Sketching Cubics and Cubic Inequalities

Q: Do I need to calculate turning points when sketching cubics for 0606?

No, for CIE IGCSE Additional Mathematics 0606, you only need to label axis intercepts and show the correct overall cubic shape when given a factored form. No turning point calculation is required for this topic.

Q: Can I solve cubic inequalities algebraically instead of graphically?

While algebraic methods exist, the 0606 syllabus expects you to use the graphical method for cubic inequalities, so practice interpreting your sketched graph to find solution regions.

CIE IGCSE Additional Mathematics· 4.4, 4.5· 18 min read

1. Sketching Factored Cubic Functions★★☆☆☆⏱ 5 min

📘 Definition

Factored cubic function

$y = k (x - a) (x - b) (x - c)$

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.

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: $x + 1 = 0 ⟹ x = - 1$ , $x - 2 = 0 ⟹ x = 2$ , $x + 3 = 0 ⟹ x = - 3$ . 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.

2. Sketching the Modulus of a Cubic Function★★★☆☆⏱ 6 min

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.

📐 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 $y \geq 0$ .

3. Solving Cubic Inequalities Graphically★★★☆☆⏱ 7 min

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.
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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 $x > 2$ .

4. Cubics with a Repeated Root (Touching the x-axis)★★★☆☆⏱ 5 min

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.

📘 Definition

Repeated (double) root

$y = a (x + b)^{2} (x + c)$

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 $(- c, 0)$ .

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.

5. Working Backwards from a Modulus Graph★★★★☆⏱ 6 min

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.

📐 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 $f (0) = k (2) (- 1) (- 3) = 6 k$ , so $∣6 k ∣ = 12$ , giving $∣ k ∣ = 2$ .
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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.

6. Cubic Inequalities with a Non-Zero Right-Hand Side★★★★☆⏱ 6 min

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.

📐 Worked Example

Solve $x^{3} - 7 x \leq - 6$ .

1
Step 1: Move every term to the left so one side is 0: $x^{3} - 7 x + 6 \leq 0$ . Let $g (x) = x^{3} - 7 x + 6$ .
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.
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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.

7. Common Pitfalls

Wrong move:

Assuming all cubic graphs have a positive leading coefficient, drawing the end behaviour backwards for negative $k$ .

Why:

A negative leading coefficient reverses the end shape of the cubic, leading you to identify the wrong regions for inequalities.

Correct move:

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 move:

Forgetting to reflect the y-intercept when sketching the modulus of a cubic.

Why:

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 move:

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 move:

Including or excluding intercepts incorrectly in inequality solutions (e.g., using $<$ instead of $\leq$ when the inequality is $\geq 0$ ).

Why:

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 move:

Check the inequality symbol: use closed bounds or $\leq$ / $\geq$ for inclusive inequalities, open bounds or $<$ / $>$ for exclusive ones.

Wrong move:

Trying to calculate turning points or inflexion points for the cubic sketch.

Why:

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 move:

Only draw the general S-shaped cubic passing through the intercepts, with correct end behaviour, no turning point calculation is needed.

Wrong move:

Writing overlapping or incorrectly ordered regions for inequality solutions.

Why:

Cubic inequalities often have two separate solution regions, and mixing up the order of the bounds leads to invalid solutions.

Correct move:

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 move:

Drawing the curve crossing the x-axis at a repeated (squared) root instead of just touching it.

Why:

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 move:

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 move:

Solving $f (x) \geq d$ (with $d \neq = 0$ ) by reading off the roots of $f (x)$ instead of rearranging first.

Why:

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 move:

Rearrange to $g (x) = f (x) - d$ , compare it with 0, factorise $g$ , and use the roots of $g$ to find the sign regions.

8. Quick Reference Cheatsheet

Task	Steps	Key Reminder
Sketch factored cubic	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	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	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 $f$

9. Frequently Asked

Do I need to calculate turning points when sketching cubics for 0606?

No, for CIE IGCSE Additional Mathematics 0606, you only need to label axis intercepts and show the correct overall cubic shape when given a factored form. No turning point calculation is required for this topic.

Can I solve cubic inequalities algebraically instead of graphically?

While algebraic methods exist, the 0606 syllabus expects you to use the graphical method for cubic inequalities, so practice interpreting your sketched graph to find solution regions.

Going deeper

syllabusCIE 0606 2025-2027 SyllabusRefer to sections 4.4 and 4.5 for official topic guidance

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.