TITLE: Why are polynomials tangential to the $x$ axis at real double roots?
QUESTION [6 upvotes]: If $(x-a)^2$ is a root of a polynomial, then the graph will be tangent to the $x$ axis at $x=a$ but why? I know this is always the case for real double roots however I do not know the explanation for this behavior. Does this behavior apply to all real polynomial roots with even index multiplicities?

REPLY [7 votes]: First a comment
Saying that $(x-a)^2$ is a root of a polynomial $p(x)$ is an improper wording. You should say that $(x-a)^2$ divides $p(x)$ or that $a$ is a root of $p(x)$ of multiplicity at least equal to $2$.
Regarding your question
So you suppose that $p(x) = (x-a)^2q(x)$. Then taking the derivative you get
$$p^\prime(x) = 2(x-a)q(x) + (x-a)^2 q^\prime(x).$$
Therefore you have $p(a)=p^\prime(a)=0$. Proving that the graph of the function $x \ \mapsto p(x)$ is tangent to the $x$-axis at $x=a$.