Tangents and Normals (AA SL)
Once you can differentiate a function, the derivative unlocks the gradient of any tangent line at any point on the curve - and from there, the normal line too. This page covers the single method that answers almost every tangent-and-normal question: differentiate, evaluate, substitute. It's part of the broader Differentiation topic.
23 questions on this sub-topic.
Finding the line
Covered under IB syllabus reference SL5.4: tangents and normals at a given point, and their equations, found using both analytic methods and technology.
Tangent line
\(y-y_1=m(x-x_1)\), where \(m=f'(x_1)\) is the derivative evaluated at the point.
Differentiate first, then plug the x-value into the derivative - not into the original function.
Normal line
\(y-y_1=-\dfrac{1}{m}(x-x_1)\), using the negative reciprocal of the tangent gradient.
The normal is perpendicular to the tangent at the same point, so its gradient flips sign and inverts.
Need the full differentiation syllabus and a GDC walkthrough for graphing tangent lines? See Differentiation.
Worked examples
Find the equation of the tangent to \(y = x^2 - 4x\) at \(x = 3\).
Worked solution
The tangent touches the curve at \(x=3\), so we need that \(y\)-value: \(y=3^2-4(3)=-3\), giving \((3,-3)\). A1
The gradient of the tangent equals the derivative at that point. \(\dfrac{dy}{dx}=2x-4\) M1 ; at \(x=3\), \(m=2(3)-4=2\). A1
Using \(y-y_1=m(x-x_1)\): \(y+3=2(x-3)\;\Rightarrow\;y=2x-9.\) M1 A1
Consider \(y = x^2 + 1\) at the point where \(x = 2\).
(a) Find the gradient of the tangent.
(b) Find the equation of the normal.
Worked solution
(a) \(\dfrac{dy}{dx}=2x\) M1
Step 2 - Evaluate at x = 2.
At \(x=2\), gradient \(=4.\) A1
(b) \(y(2)=2^2+1=5\), so the point is \((2,5).\) A1
Step 2 - Gradient of the normal.
The normal is perpendicular to the tangent, so its gradient is the negative reciprocal \(-\tfrac14.\) M1
Step 3 - Form the equation.
\(y-5=-\tfrac14(x-2)\) M1
Step 4 - Simplify.
\(y=-\tfrac14 x+\tfrac{11}{2}.\) A1
Common mistakes
- Substituting into \(f(x)\) instead of \(f'(x)\) for the gradient. The gradient of the tangent is the value of the derivative at that point, not the value of the original function - mixing the two gives a nonsense line.
- Forgetting to find the \(y\)-coordinate. The point-gradient form \(y-y_1=m(x-x_1)\) needs a full point \((x_1,y_1)\); skipping the substitution back into the original function leaves the equation incomplete.
- Using \(m\) instead of \(-1/m\) for the normal. The normal is perpendicular to the tangent, so its gradient is the negative reciprocal - reusing the tangent's own gradient by mistake is a very common slip.
Ready to practise properly?
23 tangent-and-normal questions, marked instantly like the real exam.
Quick answers
How do I find the equation of a tangent line?
Differentiate to get \(f'(x)\), evaluate it at the given \(x\)-value to get the gradient \(m\), find the \(y\)-coordinate at that point, then substitute both into \(y-y_1=m(x-x_1)\).
How is the normal different from the tangent?
The normal is perpendicular to the tangent at the same point, so its gradient is the negative reciprocal of the tangent gradient, \(-1/m\).