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

2792 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

How to express (4x)/(x^2-9)-2/(x+3)as a single fraction in its simplest form.


Use the formula 5p + 2q = t to find the value of q when p = 4 and t = 24. 6


Find the antiderivative of the function f(x)=(6^x)+1


Integrate sec^2(x)tan(X)dx


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