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

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