# Sketching Cubics and Cubic Inequalities

> CIE IGCSE Additional Mathematics · 2025-2027
> Source: https://www.owlsprep.com/study/cie-0606-u4-sketching-cubics-and-cubic-inequalities/

This guide teaches you to sketch factored cubics, draw their modulus graphs, and solve cubic inequalities using sketches, exactly as required for CIE IGCSE Add Maths 0606.

**Prerequisites:** Solving linear and quadratic inequalities; Factoring linear expressions

## Learning objectives

- Sketch a cubic given as a product of three linear factors, correctly labelling all x and y intercepts
- Sketch the modulus of a factored cubic function
- Solve cubic inequalities by interpreting the shape and intercepts of the corresponding cubic graph
- Identify regions where a cubic function lies above or below the x-axis for inequality solutions
- Sketch cubics with a repeated root, distinguishing where the curve touches the x-axis from where it crosses
- Work backwards from a graph of the modulus of a cubic to find a possible expression for the original cubic
- Solve cubic inequalities with a non-zero right-hand side by rearranging to compare with zero

## Sketching Factored Cubic Functions

**Factored cubic function** — A cubic function where $k$ is the leading coefficient (determines end behaviour: positive $k$ means right end points up, left end down; negative $k$ means right end down, left end up), and $a, b, c$ are the x-intercepts (roots) of the function.

*Notation:* $y = k(x-a)(x-b)(x-c)$

*Example:* For $y = 2(x-1)(x+2)(x-3)$, x-intercepts are (1, 0), (-2, 0), (3, 0), y-intercept is 12, so (0, 12).

To sketch a factored cubic, follow 3 simple steps: 1. Find all x-intercepts by setting each linear factor equal to zero and plotting them on the x-axis. 2. Calculate the y-intercept by substituting $x=0$ into the function, plot this on the y-axis. 3. Use the sign of the leading coefficient $k$ to draw the correct S-shaped cubic curve passing through all intercepts.

**Worked example:** Sketch the cubic function $y = (x+1)(x-2)(x+3)$, labelling all axis intercepts.

1. Step 1: Find x-intercepts: set each factor to 0: $x+1=0 \implies x=-1$, $x-2=0 \implies x=2$, $x+3=0 \implies x=-3$. Plot (-3, 0), (-1, 0), (2, 0) on the x-axis.
2. Step 2: Calculate y-intercept: substitute $x=0$: $y = (0+1)(0-2)(0+3) = (1)(-2)(3) = -6$. Plot (0, -6) on the y-axis.
3. Step 3: Leading coefficient is 1 (positive), so right end points up, left end points down. Draw a smooth S-shaped curve passing through all 4 intercepts, following the end behaviour.

> **Exam tip:** You do not need to draw the graph to scale, just ensure intercepts are correctly labelled and the shape matches the sign of the leading coefficient. Examiners only check for correct intercepts and shape for this topic.

## Sketching the Modulus of a Cubic Function

The modulus of a cubic function $|f(x)|$ reflects all parts of the original cubic graph that lie below the x-axis (where $f(x) < 0$) across the x-axis, so every point on the modulus graph has a non-negative y-value.

> **tip**
>
> The x-intercepts of the original cubic are exactly the x-intercepts of the modulus function, since $|0| = 0$, so these points stay unchanged when you reflect the graph.

**Worked example:** Sketch $y = |(x+1)(x-2)(x+3)|$, using the cubic you sketched in the previous example, label all intercepts.

1. Step 1: Start with the original cubic sketch from the previous example. It lies below the x-axis for $x < -3$ and for $-1 < x < 2$ (for instance $f(-4) = -18$ and $f(0) = -6$ are below the axis, while $f(-2) = 4$ is above it). These two portions, including the unbounded left-hand tail, are the parts to be reflected up across the x-axis.
2. Step 2: Reflect all parts of the curve that are below the x-axis across the x-axis, so those regions now lie above the x-axis.
3. Step 3: Label the x-intercepts at (-3, 0), (-1, 0), (2, 0), and the y-intercept is now $|-6| = 6$, so (0,6). All points on the curve have $y \geq 0$.

