IB Maths AA HL Topic 2 — Functions Paper 1 & 2 HL only ~7 min read

Modulus Functions

The modulus (or absolute value) function strips the sign off any input — so |3| = 3 and |−2| = 2. Its graph is a clean V-shape with vertex at the origin. Once you know how the basic y = |x| transforms via the standard form y = a|x + p| + q, you can sketch any modulus graph by reading off the vertex and the steepness directly from the equation.

📘 What you need to know

Definition: |x| = x if x ≥ 0; |x| = −x if x < 0. Equivalently, |x| = √(x²).
Domain and range: domain is all real x; range is y ≥ 0.
Basic graph: y = |x| is a V-shape with vertex at (0, 0). Continuous everywhere; not differentiable at x = 0.
Standard form: y = a|x + p| + q has vertex at (−p, q).
Shape: a > 0 → V-shape (opens upward); a < 0 → ∧-shape (opens downward). Larger |a| means steeper sides.
Two key identities for rearranging: |ab| = |a||b| and |a − b| = |b − a|.
Watch the sign of p: vertex x-coordinate is −p, not p. Sign-flip is the most common slip.

The modulus function and its graph

Definition |x| = { x if x ≥ 0; −x if x < 0 }

The modulus gives the unsigned distance of x from zero. Geometrically, the graph of y = |x| is the line y = x reflected upward whenever it would otherwise dip below the x-axis.

The basic graph y = |x|

Why the V? When x < 0, the modulus flips the sign — so the graph below the x-axis is reflected upward. The two half-lines meet sharply at the origin, where the gradient changes from −1 to +1 instantly.

The standard form: y = a|x + p| + q

Vertex from standard form y = a|x + p| + q → vertex at (−p, q)

Reading off the vertex is just inspection — flip the sign of the number inside the modulus, and pair it with the constant added at the end.

a > 0

V-shape (opens up)

vertex is the minimum; range is y ≥ q

a < 0

∧-shape (opens down)

vertex is the maximum; range is y ≤ q

y = a|x + p| + q — vertex at (−p, q)

Rearranging into standard form

Equations like y = |10 − 5x| + 2 don’t look like the standard form, but two short identities convert them in seconds:

Product identity

|ab| = |a| · |b|

factor out coefficients of x

Symmetry identity

|a − b| = |b − a|

flip the order to get x first

🧭 Recipe — convert to standard form

Make sure x appears first inside the modulus, using |a − b| = |b − a| if needed.
Factor out the coefficient of x inside the modulus.
Apply |ab| = |a||b| to bring the coefficient outside as a positive number.
Read off vertex (−p, q) and shape from the standard form.

Worked examples

WE 1

Identify vertex, shape, and intercepts from standard form

State the vertex, shape, y-intercept, and x-intercepts of y = |x + 5| − 3.

Step 1: Match standard form y = a|x + p| + q a = 1, p = 5, q = −3 Step 2: Vertex is (−p, q) vertex: (−5, −3) Step 3: Shape — a > 0 → V-shape Step 4: y-intercept (sub x = 0) y = |0 + 5| − 3 = 5 − 3 = 2 → (0, 2) Step 5: x-intercepts (set y = 0) |x + 5| = 3 → x + 5 = ±3 → x = −2 or x = −8 vertex (−5, −3); V-shape; y-int (0, 2); x-ints (−8, 0) and (−2, 0) flip the sign of p to get the vertex’s x-coordinate — the most common slip is keeping +5

WE 2

Modulus graph with a negative leading coefficient

State the vertex, shape, y-intercept, and x-intercepts of y = −2|x − 1| + 4.

Step 1: Match standard form a = −2, p = −1, q = 4 Step 2: Vertex is (−p, q) vertex: (1, 4) Step 3: Shape — a < 0 → ∧-shape (steeper than basic since |a| = 2) Step 4: y-intercept y = −2|0 − 1| + 4 = −2 + 4 = 2 → (0, 2) Step 5: x-intercepts (set y = 0) −2|x − 1| + 4 = 0 → |x − 1| = 2 → x = −1 or x = 3 vertex (1, 4); ∧-shape; y-int (0, 2); x-ints (−1, 0) and (3, 0) ∧-shape with a vertex above the x-axis always has two x-intercepts — symmetric about the vertex’s x-coordinate

WE 3

Rearrange into standard form

Express y = |10 − 5x| + 2 in the form y = a|x + p| + q, and state the vertex.

