Finding a polynomial with a turning point at y=0 that cannot be solved analytically

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timhunderwood
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Hi I am working on a project and have been slightly side tracked. I want to examine a graph which has a turning point when y=0 (like y=x^2) however it most be insoluble analytically (i.e. a quintic or above).

Can someone give me an example of one or tell me a method to generate one.

I tried a method of saying : f(x) = 0 and f ' (x) = 0 for the same x but couldn't crack it.

help appreciated.

Thanks
 

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Well, don't use "x" if it a specific value. Since you are apparently tryng to make up a specific example. Since you want "a quintic or above", let's make it quintic for simplicity. Since you want it to be "insoluble analytically" you will also have to assume roots that are not rational numbers. What I recommend is that you start by ignoring the "turning point when y= 0" part and just pick some quintic that does not have rational roots. Then determine (possibly by graphing) where a turning point is and subtract that y value from your quintic.
 
for any graph to have a turning point on the x-axis means that the x-axis is a tangent to it...and so i believe that a perfect square/cube/etc should be in the function