IB DP Mathematics: Analysis and Approaches – Straight lines, functions, inverses and quadratics Revision Notes

For full explanations and longer worked examples, read the study guide first. These notes are for recall in the final weeks.

They cover IB Diploma Programme Mathematics: Analysis and Approaches, aligned to the IB Mathematics: analysis and approaches guide, first assessment 2021, syllabus sections 2.1–2.7. All seven sections are common content for SL and HL. The notes follow the IB guide for first assessment 2021, which remains the examined syllabus until the new course is first assessed in May 2029, so they apply to the May and November 2026, 2027 and 2028 sessions.

Test yourself with the practice questions. For the whole strand in one page, see the Functions strand overview. The IB DP Maths AA course hub and the printable syllabus checklist let you tick off each section.

Definitions

• Gradient: m = (y₂ − y₁)/(x₂ − x₁). In context: vertical rise ÷ horizontal distance.
• Function: a rule that gives exactly one output for each input in its domain.
• Domain: the allowed inputs. By default, the largest set of real numbers for which the rule works.
• Range: the set of outputs produced from the domain.
• One-to-one: each output comes from exactly one input. Only one-to-one functions have inverses.
• Inverse f⁻¹: reverses f. Its graph is the reflection of y = f(x) in y = x. Domain of f⁻¹ = range of f.
• Composite: (f ∘ g)(x) = f(g(x)). Apply g first.
• Identity function: x ↦ x. (f ∘ f⁻¹)(x) = (f⁻¹ ∘ f)(x) = x.
• Parabola: the graph of a quadratic function.
• Roots / zeros: solutions of f(x) = 0; the x-intercepts of y = f(x).

Formulas and forms

Item	Result
Gradient-intercept form	y = mx + c
General form	ax + by + d = 0 (gradient −a/b)
Point-gradient form	y − y₁ = m(x − x₁)
Parallel lines	m₁ = m₂
Perpendicular lines	m₁ × m₂ = −1
Quadratic, expanded	f(x) = ax² + bx + c, y-intercept (0, c), axis x = −b/(2a)
Quadratic, factorised	f(x) = a(x − p)(x − q), x-intercepts (p, 0), (q, 0)
Quadratic, vertex form	f(x) = a(x − h)² + k, vertex (h, k), axis x = h
Quadratic formula	x = (−b ± √(b² − 4ac))/(2a)
Discriminant	Δ = b² − 4ac

Method in steps

Equation of a line through a point, parallel or perpendicular to a given line

1. Rearrange the given line to y = … to read its gradient.
2. Parallel: same gradient. Perpendicular: negative reciprocal.
3. Substitute into y − y₁ = m(x − x₁).
4. Rearrange to the form the question asks for (for example ax + by + d = 0 with integers).

Finding f⁻¹(x)

1. Check f is one-to-one on its domain (sketch it).
2. Write y = f(x) and make x the subject.
3. Choose the correct sign if there is a square root (use the domain of f).
4. Swap x and y. State the domain of f⁻¹ as the range of f.

Completing the square for ax² + bx + c

1. Take out a from the x² and x terms.
2. Halve the coefficient of x inside the bracket: x² + Bx = (x + B/2)² − (B/2)².
3. Multiply back by a and collect the constant.

Solving a quadratic inequality

1. Rearrange so one side is 0.
2. Find the critical values (roots).
3. Sketch the parabola. Read off where it is above (> 0) or below (< 0) the x-axis.
4. Write the answer as inequalities. Use ≤ or ≥ if the roots are included.

Discriminant problems with a parameter k

1. Identify a, b, c in terms of k.
2. Form Δ and simplify it.
3. Set Δ > 0, = 0 or < 0 and solve for k.
4. If a contains k, exclude the value that makes a = 0.

Changing between the three quadratic forms

From	To	How
a(x − p)(x − q)	ax² + bx + c	Expand the brackets
a(x − h)² + k	ax² + bx + c	Expand (x − h)², multiply by a, add k
ax² + bx + c	a(x − p)(x − q)	Factorise, or find the roots with the formula
ax² + bx + c	a(x − h)² + k	Complete the square
roots p, q and one point	a(x − p)(x − q)	Substitute the point to find a
vertex (h, k) and one point	a(x − h)² + k	Substitute the point to find a

Finding a domain and range

1. Domain: look for what is not allowed. A square root needs the expression inside to be ≥ 0. A fraction needs the denominator ≠ 0.
2. Range: sketch the graph over the domain. Find the lowest and highest y-values reached, using the vertex or an end point.
3. Write the range with f(x) or y, not x.

