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# Turbine points watch

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1. Can someone please explain to me how you find the second turning point in a cubic?? I know how to find the first by using completing the square but the second I'm stuck on !!
2. (Original post by kyra.wilson)
Can someone please explain to me how you find the second turning point in a cubic?? I know how to find the first by using completing the square but the second I'm stuck on !!
I'm unclear how you would complete the square on a cubic. Can you show an example?

To find turning points you are looking for points on the graph with a gradient of 0
Differentiate the cubic and solve when dy/dx=0

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3. (Original post by gdunne42)
I'm unclear how you would complete the square on a cubic. Can you show an example?

To find turning points you are looking for points on the graph with a gradient of 0
Differentiate the cubic and solve when dy/dx=0

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Yep sorry I made a mistake in saying that! I might be making mistakes as when I check it with a graph plotter the results come out different so can you please explain the steps I need to do for 12a ?
4. (Original post by kyra.wilson)
Yep sorry I made a mistake in saying that! I might be making mistakes as when I check it with a graph plotter the results come out different so can you please explain the steps I need to do for 12a ?
To answer 12a the exact coordinates of the turning points don't matter. You just have to get the general shape of the curve right and find the points where it crosses or touches the x or y axes. You only need to get the turning points roughly right in the sketch.

If you were to expand the brackets you would see the maximum power of x is a cubic. Even without fully expanding you should be able to see whether its a positive or negative x^3 and you should know the general shape of each.

Since 12a is already factorised you can instantly see the x solutions for y=0. Whenever you have two identical factors, the curve touches the axis at that solution rather than crossing the axis. To find where it crosses the y axis you just need to evaluate the value when x=0.

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Updated: November 9, 2015
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