How Do You Find the Equation of a Tangent or Normal Line at a Point?

Find the point on the curve, evaluate the derivative there, and substitute the resulting gradient into point-gradient form. For a normal line, use the negative reciprocal of the tangent gradient.

This is SL 5.4: Tangents and normals. The explanation assumes an SL Paper 1 short-response question, although the same method applies on Paper 2 and to more complicated functions using a GDC.

Method

For a curve $y=f(x)$ at $x=a$:

1. Find the point of contact: ((a,f(a))).
2. Differentiate to obtain the gradient function (f'(x)).
3. Calculate the tangent gradient: $m_{\mathrm{tangent}}=f'(a)$.
4. For the normal, calculate $m_{\mathrm{normal}}=-\frac{1}{m_{\mathrm{tangent}}}$.
5. Substitute the point and gradient into $y-y_1=m(x-x_1)$.
6. Simplify only if the question requires another form, such as $y=mx+c$.

Special cases matter. If $f'(a)=0$, the tangent is horizontal with equation $y=f(a)$, while the normal is the vertical line $x=a$. If the tangent is vertical, the normal is horizontal. In these cases, do not try to calculate a negative reciprocal of zero or an undefined gradient; state the line directly from its geometric orientation.

Worked Example

Find the tangent and normal to $f(x)=x^2+1$ at $x=2$.

First, $f(2)=5$, so the point of contact is $(2,5)$. Since $f'(x)=2x$, the tangent gradient is .

Therefore, the tangent is

$$y-5=4(x-2),$$

so $y=4x-3$.

The normal gradient is $m_n=-\frac{1}{4}$. Thus, the normal is

$$y-5=-\frac{1}{4}(x-2),$$

or $y=-\frac{1}{4}x+\frac{11}{2}$.

A common misconception is using $-m_t$ as the normal gradient. This is incorrect: perpendicular gradients satisfy $m_tm_n=-1$, so the normal gradient must be the . Also, do not use as the gradient; it is the point's -coordinate.

Exam Technique

Show the point, derivative, evaluated gradient, and line equation. For a complicated function, a GDC may evaluate (f'(a)), but record the derivative value or calculator setup instead of giving only the final equation. Keep full precision until the final answer and check that both lines pass through the stated point.