I need help with Precalc on parabolas

Find the equation of the following parabola.
Directrix is the line x=-1/2
Vertex (0,0)

How do I do this?

If the directrix is the line x=-1/2, then you know that the focus is (1/2, 0). Since the focus is on the x-axis and is positive, you know that the equation is y^2=(something)x. The standard form for an equation of a parabola with vertex (0,0) is y^2=4px, where p is the focus, so the equation is y^2=4(1/2)x, which is y^2=2x.

x = -1/2 is a VERTICAL line, so the parabola is HORIZONTAL.
The vertex lies to the right of the directrix, so the parabola opens to the right.

equation of a right-opening parabola:
     x = a(y−k)² + h
     a > 0
     vertex is (h, k)
     p = 1/(4a)
     focus (h+p, k)
     directrix x = h−p

Apply your data
vertex (0, 0) ⇒ h = k = 0
directrix x = -1/2 ⇒ p = 1/2 ⇒ a = 1/
The equation becomes
     x = (1/2)y²