Key Anatomical Features of a Function Graph
- Axial Intercepts:
- $y$-intercept: The point where the graph crosses the $y$-axis. Evaluate $f(0) \implies (0, f(0))$. A function has at most one $y$-intercept.
- $x$-intercepts (roots / zeros): The points where the graph crosses or touches the $x$-axis. Solve $f(x) = 0 \implies (x_i, 0)$.
- Turning Points (Extrema):
- Local Maximum: A point where the curve changes from increasing to decreasing. The function value is greater than or equal to all surrounding points in an open interval.
- Local Minimum: A point where the curve changes from decreasing to increasing. The function value is less than or equal to all surrounding points in an open interval.
- End-Behavior: How $f(x)$ behaves as $x \to \infty$ and $x \to -\infty$. For polynomials, this is determined completely by the degree and sign of the leading coefficient.
(a) Find the $y$-intercept.
(b) Factor $f(x)$ completely to determine all $x$-intercepts.
(c) Given that the local extrema occur at $x = -1$ and $x = 3$, find their coordinates and classify each as a maximum or minimum.
Part (a): $y$-intercept
$$f(0) = 0^3 - 3(0)^2 - 9(0) + 27 = 27 \implies (0, 27)$$Part (b): $x$-intercepts via grouping
$$f(x) = x^2(x - 3) - 9(x - 3) = (x^2 - 9)(x - 3) = (x + 3)(x - 3)(x - 3) = (x + 3)(x - 3)^2$$Setting $f(x) = 0$ yields roots at $x = -3$ and $x = 3$ (repeated root):
$$\text{Intercepts: } (-3, 0) \text{ and } (3, 0)$$(Notice: at $x = -3$ the curve crosses the $x$-axis; at $x = 3$ the curve is tangent to the axis because of the even multiplicity).
Part (c): Coordinates and classification
$$f(-1) = (-1)^3 - 3(-1)^2 - 9(-1) + 27 = -1 - 3 + 9 + 27 = 32 \implies (-1, 32) \text{ is a Local Maximum}$$ $$f(3) = 3^3 - 3(3)^2 - 9(3) + 27 = 27 - 27 - 27 + 27 = 0 \implies (3, 0) \text{ is a Local Minimum}$$Symmetry: Even and Odd Functions
- Even Function ($y$-Axis Reflectional Symmetry): $$f(-x) = f(x) \quad \text{for all } x \in \text{dom } f$$ The graph is perfectly symmetric across the vertical line $x = 0$ ($y$-axis). If $(a, b)$ is on the graph, then $(-a, b)$ must also be on the graph. Examples: $f(x) = x^2, \cos x, |x|, x^4 - 3x^2 + 5$.
- Odd Function (Origin $180^\circ$ Rotational Symmetry): $$f(-x) = -f(x) \quad \text{for all } x \in \text{dom } f$$ The graph has half-turn rotational symmetry about the origin $(0, 0)$. If $(a, b)$ is on the graph, then $(-a, -b)$ must also be on the graph. If $0 \in \text{dom } f$, then $f(0) = 0$. Examples: $f(x) = x^3, \sin x, \frac{1}{x}, x^5 - 4x$.
- Rigorous IB Algebraic Test: Always test by substituting $-x$ directly into $f(x)$ and simplifying until you prove whether it equals $f(x)$, $-f(x)$, or neither. Never test only a single numerical value!
(a) $g(x) = \frac{x^3}{x^2 + 1}$
(b) $h(x) = x^4 - 2x^2 + 3$
Part (a): Evaluate $g(-x)$
$$g(-x) = \frac{(-x)^3}{(-x)^2 + 1} = \frac{-x^3}{x^2 + 1} = -\left(\frac{x^3}{x^2 + 1}\right) = -g(x)$$Since $g(-x) = -g(x)$ for all $x \in \mathbb{R}$, $g(x)$ is an odd function (symmetric about origin).
Part (b): Evaluate $h(-x)$
$$h(-x) = (-x)^4 - 2(-x)^2 + 3 = x^4 - 2x^2 + 3 = h(x)$$Since $h(-x) = h(x)$ for all $x \in \mathbb{R}$, $h(x)$ is an even function (symmetric about $y$-axis).
