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Equations, Inequalities & Graphs

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Summary — What This Chapter Covers

Modulus function — makes any input positive. |7| = 7 and |-7| = 7.
Sketching modulus graphs — draw the original graph, then reflect anything below the x-axis upward.
Solving modulus equations — sketch both sides, find intersections, work out which "piece" (normal or reflected) applies at each one.
Solving modulus inequalities — same sketching approach, but shade the region satisfying the inequality instead of just finding points.
Cubic polynomials — degree-3 functions, always in factorised form for sketching, four possible shapes depending on the sign of a and the roots.
Modulus cubic graphs — apply the same reflection rule to cubics; the graph will only touch (never cross) the x-axis.
Finding a cubic from its modulus graph — read off roots and the y-intercept, then solve for the leading coefficient.
Solving cubic inequalities graphically — rearrange to zero, factorise, sketch, then read off the x-values where the curve sits above/below the axis.

1. Modulus Functions

1.1 What is a modulus function?

Think of the modulus (also called absolute value) as a machine that strips away the sign of whatever goes in and just tells you the size — the distance from zero. It doesn't care whether the number was positive or negative to begin with; it only cares "how far from zero is this?"

It's written with a pair of vertical bars around the input — similar to how brackets group things together:

Definition

|7| = 7 and |−7| = 7

Whatever goes in, whatever comes out is positive (or zero). The modulus of a positive number is itself; the modulus of a negative number is its positive twin.

Two forms you'll meet in this course:

Linear form

|ax + b|
e.g. |3x − 3|

Quadratic form

|ax² + bx + c|
e.g. |x² + x − 6|

1.2 Sketching y = |f(x)|

Here's the trick that makes this whole topic click: you never need to sketch a modulus graph from scratch. You sketch the ordinary graph first (the "before" picture), and then apply one simple rule.

The Two-Step Method

1 Pencil in the graph of y = f(x) as normal.

2 Reflect anything below the x-axis upward, in the x-axis, to get y = |f(x)|.

y y │ / │ / │ / y=f(x) │ / y=f(x) │ / │ / ───┼───●─────── x ──► ───┼───●─────── x │ / │ ╲ (reflected up) │ / (this dips │ ╲ y=|f(x)| │/ below axis) │ ╲

Two things to notice

1. If the original y-intercept was negative, it becomes positive on the modulus graph.
2. At every x-axis crossing, the modulus graph gets a sharp "V" shape (not a smooth curve like a turning point) — because reflecting creates a corner, not a curve.

Watch out

If the graph is y = −|f(x)| instead, the whole thing is now wholly negative — every crossing point becomes a sharp "Λ" (upside-down V) instead. Don't mix these two up!

Practice Question

Sketch y = |x − 4|. State the coordinates of the vertex and the y-intercept.

1.3 Solving modulus equations

Why bother sketching to solve an equation? Because with a modulus involved, there can be more than one solution — and graphing is the safest way to see exactly how many, and where.

Compare: two ordinary straight lines like y = x + 1 and y = 2x cross once. But y = |2x| and y = |x + 1| can cross twice, because the modulus creates extra "folded" pieces of the graph that can intersect the other line in more places.

Method: Solving |f(x)| = |g(x)| type equations

1 Sketch both graphs, including any reflected (modulus) parts.

2 Locate the graph intersections.

3 At each intersection, figure out which piece of each graph is involved — the "normal" piece or the "reflected" piece — then solve the corresponding equation.

Worked Example — Solve |2x − 4| = |x − 1|

Sketching both graphs shows two intersections, A and B.

At A: the "normal" part of x − 1 meets the "reflected" part of 2x − 4:

x − 1 = −(2x − 4)

3x = 5 → x = 5/3

At B: the "normal" parts of both graphs meet each other:

x − 1 = 2x − 4 → x = 3

So the two solutions are x = 5/3 and x = 3.

A classic trap

Solving algebraically without a sketch can hand you invalid solutions. For example, x = 1 solves x − 4 = 2x − 5, but if you sub x = 1 into |x − 4| = 2x − 5, it doesn't actually work (try it: left side is 3, right side is −3). The sketch tells you which pieces genuinely intersect — trust it over blind algebra.

