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This item asks that you write equations for several polynomial functions of odd degree in graph each function.
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And then upon doing so, they'd like you to answer the question, is it possible for the graph to have no real zeros? explain.
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And then they'll like you to repeat doing the same procedure for polynomial functions of even degree.
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So let's take a look.
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Let's start with the odd case, odd functions, and let's just write three odd functions.
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So we can do f of x equals 4x minus 7.
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Let's do 6x to the 5th plus x minus 1, and we can do x cubed plus 4.
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So we'll look at those three, and let's see if we find any evidence of it being possible for the graph to have no real zeros.
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And then we'll talk about y.
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All right, so we've got 4x minus 7 first.
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Okay, there's our linear, but it is an odd function.
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We see that we do have an x intercept.
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The next one we have is negative 6x to the fifth.
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Plus x minus 1.
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Again, we've got one, real zero.
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We didn't get three, but we've got one.
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And this was a fifth degree.
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So we had the potential to have five x intercepts.
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And then finally, what was the other one i did? i did, y equals x cubed plus four.
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Again, we have one.
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So it's interesting that for of these for none of these did we get three we didn't get five for the second one for all of them we got one and let's just note that the reason for that is so we had a degree one we had a degree five and we had a degree three and you know well through working this unit that we could have used descartes rule of signs to check how possible positive negative or complex imaginary roots and here the possible number of zeros or solutions was one five and three you know it matches up with the degree and then the zeros they always differ by twos like as far as whether they're positive or negative because we always have those imaginaries lurking out here so here our positive negative possible zero was one.
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We would have never gotten an imaginary here.
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So this one, positive, negative, zero, we had certainty on that one.
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In the second case, we could have had five or three or one, but again, we at least had one.
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In that case, if we had had five, three, or one, we would have had zero, two, or four on the imaginaries.
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And then finally, in this case, here we go again, we've got three or one...