Key Idea: A tangent touches a curve at a point and a normal cuts it at right angles there. The derivative hands you the gradient, and you finish with the straight-line equation — a staple Paper 1 by-hand task, with the occasional Paper 2 value to read off.
📐 The two lines at a point
- the gradient of the line you want
- the point of contact (a, f(a)) — same for tangent and normal
| Tangent | Normal | |
|---|---|---|
| Gradient | f′(a) | −1 / f′(a) (negative reciprocal) |
| Direction | touches the curve | perpendicular to the tangent |
| Point used | (a, f(a)) | (a, f(a)) — the same point |
The normal gradient is −1/f′(a): turn the fraction upside down and swap the sign. Gradient 2 → normal −½; gradient −¾ → normal +⁴⁄₃. The two gradients always multiply to −1.
(1) Gradient — differentiate, evaluate f′(a). (2) For a normal, take the negative reciprocal −1/f′(a). (3) Point — substitute into the original f(x) for y₁ = f(a). (4) Line — put m and (a, f(a)) into y − y₁ = m(x − x₁) and simplify.
Important: If f′(a) = 0 (a stationary point): the tangent is horizontal, y = f(a), and the normal is vertical, x = a (you can't do −1/0 — read it geometrically). For a tangent parallel to a line, set f′(x) equal to that line's gradient.
✏️ IB-style worked examples
IB-style question — equation of a tangent
Find the equation of the tangent to y = x² − 5x + 4 at the point where x = 4, without a calculator.
Step by step:
Gradient: differentiate and evaluate f′(4).
Point: substitute x = 4 into the original f(x).
Line through (4, 0) with gradient 3.
y = 3x − 12.
IB-style question — horizontal tangents
Find the equations of the horizontal tangents to y = x³ − 27x.
Step by step:
Horizontal → gradient 0, so solve f′(x) = 0.
Find the y-coordinate at each x from the original function.
A horizontal tangent is just y = (that y-value).
y = −54 (at x = 3) and y = 54 (at x = −3).
IB-style question — equation of a normal
Find the equation of the normal to y = x² at the point where x = 2, without a calculator.
Step by step:
Tangent gradient first: f′(2).
Normal gradient is the negative reciprocal; point is (2, 4).
Use y − y₁ = m(x − x₁) and simplify.
y = −¼x + 9/2.
IB-style question — tangent and normal at a vertex
Find the equations of the tangent and the normal to y = x² − 6x + 11 at its vertex, where x = 3.
Step by step:
Gradient at the vertex is 0 → the tangent is horizontal.
Point: f(3) = 9 − 18 + 11 = 2, so the tangent is y = 2.
The normal is perpendicular to a horizontal line → vertical.
Tangent y = 2 (horizontal); normal x = 3 (vertical).
Important: Don't reuse f′(a) for the normal, and don't just flip the sign. The normal gradient is −1/f′(a) — flip the fraction and change the sign. f′(a) = 2 gives −½, not −2 and not ½. Always substitute into the original f(x) for the y-coordinate, never into f′(x).
Tap each card to reveal the answer.
f′(a) = 5. What is the normal gradient? −1/5 — flip and change the sign of the tangent gradient.
Tangent gradient is −⅔. What is the normal gradient? +³⁄₂ — negative reciprocal: flip −⅔ and swap the sign.
y = x² + 1 at x = 3: what is the tangent gradient? 6 — f′(x) = 2x, so f′(3) = 6.
A horizontal tangent has what form? y = constant — the y-coordinate of the point (gradient 0).
At a stationary point, what is the normal? Vertical, x = a — the tangent is horizontal, so the normal is vertical.
Tangent to y = x² − x at x = 2 passes through which point? (2, 2) — f(2) = 4 − 2 = 2; use the original f(x) for y₁.
Exam Tips
- Tangent gradient = f′(a); always get y₁ from the ORIGINAL f(x), not f′(x).
- Normal gradient = −1/f′(a): flip the fraction AND change the sign.
- Tangent and normal share the same point of contact (a, f(a)).
- Horizontal tangent → solve f′(x) = 0; the answer is y = constant.
- Stationary point → tangent y = f(a) (horizontal), normal x = a (vertical).