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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?

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  • $\begingroup$ The equation $x^b$ crosses the $x$-axis exactly once - at $x=0$. On a pocket graphing calculator it can appear like $x^b$ crosses infinitely many times near $x=0$ when $b$ is odd simply because the resolution of a pocket graphing calculator isn't good enough to show all the details of a graph $\endgroup$
    – Rob Bland
    Sep 14, 2016 at 0:47
  • $\begingroup$ Also, you may be remembering a mangled version of the fundamental theorem of algebra - a polynomial of degree $n$ always has $n$ complex roots (which come in conjugate pairs) and so a polynomial of odd degree is guaranteed to have at least one real root. $\endgroup$
    – Rob Bland
    Sep 14, 2016 at 0:49
  • $\begingroup$ Is “parent function” = “polynomial function”? $\endgroup$
    – Winther
    Sep 14, 2016 at 1:26
  • $\begingroup$ It would be easier to give a good answer if you had a plot of what you see, or at least a description of what you plot and what the axes are: e.g. "I plot $x^5$ from $x=-1$ to $x=1$ with the $y$-range being $[-2,2]$". $\endgroup$
    – Winther
    Sep 14, 2016 at 1:32

2 Answers 2

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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.

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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.

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