Practice Question

Solve the equation |½x + 1| = |5 − x|.

1.4 Solving modulus inequalities

Same sketching process as equations — the only difference is your final step. Instead of just stating the x-values at the intersections, you shade the part(s) of the graph that satisfy the inequality, then read off the x-range underneath that shading.

Method

1 Sketch the graph(s), including reflected parts.

2 Locate the intersections (or x-intercepts if one side is zero).

3 Determine which part(s) of the graph satisfy the inequality — highlight it. Find the intersection x-values.

4 Write the final answer(s), being careful with <, >, ≤, ≥.

Worked Example — Solve |x² − 9x + 14| ≤ 6

Sketching y = |x² − 9x + 14| against the line y = 6 gives four intersection points A, B, C, D.

Points A and D (outer ones) use the "normal" part of the quadratic:

x² − 9x + 14 = 6 → x² − 9x + 8 = 0 → (x − 8)(x − 1) = 0

x = 1, x = 8

Points B and C (inner ones) use the "reflected" part of the quadratic:

−(x² − 9x + 14) = 6 → x² − 9x + 20 = 0 → (x − 4)(x − 5) = 0

x = 4, x = 5

The region below the line y = 6 gives the final answer:

1 ≤ x ≤ 4, 5 ≤ x ≤ 8

Practice Question

Using your answer to the earlier equation |½x + 1| = |5 − x| (x = 8/3 and x = 12), solve the inequality |½x + 1| ≥ |5 − x|.

2. Graphs of Cubic Polynomials

2.1 What is a cubic polynomial?

A cubic polynomial is a function of the form ax³ + bx² + cx + d, where a, b, c, d are constants — it's "degree 3" because the highest power of x is 3. Note that b, c, and/or d could be zero, but a can never be zero (otherwise it wouldn't be a cubic anymore).

To actually sketch a cubic, you need it in factorised form — that's the form that reveals the roots (x-intercepts) directly.

Example

2x³ − 3x² − 11x + 6 = (2x − 1)(x + 2)(x − 3)

The right-hand side is the factorised form — from it you can immediately read off the roots: x = ½, x = −2, x = 3.

2.2 What does a cubic graph look like?

There are essentially two "families" of cubic shape, and each family has two variations depending on how many turning points there are:

Positive cubic (a > 0)

"Starts" in the third quadrant (bottom-left)
"Ends" in the first quadrant (top-right)
Overall direction: bottom-left → top-right

Negative cubic (a < 0)

"Starts" in the second quadrant (top-left)
"Ends" in the fourth quadrant (bottom-right)
Overall direction: top-left → bottom-right

Positive cubic (a > 0) Negative cubic (a < 0) │ ╱ ╲ │ │ ╱ ╲ │ ─────┼────╱──── ─────╲────┼───── ╱ │ │ ╲ ╱ │ │ ╲

All cubic curves are smooth — no sharp corners, unlike modulus graphs. They can have either 0 or 2 turning points (a smooth "wiggle" in the middle), and every real root shows up as a place where the curve crosses (or touches) the x-axis.

2.3 How to sketch a cubic polynomial

5-Step Method

1 Find the y-axis intercept by setting x = 0.

2 Find the x-axis intercepts (roots) by setting y = 0. A repeated root means the curve touches the x-axis there instead of crossing it.

3 Decide the shape — positive or negative cubic? Where does it "start" and "end"?

4 Work out roughly where the turning point(s) should sit.

5 Sketch a smooth curve, labelling every axis intercept.

Worked Example — Sketch y = (2x − 1)(x − 3)²

Step 1 — y-intercept: y = (−1)(−3)² = −9, so it passes through (0, −9).

Step 2 — x-intercepts: 2x − 1 = 0 → x = ½; x − 3 = 0 → x = 3 (repeated).

Step 3 — shape: a = 2 > 0, so it's a positive cubic. Since x = 3 is a repeated root, the graph touches the x-axis there (doesn't cross).

