Syllabus map

AA SL Functions syllabus

The checklist below names the Standard Level syllabus points for this topic. Higher Level extensions are not included on this page.

  1. 2.1Gradients and equations of straight lines
  2. 2.2Introduction to functions
  3. 2.3Graphing functions
  4. 2.4Key features of graphs
  5. 2.5Composite, identity and inverse functions
  6. 2.6Quadratic functions
  7. 2.7Solving quadratic equations and inequalities; the discriminant
  8. 2.8Reciprocal and rational functions
  9. 2.9Exponential and logarithmic functions
  10. 2.10Graphical and analytical solutions
  11. 2.11Transformations

Study by topic

Functions lessons

Work through focused lessons mapped to the syllabus points. Each lesson includes concise notes, an original worked example, practice and common mistakes.

2.1, 2.2Straight Lines and Function NotationA linear rule describes a constant rate of change. Function notation makes the input-output relationship explicit and helps track domain restrictions.Open lesson 2.3, 2.4, 2.10Function Graphs, Features and IntersectionsA graph displays how a function behaves. Intercepts, turning points, symmetry and asymptotes can help answer questions and check algebraic solutions.Open lesson 2.5Composite, Identity and Inverse FunctionsComposition applies one function after another. An inverse reverses a function’s mapping on a domain where each output comes from only one input.Open lesson 2.6, 2.7Quadratic Functions, Equations and DiscriminantEquivalent quadratic forms reveal roots, intercepts or the vertex. The discriminant predicts the number of real roots before solving.Open lesson 2.8Reciprocal and Rational FunctionsA rational function is a quotient of polynomials. Denominator restrictions shape its domain and may create vertical asymptotes or removable holes.Open lesson 2.9Exponential and Logarithmic FunctionsExponential and logarithmic functions are inverse families. Their domains, ranges and growth patterns explain how they model changing quantities.Open lesson 2.11Function TransformationsTransformations connect changes in an equation to changes in its graph. Read inside and outside changes separately to avoid reversing the horizontal direction.Open lesson

Learn the ideas

Key ideas and methods

Inputs, outputs and domain

  • A function assigns exactly one output to each input in its domain. Use function notation such as $f(3)$ to denote the output when the input is $3$.
  • The domain is the set of allowed inputs; the range is the set of resulting outputs. For rational functions, exclude inputs where the denominator is zero. For real logarithmic functions $\log_a(u)$, require the argument $u > 0$.
  • A straight line with gradient $m$ through $(x_1, y_1)$ can be written in point-gradient form $y - y_1 = m(x - x_1)$.

Compositions, identity and inverse functions

  • A composite function applies one function after another: $(f \circ g)(x) = f(g(x))$. Order is critical; $f(g(x)) \neq g(f(x))$ in general.
  • The identity function maps $x \mapsto x$. An inverse reverses the mapping: $f^{-1}(f(x)) = x$ on the restricted domain. To find an inverse algebraically, set $y = f(x)$, interchange $x$ and $y$, then isolate $y$.
  • A function must be one-to-one (bijective on its restricted domain) to possess an inverse function. Graphically, it must pass the horizontal line test. The graphs of $y = f(x)$ and $y = f^{-1}(x)$ are reflections of each other across $y = x$.

Quadratics and other graph families

  • A quadratic can be analyzed in standard form $y = ax^2 + bx + c$, factorized form $y = a(x - p)(x - q)$, or vertex form $y = a(x - h)^2 + k$.
  • The discriminant $\Delta = b^2 - 4ac$ dictates the nature of roots: $\Delta > 0$ yields two distinct real roots, $\Delta = 0$ yields one repeated root (tangent to $x$-axis), and $\Delta < 0$ yields no real roots.
  • Reciprocal functions $y = \frac{ax + b}{cx + d}$ have vertical asymptote $x = -\frac{d}{c}$ and horizontal asymptote $y = \frac{a}{c}$. Exponential $y = a^x$ and logarithmic $y = \log_a x$ are inverse families.
  • Transformations: $y = f(x - h) + k$ translates by vector $\begin{pmatrix} h \\ k \end{pmatrix}$. Multiplier $a$ outside scales vertically by $|a|$ (and reflects across $x$-axis if $a < 0$); multiplier $b$ inside scales horizontally by $\frac{1}{|b|}$ (and reflects across $y$-axis if $b < 0$).

