The function f(x) = (x-1)³(x+4) has a zero at x=1. What is the behavior of the graph of f at this zero?

The factor $(x - 1)^3$ has odd multiplicity, 3, at the zero $x = 1$. Zeros of odd multiplicity are crossings: the graph passes through the $x$-axis, flattening out near the zero because of the multiplicity 3. That makes A correct. Choice B describes even multiplicity, as in $(x-1)^2$, where the graph touches and turns back. Choice C is wrong, because a vertical asymptote needs a factor in the denominator and this function is a polynomial. Choice D is wrong because $f(1) = 0$ is defined; polynomials are defined for every real number.