Vertical and Horizontal Asymptotes
- Vertical Asymptote ($x = a$): A vertical line that the graph approaches as $y$ heads towards $\pm\infty$: $$\lim_{x \to a^+} f(x) = \pm\infty \quad \text{or} \quad \lim_{x \to a^-} f(x) = \pm\infty$$ For a rational function $\frac{P(x)}{Q(x)}$ in lowest terms, vertical asymptotes occur at the zeros of the denominator ($Q(x) = 0$).
- Horizontal Asymptote ($y = b$): A horizontal line that the graph approaches as $x$ heads towards positive or negative infinity: $$\lim_{x \to \infty} f(x) = b \quad \text{or} \quad \lim_{x \to -\infty} f(x) = b$$ For a linear rational function $f(x) = \frac{ax + b}{cx + d}$ ($c \ne 0$), the horizontal asymptote is $y = \frac{a}{c}$.
(a) State the equations of the vertical and horizontal asymptotes.
(b) Find the coordinates of all axial intercepts.
Part (a): Asymptote equations
Vertical asymptote occurs where denominator equals zero:
$$x + 3 = 0 \implies x = -3$$Horizontal asymptote as $x \to \pm\infty$:
$$\lim_{x \to \pm\infty} \frac{2x - 4}{x + 3} = \frac{2}{1} = 2 \implies y = 2$$Part (b): Axial intercepts
$y$-intercept (set $x = 0$):
$$f(0) = \frac{2(0) - 4}{0 + 3} = -\frac{4}{3} \implies \left(0, -\frac{4}{3}\right)$$$x$-intercept (set $f(x) = 0$ $\implies$ numerator equals zero):
$$2x - 4 = 0 \implies 2x = 4 \implies x = 2 \implies (2, 0)$$Solving Simultaneous Equations & Intersections
- Intersection Principle: The points where two graphs $y = f(x)$ and $y = g(x)$ intersect are the solutions to the equation: $$f(x) = g(x)$$ Each solution $x = x_0$ gives an intersection point $(x_0, f(x_0))$.
- Inequalities from Graphs: Solving $f(x) > g(x)$ graphically means finding the $x$-intervals where the graph of $f(x)$ lies strictly above the graph of $g(x)$.
- GDC Best Practices (Paper 2):
- Enter $Y_1 = f(x)$ and $Y_2 = g(x)$.
- Adjust window boundaries ($X\min, X\max, Y\min, Y\max$) so all turning points and intersections are clearly visible.
- Use the calculator's `CALC` $\to$ `Intersect` function. Report answers to 3 significant figures unless exact values are required.
(a) Find the exact coordinates of the points of intersection of the two graphs.
(b) Hence, solve the inequality $f(x) \le g(x)$.
Part (a): Set $f(x) = g(x)$ and solve algebraically.
$$x^2 - 2x - 3 = 2x + 2$$ $$x^2 - 4x - 5 = 0$$ $$(x - 5)(x + 1) = 0 \implies x = 5 \quad \text{or} \quad x = -1$$Find the corresponding $y$-values using $g(x) = 2x + 2$:
$$\text{At } x = -1: \quad y = 2(-1) + 2 = 0 \implies (-1, 0)$$ $$\text{At } x = 5: \quad y = 2(5) + 2 = 12 \implies (5, 12)$$Intersection points: $(-1, 0)$ and $(5, 12)$.
Part (b): Solve $f(x) \le g(x)$.
Since $f(x)$ is an upward-opening parabola, it lies below the line $g(x)$ between their two intersection points.
Solution: $-1 \le x \le 5$ (or $[-1, 5]$).
Graphical Intersections & Parameter Extraction
(a) Find the coordinates of the local maximum and local minimum of $f(x)$ on this interval. [3 marks]
(b) Find the coordinates of all points of intersection between the curves $y = f(x)$ and $y = g(x)$. [4 marks]
(c) State the range of $f(x)$ for the domain $-2 \le x \le 4$. [2 marks]
Part (a): Find turning points using derivative $f'(x) = 3x^2 - 6x - 1$:
$$3x^2 - 6x - 1 = 0 \implies x = \frac{6 \pm \sqrt{36 - 4(3)(-1)}}{6} = \frac{6 \pm \sqrt{48}}{6} = 1 \pm \frac{2\sqrt{3}}{3} \quad [M1]$$Local maximum at $x \approx -0.155 \implies f(-0.155) \approx 3.08$ [A1].
