Let $\mathbb{A}^n = \mathbb{A}^n_{\mathbb{C}}$ be n-affine over the field of complex numbers.
Let $f \in \mathbb{C}[x_1, \cdots, x_n]$ be an irreducible, non-constant polynomial, and let $X = \mathbb{V}(f)$ be the vanishing set of $f$.
For a point $p = (p_1, \cdots, p_n)\in X$, define the tangent space $$T_pX := \{q = (q_1, \cdots, q_n) \in \mathbb{A}^n \ |\ \sum_{i=1}^n\frac{\partial f}{\partial x_i}|_p (q_i - p_i) = 0\}.$$
A point $p \in X$ is singular if $\frac{\partial f}{\partial x_i}|_p = 0$ for all $1 \leq i \leq n$, and non-singular otherwise.
Why is it that if $p\in X$ is non-singular, then $T_pX \simeq \mathbb{A}^{n-1}$?
I assume you are not using Schemes. If $p$ is non-singular then the matrix $(\frac{\partial f}{\partial x_1} ~\dots~ \frac{\partial f}{\partial x_n})$ has full rank, so its kernel (which is exactly $T_p X - p$ is a $n-1$ dimensional subspace of $\mathbb{C}^n = \mathbb{A}^n$ (and therefore isomorphic to $\mathbb{A}^{n-1}$).