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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 $-4,3,5$. – copper.hat Nov 5 '12 at 7:36
Nope, not at all. The zeros are $5,3$ and $-4$. – Cameron Buie Nov 5 '12 at 7:36
@copper.hat ah sorry, zeros are only where the line touches the x-axis correct? – Tyler Zika Nov 5 '12 at 7:37
Yes, zeros are the points where the function has value zero. – copper.hat Nov 5 '12 at 7:38
up vote 7 down vote accepted

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? – Tyler Zika Nov 5 '12 at 8:28
That works! You got the 1/20 by dividing the 3 into 60? – Tyler Zika Nov 5 '12 at 8:37
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? – Cameron Buie Nov 5 '12 at 8:38
Precisely, Tyler! (Sorry for the deletion and repost of my comment. I ran out of time to edit it.) – Cameron Buie Nov 5 '12 at 8:38

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? – Cameron Buie Nov 5 '12 at 7:43
Using my super glasses, I can see that the local extrema are at $\frac{1}{3}(4 \pm \sqrt{67})$. – copper.hat Nov 5 '12 at 7:45
I see your point. The solution was much more simple than I realized, and your answer is (obviously) correct. – Charles Boyd Nov 5 '12 at 7:45
Well, it isn't necessarily correct. It's just the best we can do with the information we can clearly discern. – Cameron Buie Nov 5 '12 at 7:47
Well, we can pick another point to convince ourselves. $p(-5)=-4$. – copper.hat Nov 5 '12 at 7:48

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