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.

Answered by Rafe L. Maths tutor

8926 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Solve (3x+6)/4 - (2x-6)/5 = (x+7)/8.


Express the following in partial fractions: (1+2x^2)/(3x-2)(x-1)^2


The equation of curve C is 3x^2 + xy + y^2 - 4x - 6y + 7 = 0. Use implicit differentiation to find dy/dx in terms of x and y.


Find the gradient at the point (0, ln 2) on the curve with equation e^2y = 5 − e^−x . [4]


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences