IB DP Mathematics: Analysis and Approaches – Polynomials, rational functions, odd/even functions and modulus graphs (HL) Study Guide

This study guide teaches the HL further functions unit of IB Diploma Programme Mathematics: Analysis and Approaches from scratch. It is aligned to the IB Mathematics: analysis and approaches guide, first assessment 2021, and covers syllabus sections 2.12–2.16. All of this content is HL only (AHL). It follows the IB guide for first assessment 2021, which remains the examined syllabus until the new course is first assessed in May 2029, so it applies to the May and November 2026, 2027 and 2028 HL sessions.

When you have worked through it, use the further functions revision notes for final-weeks recall and the further functions practice questions to test yourself. The SL groundwork (domain, range, composite functions, transformations, the reciprocal function) is in the functions study guide. For the whole course, see the IB DP Maths AA course hub and the printable syllabus checklist.

What this unit covers

Syllabus section	What you must be able to do	SL/HL
2.12	Work with polynomial graphs, zeros, roots and factors; use the factor and remainder theorems; use the sum and product of the roots	HL only
2.13	Sketch rational functions (ax + b)/(cx² + dx + e) and (ax² + bx + c)/(dx + e) with all asymptotes (horizontal, vertical, oblique) and intercepts	HL only
2.14	Test for odd and even functions, including periodic ones; find f⁻¹(x) with a domain restriction; recognise self-inverse functions	HL only
2.15	Solve g(x) ≥ f(x) graphically and analytically (by hand for simple polynomials up to degree 3; with technology otherwise)	HL only
2.16	Sketch y = |f(x)|, y = f(|x|), y = 1/f(x), y = f(ax + b) and y = [f(x)]²; solve modulus equations and inequalities	HL only

2.12 Polynomials

A polynomial of degree n is p(x) = aₙxⁿ + … + a₁x + a₀ with aₙ ≠ 0. Its graph has at most n x-intercepts and at most n − 1 turning points.

• A zero of p is where p(x) = 0; a root solves p(x) = 0. They are the same numbers.
• A single factor (x − r) means the graph crosses the x-axis at r; (x − r)² means it touches; (x − r)³ means it crosses with a point of inflection.

Factor and remainder theorems

• Remainder theorem: when p(x) is divided by (x − r), the remainder is p(r). Dividing by (ax − b) gives remainder p(b/a).
• Factor theorem: (x − r) is a factor of p(x) if and only if p(r) = 0.

Worked example 1

p(x) = 2x³ + ax² + bx − 6. (x − 2) is a factor, and the remainder on dividing by (x + 1) is 6. Find a and b and factorise p(x) fully.

p(2) = 0:   16 + 4a + 2b − 6 = 0   →  2a + b = −5
p(−1) = 6:  −2 + a − b − 6 = 6     →  a − b = 14
Add:        3a = 9                 →  a = 3, b = −11

So a = 3 and b = −11.

Divide by (x − 2): 2x³ + 3x² − 11x − 6 = (x − 2)(2x² + 7x + 3) = (x − 2)(2x + 1)(x + 3).

Check: p(−1) = −2 + 3 + 11 − 6 = 6. Correct.

Sum and product of the roots

For the polynomial equation aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀ = 0, counting complex roots and repeated roots:

• sum of the roots = −aₙ₋₁/aₙ
• product of the roots = (−1)ⁿa₀/aₙ

For a cubic ax³ + bx² + cx + d = 0 this gives sum −b/a and product −d/a. For a quartic the product is +a₀/a₄. Complex roots of real polynomials come in conjugate pairs (section 1.14 — see the complex numbers study guide).

Worked example 2

The roots of x³ − 6x² + kx − 6 = 0 form an arithmetic sequence. Find k and the roots.

Let the roots be m − d, m, m + d. Their sum is 3m = −(−6)/1 = 6, so m = 2. Since 2 is a root, 8 − 24 + 2k − 6 = 0, so k = 11. Then x³ − 6x² + 11x − 6 = (x − 2)(x² − 4x + 3) = (x − 1)(x − 2)(x − 3), so the roots are 1, 2 and 3. Check the product: (−1)³(−6)/1 = 6 = 1 × 2 × 3.

