1. ## Parabolas- Need help

Graph the parabola with the given equation.

(y-2)^2=6(x-1)
I know the vertex would be (1,2)

(x+1)^2=6(y-1)
The vetex would be (-1,1)

2. Originally Posted by soly_sol

Graph the parabola with the given equation.

(y-2)^2=6(x-1)
I know the vertex would be (1,2)
yes, and where should be the focus?
note that the distance of the focus from the vertex of the parabola $(y-k)^2 = 4p(x-h)$ is p. what is your p in your equation?
also note that the distance of the endpoints of the latus rectum is 4p, or teh distance of one endpoint from the focus is 2p.. (by the way, remember that the latus rectum is the segment perpendicular to the axis of symmetry of the parabola passing through the focus)

with this information, i hope you could sketch the graphs..

Originally Posted by soly_sol
(x+1)^2=6(y-1)
The vetex would be (-1,1)

3. Graph the parabola with the given equation.

(y-2)^2=6(x-1)
I know the vertex would be (1,2)
Here else is what I got:
I got p=24
foci: (25,2)
directrix: x=-23
axis:y=2

(x+1)^2=6(y-1)
The vetex would be (-1,1)
foci: (-1, 25)
directrix: y=-23
axis: -1

4. Originally Posted by soly_sol
Graph the parabola with the given equation.

(y-2)^2=6(x-1)
I know the vertex would be (1,2)
Here else is what I got:
I got p=24
foci: (25,2)
directrix: x=-23
axis:y=2
4p = 6 $\implies p = \frac{3}{2}$
from here, do the rests...