The gradient of a curve is given by dy/dx = 3 - x^2. The curve passes through the point (6,1). Find the equation of the curve.

Since we differentiate a function to find the gradient of a curve at any point, we need to reverse that to find the equation of the curve. We do this by integrating with respect to x:If you have a constant (a number without x), it becomes (constant)x. In this case, 3 becomes 3xThen, if you do have an x, you add one to the power and divide by the new power. So, here, -x^2 will become (-x^3)/3If you're given a point and told to find the equation of the curve, you have to find the constant, c. This is because when you a constant, it becomes zero. To do this, you substitute the coordinates into your integrated form: y = 3x - (x^3)/3 + c. This leads to 1 = 3(6) - (6^3)/3 + c. Solve for c and you'll get 55.So the equation of the curve is y = 3x - (x^3)/3 + 55.Never forget +c!!

DN
Answered by Darya N. Maths tutor

9993 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Find the equation of the tangent at x=1 for the curve y=(4x^2+1)^3


Simplify √32 + √18 giving your answer in the form of a√2.


Find the turning points of the equation y=4x^3-9x^2+6x?


How do I integrate and differentiate 1/(x^2)?


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:

© 2025 by IXL Learning