Tangents and normalsEdexcel International A Level Maths: Revision notes

Section 1

Gradient of a tangent

The gradient of the tangent to the curve $y=f(x)$ at the point where $x=a$ is the value of the derivative there, $f'(a)$ . To find it, differentiate and substitute the $x$ -coordinate. Example: $y=x^2-3x+4$ at $P(3,4)$ . $\frac{\mathrm{d}y}{\mathrm{d}x}=2x-3$ , so the gradient at $P$ is $2(3)-3=3$ .

Key termstangent gradient

Common mistake

Substituting into the original equation instead of the derivative. That gives the $y$ -coordinate, not the gradient.

Section 2

Equation of a tangent

A straight line through $(x_1,y_1)$ with gradient $m$ has equation $y-y_1=m(x-x_1).$ For a tangent, $(x_1,y_1)$ is the point on the curve and $m=f'(x_1)$ . If only the $x$ -coordinate is given, find $y_1$ from the curve first. Example: at $P(3,4)$ with $m=3$ : $y-4=3(x-3)$ , so $y=3x-5$ .

Key termspoint-gradient form

Exam tip

Find all three things in order: the point, the gradient, then the equation. Rearrange to the form the question asks for, for example $y=mx+c$ or $ax+by+c=0$ .

Section 3

Normals

The normal at a point is the straight line through that point perpendicular to the tangent. If the tangent has gradient $m$ , the normal has gradient $-\frac{1}{m}$ , because the product of the gradients of perpendicular lines is $-1$ . Example: for $y=2x^2-\frac8x$ at $A(2,4)$ the tangent gradient is $4x+8x^{-2}=10$ , so the normal gradient is $-\frac1{10}$ and its equation is $y-4=-\frac1{10}(x-2)$ , i.e. $x+10y-42=0$ . It meets the $x$ -axis where $y=0$ , at $(42,0)$ .

Key termsnormal

Common mistake

Only changing the sign of the gradient ( $10\to-10$ ), or only taking the reciprocal ( $10\to\frac1{10}$ ). You need both: $-\frac{1}{10}$ .

Section 4

Finding points with a given gradient

To find where the tangent has a given gradient $k$ , solve $\frac{\mathrm{d}y}{\mathrm{d}x}=k$ for $x$ and then find $y$ from the curve. Tangents are parallel when their gradients are equal; a horizontal tangent has gradient $0$ . Example: for $y=x^3-6x$ the gradient at $x=2$ is $6$ . Solving $3x^2-6=6$ gives $x=\pm2$ , so the tangent at $(-2,4)$ is parallel to the tangent at $(2,-4)$ .

Key termsparallel tangents

Common mistake

Giving only the positive root of $x^2=4$ . Both $x=2$ and $x=-2$ can be valid; use the question to decide which point is wanted.

Section 5

Where a tangent meets the curve again

To find where a line meets a curve, set the two expressions for $y$ equal and solve. If the line is a tangent at $x=a$ , then $x=a$ is a repeated root, so $(x-a)^2$ is a factor. This lets you find the remaining root. Example: the tangent to $y=x^3-3x^2-x+7$ at $x=2$ is $y=-x+3$ . Equating: $x^3-3x^2+4=0$ , which factorises as $(x-2)^2(x+1)=0$ . So the tangent meets the curve again where $x=-1$ , at $(-1,4)$ .

Key termsrepeated root

Exam tip

Use the repeated root as a check: after equating, $(x-a)^2$ must divide your cubic exactly.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Tangents and normals

The curve $C$ has equation $y=x^2-3x+4$ . The point $P(3,4)$ lies on $C$ .
Find an equation of the tangent to $C$ at $P$ .2 marks
A curve has equation $y=x^3-6x$ .
The tangent to the curve at the point where $x=2$ is parallel to the tangent at another point $B$ . Find the coordinates of $B$ .2 marks
The curve $C$ has equation $y=2x^2-\dfrac{8}{x}$ for $x\neq0$ . The point $A(2,4)$ lies on $C$ .
Find an equation of the tangent to $C$ at $A$ , giving your answer in the form $y=mx+c$ .3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).