Give the coordinates of the vertex for the graph of each equation

1. y = x² + 6x + 9

2. y = -x² + 9

x coord of vertex is always -b/2a
1. (-3,0)
2.(0,9)

We find the vertex by completing the square:

1. y = x² + 6x + 9
This is a perfect square: we can simplify it as is:
y = (x + 3)² + 0.
Now it is in vertex form:
y=a(x - p)² + q., where the vertex is at (p,q).

p is -3, and q is 0.
(remember that p is subtracted, so we are subtracting -3).

Therefore the vertex is at (-3,0).

2. y = -x² + 9
In this case, we realise that this is secretly also in vertex form.
Let me rewrite it as:

y = -(x + 0)² + 9

So (p,q) = (0,9).

Therefore, the vertex happens at (0,9).

If the parabola is in the form
y = a(x-h)^2 + k
Then the vertex is (h,k)

1. y = (x+3)^2 + 0
Vertex - (-3,0)

2. y = -x^2 + 9
y = -(x+0)^2 + 9
Vertex - (0,9)