Find the turning points of the curve y = x^3 +5x^2 -6x +4
y= x3 +5x2 -6x +4
dy/dx = 3x2 +10x -6
at turning points dy/dx = 0 therefore
3x2 +10x -6 = 0
This quadratic is factorisable. When factorised you get:
(3x -2)(x +4) = 0
therefore x = 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 -6x +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
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