Tangents and Normals (AI HL)
A curve's gradient changes from point to point, but at any single point you can draw two straight lines that describe it: the tangent, which just touches the curve and shares its gradient there, and the normal, which is perpendicular to the tangent. Both are found from the same derivative - only the gradient step differs between them. It's part of the broader Differentiation topic.
19 questions on this sub-topic.
Finding each line
Covered under IB syllabus reference SL5.4: tangents and normals at a given point, and their equations, using both analytic approaches and technology.
Tangent
Differentiate \(f(x)\) and substitute the \(x\)-value to get the gradient \(m\) at that point, then use \(y - y_1 = m(x - x_1)\).
Not in the formula booklet as a named formula - it's the point-gradient line equation applied with the derivative's value.
Normal
Find the tangent's gradient \(m\) as above, then use the perpendicular gradient \(-\dfrac{1}{m}\) in \(y - y_1 = -\dfrac{1}{m}(x - x_1)\).
Not in the formula booklet - built from the tangent gradient and the perpendicular-gradient relationship.
Need the surrounding derivative rules and formula-booklet reference table? See Differentiation.
Worked examples
Find the equation of the tangent to \(y = x^2 - 3x + 4\) at the point where \(x = 1.\)
Worked solution
At \(x=1\): \(y\) M1
\(= 2.\) A1
\(\dfrac{dy}{dx} = 2x - 3\), gradient \(= -1.\) M1
\(-1.\) A1
Tangent: \(y = -x + 3.\) A1
Find the equation of the normal to \(y = x^2\) at the point \((2, 4).\)
Worked solution
\(\dfrac{dy}{dx} = 2x\), tangent gradient at \(x\) M1
\(=2\) is 4. A1
Normal gradient \(= -\tfrac14.\) M1
\(-\tfrac14.\) A1
Normal: \(y = -\tfrac14 x + 4.5.\) A1
Common mistakes
- Using the negative reciprocal for a tangent instead of a normal. The tangent's gradient is just \(f'(x)\) at the point - the perpendicular flip only applies when you're asked for the normal.
- Substituting into \(f'(x)\) instead of \(f(x)\) for the point. You need both the \(y\)-coordinate from the original function and the gradient from the derivative - mixing them up gives a line through the wrong point.
- Leaving the answer as a gradient instead of a full line equation. The question asks for the equation of the tangent or normal, so finish with \(y - y_1 = m(x - x_1)\) rearranged, not just the value of \(m\).
Ready to practise properly?
19 tangent and normal questions, marked instantly like the real exam.
Quick answers
How do you find the equation of a tangent to a curve?
Find the \(y\)-coordinate at the given \(x\)-value, differentiate to get the gradient function, substitute in the \(x\)-value to get the tangent's gradient, then use \(y - y_1 = m(x - x_1)\) with the point and gradient.
How do you find the equation of a normal to a curve?
Find the tangent's gradient at the point in the usual way, then take the negative reciprocal of it, since the normal is perpendicular to the tangent, and use \(y - y_1 = m(x - x_1)\) with that gradient and the point. See the parent Differentiation page's GDC guidance for graphing tangent/normal lines to check your answer.