SOLUTION: Find a) vertex, b ) focus, c) directrix, d) symmetry axis, e) focal distance of the parabola of this equation, and sketch it: y^2+y-x-1=0

Question 738653: Find a) vertex, b ) focus, c) directrix, d) symmetry axis, e) focal distance of the parabola of this equation, and sketch it:
y^2+y-x-1=0
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Find a) vertex, b ) focus, c) directrix, d) symmetry axis, e) focal distance of the parabola of this equation, and sketch it:
y^2+y-x-1=0
complete the square:
(y^2+y+(1/4))=x+1+1/4
(y+1/2)^2=(x+5/4)
This is an equation of a parabola that opens rightward.
Its basic form: (y-k)^2=4p(x-h), (h,k)=(x,y) coordinates of the vertex
For given equation:(y+1/2)^2=(x+5/4)
(a)vertex: (-5/4,-1/2)
(d)axis of symmetry: y=-1/2
4p=1
p=1/4
(b)focus: (-1,-1/2) (p-distance to the right of the vertex on the axis of symmetry)
(c)directrix: x=-3/2 (p-distance to the left of the vertex on the axis of symmetry)
(e)focal distance=4p=1
see graph below as a visual check:
y=�(x+5/4)^.5-(1/2)

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