maths question -- - help

• valentinoo

  15 Jan 07, 21:18

  Find the co-ordinates of the stationary points on the curve y = x^3 - 3x^2 + 4
  
  Anyone remember?
  
• kenn3th

  15 Jan 07, 21:21

  do the cross method?
  
  this one is what level de maths?
• snop_coolz

  15 Jan 07, 21:24

  omg...dy/dx = 0 lor...
  
• omgukilledkenny

  15 Jan 07, 21:25

  differentiate that equation,
  then put dy/dx = 0
  u can find the x coordinates already.
  sub the x coords into the y-equation to get the y coords
• CoolMyth

  15 Jan 07, 21:25

  Originally posted by valentinoo:

  Find the co-ordinates of the stationary points on the curve y = x^3 - 3x^2 + 4
  
  Anyone remember?
  

  Find max, min and points of inflection.
• snop_coolz

  15 Jan 07, 21:25

  hey..i faster
  
  3X^2-6X =0
• valentinoo

  15 Jan 07, 21:25

  dy/dx = 3x^2 - 6x
  dy/dx = 0 for stationary points
  
  3x^2 = 6x
  3x=6
  x= 2
  
  f(2) = 8 - 12 + 4 = 0
  (2, 0)
  
  correct or not
• snop_coolz

  15 Jan 07, 21:27

  yup.. correct
• CoolMyth

  15 Jan 07, 21:30

  Where's your point (0, 4)?
• omgukilledkenny

  15 Jan 07, 21:34

  there should be more than one stationary pts
• snowiesnow

  15 Jan 07, 21:36

  simultaneous equation also can ma
  
• HyuugaNeji

  15 Jan 07, 21:40

  Originally posted by valentinoo:

  dy/dx = 3x^2 - 6x
  dy/dx = 0 for stationary points
  
  3x^2 = 6x
  3x=6
  x= 2
  
  f(2) = 8 - 12 + 4 = 0
  (2, 0)
  
  correct or not

  only half correct.
  
  Is 3x^2-6x=0
  so x(3x-6)=0
  x=0 or x=2
  
  Then look for y.
• valentinoo

  15 Jan 07, 21:45

  Originally posted by CoolMyth:

  Where's your point (0, 4)?

  the other point is ths one
• snop_coolz

  15 Jan 07, 21:47

  oh ya..x^3 graph..