Hi! So I've been working on this problem for a while now but I can't figure it out. The question is "Find the vertex, directrix and focus of the parabola and sketch it's graph. y^2=3x" I'm confused on how to start it. If you could help, that would be awesome! :)
let's start with a tiny bit of rewriting so
(y-0)^2 = 3(x-0)
this tells us the vertex by looking at what's been added to y and x so
vertex is at (0,0)
the directrix is found by setting x=0-3/4 so directrix is x=-3/4
the focus is then found by (0+3/4, 0) so the focus is at (3/4, 0)
by plotting each of these it should be fairly easy to draw remembering the parabola should be equal distance away from the focus and directrix at all times (and the fact it's y squared means it's a sideways parabola)
remember in general you want the eqn in this form
(y-k)^2 = 4p(x-h)
so the vertex is at (h,k), the focus at (h+p,k) and the directrix at x=h-p
