Explorations with
y = (x - d)^2 - 2

Let's examine equations of the form:

y = (x - d)^2 - 2

For simplicity's sake when viewing the graphs, we shall restrict d so that

d is an element of the set of integers

but we maintain that there is no loss of generality in doing so. Thus, the simplest form of our equation is when d = 0, and results in the following graph.

Graph
y = x^2 - 2

We see that the graph is a parabola. Why is this? If we multiply the right side out, we obtain the equation:

y = x^2 + 2dx + d^2 - 2
Which is of the form:
y = ax^2 + bx + c
where
a = 1
b = 2d
c = d^2 - 2
since d is any integer (also holds for d any real number).

Let's observe some graphs to garner some intuition as to what changing d will do the graph. For a reference, I have included our original graph in each of the following.

When d = 1, we have

Graph
y = x^2 - 2 y = (x - 1)^2 - 2

When d = -1, we have

Graph
y = x^2 - 2 y = (x + 1)^2 - 2

Now let's try, d = 2 ...

Graph
y = x^2 - 2 y = (x - 2)^2 - 2

and d = -2

Graph
y = x^2 - 2 y = (x + 2)^2 - 2

Noticing a pattern? Let's look at two more graphs and then decide...
When d = 3, our graph is

Graph
y = x^2 - 2 y = (x - 3)^2 - 2

Then for d = -3, our graph becomes

Graph
y = x^2 - 2 y = (x + 3)^2 - 2

So were you able to decide? Does changing d effect the shape of the graph?
We can see that it does not. The parabola always opens upwards, and it's vertex always lies on the line y = -2. The only change in the graph is where the vertex is in relation to x. Notice that the vertex can be found at the coordinates (d, -2) for each graph. So changing d simply moves the graph right when d is positive, and left when d is negative.
The following graphs further illustrate this relationship.
For d = 7

Graph
y = x^2 - 2 y = (x - 7)^2 - 2

For d = -7

Graph
y = x^2 - 2 y = (x + 7)^2 - 2

For d = 10

Graph
y = x^2 - 2 y = (x - 10)^2 - 2

For d = -10

Graph
y = x^2 - 2 y = (x + 10)^2 - 2

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Write-up by Robert Childres.