Solving graphically and analytically

  • Intersections of $y = f(x)$ and $y = g(x)$ solve the equation $f(x) = g(x)$. Use GDC intersect features in Paper 2 and exact algebraic factorisation or quadratic formulas in Paper 1.
  • Always sketch functions with labeled axes, intercepts, turning points, and clearly indicated asymptote dashed lines with equations.

Follow the reasoning

Worked examples

These original examples show common reasoning steps. They are not copied from IB examination papers.

Worked example 1

Read a quadratic in vertex form

Question. For $f(x) = (x - 2)^2 - 3$, state the vertex, axis of symmetry, and exact roots.

  1. The vertex form $a(x - h)^2 + k$ gives vertex $(h, k) = (2, -3)$.
  2. The axis of symmetry is the vertical line $x = 2$.
  3. Set $f(x) = 0 \implies (x - 2)^2 = 3 \implies x - 2 = \pm\sqrt{3} \implies x = 2 \pm \sqrt{3}$.

Answer: Vertex $(2, -3)$, axis $x = 2$, roots $x = 2 \pm \sqrt{3}$.

Worked example 2

Find an inverse function

Question. Find the inverse of $f(x) = 3x - 5$.

  1. Write $y = 3x - 5$.
  2. Interchange $x$ and $y$: $x = 3y - 5$.
  3. Solve for $y$: $3y = x + 5 \implies y = \frac{x + 5}{3}$.

Answer: $f^{-1}(x) = \frac{x + 5}{3}$.

Worked example 3

Solve a quadratic inequality

Question. Solve $x^2 - 4 \le 0$.

  1. Factor: $(x - 2)(x + 2) \le 0$. Critical values are $x = -2$ and $x = 2$.
  2. The parabola opens upward ($a = 1 > 0$), so it is on or below the $x$-axis between and including the roots.

Answer: $-2 \le x \le 2$.

Try it yourself

Practice with worked answers

  1. Question 1. Find the gradient of the line through $(-1, 4)$ and $(3, 12)$, then write its equation in slope-intercept form.

    Show worked answer

    $m = \frac{12 - 4}{3 - (-1)} = \frac{8}{4} = 2$. Using $(-1, 4)$: $y - 4 = 2(x + 1) \implies y = 2x + 6$.

  2. Question 2. For $g(x) = -(x + 1)^2 + 4$, give the coordinates of the vertex and state the maximum value.

    Show worked answer

    Vertex form gives $(-1, 4)$. Because $a = -1 < 0$, the parabola opens downward, giving a global maximum value of $4$.

  3. Question 3. Solve $x^2 - 5x + 6 = 0$ and state the discriminant.

    Show worked answer

    Factoring: $(x - 2)(x - 3) = 0 \implies x = 2$ or $x = 3$. The discriminant is $\Delta = (-5)^2 - 4(1)(6) = 25 - 24 = 1 > 0$, confirming two distinct real roots.

Avoid these slips

Common mistakes

  • Confusing $f^{-1}(x)$, the inverse function, with $\frac{1}{f(x)}$, the reciprocal function.
  • Reversing the horizontal translation direction: $f(x - h)$ shifts right by $h$, not left.
  • Neglecting domain restrictions when stating roots or inverse ranges.
  • Using a graphical rounded estimate when an exact analytical radical or fraction is required on Paper 1.

Before you move on

Revision checklist

  • Can I move between a function rule, a table and a labelled graph?
  • Can I identify domain, range, intercepts, turning points and asymptotes where relevant?
  • Can I compose functions and find an inverse on a stated domain?
  • Can I explain whether an equation has zero, one or two real roots using the discriminant?

About this guide

Independent SJMaths resource; not an official IB publication. Follows the AA syllabus first assessed in 2021 (final assessment: November 2028). Use the separate 2029 guide for the new course.

Check the IB Mathematics programme page and the official AA subject guide for the authoritative syllabus and examination-session requirements.