# Why does an $n$th degree polynomial have at most $n-1$ turning points?

How can one explain that polynomial of degree $n$ can have up to $n-1$ turning points and $n$ intersections with the $x$-axis?

If it is easier to explain, why can't a cubic function have three or more turning points?

Usually, these two phenomenons are just given, but I couldn't find an explanation for such polynomial function behavior.

• So the title should be "at most" $n-1$ turning points. – megas Dec 25 '14 at 23:31

The number of turning points is the number of solutions to the equation $${dy\over dx}=0$$ and the derivative of a polynomial has one less power of $x$ than the polynomial itself. The number of intersections with the x axis is the number of roots of the polynomial. The statement that a polynomial of order $n$ has $n$ roots is called the fundamental theorem of algebra.
• I would say solutions $x_0\in\mathbb R$ to the equation $\frac{dy}{dx}\small{(x_0)} = 0$. Also the fact that a polynomial can have at most $n$ roots is just basic algebra. – GPerez Dec 26 '14 at 0:53
• To be specific, my first remark is to avoid confusion with differential equations, and the second is to say two things: you don't need to go as far as the fundamental theorem of algebra, and in fact it doesn't depend on being in $\mathbb C$ or even $\mathbb Q$, for that matter. – GPerez Dec 26 '14 at 1:18
A degree $n$ polynomial has at most $n$ roots (intersections with the $x$ axis). A "turning point" is a place where the derivative of the polynomial is zero (though not every place the derivative vanishes is a turning point), and since the derivative of a degree $n$ polynomial is a degree $n-1$ polynomial, there are at most $n-1$ turning points.
• It'd be more accurate/clear to say "The derivative of a polynomial is $0$ at a turning point" - as it's written now, it looks like "derivative is 0" and "turning point" are equivalent, rather than just the latter implying the former. – Milo Brandt Dec 25 '14 at 23:51