In describing part of the geometry of the cone of curves for an algebraic surface $S$, we need to find $(-1)$ curves within $S$. Once we've done that, then we can say that the "negative" part of the cone of curves has as many extremal rays as $(-1)$ curves. Here I am using the cone theorem: $$\overline{\mathrm{NE}}(S)=\overline{\mathrm{NE}}(S)_{K_S\geq 0}+\sum_i \mathbb{R}^+[C_i]$$ where $C$ are such negative self-intersection rational curves.
However, is there an intuitive argument showing that if we have a curve of negative self-intersection, then such a curve is going to generate a extremal ray in the cone of curves?
What is known about this "positive" part $\overline{\mathrm{NE}}(S)_{K_S\geq 0}$ of the cone describe in the theorem?
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EDIT: We may assume that the Picard number is at least two, as otherwise the cone is simply a ray generated by any effective curve. In particular, every effective curve is extremal. I will also assume that "curve" means "effective curve". (This edit was prompted by Damiano's comment that is now (sadly) deleted. It was a useful contribution.)
A curve on a surface is simultaneously a curve and a divisor and assuming the surface is smooth or at least $\mathbb Q$-factorial, then the curve, as a divisor, induces a linear functional on $1$-cycles. This works better if the surface is proper, so let's assume that.
So, if $C$ is such a curve, then the corresponding linear function on the space where $NE(S)$ lives is best represented by the hyperplane on which it vanishes and remembering which side is positive and which one is negative.
If $C$ is reducible, then it may have negative self-intersection, but it is not extremal. For an example, blow up two separate points on a smooth surface and take the sum of the exceptional divisors. My guess is that you meant irreducible, so let's assume that.
Now we have $3$ cases:
$C^2>0$. In this case $C$ is in the interior of the cone and it cannot be extremal, can't even be on the boundary (Use Riemann-Roch to prove this).
$C^2=0$. Since $C$ is irreducible, it follows that it is nef and hence a limit of ample classes, so it is effective, but as Damiano pointed out I have already assumed that. (It is left to the reader to rephrase this if $C$ is assumed to be nef instead of effective). In this case the hyperplane corresponding to $C$ as a linear functional is a supporting hyperplane of the cone, intersecting it at least in the ray generated by $C$. So $C$ is definitely on the boundary, but it may or may not be extremal depending on the surface. For example any curve of self-intersection $0$ on an abelian surface is extremal, but for instance a member of a fibration that also has reducible fibers is not extremal despite being irreducible. For the latter think of a K3 surface with an elliptic fibration that has some $(-2)$-curves contained in some fibers.
$C^2<0$. If $C$ is effective, then $C\cdot D>0$ for any irreducible curve $D\neq C$. This means that $C$ and all other irreducible curves lie on different sides of the hyperplane corresponding to $C$ as a linear functional, so the convex cone they generate must have $C$ generating an extremal ray.
Observe that we did not use the Cone Theorem. In fact one gets a different "cone theorem" this way:
There is also one for $K3$'S, using the above notation:
For proofs and more details, see The cone of curves of a K3 surface. (There is also a newer version which is less detailed, but works in arbitrary characteristic. See here or here.)