The big idea: At any point on a curve, two special lines can be drawn: Tangent — touches the curve and goes in the same direction as the curve. Normal — perpendicular to the tangent at the same point. It goes in the direction 90° away from the tangent.
| Line | Gradient | Relationship to curve |
|---|---|---|
| Tangent | mt = f′(x₁) | Goes in the same direction as the curve at that point |
| Normal | mn = −1 / f′(x₁) | Perpendicular to the tangent at the same point |
Perpendicular lines — the gradient rule: Two lines are perpendicular if and only if their gradients multiply to give −1: mt × mn = −1 Rearranging: mn = −1 / mt This is the only formula you need for the normal gradient.
- gradient of the tangent — found from f′(x)
- gradient of the normal — negative reciprocal of m_t
[Diagram: math-derivative-tangent] - Available in full study mode
The four steps: Step 1 — Differentiate: Find f′(x). Step 2 — Find the tangent gradient: mt = f′(x₁). Step 3 — Find the normal gradient: mn = −1/mt. Step 4 — Write the equation: y − y₁ = mn(x − x₁).
Worked example 1
Find the equation of the normal to y = x² + 1 at x = 2. Give the answer in the form y = mx + c.
Step by step
- Step 1: Differentiate.
- Step 2: Tangent gradient at x = 2.
- Step 3: Normal gradient.
- Find y₁: y = (2)² + 1 = 5. Point (2, 5).
- Rearrange to y = mx + c.
Final answer
y = −¼x + 11/2
The most common error in normal-line questions: Using the tangent gradient when asked for the normal equation. The question will say 'normal' — do not skip the negative-reciprocal step.
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The big idea: The negative-reciprocal rule works whatever the tangent gradient is — positive, negative, or fractional. Take care with the signs.
| Tangent gradient mt | Normal gradient mn = −1/mt | Key step |
|---|---|---|
| 4 | −1/4 | Flip and negate |
| −3 | 1/3 | Flip: 1/3, negate: 1/3 (two negatives cancel) |
| 1/2 | −2 | Flip: 2, negate: −2 |
| −2/5 | 5/2 | Flip: 5/2, negate: 5/2 (two negatives cancel) |
Worked example 2 — negative tangent gradient
Find the normal to f(x) = 6 − x² at x = 3.
Step by step
- Differentiate.
- Tangent gradient at x = 3.
- Normal gradient.
- y₁: f(3) = 6 − 9 = −3. Point (3, −3).
- Rearrange.
Final answer
y = (1/6)x − 7/2
The big idea: If f′(x₁) = 0, the tangent is horizontal. You cannot divide −1 by 0, so the normal is vertical — a line of the form x = x₁.__LINEBREAK__IB rarely asks this exact case, but recognising it shows understanding.
| Tangent is ... | Normal is ... | Normal equation |
|---|---|---|
| Horizontal (mt = 0) | Vertical | x = x₁ |
| Vertical (undefined slope) | Horizontal | y = y₁ |
| Any other gradient mt | Perpendicular slope | mn = −1/mt |
IB exam context: For polynomials in AI SL, the tangent gradient is rarely 0 at the given point unless the question is specifically about stationary points. Most normal-line questions use a well-defined non-zero gradient.
Full comparison example
For f(x) = x³ − 3x at x = 1, find (a) the tangent equation and (b) the normal equation.
Step by step
- Differentiate.
- Tangent gradient at x = 1.
- f(1) = 1 − 3 = −2. Point (1, −2).
- (a) Tangent: horizontal line through (1, −2).
- (b) Normal: vertical line through (1, −2).
Final answer
(a) y = −2 (b) x = 1