Question 201601
y= x^3 -15x^2 +12x +6
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y' = 3x^2 -30x +12
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to  find vertex,  set  slope(y') =0
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3x^2 -30x +12 =0
3(x^2 -10x +4) =0
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solve  using  quadratic  eqn,,,x= 9.58,,,.417
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to  find  y,,,substitute  in  original  eqn,,,,(9.58,-376.5),,,(.417,,8.47)
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since  this  is  a  positive  cubic,,rises  to  right,,
1st  vertex,,,(.417,8.47) is  concave  down,,,and
  second  vertex (9.58, -376.5)  is  concave  up
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checking,,,(.5,8.375) (.3,8.277),,,ok  cc down
(9.5,-376.4),(9.7, -376.3) ,,,,ok  cc  up