Key Idea: The derivative f'(x) gives the gradient of the tangent line to a curve at any point. Topic 5.4 uses derivatives to write the equation of the tangent (touching the curve) and the equation of the normal (perpendicular to the tangent) at a specific point. These are standard IB exam question types.
✅ Tangent and normal formulas
| Quantity | Formula | Note |
|---|---|---|
| Tangent gradient | f'(a) | Differentiate f(x), substitute x = a |
| Tangent equation | y − f(a) = f'(a)(x − a) | Point-slope form through the point (a, f(a)) |
| Normal gradient | −1/f'(a) | Perpendicular: m₁ × m₂ = −1 |
| Normal equation | y − f(a) = [−1/f'(a)](x − a) | Same point (a, f(a)), perpendicular gradient |
| Special case | If f'(a) = 0, normal is vertical: x = a | Tangent is horizontal; normal has undefined gradient |
Example: f(x) = x³ − 3x + 2. Find the tangent and normal at x = 2. f(2) = 8 − 6 + 2 = 4. Point: (2, 4). f'(x) = 3x² − 3. f'(2) = 12 − 3 = 9. Tangent: y − 4 = 9(x − 2) → y = 9x − 14 Normal gradient = −1/9. Normal: y − 4 = −(1/9)(x − 2) → y = −x/9 + 2/9 + 4 → y = −(1/9)x + (38/9)
The point (a, f(a)) lies on both the tangent and the normal. Always calculate f(a) — it gives the y-coordinate of the point of contact. The tangent touches the curve at one point; it does not cross it (at that point). If the question asks where else the tangent meets the curve, set the tangent equation equal to f(x) and solve.
Paper 1 (GDC allowed): Show all differentiation steps. Write f'(x) first, then substitute x = a. Then write the point-slope equation. Paper 2 (GDC allowed): Use GDC to evaluate f(a) and f'(a) if the function is complex. But still write the equation construction steps in your working.
IB-style question [5 marks]
Find the equation of the tangent to the curve y = x³ − 2x at the point where x = 1.
Step by step:
Find the y-coordinate of the point.
Differentiate and find the gradient at x = 1.
Use y − y₁ = m(x − x₁).
Final answer:
y = x − 2.