Step 4 — turning points: One minimum sits right where the curve touches at x = 3; a maximum sits somewhere between x = ½ and x = 3.

Step 5: the resulting curve rises from bottom-left, passes through (0, −9), crosses at (0.5, 0), rises to a local max, dips back down to touch (3, 0), then shoots up to the top-right.

Exam tip

In the exam, the cubic will always be given to you already factorised (as a product of three linear factors, possibly with a repeated one). Build your sketch up gradually as you complete each step — don't try to picture the whole thing at once. If it looks messy, redraw it cleanly for your final answer.

Practice Question

Sketch y = −(x + 2)(x − 1)(x − 4). State the y-intercept and describe the overall shape.

2.4 Modulus cubic graphs

Same reflection idea as before, just applied to a cubic. Because any real cubic f(x) = a(x−b)(x−c)(x−d) must cross the x-axis at least once, y has to take both positive and negative values at some point. So y = |f(x)| will always have something to reflect.

Example of the reflection rule

f(x) = x³ → f(−1) = −1, but |f(−1)| = |−1| = 1

Because a modulus cubic graph can never dip below the x-axis:

the graph will not cross the x-axis
the graph will touch the x-axis (at least once)

How to sketch y = |f(x)| for a cubic

1 Sketch the original cubic y = f(x) first — don't try to jump straight to the modulus version.

2 Reflect any part below the x-axis upward.

Smooth vs. pointy — how to tell them apart

Where a modulus graph touches the x-axis, it's sharp/pointy (V-shaped) — that's a reflection corner. Where the curve has an ordinary turning point away from the axis, it stays smooth (U-shaped) — that part wasn't reflected.

2.5 Finding a cubic function from its modulus graph

This is the reverse process — you're given the picture of y = |f(x)| and need to work backwards to find f(x).

Method

1 Decide whether the original (un-reflected) graph would be positive or negative — sketch a rough "before" version without worrying about labels yet.

2 Read off the roots from where the modulus graph touches the x-axis; write the factors (x − b)(x − c)(x − d).

3 Use the y-intercept of the original (not the modulus) graph to solve for the leading coefficient a, since a(x−b)(x−c)(x−d) at x = 0 gives a × (−b)(−c)(−d) = −abcd.

Worked Example

Given a graph of y = |f(x)| where f(x) is a positive cubic, touching the x-axis at x = −4, x = −½, x = 3, with the modulus graph's y-intercept at 12.

Step 1: Since f(x) is positive cubic, its original graph starts bottom-left and ends top-right. Sketching this rough shape shows the y-intercept of the original graph would actually be negative (−12), even though the modulus graph shows +12.

Step 2 — factors from the roots:

(x + 4)(x + ½)(x − 3)

Step 3 — solve for a using the original y-intercept of −12:

a × 4 × ½ × (−3) = −12

−6a = −12 → a = 2

To get integer coefficients, multiply the 2 into the (x + ½) factor:

f(x) = (x + 4)(2x + 1)(x − 3)

Don't get caught out

The y-intercept you're given on the modulus graph is always positive (or zero) — but the y-intercept of the underlying f(x) might actually be negative! Always sketch the rough "before" picture to check whether you need to flip the sign before solving for a.

Practice Question

A modulus graph y = |f(x)|, where f(x) is a negative cubic, touches the x-axis at x = −1, x = 2, and x = 5, and has a y-intercept of 10. Find f(x).

2.6 Solving cubic inequalities graphically

A cubic inequality is any of f(x) > 0, f(x) ≥ 0, f(x) < 0, or f(x) ≤ 0. If it's not already in one of these forms, rearrange first so one side is zero.

Method

1 Rearrange so one side is zero. Factorise the cubic if it isn't already.

2 Sketch the graph — the x-intercepts (roots) are what actually matter here; the y-intercept isn't needed for this type of question.

3 Highlight the part(s) of the curve that satisfy the inequality (above the axis for "> 0", below for "< 0", etc.).

4 Write down the x-values under the highlighted region(s) — decide whether to include (≤, ≥) or exclude (<, >) the roots themselves.