Local minimum at $x \approx 2.155 \implies f(2.155) \approx -3.08$ [A1].
Part (b): Equate functions $f(x) = g(x)$ (Paper 2 GDC Intersect):
$$x^3 - 3x^2 - x + 3 = 2x - 1 \implies x^3 - 3x^2 - 3x + 4 = 0 \quad [M1]$$Using GDC Polynomial Root Finder / Intersect tool on $[-2, 4]$ [M1]:
$$x_1 \approx -1.36 \implies y_1 = 2(-1.36) - 1 \approx -3.71 \implies (-1.36, -3.71) \quad [A1]$$ $$x_2 \approx 0.867 \implies y_2 = 2(0.867) - 1 \approx 0.735 \implies (0.867, 0.735) \quad [A1]$$ $$x_3 \approx 3.49 \implies y_3 = 2(3.49) - 1 \approx 5.98 \implies (3.49, 5.98) \quad [A1]$$Part (c): Range on $[-2, 4]$:
Evaluate endpoints: $f(-2) = (-2)^3 - 3(-2)^2 - (-2) + 3 = -8 - 12 + 2 + 3 = -15$.
$$f(4) = 4^3 - 3(4^2) - 4 + 3 = 64 - 48 - 4 + 3 = 15.$$Comparing endpoints with local extrema $(-3.08 \text{ and } 3.08)$: absolute minimum is $-15$ and absolute maximum is $15$.
$$\text{Range: } -15 \le y \le 15 \quad [A1]$$Answers: (a) Max $(-0.155, 3.08)$, Min $(2.155, -3.08)$; (b) $(-1.36, -3.71)$, $(0.867, 0.735)$, $(3.49, 5.98)$; (c) $-15 \le y \le 15$
10 Progressive Practice Exercises
Tier 1 tests polynomial intercepts and symmetry; Tier 2 introduces asymptotes, turning points, and tangent conditions; Tier 3 challenges multi-intersection systems and physical modeling.
Tier 1 Core Fluency (Problems 1–3)
Find all $x$-intercepts and the $y$-intercept of the polynomial function $f(x) = 2x^3 - 8x$.
View Full Worked Solution
$y$-intercept: Set $x = 0$:
$$f(0) = 2(0)^3 - 8(0) = 0 \implies (0, 0)$$$x$-intercepts: Set $f(x) = 0$ and factor completely:
$$2x(x^2 - 4) = 0 \implies 2x(x - 2)(x + 2) = 0$$ $$x = 0, \quad x = 2, \quad x = -2$$Answer: $x$-intercepts are $(-2, 0)$, $(0, 0)$, and $(2, 0)$; $y$-intercept is $(0, 0)$.
Using algebraic substitution, classify each function as even, odd, or neither:
(a) $f(x) = 3x^4 - 5x^2 + 1$
(b) $g(x) = 2x^3 - 4x$
View Full Worked Solution
Part (a):
$$f(-x) = 3(-x)^4 - 5(-x)^2 + 1 = 3x^4 - 5x^2 + 1 = f(x)$$Since $f(-x) = f(x)$, $f$ is an even function.
Part (b):
$$g(-x) = 2(-x)^3 - 4(-x) = -2x^3 + 4x = -(2x^3 - 4x) = -g(x)$$Since $g(-x) = -g(x)$, $g$ is an odd function.
Find the coordinates of the points of intersection between the parabola $y = x^2 - 4x + 3$ and the line $y = x - 1$.
View Full Worked Solution
Step 1: Set equations equal to each other:
$$x^2 - 4x + 3 = x - 1 \implies x^2 - 5x + 4 = 0$$Step 2: Factor the quadratic:
$$(x - 4)(x - 1) = 0 \implies x = 1 \quad \text{or} \quad x = 4$$Step 3: Find corresponding $y$-values:
$$\text{At } x = 1: \quad y = 1 - 1 = 0 \implies (1, 0)$$ $$\text{At } x = 4: \quad y = 4 - 1 = 3 \implies (4, 3)$$Answer: $(1, 0)$ and $(4, 3)$.
