Hypersurfaces have no embedded points (Vakil 5.5.I)

Let $R:=k[x_1,\ldots,x_n]$ just for ease of notation, let $\pi:R\to R/(f)$ be the projection and let $P\subseteq R/(f)$ an associated prime ideal, i.e. there is some $g\in R$ such that $P=\operatorname{ann}(g)=\{ h\in R/(f) \mid gh=0 \}$. Then, $Q:=\pi^{-1}(P)=\{ h\in R \mid gh\in (f)\}$, so you know that $g$ and $f$ have a (largest) common divisor $d$, say $f=ad$ and $g=bd$. Then, $bdh=gh\in(f)$ is equivalent to $a$ being a factor of $h$, so $Q=(a)\supseteq(f)$. Since $Q$ is a minimal prime ideal over $(f)$, the ideal $P=(\pi(a))$ is a minimal prime ideal of $R/(f)$ - hence a generic point.