In class we stated and proved the implicit function theorem in the case where we have an open set $A \subset R^2$ a function $f:A \rightarrow R, \ f \in C^1_A$ and a point $ (x_0, y_0) \in A$ s.t. $f(x_0,y_0) = 0$ and $\partial f (x_0, y_0) / \partial y \ne 0$.
(by $f \in C^1_A$ I mean that $f$ is continuous on $A$ and also all it's partial derivatives exist and are continuous on $A$.)
Then we have stated, but not proven, the implicit function theorem for functions of more variables:
Given an open set $A \subset R^n$, and given $V$ a $q$-manifold of class $C^k$ where $k > 1$ and, $1 \le q \le n$,then $V$ is locally the graph of a function $\phi : I \rightarrow R^{n-q}$ where $I = [a_1, b_1] \times \dots \times [a_q, b_q] \subset R^q$.
I don't understand this statement and how it generalizes the previous one. Could somebody help me understand this generalization? as a start what do we mean with "$V$ is locally the graph of a function"? What is meant with graph of a function?
I can see your confusion Monolite. The second statement about a graph does not directly generalize from the former statement, but must be proven to be equivalent.
The direct generalization you are looking for is here: http://mathworld.wolfram.com/ImplicitFunctionTheorem.html
To get from the direct generalization, to the whole idea about "graph of a function", you should first show that the solution to the implicit function gives you a sufficiently differentiable manifold, then show that that manifold may be represented as a graph of a function. In my opinion that's the clearest way to see what is going on.