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

Answer 1

\[x^3+ax^2+bx+c\]

Answer 2

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 :)