The function f(x) = (x-2)²(x+3) has a zero at x=2. What is the behavior of the graph of f at this zero?

The factor $(x-2)^2$ has even multiplicity, 2, at the zero $x = 2$. When a zero has even multiplicity, the graph touches the $x$-axis and turns around, like the vertex of a parabola, without crossing. That is choice B. The other factor $(x+3)$ has multiplicity 1, so the graph crosses at $x = -3$. Choice A would be correct only for an odd multiplicity of 1. Choice C is wrong because a vertical asymptote needs a factor in the denominator. Choice D is wrong because $f(2) = 0$ is perfectly defined, since the function is a polynomial.