Key features from a GDC (Paper 2)

1. Graph the function in a window that shows all turning points and intercepts.
2. Use the zero, maximum, minimum and intersect tools. Read the y-intercept from f(0).
3. For asymptotes, trace towards the gap or the end of the curve and write the line as an equation (x = a or y = b).
4. Record values to 3 significant figures and give points as coordinates.

Small worked reminders

• Line through (2, −1) perpendicular to y = 4x + 3: m = −1/4, so y + 1 = −(1/4)(x − 2), giving x + 4y + 2 = 0.
• f(x) = 7 − x²: domain x ∈ ℝ, range f(x) ≤ 7.
• f(x) = 2x + 3, g(x) = x − 4: (g ∘ f)(x) = 2x − 1, but (f ∘ g)(x) = 2x − 5.
• x² − 10x + 18 = (x − 5)² − 25 + 18 = (x − 5)² − 7, vertex (5, −7).
• 2(x + 1)(x − 5) has roots −1 and 5, so the axis is x = 2.

Using technology (Paper 2)

Paper 1 allows no technology; Paper 2 requires it. With a GDC, find zeros, maximum and minimum points, intersections of two graphs, and vertical and horizontal asymptotes. Give values to 3 significant figures and points as coordinates. Everything in the method boxes above must also work by hand for Paper 1.

Must-know distinctions

• Draw vs sketch. Draw: accurate, to scale, plotted points. Sketch: correct shape with key features labelled.
• f⁻¹(x) vs 1/f(x). f⁻¹ is the inverse function, not the reciprocal.
• (f ∘ g)(x) vs (g ∘ f)(x). Usually different; the inner function is applied first.
• Domain vs range. Domain restricts inputs; range describes outputs. They swap for f and f⁻¹.
• Δ = 0 vs Δ ≥ 0. Δ = 0 gives two equal roots; Δ ≥ 0 means “real roots” (equal or distinct).
• x-intercepts vs vertex. Factorised form gives intercepts; vertex form gives the turning point.

Quick self-test

1. Find the gradient of the line through (3, −1) and (7, 11).
2. State the gradient and y-intercept of 4x − 2y + 7 = 0.
3. A line has gradient −4/5. State the gradient of any line perpendicular to it.
4. State the range of f(x) = 7 − x², x ∈ ℝ.
5. f(x) = 2x + 3 and g(x) = x − 4. Find (f ∘ g)(2).
6. Find f⁻¹(x) for f(x) = (x − 6)/4.
7. f(x) = 5x − 3. Find f⁻¹(12) without finding f⁻¹(x).
8. Write down the vertex of y = −3(x − 2)² + 7 and say whether it is a maximum or a minimum.
9. Write down the x-intercepts and the axis of symmetry of y = 2(x + 1)(x − 5).
10. Use the discriminant to state the nature of the roots of 2x² − 3x + 5 = 0.
11. Solve x² + 2x − 15 ≤ 0.
12. Write x² − 10x + 18 in the form (x − h)² + k.

Answers

1. m = 12/4 = 3
2. y = 2x + 7/2: gradient 2, y-intercept (0, 7/2)
3. 5/4
4. f(x) ≤ 7
5. g(2) = −2, f(−2) = −1
6. y = (x − 6)/4 → x = 4y + 6, so f⁻¹(x) = 4x + 6
7. Solve 5x − 3 = 12: f⁻¹(12) = 3
8. (2, 7), maximum (a = −3 < 0)
9. (−1, 0) and (5, 0); x = 2
10. Δ = 9 − 40 = −31 < 0: no real roots
11. (x + 5)(x − 3) ≤ 0: −5 ≤ x ≤ 3
12. (x − 5)² − 7

Where marks are usually lost

• Giving a perpendicular line the same gradient, or using the reciprocal without changing the sign.
• Reading the gradient of 4x − 2y + 7 = 0 as 4 or −2 instead of rearranging to get 2.
• Leaving a line in a form other than the one the question asks for (for example not giving integer a, b, d).
• Stating a range with the wrong inequality sign, or as an interval that includes values the function never reaches.
• Composing in the wrong order: (f ∘ g)(x) means f of g.
• Finding f⁻¹(x) with “±√” left in, when the domain of f fixes the sign.
• Not stating the domain of f⁻¹ when asked, or stating the domain of f instead.
• Writing the vertex of a(x + h)² + k as (h, k) instead of (−h, k).
• Giving the region between the roots of a quadratic inequality when the sketch shows it is outside them (or the reverse).
• Forgetting that k making the x² coefficient zero turns the quadratic into a linear equation.

Official syllabus

International Baccalaureate Organization, Diploma Programme, Mathematics: analysis and approaches guide, first assessment 2021 (published February 2019, updated November 2020), syllabus sections SL 2.1–2.7.