Can you help me identify the vertex, focus and directrix of the graph of each of these equations??
1) y+1=1/4(x3)^2
2) x=2y^2
3) y^24x2y=3
I have no clue how to do this..my teacher never taught us but its on the homework >:( Anyway if you could just help me through solving for one of those equations I would appreciate it SOOO much...just figured I'd post them all in case anybody had some serious time on their hands ;)
Thanks!!
1) y+1=1/4(x3)^2
2) x=2y^2
3) y^24x2y=3
I have no clue how to do this..my teacher never taught us but its on the homework >:( Anyway if you could just help me through solving for one of those equations I would appreciate it SOOO much...just figured I'd post them all in case anybody had some serious time on their hands ;)
Thanks!!

Some basic equations:
(xh)^2 = 4*p*(yk)...for parabolas that open up & down. The vertex is at (h,k). P is positive for opens up, P is negative for opens down. The focus is at (h, k+p), which is "inside" the parabola. The directrix is at y = kp, which is above or below the parabola (depending on the value of "p")
(yk)^2 = 4*p*(xh)...for parabolas that open left & right. The vertex is at (h,k). P is positive for opens right, P is negative for opens left. Focus is at (h+p, k). The directrix is at x = kp.
For #1:
Vertex (3, 1)
Fits the first case...
multiply both sides by 4 on both sides.
4(y+1) = (x3)^2
4 = 4p
p = 1
Focus is at (2,1)
Directrix is at x = 4
#2
This one is a little more straight forward.
Vertex at (0,0)
0.5x = y^2
0.5 = 4p
p = 0.125
Focus is at (0.125 , 0)
Directrix is at x = 0.125
#3
You need to complete the square
y^2  2y +1 4x = 3+1
(y1)^2 = 4x+4
(y1)^2 = 4(x+1)
so the vertex is at (1, 1)
4 = 4p
p = 1
so the focus is at (0,1)
the directrix is at x = 2.
(xh)^2 = 4*p*(yk)...for parabolas that open up & down. The vertex is at (h,k). P is positive for opens up, P is negative for opens down. The focus is at (h, k+p), which is "inside" the parabola. The directrix is at y = kp, which is above or below the parabola (depending on the value of "p")
(yk)^2 = 4*p*(xh)...for parabolas that open left & right. The vertex is at (h,k). P is positive for opens right, P is negative for opens left. Focus is at (h+p, k). The directrix is at x = kp.
For #1:
Vertex (3, 1)
Fits the first case...
multiply both sides by 4 on both sides.
4(y+1) = (x3)^2
4 = 4p
p = 1
Focus is at (2,1)
Directrix is at x = 4
#2
This one is a little more straight forward.
Vertex at (0,0)
0.5x = y^2
0.5 = 4p
p = 0.125
Focus is at (0.125 , 0)
Directrix is at x = 0.125
#3
You need to complete the square
y^2  2y +1 4x = 3+1
(y1)^2 = 4x+4
(y1)^2 = 4(x+1)
so the vertex is at (1, 1)
4 = 4p
p = 1
so the focus is at (0,1)
the directrix is at x = 2.