thisgirl0709
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The curve y=x^3 +10x^2 -kx -2 has a stationary point at x=-8.
Find the x coordinates of the other stationary points
Oh don’t worry I found out how😂
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vbzl
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(Original post by thisgirl0709)
The curve y=x^3 +10x -kx -2 has a stationary point at x=-8.
Find the x coordinates of the other stationary points
What have you done so far?
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hexaflexagon
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Do you know how to differentiate? Differentiation is the rate of change on the graph. A stationary point is when the rate of change is 0. To find the stationary points, you diffrentiate the equation and equate it to 0. This gives the x coordinates of the other stationary points.
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(Original post by hexaflexagon)
Do you know how to differentiate? Differentiation is the rate of change on the graph. A stationary point is when the rate of change is 0. To find the stationary points, you diffrentiate the equation and equate it to 0. This gives the x coordinates of the other stationary points.
(Original post by vbzl)
What have you done so far?Name:  image.jpg
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ThisName:  image.jpg
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Size:  86.1 KBi misread the question but the answer is telling me x= 4/3
Last edited by thisgirl0709; 5 days ago
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(Original post by hexaflexagon)
Do you know how to differentiate? Differentiation is the rate of change on the graph. A stationary point is when the rate of change is 0. To find the stationary points, you diffrentiate the equation and equate it to 0. This gives the x coordinates of the other stationary points.
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vbzl
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(Original post by thisgirl0709)
This
Yes it is good. No need to find y coordinate of stationary point since it is not required. Once you find k, plug it in dy/dx and solve for x. So, your answer becomes x= 8
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hexaflexagon
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Looks correct to me. x = ±8
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