medal!! :)
Using a directrix of y = -2 and a focus of (2, 6), what quadratic function is created?

A. f(x) = -(1/8) (x-2)^2 - 2
B. f(x) = 1/16 (x-2)^2 + 2
C. f(x) = 1/8 (x-2)^2 - 2
D. f(x) = -(1/16) (x+2)^2 - 2

Question

medal!! :)
Using a directrix of y = -2 and a focus of (2, 6), what quadratic function is created?

A. f(x) = -(1/8) (x-2)^2 - 2
B. f(x) = 1/16 (x-2)^2 + 2
C. f(x) = 1/8 (x-2)^2 - 2
D. f(x) = -(1/16) (x+2)^2 - 2

perl · Answer

this a vertical parabola, opens upwards
y - k = 1/(4p) ( x - h)^2

perl · Answer

the vertex (h,k) is midway between the focus (2,6) and the directrix y = -2

anonymous · Answer

wait soo..how do i solve it lol @perl

perl · Answer

it might help to make a rough graph first

anonymous · Answer

i'm really bad at math so idk how to do or make anything besides a quadratic formula lol

perl · Answer

|dw:1419293856674:dw|

perl · Answer

so midway between the focus and the directrix is the vertex (h,k)

anonymous · Answer

so 2 is h and 6 is k?

perl · Answer

|dw:1419293941681:dw|