# Why are polynomials tangential to the $x$ axis at real double roots?

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?

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 $$2$$.

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 $$x=a$$.

You can prove this easily using calculus, but I'll give an even more elementary argument.

The graph of $$f(x)=(x-a)^{2}$$ is an upward facing parabola. In particular, it is the graph of $$f(x)=x^{2}$$ shifted to the right by $$a$$ units (if a is positive).

Note that, being a square, $$(x-a)^{2}$$ is always greater than or equal to $$0$$. And it is only equal to $$0$$ at $$x=a$$.

Combining the above two observations, the graph of $$f(x)$$ is tangent to the $$x$$-axis at $$x=a$$.

• The OP is asking about any polynomial that has $(x-a)^2$ as a factor, not just $(x-a)^2$. – amd Jan 16 '19 at 18:24