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

9444 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Can you please explain how to expand two brackets, eg. (3x-7)(5x+6)


Show that the cubic function f(x) = x^3 - 7x - 6 has a root x = -1 and hence factorise it fully.


How do we know the derivative of x^n


What is the normal distribution and how do I use it?


We're here to help

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

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences