(I'm assuming that you can get Sage running and evaluate the sum 1+1, say.) Sage comes with many graphs preinstalled. Thus the command

K = graphs.CompleteGraph(5); K
Complete graph: Graph on 5 vertices

sets $K$ equal to the complete graph on 5 vertices. Now produces a drawing of the graph in a separate window.   # not tested

The command

[0, 1, 2, 3, 4]

displays the vertices of our graph and

[(0, 1, None), (0, 2, None), (0, 3, None), (0, 4, None), 
(1, 2, None), (1, 3, None), (1, 4, None), 
(2, 3, None), (2, 4, None), (3, 4, None)]

displays the edges. To avoid the empty labels, try

[(0, 1), (0, 2), (0, 3), (0, 4), (1, 2), 
(1, 3), (1, 4), (2, 3), (2, 4), (3, 4)]

The command produces a list of the vertex degrees, while K.deg(u) gives the degree of the vertex $u$. Further K[i] returns the neighbors of $i$ in $K$. If your graph is regular (K.is_regular() returns True) then its valency is given by[0]

There are many built-in graphs in Sage. Thus:

C = graphs.CycleGraph(8); C
Cycle graph: Graph on 8 vertices

gives us the cycle on 8 vertices. To get the list of possible graphs, type graphs. at the prompt, followed by a tab. (That's 8 keypresses---6 letters, a period and a tab.)

There are many graph operations we can use. We can get the complement of our cycle by

CC = C.complement(); CC
complement(Cycle graph): Graph on 8 vertices

and its line graph by

LC = C.line_graph(); LC
Graph on 8 vertices

Of course a cycle is isomorphic to its line graph, which we can verify:


Another way to verify that $LC$ is the cycle $C_8$ is to verify that it is connected


and that it is regular of degree two
[2, 2, 2, 2, 2, 2, 2, 2]

Sage supplies the Petersen graph

P = graphs.PetersenGraph(); P
Petersen graph: Graph on 10 vertices

and we can verify this is correct by computing is diameter, girth and the number of vertices since these three properties characterize the graph.

P.diameter(), P.girth(), P.num_verts()
(2, 5, 10)   

In practice it is very important to check that any graph you construct is the one you wanted. It may be hard to prove that your graph is correct, but simple checks can still be useful. One common source of error is an incomplete understanding of the commands you use. For example

K2 = graphs.CompleteGraph(2)
K3 = graphs.CompleteGraph(3)
M = K2.union(K3)
Graph on 3 vertices

produces a graph on three vertices!

I should have used K2.disjoint_union(K3).

Complete graph disjoint_union Complete graph: Graph on 5 vertices