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

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#1
Help much appreciated!
0
5 years ago
#2
What have you tried?
0
#3
(Original post by james22)
What have you tried?
I just don't know how to go about it...
0
5 years ago
#4
(Original post by Venus fly trap)
Help much appreciated!
Do you know the properties of a stationary point?
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5 years ago
#5
Try using the product rule+the fact that at a stationary point dy/dx=0
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#6
(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!
0
5 years ago
#7
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.
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5 years ago
#8
(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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