Gradients, Tangents and Normals

Gradients, Tangents and Normals

A LevelAQAEdexcelOCRAQA 2022OCR 2022

Gradients, Tangents and Normals

As mentioned in the Differentiation section, we can find a derivative to give the gradient of a graph at any given point.

From that information, we can create a tangent and a normal.

A Level AQA Edexcel OCR

How to Compare the Tangent and Normal

Let’s denote the gradient of the tangent m_T and gradient of the normal m_N.

For a pair of tangent and normal lines at one point, we have one rule:

The two must be perpendicular.

This means that we must have

m_T\cdot m_N = -1

A LevelAQAEdexcelOCR
A Level AQA Edexcel OCR

Example 1: Finding the Tangent

tangent is a straight line which touches our graph, but doesn’t pass through it at the meeting point. By definition, a straight line graph (i.e. y = x) cannot have a tangent – only a curved graph can have a tangent.

So, as an example, here’s the graph of \textcolor{blue}{y = (x - 1)^2 - 2}. Find the equation of the tangent line at x = \dfrac{3}{2}.

[4 marks]

To find the tangent, we first find the gradient at our point of interest.

Since we have

\textcolor{blue}{y = (x - 1)^2 - 2}

= \textcolor{blue}{x^2 - 2x - 1}

we have a gradient of

\dfrac{dy}{dx} = 2x - 2

at any point, and \dfrac{dy}{dx} = 1 when x = \dfrac{3}{2}.

Since our gradient is a straight line, it must have the general equation

y = mx + c

When x = \dfrac{3}{2}, y = -\dfrac{7}{4} and \dfrac{dy}{dx} = m_T = 1.

Then we can plug in these values to find c:

-\dfrac{7}{4} = \dfrac{3}{2} + c

c = -\dfrac{7}{4} - \dfrac{3}{2} = -\dfrac{13}{4}

Therefore, our tangent to y = (x - 1)^2 - 2 at x = \dfrac{3}{2} is given by the equation \textcolor{red}{y = x - \dfrac{13}{4}}.

A LevelAQAEdexcelOCR

Example 2: Finding the Normal

The normal is a straight line which is exactly perpendicular to the tangent.

Say we have the same graph from Example 1, but now we wish to find the normal at the point x = \dfrac{3}{2}.

[3 marks]

Since we already know that the gradient of the tangent, m_T = 1, we can conclude that the gradient of the normal, m_N = -1.

Given, also, that the point of interest is at \left( \dfrac{3}{2}, -\dfrac{7}{4} \right), we can form a straight line equation for the normal:

-\dfrac{7}{4} = \left(-1 \times \dfrac{3}{2} \right) + c

c = -\dfrac{7}{4} + \dfrac{3}{2} = -\dfrac{1}{4}

The equation of the normal is given by \textcolor{limegreen}{y = -x - \dfrac{1}{4}}.

A LevelAQAEdexcelOCR

Example Questions

f(x) = -x^2 gives f'(x) = -2x

When x = 2,

m_T = -4 and m_N = \dfrac{-1}{-4} = \dfrac{1}{4}

y = x^3 gives a gradient of

\dfrac{dy}{dx} = 3x^2

When x = 1,

\dfrac{dy}{dx} = 3 and y = 1


1 = (3 \times 1) + c

c = 1 - 3 = -2

The equation of the tangent at x = 1 is y = 3x - 2.

At x = -1, \dfrac{dy}{dx} = 3 \times (-1)^2 = 3.

Therefore, we can see that the gradients of the tangents at x = 1 and x = -1 are the same, so the two must run parallel.

When x = \dfrac{-1}{2}, y = \sqrt{\dfrac{7}{2}}.

We have m_T = \dfrac{1}{2\sqrt{\dfrac{7}{2}}} = \dfrac{1}{\sqrt{14}}. Then m_N = -2\sqrt{\dfrac{7}{2}} = -\sqrt{14}.

y = mx + c gives

\sqrt{\dfrac{7}{2}} = \left( -\sqrt{14} \times \dfrac{-1}{2} \right) + c


\sqrt{\dfrac{7}{2}} = \sqrt{\dfrac{14}{4}} + c


\sqrt{\dfrac{7}{2}} = \sqrt{\dfrac{7}{2}} + c


c = 0

So, the normal intercepts the y-axis at y = 0, or, more appropriately, the origin.

Additional Resources


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