Can you please help me with this maths question

1. Consider the cubic curve y = (x-1)(x-2)(x+4)
a) Find the stationary points

Can you please help

stationary points → where the derivative is 0

y = (x - 1)(x - 2)(x + 4)

dy/dx = (x - 2)(x + 4) + (x - 1)(x + 4) + (x - 1)(x - 2) =
= x² + 2x - 8 + x² + 3x - 4 + x² - 3x + 2 =
= 3x² + 2x - 10

3x² + 2x - 10 = 0
Δ = 4 + 120 = 124; √Δ = 2√31

⇒ x1 = (-2 + 2√31)/6 = 1/3 (-1 + √31) --- stationary point #1
⇒ x2 = (-2 - 2√31)/6 = 1/3 (-1 - √31) ----- stationary point #2

I forgot how to work out stationary points or what ever its called.



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