## Solving Cubic Inequalities Graphically

To solve a cubic inequality, first rearrange it so one side is 0, then sketch the corresponding factored cubic graph, and identify the regions of the x-axis where the graph satisfies the inequality (above the x-axis for $> 0$, below for $< 0$, including intercepts for $\geq$ or $\leq$).

**Worked example:** Solve the inequality $(x+1)(x-2)(x+3) > 0$ using the cubic graph you sketched earlier.

1. Step 1: The corresponding cubic is $y = (x+1)(x-2)(x+3)$, which we already sketched, with intercepts at x=-3, x=-1, x=2, positive leading coefficient.
2. Step 2: We need regions where $y > 0$, i.e., the curve lies above the x-axis.
3. Step 3: Looking at the graph: the curve is above the x-axis between x=-3 and x=-1, and for $x > 2$. So the solution is $-3 < x < -1$ or $x > 2$.

**Exam command terms**

Common command terms for this topic:

- **Sketch** — Draw the general shape of the graph, label all required points (intercepts) clearly, no scale is needed unless specified.

- **Solve** — Find the set of x-values that satisfy the given inequality, write your solution using inequality notation or set notation as required.

- **Label** — Write the coordinates of all intercepts next to the corresponding points on your graph, do not leave them unmarked.

## Cubics with a Repeated Root (Touching the x-axis)

Not every cubic has three different roots. When a factor is squared, as in $y = a(x + b)^2(x + c)$, the value $x = -b$ is a repeated (double) root. At a double root the curve does not cross the x-axis; instead it comes down, touches the axis, and turns back on the same side, just like the vertex of a parabola. At the single root $x = -c$ the curve crosses straight through the axis as usual.

> **tip**
>
> Quick rule: a squared factor $(x + b)^2$ gives a touch at $x = -b$ (the curve stays on one side of the axis there), while a plain factor $(x + c)$ gives a crossing at $x = -c$ (the curve changes side).

**Repeated (double) root** — A root that comes from a squared factor. The factor $(x + b)^2$ is never negative, so it cannot change the sign of $y$ as $x$ passes through $x = -b$; the curve therefore touches the x-axis at $(-b, 0)$ rather than crossing it. The single factor $(x + c)$ still gives an ordinary crossing at $(-c, 0)$.

*Notation:* $y = a(x + b)^2(x + c)$

*Example:* $y = (x - 2)^2(x + 1)$ touches the x-axis at $(2, 0)$ and crosses it at $(-1, 0)$.

**Worked example:** Sketch $y = (x - 2)^2(x + 1)$, showing clearly where the curve touches and where it crosses the x-axis, and labelling all axis intercepts.

1. Step 1: Read off the roots. The squared factor $(x - 2)^2$ gives a double root at $x = 2$, so the curve touches the x-axis at $(2, 0)$. The single factor $(x + 1)$ gives a simple root at $x = -1$, so the curve crosses at $(-1, 0)$.
2. Step 2: Find the y-intercept: substitute $x = 0$: $y = (0 - 2)^2(0 + 1) = (4)(1) = 4$, so the y-intercept is $(0, 4)$.
3. Step 3: The leading coefficient is $+1$ (positive), so the left end points down and the right end points up. Coming from the bottom left, the curve rises and crosses the axis at $(-1, 0)$, climbs to a local maximum, comes back down to just touch the axis at $(2, 0)$, then turns and rises again on the right. Check a point between the roots: $f(1) = (1 - 2)^2(1 + 1) = (1)(2) = 2 > 0$, confirming the curve stays above the axis between $x = -1$ and $x = 2$.

> **Exam tip:** A squared factor means the curve touches the x-axis and turns back; a plain factor means the curve passes straight through. Show this difference clearly in your sketch.

## Working Backwards from a Modulus Graph

Some questions show you the graph of $y = |f(x)|$, where $f(x)$ is a cubic, and ask you to find a possible expression for $f(x)$. You reverse the earlier process: read the x-intercepts to get the factors, read the shape at each intercept to decide whether the root is single or repeated, and use the y-intercept to fix the constant.

> **tip**
>
> Reading a modulus graph: a sharp corner (cusp) where the graph meets the x-axis means the original cubic crossed there, so that is a single root; a smooth, rounded touch means the original cubic only touched there, so that is a repeated (double) root.

