Build tangent and normal equations at a curve point
Evaluate the curve and its derivative at the same input, use a point-gradient line and handle a horizontal tangent's vertical normal.
Section 1.7 · Differentiation
Pure Mathematics 1 (Paper 1) · AS & A Level Mathematics 9709, 2026–2027. Use an informal understanding of limits; formal general differentiation from first principles and points of inflexion are not required in section 1.7. Written practice is self-checked.
Before you begin
- Differentiate powers and composite functions.
- Use point-gradient line equations and perpendicular gradients.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For y = x² + 1 at x = 1, what is the curve point?
What you will learn
- Find tangent and normal equations using a differentiated curve.
- Handle zero tangent gradients without dividing by zero.
The point comes from the curve; the gradient comes from its derivative
At input x = a on y = f(x), the curve point is (a, f(a)) and its finite tangent gradient is f′(a). The tangent equation is y − f(a) = f′(a)(x − a). Substituting a into the derivative gives a gradient, not the point's height. Find both quantities before building the line.
For y = x² − 3x + 2 at x = 3, f(3) = 2 and f′(3) = 2(3) − 3 = 3. The tangent is y − 2 = 3(x − 3), or y = 3x − 7. It passes through (3, 2) and its coefficient of x is the calculated gradient 3; these two checks distinguish it from a line with the correct gradient but wrong intercept.
A normal is perpendicular to the tangent
When the tangent gradient m is finite and non-zero, the normal has gradient −1/m through the same curve point. For the example, the normal is y − 2 = −(x − 3)/3, equivalently x + 3y = 9. The product of the gradients is −1 and both lines contain the point.
If m = 0, the tangent is horizontal and the normal is vertical. For y = (x − 2)⁴ + 1 at x = 2, the point is (2, 1), the tangent is y = 1 and the normal is x = 2. Do not try to turn −1/0 into a finite gradient or force the normal into y = mx + c. Where a finite derivative does not exist, the usual point-gradient procedure needs separate geometric consideration.
Pause and explain
A curve's finite tangent gradient is zero at (2, 1). What is its normal equation?
Worked example
Find the tangent and normal to y = x² − 3x + 2 at x = 3.
Show the worked solution
- Evaluate the curve to get (3, 2), and differentiate to get y′ = 2x − 3, hence tangent gradient 3.
- The tangent is y − 2 = 3(x − 3), or y = 3x − 7.
- The normal gradient is −1/3, so y − 2 = −(x − 3)/3, or x + 3y = 9; check both at (3, 2).
Answer Tangent y = 3x − 7; normal x + 3y = 9.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find tangent and normal equations to y = x² + 1 at x = 1.
Give me a hint
Find the point and tangent gradient separately, then use the same point for both lines.
Compare my reasoning
- The point is (1, 2), and y′ = 2x gives tangent gradient 2.
- The tangent is y − 2 = 2(x − 1), hence y = 2x.
- The normal gradient is −1/2, so y − 2 = −(x − 1)/2, giving y = −x/2 + 5/2.
Tangent y = 2x; normal y = −x/2 + 5/2.
Look for these in your work
- Use the original curve and derivative for their different roles.
- Verify both lines pass through the same point and have perpendicular gradients.
Find the tangent and normal to y = √(4x + 1) at x = 2.
Give me a hint
Use the chain rule for the gradient and verify the input lies inside the finite-derivative domain.
Compare my reasoning
- The point is (2, 3), and y′ = 2/√(4x + 1) gives gradient 2/3.
- The tangent y − 3 = (2/3)(x − 2) simplifies to y = 2x/3 + 5/3.
- The normal gradient is −3/2, so its equation is y − 3 = −3(x − 2)/2, or y = −3x/2 + 6.
Tangent y = 2x/3 + 5/3; normal y = −3x/2 + 6.
Look for these in your work
- Retain the inner derivative factor in the root calculation.
- Use a negative reciprocal and the actual curve point.
For y = (x − 2)⁴ + 1, a learner tries to use −1/0 as a normal gradient at x = 2. Give both correct line equations and explain the geometric correction.
Give me a hint
Differentiate the curve and identify the perpendicular line from its direction.
Compare my reasoning
- The curve point is (2, 1), and y′ = 4(x − 2)³ gives tangent gradient zero.
- The horizontal tangent through the point is y = 1.
- Its perpendicular normal is vertical, x = 2, with undefined finite gradient; it should not be represented by division by zero.
Tangent y = 1; normal x = 2; use perpendicular directions rather than −1/0.
Look for these in your work
- Identify the zero-gradient exception correctly.
- Write the vertical normal in an appropriate line form.
Common mistakes
- Evaluate the original curve for the point's height, not the derivative.
- Use the negative reciprocal for a non-zero tangent gradient.
- A horizontal tangent has a vertical normal with an x = constant equation.
What can you explain now?
At the point (2, 1) on y = (x − 2)⁴ + 1, find tangent and normal equations.
Compare with the explanation
Tangent y = 1; normal x = 2.
The derivative is 4(x − 2)³, so it is zero at the point; perpendicularity therefore gives a vertical normal.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Explain why the original function and its derivative are both needed to build a tangent. Then describe how the normal changes when the tangent gradient is zero.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.