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ilovemattc · Mar 17, 2009, 10:51 PM

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

galactus · Mar 18, 2009, 10:01 AM

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.

galactus · Mar 19, 2009, 10:17 AM

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.