**Worked example:** The diagram shows $y = |f(x)|$, where $f(x)$ is a cubic. The graph meets the x-axis at $x = -2$, $x = 1$ and $x = 3$, forming a sharp corner (cusp) at each of these points, and it passes through $(0, 12)$. Find a possible expression for $f(x)$.

1. Step 1: The x-intercepts of $|f(x)|$ are the roots of $f(x)$, so $f$ has roots at $x = -2$, $x = 1$ and $x = 3$. A cusp at each intercept means each is a single root (the cubic crosses there), so $f(x) = k(x + 2)(x - 1)(x - 3)$.
2. Step 2: Use the y-intercept. On the modulus graph the y-intercept is $12$, so $|f(0)| = 12$. Now $f(0) = k(2)(-1)(-3) = 6k$, so $|6k| = 12$, giving $|k| = 2$.
3. Step 3: Since $|f(x)| = |-f(x)|$, both $k = 2$ and $k = -2$ produce the same modulus graph. Taking the simplest, a possible expression is $f(x) = 2(x + 2)(x - 1)(x - 3)$. (The choice $f(x) = -2(x + 2)(x - 1)(x - 3)$ is equally valid.)

> **Exam tip:** Because $|f(x)|$ and $|-f(x)|$ give exactly the same picture, the question asks for a possible $f(x)$: usually give the simplest one (for example with a positive leading coefficient) unless extra information fixes the sign.

## Cubic Inequalities with a Non-Zero Right-Hand Side

So far each inequality already compared the cubic with 0. When instead you meet something like $f(x) \geq d$ with $d \neq 0$, you cannot use the roots of $f$ directly. First move everything to one side to form $g(x) = f(x) - d$, so the inequality becomes $g(x) \geq 0$ (or the matching sign). Then factorise $g(x)$ and analyse the sign of this new cubic exactly as before, using its roots.

> **tip**
>
> A common slip is to solve $f(x) \geq d$ by reading off where $f(x) = 0$. Those are the wrong boundary points. Rearrange to $g(x) = f(x) - d$ first; the roots of $g$ are the correct boundaries.

**Worked example:** Solve $x^3 - 7x \leq -6$.

1. Step 1: Move every term to the left so one side is 0: $x^3 - 7x + 6 \leq 0$. Let $g(x) = x^3 - 7x + 6$.
2. Step 2: Factorise $g(x)$ using the factor theorem. Testing $x = 1$: $1 - 7 + 6 = 0$, so $(x - 1)$ is a factor. Dividing gives $g(x) = (x - 1)(x^2 + x - 6) = (x - 1)(x + 3)(x - 2)$, so the roots are $x = -3$, $x = 1$ and $x = 2$.
3. Step 3: Sketch $g(x)$: three single roots at $-3, 1, 2$ with a positive leading coefficient. The curve is on or below the x-axis (so $g(x) \leq 0$) for $x \leq -3$ and for $1 \leq x \leq 2$. Check $x = 0$: $g(0) = 6 > 0$, which correctly lies outside the solution.
4. Step 4: Because the inequality is $\leq$, include the boundary points. The solution is $x \leq -3$ or $1 \leq x \leq 2$.

> **Exam tip:** The boundary points of the solution are the roots of $g(x) = f(x) - d$, not the roots of $f(x)$. Always rearrange to compare with 0 before reading any roots.

