Question

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

Answer 1

In summary, find the vertex and x-intercepts, and plug and chug for additional points. Finally, connect 'em all together with a neat curve.

Explanation:

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]}