Problem: Below is the graph of a polynomial with real coefficients
What can you say about the degree of the polynomial and about the sign of the first three and last three coefficients when written in the usual manner.
My attempt: let $P(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+ \cdots a_2x^2+a_1x+a_0$
Clearly the degree is even as the polynomial has even (8) number of roots.
Also $P(0)>0$ and $P'(0)=0$ and $P''(0) <0$ (as we can see that 0 is a locally maxima and also the tangent is above the graph at 0).
This is wrong as pointed out in the answer:Also, its obvious that sum of the roots of is positve (from the figure). $a_{n-1} >0$ as $a_n < 0$ since the polynomial has a global maxima and no global minima.
But I am unable to comment on the sign of $a_{n-2}$.
So, I would request someone to help me.
You're right about $a_0$, $a_1$, $a_2$.
It doesn't seem to be stated that all of the real zeroes are in the $x$-range shown on the graph, but if you do assume that, you can certainly also conclude that the degree is even, and that $a_n$ is negative.
You can say more about the degree than that, though. Clearly it must be at least $8$ in order to allow for all of the zeroes. But if you count stationary points, you can see it must be at least $12$ -- and if you count inflection points it must be at least $18$.
You can't say anything about $a_{n-1}$ and $a_{n-2}$, though, even with the assumption that you can see all of the zeroes.
For example, if the degree of the polynomial you see is $2n$, then adding a sufficiently small multiple of $-x^{2n+4}\pm x^{2n+3} \pm x^{2n+2}$ will not modify the graph visibly, and will not create any new zeroes -- so you cannot see whether this has already happened.
It is true that $-\frac{a_{n-1}}{a_n}$ is the sum of the roots, but you don't know what the sum of the roots is. All you can judge from the graph is the sum of the real roots; you can't see where there might be complex roots, whose real parts can easily dominate the sum of the roots.