SOLUTION: 2y2-4y-2=x how do i graph this parabola, and identify the vertex, focus, andd directrix

Question 722060: 2y2-4y-2=x
how do i graph this parabola, and identify the vertex, focus, andd directrix
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
2y^2-4y-2=x
how do i graph this parabola, and identify the vertex, focus, andd directrix
2y^2-4y-2=x
complete the square
2(y^2-2y+1)-2-2=x
2(y-1)^2=x+4
(y-1)^2=(1/2)(x+4)
This is an equation of a parabola that opens rightward.
Its basic equation: (y-k)^2=4p(x-h), (h,k)=(x,y) coordinates of the vertex
For given parabola:
vertex: (-4,1)
axis of symmetry: y=1
4p=1/2
p=1/8
focus: (-31/8,1) (p-distance to the right of the vertex on the axis of symmetry)
directrix:x=33/8 (p-distance to the left of the vertex on the axis of symmetry)
Graphing:
y-intercept:
set x=0
(y-1)^2=2
y-1=�√2
y=1�√2
..
x-intercept
set y=0
x=2 (from original equation)
..
y=((x+4)/2)^.5+1
see graph below as a visual check:

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