Tangents and normals

Tangent

1. Find the point: substitute x into y.

2. Find the gradient m: substitute x into dy/dx.

3. Use y - y_1 = m(x - x_1).

Normal

The normal is perpendicular to the tangent, so its gradient is -1/m. Then use y - y_1 = (-1/m)(x - x_1).

Working backwards

To find where the gradient has a given value k, solve dy/dx = k.

Worked example

Find the tangent and normal to y = x² - 3x + 4 at x = 2.

1. y = 4 - 6 + 4 = 2, so the point is (2, 2)
2. dy/dx = 2x - 3 = 1
3. Tangent: y - 2 = 1(x - 2), so y = x
4. Normal gradient -1: y - 2 = -(x - 2), so y = -x + 4

Answer: Tangent y = x; normal y = -x + 4

Check yourself

1. Find the gradient of the tangent to y = x³ - 2x at x = 2.
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   10. dy/dx = 3x² - 2 = 10 at x = 2.

2. Find the y-intercept of the tangent to y = x² + 1 at (2, 5).
   Show answer

   -3. dy/dx = 2x = 4. y - 5 = 4(x - 2) gives y = 4x - 3, so the intercept is -3.

3. Find the gradient of the normal to y = x² + 3x at x = 1. Give your answer as a fraction in its simplest form.
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   -1/5. dy/dx = 2x + 3 = 5, so the normal gradient is -1/5.

4. The normal to y = x² at (1, 1) crosses the y-axis. Find the y-coordinate.
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   1.5. The tangent gradient is 2, so the normal gradient is -1/2. y - 1 = -1/2 (x - 1) gives y = -x/2 + 3/2, so the y-intercept is 1.5.

5. Find the x-coordinate of the point on y = x² - 6x + 1 where the gradient is 4.
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   5. 2x - 6 = 4, so x = 5.

6. Find the equation of the tangent to y = 4/x at x = 2.
   • y = x
   • x + y = 4
   • x + y = 2
   • y = -4x + 10
   Show answer

   x + y = 4. The point is (2, 2). dy/dx = -4x^-2 = -1 at x = 2. y - 2 = -(x - 2), so x + y = 4.

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Sources

• Pearson Edexcel International Advanced Level in Mathematics specification (issue 3)
• Pearson Edexcel IAL Pure Mathematics 1 scheme of work

Last updated 7 November 2026. Spotted a mistake? Tell us.