How to find the equation of a tangent to a curve at a specific point.

The first thing to remember is that the tangent to the curve at a given point has the same gradient as the curve at that point.

Let's consider the general case y=f(x) at a point c=(x1,y1).

Step 1:We have to differentiate to find dy/dx.

Step 2: Calculate dy/dx when x=x1. This is the gradient at the point c.

Step 3: Use the equation of a general straight line with gradient m at a point c 'y-y1=m(x-x1)', where m is the value of the gradient calculated in Step 2.

Here is an example. Find the equation of the tangent to the curve y=x^3-4x at the point (1,-3). 

In this example x1=1 and y1=-3

Step 1: dy/dx=3x^2-4

Step 2: When x=1, dy/dx =-1, so m=-1

Step 3: Use the equation y-y1=m(x-x1) to obtain y--3=-1(x-1)

We can then rearrange this to the nicer form of y=-x-2.

RL
Answered by Rafe L. Maths tutor

10784 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

f(x) = (4x + 1)/(x - 2). Find f'(x)


A curve C has equation y = (2 - x)(1 + x) + 3 . A line passes through the point (2, 3) and the point on C with x-coordinate 2 + h . Find the gradient of the line, giving your answer in its simplest form.


Maths C1 2017 1. Find INT{2x^(5) + 1/4x^(3) -5}


Does the equation x^2 + 2x + 5 = 0 have any real roots?


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning