubics and cubic inequalities

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