Find the x and y coordinates of the turning points of the curve 'y = x^3 - 3x^2 +4'. Identify each turning point as either a maximum or a minimum.

The first part of the problem is solved by differentiating once and equating this to zero:
y = x^3 - 3x^2 +4 .dy/dx = 3x^2 - 6x .dy/dx = x(3x - 6) .
At the turning points;
x(3x - 6) = 0 (turning points occur where the gradient, dy/dx, equals zero) .
Hence, x = 0 or 2.
Inputting these x-values into the original equation yields the respective y-coridnates or the turning points. The locations are (0, 4) and (2, 0).
The nature of the turning points can be determined by finding the second derivative of the original equation:
d^2y/dx^2 = 6x - 6 .
At (0, 4), d^2y/dx^2 = -6 .At (2, 0), d^2y/dx^2 = 6 .
Therefore (0, 4) is a maximum and (2, 0) is a minimum (positive second derivative --> minimum, negative second derivative --> maximum).

RM
Answered by Robbie M. Maths tutor

12425 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

x^2 + y^2 + 10x + 2y - 4xy = 10. Find dy/dx in terms of x and y, fully simplifying your answer.


The curve C has a equation y=(2x-3)^5; point P (0.5,-32)lies on that curve. Work out the equation to the tangent to C at point P in the form of y=mx+c


Why is the derivative of sin(x), cos(x)?


What is the equation of the curve that has gradient dy/dx=(4x-5) and passes through the point (3,7)?


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