Find the turning points of the curve y = x^3 +5x^2 -6x +4

y= x3 +5x2 -6+4

dy/dx = 3x2 +10-6

at turning points dy/dx = 0 therefore 

3x2 +10-6 = 0

This quadratic is factorisable. When factorised you get:

(3-2)(+4) = 0

therefore = 2/3 and -4 at the turning points

to find the y co-ordinates, substitue these values of x into the original equation of y= x^3 +5x^2 -6+4

y = (-4)3 +5(-4)2 -6(-4) +4 = 44

y = (2/3)3 +5(2/3)2 -6(2/3) +4 = 68/27

thw turning points of the curve are at the points (-4,44) and (2/3,68/27)

  

 

AB
Answered by Arshan B. Maths tutor

19429 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

find the coordinates of the turning points of the curve y = 2x^4-4x^3+3, and determine the nature of these points


How come x^2 = 25 has 2 solutions but x=root(25) only has one? Aren't they the same thing?


Find the indefinite integral tan(5x)tan(3x)tan(2x)


(A) express 4^x in terms of y given that 2^x = y. (B) solve 8(4^x ) – 9(2^x ) + 1 = 0


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