maths question -- - help

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

Anyone remember?

do the cross method?

this one is what level de maths?

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

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

Find max, min and points of inflection. Laughing

hey..i faster Laughing

3X^2-6X =0

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

yup.. Laughing correct

Where's your point (0, 4)? Laughing

there should be more than one stationary pts

simultaneous equation also can ma

only half correct.

Is 3x^2-6x=0
so x(3x-6)=0
x=0 or x=2

Then look for y.

the other point is ths one Laughing

oh ya..x^3 graph.. Laughing