## Common pitfalls

- **Wrong:** Assuming all cubic graphs have a positive leading coefficient, drawing the end behaviour backwards for negative $k$.
  - Why it fails: A negative leading coefficient reverses the end shape of the cubic, leading you to identify the wrong regions for inequalities.
  - Correct: Always check the sign of $k$ before sketching: positive $k$ = right end up, left end down; negative $k$ = right end down, left end up.
- **Wrong:** Forgetting to reflect the y-intercept when sketching the modulus of a cubic.
  - Why it fails: If the original y-intercept is negative, the modulus y-intercept is its absolute value, not the original value, leading to an incorrect graph.
  - Correct: Calculate the modulus of the original y-intercept when sketching $|f(x)|$, or reflect the entire part of the graph below the x-axis including the y-intercept if it is negative.
- **Wrong:** Including or excluding intercepts incorrectly in inequality solutions (e.g., using $<$ instead of $\leq$ when the inequality is $\geq 0$).
  - Why it fails: Inequalities with $\geq$ or $\leq$ include the x-intercepts where the function equals 0, while $<$ or $>$ exclude them, leading to lost marks for incorrect bounds.
  - Correct: Check the inequality symbol: use closed bounds or $\leq$/$\geq$ for inclusive inequalities, open bounds or $<$/$>$ for exclusive ones.
- **Wrong:** Trying to calculate turning points or inflexion points for the cubic sketch.
  - Why it fails: Calculating turning points is out of scope for this 0606 topic, and wasting time on this will lose you time for other questions in the exam.
  - Correct: Only draw the general S-shaped cubic passing through the intercepts, with correct end behaviour, no turning point calculation is needed.
- **Wrong:** Writing overlapping or incorrectly ordered regions for inequality solutions.
  - Why it fails: Cubic inequalities often have two separate solution regions, and mixing up the order of the bounds leads to invalid solutions.
  - Correct: List the roots in ascending order on the x-axis, then test each region between the roots, or use the graph shape to identify the correct regions, writing them as separate non-overlapping intervals.
- **Wrong:** Drawing the curve crossing the x-axis at a repeated (squared) root instead of just touching it.
  - Why it fails: A squared factor cannot change the sign of $y$, so the curve must touch and turn back at that root; drawing a crossing gives the wrong shape and the wrong inequality regions.
  - Correct: Treat a squared factor $(x+b)^2$ as a touch at $x=-b$ (curve stays on one side) and a plain factor as a crossing.
- **Wrong:** Solving $f(x) \geq d$ (with $d \neq 0$) by reading off the roots of $f(x)$ instead of rearranging first.
  - Why it fails: The boundary points of the solution are where $f(x) = d$, that is the roots of $g(x) = f(x) - d$, not where $f(x) = 0$.
  - Correct: Rearrange to $g(x) = f(x) - d$, compare it with 0, factorise $g$, and use the roots of $g$ to find the sign regions.

## Cheatsheet

| Task | Steps | Key Reminder |
| --- | --- | --- |
| Sketch factored cubic | 1. Plot x-intercepts (set factors = 0) 2. Calculate & plot y-intercept (x=0) 3. Draw S-shape matching $k$ sign | No turning point calculation needed |
| Sketch modulus of cubic | 1. Sketch original cubic 2. Reflect all parts below x-axis across x-axis | x-intercepts stay the same, y-values all $\geq 0$ |
| Solve cubic inequality | 1. Rearrange to $f(x)$ [inequality] 0 2. Sketch $f(x)$ 3. Identify regions matching inequality | Include intercepts for $\geq$/$\leq$, exclude for $>$/$<$ |
| Sketch cubic with a repeated root | Squared factor $(x+b)^2$ = touch at $x=-b$; plain factor = crossing. Plot the y-intercept, then draw the shape | The curve stays on one side of the axis at a touch |
| Find $f(x)$ from a graph of $\|f(x)\|$ | x-intercepts give the roots; cusp = single root, smooth touch = double root; y-intercept fixes $\|k\|$ | Answer is a possible $f(x)$: $\|f\| = \|-f\|$ so the sign of $k$ is not unique |
| Solve $f(x)$ [inequality] $d$ with $d \neq 0$ | Rearrange to $g(x)=f(x)-d$ [inequality] 0, factorise, use the roots of $g$ for the sign regions | Boundaries are the roots of $g$, not of $f$ |

## What's next

Now that you have mastered sketching factored cubics and solving cubic inequalities, you can move on to more advanced graph topics in the CIE IGCSE Additional Mathematics 0606 syllabus. Next, you will learn to sketch other polynomial and rational functions, which build on the intercept identification and shape analysis skills you practiced here. These graph skills are also essential for solving applied problems involving rates of change later in the course, so make sure you can quickly sketch cubics and identify inequality regions without unnecessary calculations. Practice past paper questions on this topic to get used to common exam phrasing and formats.

---

From [OwlsPrep](https://www.owlsprep.com) — free study guides for A-Level, IB, AP and IGCSE, written against the official syllabus. Canonical page: https://www.owlsprep.com/study/cie-0606-u4-sketching-cubics-and-cubic-inequalities/