Find the following parts of a parabola: vertex, focus, and directrix and show work.
(3x+6)^2=18y-36
18y=(3x+6)^2+36
18y=(3x+6)(3x+6)+36
18y=9x^2+36x+72
18y=9(x^2+4x+8)
y=(x^2+4x+8)/2

Now what? I can't get it into parabolic form with a denominator. And, do I need to complete the square? Not sure where to go from here.

Question

Find the following parts of a parabola: vertex, focus, and directrix and show work.
(3x+6)^2=18y-36
18y=(3x+6)^2+36
18y=(3x+6)(3x+6)+36
18y=9x^2+36x+72
18y=9(x^2+4x+8)
y=(x^2+4x+8)/2

Now what? I can't get it into parabolic form with a denominator. And, do I need to complete the square? Not sure where to go from here.

anonymous · Answer

$$y=(1/2)x^2 +2x+4$$
vertex is -b/2a, f(-b/2a)
            =(-2,2)

anonymous · Answer

did you see your 5th line? its 9/18=1/2

anonymous · Answer

so I got y=1/2(x+2)^2+2 by dividing and completing the square... vertex (-2,2) focus (-2,5/2), and directrix y=-3/2; hopefully it's right