Graph the following parabola (x-2)^2=(y+1) list the vertex focus and directrix

Rearranging, we have:

y=(x+2)^{2}-1

That means we have the form y=a(x-h)^{2}+k

where h and k are the vertex coordinates.

The directrix is the same distance from the vertex as the focus is from the vertex.

Pardon my typo in my first post. That should be y=(x-2)^{2}-1

Note, if we expand we get y=x^{2}-4x+3

The x coordinate of the vertex can also be easily obtained by using x=\frac{-b}{2a}

Then, sub that value back into the quadratic to find its corresponding y coordinate.

Note, you have the form y=ax^{2}+bx+c. This is a concave up parabola.

The distance from the vertex to the focus is p=\frac{1}{4a}.

Consequently, that distance is the same from the vertex to the directrix. You have found the vertex, so use that and find the directrix.

It will just be a horizontal line with coordinates y=some number. Look at the graph. That'll help.