Tier 2 Exam Modeling (Problems 4–7)
Consider the rational function $f(x) = \frac{3x + 6}{x - 2}$.
(a) State the equation of the vertical asymptote.
(b) State the equation of the horizontal asymptote.
(c) Find the coordinates where the graph crosses the coordinate axes.
View Full Worked Solution
Part (a): Vertical Asymptote
$$x - 2 = 0 \implies x = 2$$Part (b): Horizontal Asymptote
$$\lim_{x \to \pm\infty} \frac{3x + 6}{x - 2} = \frac{3}{1} \implies y = 3$$Part (c): Axial Intercepts
$y$-intercept (set $x = 0$): $f(0) = \frac{6}{-2} = -3 \implies (0, -3)$.
$x$-intercept (set numerator $= 0$): $3x + 6 = 0 \implies x = -2 \implies (-2, 0)$.
A quadratic function is given by $f(x) = -2(x - 3)^2 + 8$.
(a) State the coordinates of the maximum turning point.
(b) State the equation of the horizontal line that is tangent to the graph at its maximum point.
View Full Worked Solution
Part (a): Vertex coordinates
The vertex form is $a(x - h)^2 + k$. Here $h = 3, k = 8$, and $a = -2 < 0$.
The maximum turning point is $(3, 8)$.
Part (b): Horizontal tangent line
A horizontal line through $(3, 8)$ has slope $m = 0$, giving equation $y = 8$.
Answer: (a) $(3, 8)$; (b) $y = 8$.
The horizontal line $y = k$ intersects the curve $y = x^2 - 6x + 5$ at exactly one point.
(a) Find the value of $k$.
(b) Find the coordinates of the point of intersection.
View Full Worked Solution
Method 1: Vertex analysis
Complete the square: $y = (x - 3)^2 - 9 + 5 = (x - 3)^2 - 4$.
The parabola has its minimum value at $(3, -4)$. A horizontal line touches the parabola at exactly one point if and only if it passes through the vertex.
Thus $k = -4$, and the intersection point is $(3, -4)$.
Method 2: Discriminant analysis
$$x^2 - 6x + 5 = k \implies x^2 - 6x + (5 - k) = 0$$For exactly one solution, $\Delta = 0$:
$$\Delta = (-6)^2 - 4(1)(5 - k) = 36 - 20 + 4k = 16 + 4k = 0 \implies 4k = -16 \implies k = -4$$Then $x = \frac{-(-6)}{2(1)} = 3$, giving point $(3, -4)$.
Answer: (a) $k = -4$; (b) $(3, -4)$.
Let $f(x) = \frac{x}{x^2 + 4}$.
(a) Prove that $f(x)$ is an odd function.
(b) Find all points of intersection between the graph of $f(x)$ and the line $y = \frac{1}{5}x$.
View Full Worked Solution
Part (a): Symmetry proof
$$f(-x) = \frac{-x}{(-x)^2 + 4} = \frac{-x}{x^2 + 4} = -\left(\frac{x}{x^2 + 4}\right) = -f(x)$$Since $f(-x) = -f(x)$ for all $x \in \mathbb{R}$, $f$ is an odd function. $\blacksquare$
Part (b): Find points of intersection
$$\frac{x}{x^2 + 4} = \frac{1}{5}x \implies x\left(\frac{1}{x^2 + 4} - \frac{1}{5}\right) = 0$$Either $x = 0$ (which gives $y = 0$, point $(0, 0)$), or:
$$\frac{1}{x^2 + 4} = \frac{1}{5} \implies x^2 + 4 = 5 \implies x^2 = 1 \implies x = \pm 1$$For $x = 1$: $y = \frac{1}{5}(1) = \frac{1}{5} \implies \left(1, \frac{1}{5}\right)$.
For $x = -1$: $y = \frac{1}{5}(-1) = -\frac{1}{5} \implies \left(-1, -\frac{1}{5}\right)$.
Answer: Three points: $(-1, -0.2)$, $(0, 0)$, and $(1, 0.2)$.
Tier 3 Complex Applications & Synthesis (Problems 8–10)
Find all points of intersection between the cubic curve $y = x^3 - 3x$ and the line $y = x$, and sketch the relative positions of the two curves.
