The gradient of a curve is defined as Dy/dx = 3x^2 + 3x and it passes through the point (0,0), what is the equation of the curve

Integrate this = (3x^3)/3 + (3x^2)/2 + c So y = x^3 + (3x^2)/2 + c Using point (0,0), 0 = 0 + 0 + c so c = 0. Full equation of the curve is therefore x^3 + (3x^2)/2

LT
Answered by Laura T. Maths tutor

6243 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

A circle has equation: (x - 2)^2 + (y - 2)^2 = 16. It intersects the y-axis (y > 0) at point P and the x-axis (x < 0) at point Q. Find the equation of the line connecting P and Q and of the line perpendicular to PQ passing through the circle's centre.


Differentiate 2cos(x)sin(x) with respect to x


Find the coordinate of the stationary point on the curve y = 2x^2 + 4x - 5.


The curve C has the equation ye ^(–2x) = 2x + y^2 . Find dy/dx in terms of x and y.


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