Year 11: Tangents and normals

Tangents and normals turn derivative ideas into straight-line equations. In Year 11, you are often asked to find the gradient of the tangent at a point, write the equation of that tangent, and then find the equation of the normal line, which is perpendicular to the tangent.

The derivative gives the gradient of the tangent. The normal is perpendicular to the tangent, so its gradient is the negative reciprocal, provided the tangent gradient is not zero.

What to master

Finding the derivative and evaluating it at a point.
Using point-slope form to write the tangent equation.
Finding the normal gradient from the tangent gradient.
Interpreting tangent information on graphs and in worded questions.

The tangent line

A tangent touches the curve at a point and has the same instantaneous gradient there. If $y = f (x)$ and the point is $(a, f (a))$ , then the tangent gradient is $f' (a)$ .

Once the gradient and point are known, use point-slope form: $y - y_{1} = m (x - x_{1})$ .

Example 1: Find the tangent to

y = x^{2} + 3 x

x = 1

Differentiate: $d y / d x = 2 x + 3$ .
Substitute $x = 1$ : $m = 2 (1) + 3 = 5$ .
Find the point on the curve: $y = 1^{2} + 3 (1) = 4$ , so the point is $(1, 4)$ .
Use point-slope form: $y - 4 = 5 (x - 1)$ .
Simplify: $y = 5 x - 1$ .

Answer: the tangent is

y = 5 x - 1

The normal line

The normal is perpendicular to the tangent. If the tangent gradient is $m$ , then the normal gradient is $- 1 / m$ .

Example 2: Find the normal to the same curve at

(1, 4)

The tangent gradient is $5$ .
The normal gradient is $- 1 / 5$ .
Use point-slope form: $y - 4 = (- 1 / 5) (x - 1)$ .
If needed, expand: $y = (- 1 / 5) x + 21 / 5$ .

Answer: the normal is

y - 4 = (- 1 / 5) (x - 1)

Horizontal and vertical special cases

If the tangent gradient is zero, the tangent is horizontal. The normal is then vertical. In those cases, the normal cannot be written in the usual $y = m x + b$ form.

Example 3: Find the tangent and normal to

y = x^{2}

x = 0

Differentiate: $d y / d x = 2 x$ .
At $x = 0$ , the gradient is $0$ .
The point is $(0, 0)$ .
Tangent: horizontal line $y = 0$ .
Normal: vertical line $x = 0$ .

Interpreting a tangent on a graph

Not every question starts with an equation. Sometimes you are shown a graph and asked what the tangent means. A positive tangent gradient means the function is increasing at that point. A negative tangent gradient means it is decreasing. A zero gradient indicates a stationary point or flat turning point candidate.

Example 4: A tangent drawn to a curve slopes down from left to right.

That means the derivative at the point is negative, so the function is decreasing there.

Common mistakes

Using the tangent point incorrectly and substituting the wrong coordinates.
Forgetting that the normal gradient is the negative reciprocal, not just the negative.
Trying to write a vertical normal in slope-intercept form.
Differentiating correctly but evaluating at the wrong x-value.

Study routine

A reliable method is: differentiate, substitute to get the tangent gradient, find the point on the curve, write the tangent equation, then form the normal from the negative reciprocal. Keeping those steps separate reduces sign errors and makes your working easier to check.