View Full Worked Solution
Step 1: Set equations equal to each other:
$$x^3 - 3x = x \implies x^3 - 4x = 0$$ $$x(x^2 - 4) = 0 \implies x(x - 2)(x + 2) = 0$$The $x$-coordinates are $x = -2, 0, 2$.
Step 2: Find corresponding $y$-values using $y = x$:
- At $x = -2$: $y = -2 \implies (-2, -2)$
- At $x = 0$: $y = 0 \implies (0, 0)$
- At $x = 2$: $y = 2 \implies (2, 2)$
Answer: The three intersection points are $(-2, -2)$, $(0, 0)$, and $(2, 2)$.
A parabola $y = ax^2 + bx + c$ has its vertex at $(2, -1)$ and passes through $(0, 3)$.
(a) Find the values of $a, b,$ and $c$.
(b) Find the range of values of $m$ for which the line $y = mx - 2$ does not intersect the parabola.
View Full Worked Solution
Part (a): Find $a, b, c$
Using vertex form: $y = a(x - 2)^2 - 1$.
Passes through $(0, 3)$: $3 = a(0 - 2)^2 - 1 \implies 3 = 4a - 1 \implies 4a = 4 \implies a = 1$.
Expand: $y = 1(x^2 - 4x + 4) - 1 = x^2 - 4x + 3$.
So $a = 1$, $b = -4$, $c = 3$.
Part (b): No intersection condition
Set equal to $y = mx - 2$:
$$x^2 - 4x + 3 = mx - 2 \implies x^2 - (4 + m)x + 5 = 0$$No intersection requires strictly negative discriminant ($\Delta < 0$):
$$\Delta = [-(4 + m)]^2 - 4(1)(5) = (m + 4)^2 - 20 < 0$$ $$(m + 4)^2 < 20 \implies -\sqrt{20} < m + 4 < \sqrt{20}$$ $$-2\sqrt{5} - 4 < m < 2\sqrt{5} - 4$$Answer: (a) $a = 1, b = -4, c = 3$; (b) $-4 - 2\sqrt{5} < m < -4 + 2\sqrt{5}$.
The concentration $C(t)$ of a medication in a patient's bloodstream $t$ hours after injection is modeled by:
$$C(t) = \frac{50t}{t^2 + 9}, \quad t \ge 0$$
(a) Find the initial concentration $C(0)$.
(b) State the horizontal asymptote of $C(t)$ as $t \to \infty$ and interpret its medical meaning.
(c) Solve $C(t) = 5$ to find the two times at which the concentration equals $5\text{ mg/L}$.
View Full Worked Solution
Part (a): Initial concentration
$$C(0) = \frac{50(0)}{0^2 + 9} = 0\text{ mg/L}$$Part (b): Horizontal Asymptote
$$\lim_{t \to \infty} \frac{50t}{t^2 + 9} = 0 \implies C = 0$$Medical interpretation: Long after the injection, the medication is fully metabolized and filtered out, so the concentration in the bloodstream approaches zero.
Part (c): Solve $C(t) = 5$
$$\frac{50t}{t^2 + 9} = 5 \implies 50t = 5(t^2 + 9) \implies 10t = t^2 + 9$$ $$t^2 - 10t + 9 = 0 \implies (t - 1)(t - 9) = 0$$ $$t = 1\text{ hour} \quad \text{or} \quad t = 9\text{ hours}$$Answer: (a) $0\text{ mg/L}$; (b) Horizontal asymptote $C = 0$ (drug clears over time); (c) $t = 1\text{ hour}$ and $t = 9\text{ hours}$.
Quadratic & Function Formulas
- Axis of Symmetry of Quadratic: $$x = -\frac{b}{2a}$$ For $f(x) = ax^2 + bx + c$. Gives $x$-coordinate of the vertex.
- Quadratic Formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
Symmetry, Asymptotes & Intersections
- Even Function ($y$-Axis Symmetry): $$f(-x) = f(x)$$
- Odd Function (Origin Symmetry): $$f(-x) = -f(x)$$
- Horizontal Asymptote of Linear Rational: $$f(x) = \frac{ax + b}{cx + d} \implies y = \frac{a}{c} \quad (c \ne 0)$$
- Vertical Asymptote of Linear Rational: $$cx + d = 0 \implies x = -\frac{d}{c}$$
- Simultaneous Intersections: $$\text{Solve } f(x) = g(x)$$
Critical IB Examination Pitfalls
- Reporting Only $x$ for Points: If a question asks for the "coordinates of points of intersection", writing only $x = 1$ and $x = 4$ loses the answer mark. You must report ordered pairs $(1, y_1)$ and $(4, y_2)$.
