Tangents and Normals (AI SL)
Once you can differentiate a function, you can find the equation of the straight line that just touches its graph at a single point, the tangent, or the line perpendicular to that tangent, the normal. This page walks through both, with worked examples and the mistakes that cost marks. It's part of the wider Differentiation topic.
21 questions on this sub-topic.
Tangent and normal gradients
Covered under IB syllabus reference SL5.4. Neither line's equation is given in the formula booklet - both come from the point-gradient form of a straight line, using the derivative to supply the gradient.
Tangent
\(y - y_1 = f'(x_1)(x - x_1)\)
The gradient of the tangent at \(x=x_1\) is simply \(f'(x_1)\) - the value of the derivative at that point.
Normal
\(y - y_1 = -\dfrac{1}{f'(x_1)}(x - x_1)\)
The normal is perpendicular to the tangent, so its gradient is the negative reciprocal of \(f'(x_1)\).
Some calculators can draw the tangent or return \(f'(x_1)\) directly - see the parent topic's GDC guidance for the exact keystrokes.
Worked examples
Find the equation of the tangent to \(y = x^2 - 3x + 4\) at the point where \(x = 2.\)
Worked solution
At \(x = 2\), \(y = 4 - 6 + 4\) M1 \(= 2\), so \((2, 2).\) A1
\(\dfrac{dy}{dx} = 2x - 3\), so at \(x\) M1 \(= 2\) the gradient is \(1.\) A1
\(y - 2 = 1(x - 2) \Rightarrow y = x.\) A1
Find the equation of the normal to \(y = x^2\) at the point \((3, 9).\)
Worked solution
\(\dfrac{dy}{dx} = 2x\), so at \(x\) M1 \(= 3\) the tangent gradient is \(6.\) A1
Perpendicular, so gradient \(= -\dfrac{1}{6}.\) M1 A1
\(y - 9 = -\frac16(x - 3) \Rightarrow y = -\frac16 x + 9.5.\) A1
Common mistakes
- Using the function value instead of the derivative as the gradient. The gradient of the tangent at a point is \(f'(x_1)\), not \(f(x_1)\) - substitute into the derivative, not the original function.
- Forgetting to take the negative reciprocal for a normal. If the tangent gradient is \(m\), the normal gradient is \(-\dfrac{1}{m}\), not \(\dfrac{1}{m}\) and not \(-m\) - two separate operations, flip and negate.
- Substituting the \(x\)-coordinate into the wrong expression when writing the final line. Once the gradient and point are both found, the equation is \(y-y_1=m(x-x_1)\) using the original point \((x_1,y_1)\), not a point from an intermediate working line.
Ready to practise properly?
21 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 gradient, then use \(y-y_1=m(x-x_1)\) with that point and gradient.
How is the gradient of a normal related to the gradient of the tangent?
The normal is perpendicular to the tangent at that point, so its gradient is the negative reciprocal of the tangent gradient: if the tangent gradient is \(m\), the normal gradient is \(-\dfrac{1}{m}\).