2.13 Rational functions

HL adds two shapes to the SL reciprocal and (ax + b)/(cx + d) functions.

Type 1: f(x) = (ax + b)/(cx² + dx + e)

• Vertical asymptotes where the denominator is zero (and the numerator is not).
• Horizontal asymptote y = 0. The graph can cross it at the zero of the numerator.

Type 2: f(x) = (ax² + bx + c)/(dx + e)

• One vertical asymptote at x = −e/d.
• An oblique asymptote found by division: write f(x) = mx + k + R/(dx + e). The line y = mx + k is the asymptote, since R/(dx + e) → 0 as x → ±∞.

Give asymptotes as equations and intercepts as coordinates.

Worked example 3

Sketch y = (2x − 1)/(x² − x − 6).

• Denominator: (x − 3)(x + 2), so vertical asymptotes x = 3 and x = −2.
• Horizontal asymptote y = 0.
• x-intercept: 2x − 1 = 0 gives (1/2, 0). y-intercept: −1/−6 gives (0, 1/6).
• Signs: negative for x < −2, positive on (−2, 1/2), negative on (1/2, 3), positive for x > 3. This places each branch.

Worked example 4

Find the asymptotes and intercepts of y = (x² + x − 2)/(x − 2).

Divide: x² + x − 2 = (x − 2)(x + 3) + 4, so y = x + 3 + 4/(x − 2).

• Vertical asymptote x = 2; oblique asymptote y = x + 3.
• Numerator (x + 2)(x − 1): x-intercepts (−2, 0) and (1, 0). y-intercept (0, 1).

2.14 Odd, even, inverse and self-inverse functions

• Even: f(−x) = f(x) for all x in the domain. The graph is symmetric in the y-axis. Examples: x², x⁴ − 3x², cos x.
• Odd: f(−x) = −f(x). The graph has rotational symmetry of order 2 about the origin. Examples: x³, 1/x, sin x, tan x.
• Most functions are neither: show this with one value where both tests fail.

The guide includes periodic functions: sin 2x is odd, cos 3x is even, and sin x + cos x is neither.

Worked example 5

Show that f(x) = x³ sin x is even and g(x) = x/(x² + 1) is odd.

f(−x) = (−x)³ sin(−x) = (−x³)(−sin x) = x³ sin x = f(x), so f is even.

g(−x) = (−x)/((−x)² + 1) = −x/(x² + 1) = −g(x), so g is odd.

Inverses with a domain restriction

f⁻¹ exists only when f is one-to-one, so restrict the domain to one side of any turning point. Swap x and y, rearrange, and state the domain of f⁻¹ (the range of f).

Worked example 6

f(x) = x² − 4x + 1, x ≥ 2. Find f⁻¹(x) and its domain.

Complete the square: f(x) = (x − 2)² − 3. With x ≥ 2 the range is f(x) ≥ −3.

Let y = (x − 2)² − 3, so x − 2 = √(y + 3) (positive root, since x ≥ 2). Hence f⁻¹(x) = 2 + √(x + 3), x ≥ −3.

Self-inverse functions

f is self-inverse when f(f(x)) = x, so f⁻¹ = f and the graph is symmetric in y = x. Examples: 1/x, c − x and (ax + b)/(cx − a).

Worked example 7

Show that f(x) = (3x + 5)/(x − 3), x ≠ 3, is self-inverse.

f(f(x)) = [3(3x + 5)/(x − 3) + 5] / [(3x + 5)/(x − 3) − 3]
        = [(9x + 15 + 5x − 15)/(x − 3)] / [(3x + 5 − 3x + 9)/(x − 3)]
        = 14x / 14 = x

2.15 Solving g(x) ≥ f(x)

Move everything to one side: g(x) − f(x) ≥ 0. Find the zeros, then use a sketch or a sign table.

Worked example 8 (by hand)

Solve x³ ≥ 2x² + 5x − 6.

x³ − 2x² − 5x + 6 ≥ 0. Test x = 1: 1 − 2 − 5 + 6 = 0, so (x − 1) is a factor. Then x³ − 2x² − 5x + 6 = (x − 1)(x + 2)(x − 3).