- Writing Asymptotes as Numbers: An asymptote is a straight line equation ($x = 2$ or $y = 3$). Writing "asymptote $= 2$" is mathematically inaccurate and receives $A0$.
- Testing Single Numbers for Symmetry: Proving a function is even requires testing the general expression $f(-x)$ algebraically. Checking that $f(-2) = f(2)$ does not prove that $f(x)$ is even for all real $x$.
- Ignoring Window Bounds on the GDC: On Paper 2, default GDC zoom $[-10, 10]$ frequently conceals key turning points or second intersections that occur further out. Always adjust the window to inspect all relevant features.
IB AA SL Topic Check: Graphs & Key Features
Section A has diagnostic questions with feedback. Section B has Paper 1 and Paper 2 exam-style problems with markschemes.
Section A: Diagnostic Concept Check
Score: 0 / 4Question 1: Which of the following describes the symmetry of the function $f(x) = \frac{x^2 - 1}{x^4 + 3}$?
Question 2: What is the equation of the horizontal asymptote of $f(x) = \frac{4x - 7}{2x + 5}$?
Question 3: How many points of intersection exist between the graphs of $y = x^2 + 4$ and $y = 3$?
Question 4: What are the coordinates of the $y$-intercept of $f(x) = \frac{2x - 8}{x + 4}$?
Section B: Exam-Style Free Response Questions
Question 5: Rational Curve Features
Consider the rational function $f(x) = \frac{3x - 9}{x + 2}$.
(a) Write down the equations of the vertical and horizontal asymptotes. [2 marks]
(b) Find the coordinates of the $x$-intercept and $y$-intercept. [2 marks]
(c) Find the value of $x$ for which $f(x) = 1$. [2 marks]
View Official Markscheme & Model Solution
Part (a) [2 Marks]:
Vertical asymptote: $x = -2$ (A1)
Horizontal asymptote: $y = 3$ (A1)
Part (b) [2 Marks]:
$x$-intercept: $3x - 9 = 0 \implies x = 3 \implies (3, 0)$ (A1)
$y$-intercept: $f(0) = \frac{-9}{2} = -4.5 \implies (0, -4.5)$ (A1)
Part (c) [2 Marks]:
$$\frac{3x - 9}{x + 2} = 1 \implies 3x - 9 = x + 2 \quad \text{\bf (M1)}$$ $$2x = 11 \implies x = 5.5 \quad (\text{or } 11/2) \quad \text{\bf (A1)}$$Question 6: Quadratic-Linear Intersection & Inequality
Let $f(x) = -x^2 + 4x + 5$ and $g(x) = x + 1$.
(a) Find the exact coordinates of the points of intersection of the graphs of $f$ and $g$. [4 marks]
(b) Hence, find the set of values of $x$ for which $f(x) > g(x)$. [3 marks]
View Official Markscheme & Model Solution
Part (a) [4 Marks]:
$$-x^2 + 4x + 5 = x + 1 \quad \text{\bf (M1)}$$ $$x^2 - 3x - 4 = 0 \implies (x - 4)(x + 1) = 0 \quad \text{\bf (M1)}$$ $$x = 4 \quad \text{or} \quad x = -1 \quad \text{\bf (A1)}$$Find $y$-coordinates using $g(x) = x + 1$:
$$\text{At } x = -1: \quad y = -1 + 1 = 0 \implies (-1, 0)$$ $$\text{At } x = 4: \quad y = 4 + 1 = 5 \implies (4, 5)$$Intersection points are $(-1, 0)$ and $(4, 5)$. (A1)
Part (b) [3 Marks]:
$f(x) > g(x) \iff -x^2 + 3x + 4 > 0 \iff x^2 - 3x - 4 < 0$. (M1)
Graphically, the downward-opening parabola lies strictly above the line between its intersection points. (R1)
$$-1 < x < 4 \quad (\text{or } x \in (-1, 4)) \quad \text{\bf (A1)}$$Topic 2 Curriculum
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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.