Find the tangent to y = x^2 - 4x + 9 at the point (3,15)

First find dy/dx:

dy/dx = 4x - 4

And thus at (3,15):

dy/dx = 12 - 4 = 8 = m (as m is the gradient of a curve)

So using y - y1 = m(x - x1) where (x1,y1) = (3,15):

y - 15 = 8(x - 3)

y = 8x- 9

Answered by Scott H. Maths tutor

2796 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Differentiate the following: y=sin(x^2+2)


Explain the Principle of Moments.


I struggle with modelling with differential equation, is there an easier way of interpreting this type of wordy question?


FInd the equation of the line tangent to the graph g(x)=integral form 1 to x of cos(x*pi/3)/t at the point x=1


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