The derivative gives the gradient of the tangent at a point. With the point and the gradient, you write the line. The normal uses the perpendicular gradient through the same point.
This lesson is part of SPM Additional Mathematics differentiation. The line-equation step comes from parallel and perpendicular lines.
The method in four steps
- Find the point: substitute x into the curve to get y.
- Differentiate and substitute x to get the tangent gradient m.
- Tangent: y − y₁ = m(x − x₁).
- Normal: use gradient −1/m through the same point.
Worked example 1: tangent and normal
Find the tangent and the normal to y = x² − 3x + 4 at the point where x = 3.
Point. y = 9 − 9 + 4 = 4, so the point is (3, 4).
Gradient. dy/dx = 2x − 3. At x = 3 it is 3.
Tangent. y − 4 = 3(x − 3), so y = 3x − 5.
Normal. The gradient is −1/3, so y − 4 = −(1/3)(x − 3). Multiply by 3: 3y − 12 = −x + 3, so x + 3y = 15.
Check: (3, 4) lies on both lines. For the normal, 3 + 12 = 15.
Worked example 2: a tangent with a given gradient
Find the point on y = x² − 3x + 4 where the tangent is parallel to y = 5x − 2.
The required gradient is 5, so 2x − 3 = 5, which gives x = 4. Then y = 16 − 12 + 4 = 8, so the point is (4, 8).
Check the gradient: at x = 4, dy/dx = 8 − 3 = 5, as required.
The mistake that costs marks
Two slips cost marks: using the y-value as the gradient, and forgetting the negative when changing to the normal.
| Step | Wrong | Right |
|---|---|---|
| Gradient at x = 3 | y = 4, so m = 4 | dy/dx = 2(3) − 3 = 3 |
| Normal gradient | 1/3 or −3 | −1/3 |
The gradient always comes from dy/dx, never from the curve’s value. A quick test: the product of the tangent and normal gradients is −1, and 3 × (−1/3) = −1.
Check yourself
Find the tangent and the normal to y = x³ − 2x at x = 2.
Answer
Point: y = 8 − 4 = 4, so (2, 4).
Gradient: dy/dx = 3x² − 2 = 12 − 2 = 10.
Tangent: y − 4 = 10(x − 2), so y = 10x − 16.
Normal: gradient −1/10, so y − 4 = −(1/10)(x − 2). Multiply by 10: 10y − 40 = −x + 2, so x + 10y = 42. Check: 2 + 40 = 42.
What to study next
Go on to finding stationary points and testing their nature. For a harder version where a coefficient is unknown, try using a tangent condition to recover an unknown coefficient.
If you would like a teacher to go through your tangent and normal working, see online one-to-one Additional Mathematics tuition.