Worked Example — Solve (3x − 1)(x − 3)² > 0

This is a positive cubic with roots x = 1/3 and a repeated root at x = 3 (so the curve touches, not crosses, at x = 3).

Sketching shows the curve is below the x-axis only for x < 1/3, then rises above it for x > 1/3 — except it briefly touches zero again at x = 3 without actually crossing back below.

We want where the curve is strictly above zero, so we must exclude both roots:

x > 1/3, x ≠ 3

Repeated roots need extra care

At a repeated root, the curve touches the axis but doesn't cross — so the sign of f(x) stays the same on both sides of that root. Always double-check whether a repeated root should be included or excluded in your final answer, since it behaves differently from an ordinary crossing root.

Practice Question

Solve the inequality (x + 4)(2x + 1)(x − 3) ≤ 0.

What to Memorise

Modulus basics

|x| is always ≥ 0
Sketch original graph, then reflect below-axis parts up
y-intercepts flip sign if originally negative
x-axis crossings become sharp "V" corners

Solving modulus equations/inequalities

Sketch both sides → find intersections
Identify "normal" vs "reflected" piece at each point
Never trust pure algebra without a sketch check
Inequalities: shade region, then read off x-range

Cubic shape rules

a > 0: bottom-left → top-right
a < 0: top-left → bottom-right
Repeated root → curve touches, doesn't cross
Always factorise before sketching

Modulus cubics

Never dips below the x-axis
Touches axis at every original root
y-intercept of |f(x)| may differ in sign from f(x)
Use a × (−b)(−c)(−d) = −abcd to find a

Concepts Checklist

I can define the modulus function and evaluate |x| for any value. I can sketch y = |f(x)| by reflecting the below-axis part of y = f(x). I know that modulus graphs form sharp "V" corners at x-axis crossings, not smooth turning points. I can solve modulus equations by sketching both sides and identifying which piece applies at each intersection. I understand why algebra alone can produce invalid solutions to modulus equations. I can solve modulus inequalities by shading the correct region on a sketch. I can identify a cubic polynomial and know it must be factorised before sketching. I can determine whether a cubic is "positive" or "negative" from its leading coefficient. I can sketch a cubic using the 5-step method (y-intercept, roots, shape, turning points, smooth curve). I understand how repeated roots affect whether a cubic crosses or touches the x-axis. I can sketch y = |f(x)| for a cubic and explain why it never goes below the x-axis. I can work backwards from a modulus cubic graph to find f(x) in factorised form. I can solve cubic inequalities graphically, including deciding whether to include or exclude roots.

Exam Tips & Common Mistakes

Mistake #1 — Solving modulus equations purely algebraically

Squaring both sides or blindly manipulating a modulus equation can create solutions that don't actually satisfy the original equation. Always sketch first, and check any algebraic solution against the graph (or by direct substitution).

Mistake #2 — Forgetting to flip the y-intercept sign

When reading a y-intercept off a modulus graph, remember it could have started out negative on the original (non-modulus) graph. This trips people up constantly when working backwards to find f(x).

Mistake #3 — Treating a touch-point like a cross-point

At a repeated root, the curve doesn't change sign — it touches and bounces back. If you treat it like an ordinary root when writing an inequality's solution, you'll get the wrong range. Always sketch to check.

Mistake #4 — Mixing up positive and negative cubic shapes

A quick way to remember: positive cubic behaves like y = x³ (bottom-left to top-right); negative cubic behaves like y = −x³ (top-left to bottom-right). If you're ever unsure, plug in a very large positive x — the sign of a tells you where the curve ends up.

What examiners actually look for

1. A clearly labelled sketch with axis intercepts marked — even a "rough" sketch earns method marks if the shape and key points are correct.
2. Correct identification of which piece (normal/reflected) is used to form each equation when solving modulus problems.
3. Careful, consistent use of inequality symbols (< vs ≤) matching whether roots should be included or excluded.
4. Fully factorised final answers with integer coefficients when asked for a cubic expression.

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Exam Tips & Common Mistakes

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