3
$\begingroup$

graph

Where does one begin? I can see that the zeros are -5, -3, 0, and 4? Is that correct so far?

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    I only see zeros at $-4,3,5$.2012-11-05
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    Nope, not at all. The zeros are $5,3$ and $-4$.2012-11-05
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    @copper.hat ah sorry, zeros are only where the line touches the x-axis correct?2012-11-05
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    Yes, zeros are the points where the function has value zero.2012-11-05

2 Answers 2

7

Let's use the fact that the graph has zeros at $5,3$ and $-4$. Given the window, we may as well assume these are the only zeros of the polynomial. Note also that we don't have any "flattening" near the zeros, so the zeros must be of multiplicity $1$. Hence, the polynomial is $$p(x)=a(x-5)(x-3)(x+4)$$ for some constant $a$. We can use the fact that $p(0)=3$ to solve for $a$, and then expand if we desire.

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    I tried inputting your answer, and it is not working sadly. Could we be overlooking something?2012-11-05
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    That works! You got the 1/20 by dividing the 3 into 60?2012-11-05
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    Well, my formula $p(x)=a(x-5)(x-3)(x+4)$ gives an infinite family of polynomials, so that isn't enough (on its own) to get the answer. Using the fact that we need $p(0)=3$, we have $$3=p(0)=a(0-5)(0-3)(0+4)=a\cdot(-5)\cdot(-3)\cdot(4)=60a,$$ so it follows that $a=\frac1{20}$, so our desired polynomial is $$p(x)=\frac1{20}(x-5)(x-3)(x+4).$$ Does that fit the graph you're looking for?2012-11-05
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    Precisely, Tyler! (Sorry for the deletion and repost of my comment. I ran out of time to edit it.)2012-11-05
1

Where does one begin? I can see that the zeros are -5, -3, 0, and 4? Is that correct so far?

No, that is not correct.

The zeros are points where $y = 0$, and it appears to me that none of the points you mentioned are actually zeroes of the function.

This graph has $3$ points where $y = 0$, so by the fundamental theorem of algebra, you know that the polynomial will have degree $3$ (or higher, since there may also be complex roots).

You can use the existence of local minimum and maximum points on the graph to construct a polynomial that will have the same extrema. It is a simple application of differential calculus.

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    Based on your inspection of the graph, where ***precisely*** is the local maximum? the local minimum?2012-11-05
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    Using my super glasses, I can see that the local extrema are at $\frac{1}{3}(4 \pm \sqrt{67})$.2012-11-05
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    I see your point. The solution was much more simple than I realized, and your answer is (obviously) correct.2012-11-05
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    Well, it isn't necessarily correct. It's just the best we can do with the information we can clearly discern.2012-11-05
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    Well, we can pick another point to convince ourselves. $p(-5)=-4$.2012-11-05
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    Fair enough, @copper.hat. Of course, this could be a polynomial of arbitrarily high degree chosen to fit all the (finitely many) points in the given window that we think are "obvious".2012-11-05
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    :-). I was using Occam's sunglasses...2012-11-05