Turning points of a cubic?

Find the derivative of your function.

Set it equal to zero.

Solve for X by factoring or using quadratic formula.

Plug each value of X into original function to find the Y- coordinate of each point.

dy/dx=3x^2-6x-9

Equate this to zero

or 3x^2-6x-9=0

or x^2-2x-3=0

(x+1)(x-3)=0

or x=3 or -1

so the other turning point is when x=-1

y=(-1)^3-3(1)+9(1)+7=12 co-ordinate is(-1,12)

d/dx(x^3 - 3 x^2 - 9 x + 7) = 0

x = 3

x = -1

What you've solved for above are the roots of this function (you've positioned the criteria the position the curve crosses the x-axis). What you're searching for are the turning factors, or the position the slop of the curve is an identical as 0. you want to make certain the spinoff of the equation: y' = 3x^2 + 10x + 4 then you definately want to sparkling up for zeroes utilising the quadratic equation, yielding x = -2.9, -0.5 Plug them again into your unique equation, and also you get the criteria (-2.9,6.a million) and (-0.5,-0.9)

f '(x)=3x^2-6x-9=0 --->3(x+1)(x-3)=0 --->x=...