How do you identify the important parts of #y=x^2-1# to graph it?

The "important parts" in terms of graphing would be the x-intercepts and the vertex. From there you can just plug and chug to identify other points on the graph.

To find the x-intercepts, you set #y = 0# and solve for your #x#s:
#0 = x^2-1#
#1 = x^2# (Adding 1 to both sides)
#x = +-1# (Taking square roots)

Thus, the x-intercepts occur at #x = 1# and #x = -1#.

The vertex is the "beginning" point of a quadratic like this one. In other words, it's where the two curves meet on a parabola (not a very good definition, but it goes). To find the x-coordinate of the vertex, we use the formula #x = -b/(2a)#, where #a# and #b# are the numbers in #ax^2+bx+c# (which is the form of quadratic presented in this problem). In our equation #x^2-1#, #a = 1# and #b = 0# (since there is no middle term, we have #b = 0#). Thus, #x = -0/(2(1)) = 0#. This is the x-coordinate of the vertex; to find the y, we simple plug in: #y = x^2-1 = (0)^2-1 = -1#. The coordinates of the vertex are #(0,-1)#.

We can see everything above on the graph below. The vertex, #(0,-1)#, is the bottom of the graph; you'll want to start there and work your way up. Next are the intercepts at -1 and 1 on the x-axis; plot those points, and connect them (vertex, and both intercepts) together with a nice curve. Finally, choose x-values like -3, -2, 2, and 3 and find which y-values they produce:
#y = (-3)^2-1 = 9-1 = 8#
#y = (-2)^2-1 = 4-1 = 3#
#y = (2)^2-1 = 4-1 = 3#
#y = (3)^2-1 = 9-1 = 8#
And just like that, we've identified 4 more points to plot:
#(-3,8)#
#(-2,3)#
#(2,3)#
#(3,8)#

Finally finally, to complete the rough sketch of the graph, connect all the points together - the "special" ones (vertex and intercept(s)) and the ones we found by plugging and chugging.

graph{x^2-1 [-10, 10, -5, 5]}