find parts of a parabola and graph, show work:
3x+y^2+8y+4=0
3x=-y^2-8y-4
3x=-(y^2+8y+ )-4
3x=-(y^2+8y+16)-4+16 completed square
3x=-(y+4)^2+12
x=-1/3(y+4)^2+4 divided by 3
so vertex should be (-4,4)
focus should be (-19/4,4)
directrix should be x=-13/4

parabola should open to left...
However, when I try to find x intercept by making y=0, I get (-4/3,0) which is not on the parabola I have graphed. Where did I go wrong?

Question

find parts of a parabola and graph, show work:
3x+y^2+8y+4=0
3x=-y^2-8y-4
3x=-(y^2+8y+   )-4
3x=-(y^2+8y+16)-4+16 completed square
3x=-(y+4)^2+12
x=-1/3(y+4)^2+4 divided by 3
so vertex should be (-4,4)
focus should be (-19/4,4)
directrix should be x=-13/4

parabola should open to left...
However, when I try to find x intercept by making y=0, I get (-4/3,0) which is not on the parabola I have graphed. Where did I go wrong?