How many zeros does an odd degree function have? This is probably a stupid question, but it's kinda bugging me right now. I've noticed that on my graphical calculator that when I graph an odd degree parent function, the graph appears to be contiguous with the $x$-axis between the points $-1$ and $1$. Intuitively, I know this cannot be the case as the function can't have infinite zeros. To my knowledge --I may be wrong-- the graph $x^b$ should have $b-1$ zeros. So, can someone explain why a graph would have $b-1$ zeros (if that's the case); why does it appear as a contiguous line on my calculator (if the reason is other than scale); and finally and simply, how many zeros must and odd degree function have?
 A: A polynomial of degree $n$ will have up to $n$ distinct zeros. Note the word "distinct" - it is common for the roots of a polynomial to overlap. For example, $y = x^2$ has $0$ for both of its roots, therefore it only touches the x-axis once. By comparison, $y = x^2 - 1$ has two roots, $x = -1$ and $x = 1$. And $y = x^2 + 1$ has no (real) roots - its graph never touches the x-axis.
If you graph $y = x^b$ for a very large, odd $b$, then the graph gets very flat between $x = -1$ and $x = 1$, but the only time it actually touches the x-axis is at $x = 0$. If your graphing calculator makes it look like it's going back and forth a lot, that's just a result of either the way it draws the graph, or the way it does its calculations.
A: An odd degree polynomial has at least one (real) root and at most $n$ roots, where $n$ is the degree of the polynomial (i.e. the highest exponent of the variable). That is, if $r$ is the number of the roots of a polynomial function of odd degree $n$  then: $1\leq r\leq n$. (The "at least one real root" part, is a consequence of Bolzano's theorem, since for an odd degree polynomial function $f(x)$ we have: $\lim_{x\to\infty}f(x)=\infty$ and $\lim_{x\to-\infty}f(x)=-\infty$. The "at most $n$ real roots" part, is the fundamental theorem of algebra).
As an example: $x^3=0$ has a single real root ($x=0$) while $x^5-x^3=0$ has three real roots ($x=0,1,-1$). 
If this is not apparent on your screen, I propose you try rescaling the domain of your graph. 
