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# Stationary points Watch

1. I've been doing some reviison for a test tomorrow and have come across this question. Show that the stationary point on has coordinates (-b/2a, 4ac-b^2/4a) the cooriantes are -b all over 2a and 4ac- b^2 all over 4a

I think its the algebraic brackets that are confusing me
2. Differentiate it
When dy/dx = 0 there is no gradient therefore it is a stationary point
so put 0 = (what you get when you differentiate)
and solve as you would a simultaneous equation

You should get x = -b/2a and y=4ac-b^2/4a
3. (Original post by greeneyedgirl)
Differentiate it
When dy/dx = 0 there is no gradient therefore it is a stationary point
so put 0 = (what you get when you differentiate)
and solve as you would a simultaneous equation

You should get x = -b/2a and y=4ac-b^2/4a
I did try this origannly i differentiated it to get 2ax +bx=0 but i dont see how if i take the second dirrivative it would lead to the asnwer
4. (Original post by hazbaz)
I did try this origannly i differentiated it to get 2ax +bx=0 but i dont see how if i take the second dirrivative it would lead to the asnwer
You don't want to take the second derivative!
d^2y/dx^2 is only used for finding whether the stationary point is a max or a min, not for determining the actual stationary point.
5. (Original post by hazbaz)
I did try this origannly i differentiated it to get 2ax +bx=0 but i dont see how if i take the second dirrivative it would lead to the asnwer
I think you should check that part again.
6. (Original post by greeneyedgirl)
You don't want to take the second derivative!
d^2y/dx^2 is only used for finding whether the stationary point is a max or a min, not for determining the actual stationary point.
Opps silly me
7. (Original post by dknt)
I think you should check that part again.
2ax +b ah right yh it makes sense now thanks

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