1. Find the two ingredients
The equation of a straight line is fixed by a point on the line and its gradient. For a tangent to a differentiable curve at x = a, the derivative gives the gradient at that point.
2. Work through a complete example
Find the tangent to y = x² + 3x − 4 at x = 2.
The tangent passes through (2, 6).
Find the gradientAt x = 2: m = 2(2) + 3 = 7
y − 6 = 7(x − 2)
y = 7x − 8
Keep the point-gradient form visible before expanding. It makes the link between your point, gradient and final equation easy to check.
3. Check the result in two ways
Point check: substitute x = 2 into y = 7x − 8. It gives y = 6, so the line passes through the required point.
Gradient check: the coefficient of x is 7, which matches the derivative evaluated at x = 2.
These checks catch common errors such as calculating the derivative correctly but using the wrong y-coordinate, or expanding 7(x − 2) incorrectly.
4. If the question asks for a normal
A normal is perpendicular to the tangent at the same point. When the tangent has a finite, non-zero gradient m, the normal gradient is −1/m.
y − 6 = −(1/7)(x − 2)
A horizontal tangent has gradient zero; its normal is vertical. In that case, use a vertical-line equation such as x = a rather than dividing by zero.
Your turn: find a tangent
Find the equation of the tangent to y = x³ − 2x at x = 1. Check that your final line passes through the correct point.
Reveal the worked solution
dy/dx = 3x² − 2
At x = 1: m = 3(1)² − 2 = 1
y + 1 = 1(x − 1)
y = x − 2
Check: when x = 1, the line gives y = −1. Its gradient is 1, matching the derivative at that point.
To revise efficiently, repeat the same four prompts: point, derivative, gradient at the point, equation. Then try a question where the tangent is horizontal or where the x-coordinate must be found first.
These are original teaching examples. Your school’s syllabus and task-specific instructions take priority. Ask about personalised support.