Step 1: Get x first inside the modulus |10 − 5x| = |5x − 10| (using |a − b| = |b − a|) Step 2: Factor out the coefficient of x |5x − 10| = |5(x − 2)| Step 3: Apply |ab| = |a||b| |5(x − 2)| = 5|x − 2| Step 4: Combine with the +2 outside y = 5|x − 2| + 2 y = 5|x − 2| + 2; vertex (2, 2) a = 5 always comes out positive even if the original coefficient was negative inside — that’s what |a||b| guarantees

WE 4

Find the equation of a V-shaped graph

The graph of y = a|x + p| + q is V-shaped, with vertex at (−3, 1) and passes through (0, 4). Find the equation.

Step 1: Vertex (−p, q) = (−3, 1) → p = 3, q = 1 y = a|x + 3| + 1 Step 2: Substitute (0, 4) to find a 4 = a|0 + 3| + 1 4 = 3a + 1 → 3a = 3 → a = 1 Step 3: V-shape requires a > 0 — confirm a = 1 ✓ y = |x + 3| + 1 a = 1 is the basic V — both branches have gradients ±1

WE 5

Find the equation of a ∧-shaped graph

The graph of y = a|x + p| + q is ∧-shaped, with vertex at (2, 5) and passes through (0, 1). Find the equation.

Step 1: Vertex (−p, q) = (2, 5) → p = −2, q = 5 y = a|x − 2| + 5 Step 2: Substitute (0, 1) to find a 1 = a|0 − 2| + 5 1 = 2a + 5 → 2a = −4 → a = −2 Step 3: ∧-shape requires a < 0 — confirm a = −2 ✓ y = −2|x − 2| + 5 always check the sign of a against the shape — if it doesn’t match, you’ve slipped a sign somewhere

WE 6

Domain and range of a transformed modulus

State the domain and range of f(x) = −3|x + 4| + 7.

Step 1: Domain — modulus accepts any real input domain: x ∈ ℝ Step 2: Identify shape and vertex a = −3 → ∧-shape (max at vertex) vertex (−p, q) = (−4, 7) → maximum value y = 7 Step 3: Range follows from the vertex ∧-shape opens downward with maximum 7 → y ≤ 7 domain: ℝ; range: y ≤ 7 V-shape (a > 0) gives range y ≥ q; ∧-shape (a < 0) gives range y ≤ q

💡 Top tips

Vertex’s x-coordinate is −p, not p. Always flip the sign of whatever’s inside the modulus.
Sign of a determines shape: positive V, negative ∧. Confirm against the picture before finalising.
Modulus identities are exam shortcuts: |ab| = |a||b| and |a − b| = |b − a| handle most rearrangements.
The coefficient that comes out is always positive: |−3| = 3, never −3.
Check the y-intercept by substituting x = 0 — useful for verifying the equation when reverse-engineering.
Two x-intercepts if the vertex is on the opposite side of the x-axis to the opening; none if the vertex is on the same side; one if the vertex is on the axis.
Draw the corner sharp — the V (or ∧) point isn’t a smooth turning point; it’s a cusp where the gradient jumps.

⚠ Common mistakes

Reading p as the vertex’s x-coordinate instead of −p. Marker’s favourite slip.
Forgetting to apply the symmetry identity when the coefficient of x is negative inside the modulus.
Bringing a coefficient out as negative — |−4| = 4 always; the coefficient outside the modulus must be positive in the standard form.
Drawing a smooth curve at the vertex instead of a sharp cusp. Makes the answer look like a parabola or worse.
Confusing range with domain. Domain is always all real numbers for modulus functions; range depends on shape and vertex.
Forgetting that modulus is non-differentiable at the vertex. The gradient changes abruptly — important for calculus questions later.
Substituting x = 0 carelessly when finding the y-intercept, especially with negative p: check that |0 − 1| = 1, not 0.

Now you can sketch y = |x| and any of its standard transformations confidently. The next note, Modulus Transformations, takes a step further: applying modulus to any function — y = |f(x)| reflects the negative parts upward, and y = f(|x|) makes the graph symmetric about the y-axis. Two simple transformation rules with rich exam applications.

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