If a parabola's vertex is at (4, -2) and its focus at (4,4), write the equation of this parabola, its directrix, and its axis of symmetry. Graph your parabola with all appropriate parts labeled.
I did:
. . .vertex (h, k) = (4, -2)
. . .focus (h, k + p) = (4, 4), so k + p = 2
. . .y - k = 1/4p (x - h)^2 > equation
. . .y + 2 = 1/4(4) (x - 4)^2 > substituted
. . .y = 1/16 (x - 4)^2 - 2 > solved
. . .p < 0
. . .y = -1 5/6
. . .x = 4
ok now graph it ...how?
I did:
. . .vertex (h, k) = (4, -2)
. . .focus (h, k + p) = (4, 4), so k + p = 2
. . .y - k = 1/4p (x - h)^2 > equation
. . .y + 2 = 1/4(4) (x - 4)^2 > substituted
. . .y = 1/16 (x - 4)^2 - 2 > solved
. . .p < 0
. . .y = -1 5/6
. . .x = 4
ok now graph it ...how?