Find the gradients of y = 3x^2 − (2/3) x + 1 at x = 0

y = 3x^2 − (2/3) x + 1Derivative of 3x^2 = 6x3 * 2 = 6x^(2-1) = x^1 = xDerivative of -(2/3)x = -2/3-(2/3) * 1 = -2/3x^(1-1) = x^0 = 1dy/dx = 6x - 2/3at x = 0dy/dx = 6(0) - 2/3dy/dx = -2/3

ZB
Answered by Zaid B. Maths tutor

3676 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Find the gradient of the tangent to the curve y=4x^2 - 7x at x = 2


Find the coordinates of the turning points of the curve y = 4/3 x^3 + 3x^2-4x+1


Integrate | x^7 (ln x)^2 dx ( | used in place of sigma throughout question)


Simplify (5-2√3)/(√3-1) giving your answer in the form p +q√3, where p and q are rational numbers


We're here to help

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

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning