given x(cube)+ax (squared)+bx+c with turning points at (0;4) and (2;0)

prove tht a=-3;b=0 and c =4

Question

given x(cube)+ax (squared)+bx+c with turning points at (0;4) and (2;0)                                                                                                                             

prove tht a=-3;b=0 and c =4

anonymous · Answer

$$x^3+ax^2+bx+c$$

callisto · Answer

Turning points => dy/dx =0
let y = x^3 + ax^2 +bx +c  -(1)
So, differentiate the equation.
y' = 3x^2 +2ax + b 
Put y'=0
 3x^2 +2ax + b =0  -(2)
Put x=0 and x=2 into (2)
3(0)^2 +2a(0) + b =0  =>  b =0
3(2)^2 +2a(2) + b =0  =>  12 + 4a +b =0   => 4a = -12  => a =-3
Now, the equation becomes y = x^3 -3x^2  +c
Put (0;4) into y  = x^3 -3x^2  +c
4  = (0)^3 -3(0)^2  +c
i bet you can get c :)