The cubic is positive on (−2, 1) and for x > 3, so −2 ≤ x ≤ 1 or x ≥ 3.

Worked example 9 (technology)

Solve eˣ ≥ x + 2.

Graph y = eˣ and y = x + 2 and find the intersections: x = −1.84 and x = 1.15 (3 s.f.). The exponential is above the line outside these values, so x ≤ −1.84 or x ≥ 1.15.

2.16 Modulus and related graphs

Graph	How to get it from y = f(x)
y = |f(x)|	Reflect any part below the x-axis in the x-axis
y = f(|x|)	Keep the part for x ≥ 0, delete x < 0, reflect the kept part in the y-axis (the result is even)
y = 1/f(x)	Zeros of f become vertical asymptotes; vertical asymptotes of f become zeros; a maximum at (p, q) with q ≠ 0 becomes a minimum at (p, 1/q), and vice versa; points with y = ±1 stay put; the sign is unchanged
y = f(ax + b)	A point (p, q) moves to ((p − b)/a, q): translate by b units left, then stretch horizontally by scale factor 1/a
y = [f(x)]²	Never negative; zeros of f become points where the graph touches the x-axis; points with y = ±1 go to y = 1; values with |f| > 1 grow, values with |f| < 1 shrink

The composite f(ax + b) is HL only; SL excludes it.

Worked example 10

f(x) = x² − 2x − 3 = (x + 1)(x − 3), with minimum point (1, −4).

• y = f(2x − 4): points map to ((p + 4)/2, q). The minimum goes to (5/2, −4) and the zeros to x = 3/2 and 7/2.
• y = 1/f(x): vertical asymptotes x = −1 and x = 3; the minimum (1, −4) becomes a local maximum (1, −1/4).
• y = [f(x)]²: touches the axis at x = −1 and x = 3; local maximum (1, 16).

Modulus equations and inequalities

For |f(x)| = g(x), solve f(x) = g(x) and f(x) = −g(x), then reject any solution with g(x) < 0. For |f(x)| < |g(x)|, both sides are non-negative, so you can square both sides.

Worked example 11

Solve |2x − 3| = x + 1.

• 2x − 3 = x + 1 gives x = 4 (x + 1 = 5 > 0, valid).
• 2x − 3 = −(x + 1) gives 3x = 2, x = 2/3 (x + 1 = 5/3 > 0, valid).

So x = 4 or x = 2/3.

Worked example 12

Solve |x − 2| < |2x + 1|.

Square: x² − 4x + 4 < 4x² + 4x + 1, so 0 < 3x² + 8x − 3 = (3x − 1)(x + 3). Hence x < −3 or x > 1/3.

Using your GDC

HL Paper 1 allows no technology; Papers 2 and 3 require it. So you must factorise, find asymptotes, test odd/even, find inverses and solve polynomial inequalities up to degree 3 by hand. The guide expects technology for inequalities involving other functions (as in example 9). On a GDC, graph both sides, use the intersect tool and give endpoints to 3 s.f.

Common errors

• Using p(r) for a divisor (x + r). The remainder on dividing by (x + 1) is p(−1).
• Forgetting the (−1)ⁿ in the product of roots: for a cubic the product is −a₀/a₃, not +a₀/a₃.
• Giving y = 0 as the asymptote of (ax² + bx + c)/(dx + e). It has an oblique asymptote.
• Writing an asymptote as “3” instead of the equation x = 3.
• Testing odd/even with one number and claiming “even”. A single value can only prove “neither”.
• Leaving out the domain of f⁻¹, or choosing the wrong sign of the square root.
• Dividing an inequality by an expression that could be negative, such as (x − 1).
• Keeping modulus solutions that make the right-hand side negative.
• For f(ax + b), stretching first and then translating by b instead of b/a.

Where to go next

Use the revision notes and the practice questions next. These functions reappear in the calculus study guide. See also the syllabus guide, subject guide and exam preparation guide.

Official syllabus

International Baccalaureate Organization, Diploma Programme, Mathematics: analysis and approaches guide, first assessment 2021 (published February 2019, updated November 2020), syllabus sections AHL 2.12–2.16.