What is the equation of the tangent to the curve y=x^3+3x^2+2 when x=2

dy/dx=3x^2+6xx=2m=3(2)^2+6(2)=24at x=2 y=22(2,22)y-22=24(x-2)y-22=24x-48y=24x-26

Answered by Kieran H. Maths tutor

3156 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Find ∫(2x^5 -4x^-3 +5) dx


Differentiate y=(4x - 5)^5 by using the chain rule.


Intergrate 8x^3 + 6x^(1/2) -5 with respect to x


Differentiate y = x^3 + 2x^2 + 4x + 7


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 2013–2024

Terms & Conditions|Privacy Policy
Cookie Preferences