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The curve y=(1-x)(x^2+4x+k) has a stationary point when x=-3, find K help?

Help much appreciated!
Reply 1
Avatar for james22
james22
16
What have you tried?
Original post by james22
What have you tried?


I just don't know how to go about it...
Original post by Venus fly trap
Help much appreciated!


Do you know the properties of a stationary point?
Reply 4
Avatar for krisshP
krisshP
14
Try using the product rule+the fact that at a stationary point dy/dx=0
Original post by Khallil
Do you know the properties of a stationary point?


Not really we've just done it very briefly in class, and I'm really not sure about this question!
Reply 6
Avatar for Jonagorn
Jonagorn
A stationary point is a point on the graph where the derivative of the curve is 0. Therefore, differentiating the expanded equation will give you a quadratic equation which you can then sub in x = -3 to find K.
(edited 10 years ago)
Reply 7
Avatar for james22
james22
16
Original post by Venus fly trap
I just don't know how to go about it...


Stationary points occur when dy/dx=0, so find dy/dx and set it equal to 0. You have been told that x=-3 is a stationary point, so you should get dy/dx=0 when x=-3. Plug in this x value and you should get an equation for k.

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