parabola with vertex at (3,-2) and a directreix of x=2. how do i wrtie this as an equation....PLEASE HELPPPPPPPP

Answer 1

Let's use basic concepts for this one
let P(x,y) be any point on the parabola
the focal point will be F(4,-2), (remember the vertex is midway between the focal point and the directrix)

Then PF = distance from P to directrix

√((x-4)^2+(y+2)^2) = √(x-2)^2
square both sides and expand

x^2 - 8x + 15 + y^2 + 4y + 4 = x^2 - 4x + 4

y^2 + 4y + 20 = 4x

from here you could complete the square to get
x = 1/4(y+2)^2 + 3 to confirm that the vertex is (3,-2)