is this constant? like..
On
If by "starts", you mean the behaviour in a $(-\infty, -M)$ where $M$ some large positive real number, then yes.
On
As already noted, the answer is "yes". Here is one way to see it: factor out the leading term, so the polynomial has the form $P(x) = a x^n( 1 + \text{ expression in powers of } \frac{1}{x}).$
If $x$ is very positive or very negative, i.e. if $|x| \gg 0$, then the expression in powers of $\frac{1}{x}$ will be negligible, and so $P(x) \sim a x^n$ for $| x | \gg 0.$
So now just look at the behaviour of $a x^n$. If $n$ is odd this is very negative when $x$ is very negative, becoming more so as $x\to -\infty$, while it is very positive when $x$ is very positive, becoming more so as $x \to \infty$. If $n$ is even then it is very positive when $x$ is either very negative or very positive, and increases both as $x \to -\infty$ and as $x \to \infty$.
Edit: As Didier Piau notices, this argument with crude asymptotics is not precise enough to conclude true monotonicity for $|x| \gg 0$; Moron's argument with derivatives is better for that. However, it does give an explanation for the rough behaviour of $P(x)$ for $|x| \gg 0$.
Assuming you mean if we start drawing it from $-\infty$ and proceed towards $+\infty$, then yes that is correct.
This is because $a_m x^m$ is the dominating term in the polynomial $P(x) = \sum_{k=0}^{m} a_k x^k$.
Now for even degree $Q(x)$ (with positive leading coefficient), it's derivative $P(x) = Q'(x)$ is of odd degree and so is negative as we start out from $-\infty$, and thus $Q(x)$ is decreasing.
For odd degree $Q(x)$, the derivative has even degree and so is positive near $-\infty$ (and also $+\infty$) and is thus increasing.