Let $X = V(x_0^2 + x_1^2 + x_2^2 + x_3^2) \subseteq \mathbb{P}^3$ (a smooth quadric surface, $\cong \mathbb{P}^1 \times \mathbb{P}^1$). A general hyperplane $H = V(a_0 x_0 + \cdots + a_3 x_3)$ meets $X$ in a smooth curve of degree 2 in $H \cong \mathbb{P}^2$, i.e., a smooth conic $\cong \mathbb{P}^1$.

A special hyperplane like $H = V(x_0)$ gives $V(x_1^2 + x_2^2 + x_3^2, x_0)$, which is still a smooth conic. In fact, for this particular quadric, every hyperplane section is smooth (or empty) because $X$ itself is smooth and the dual variety has codimension $> 0$.

But if $X$ were a quadric cone $V(x_0^2 + x_1^2 - x_2^2)$ (singular at $[0:0:0:1]$), then $H = V(x_3)$ gives a smooth section, while $H = V(x_2 - x_0)$ passes through the vertex and gives a singular section.

A smooth curve $C \subseteq \mathbb{P}^n$ of degree $d$ meets a general hyperplane in $d$ distinct points (smooth = multiplicity 1 at each). Bertini says these $d$ points are all distinct (no tangencies with a general hyperplane).

For the twisted cubic $C \subseteq \mathbb{P}^3$ ($\deg C = 3$): a general plane meets $C$ in 3 distinct points.

The theorem generalizes: let $|D|$ be a linear system (family of divisors) on a smooth variety $X$. If $|D|$ is base-point free (no point lies in every member), then a general member of $|D|$ is smooth.

For $X = \mathbb{P}^2$ and $|D| = |\mathcal{O}(d)|$ (degree-$d$ curves), a general degree-$d$ curve in $\mathbb{P}^2$ is smooth. This is "obvious" for $d = 1$ (all lines are smooth) and non-trivial for higher $d$.

Consider the pencil (1-parameter family) of cubics $\{sF + tG = 0\}_{[s:t] \in \mathbb{P}^1}$ where $F$ and $G$ are two smooth cubics meeting in 9 points (by Bézout). By Bertini, a general member of the pencil is smooth. But finitely many members are singular — these correspond to special values of $[s:t]$.

For the pencil $\{s(y^2z - x^3 + xz^2) + t(y^2z - x^3 - xz^2) = 0\}$: a general member is a smooth elliptic curve, but special members degenerate (e.g., a nodal cubic, or a cuspidal cubic, or a reducible cubic).

Bertini is the algebraic side of the Lefschetz hyperplane theorem: for a smooth projective variety $X \subseteq \mathbb{P}^n$ of dimension $\geq 2$ and a general hyperplane section $Y = X \cap H$:

$\pi_i(Y) \xrightarrow{\sim} \pi_i(X) \quad \text{for } i < \dim X - 1.$

In particular, a smooth surface section of a smooth 3-fold has the same fundamental group. This is a powerful topological tool, and Bertini's smoothness is the algebraic prerequisite.

Many theorems in algebraic geometry are proved by induction on dimension, with Bertini providing the inductive step. For example:

Claim: A smooth hypersurface $X = V(F) \subseteq \mathbb{P}^n$ of degree $d \geq 2$ is connected (for $n \geq 2$).

Proof sketch: By induction on $n$. A general hyperplane section $X \cap H$ is smooth of dimension $\dim X - 1 \geq 1$ (by Bertini). If $n = 2$: $X$ is a smooth curve in $\mathbb{P}^2$, which is connected. For $n \geq 3$: $X \cap H$ is connected by induction, and Lefschetz says $\pi_0(X \cap H) \to \pi_0(X)$ is surjective, so $X$ is connected.