Question 852125
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Hi,
sketch IT (Directrix  x = 1 the clue) Parabola opening right
the vertex form of a Parabola opening right(a>0) or left(a<0), {{{x=a(y-k)^2 +h}}}
 where(h,k) is the vertex and  y = k  is the Line of Symmetry
{{{x=a(y+1)^2 + 2}}}  V(2,-1)   Finding value of a
p(distance of the directrix and focus left 0r right of vertex) = 1, a = 1/4p = 1/4
{{{x=(1/4)(y+1)^2 + 2}}}  
{{{drawing(300,300,   -6, 6, -6, 6, grid(1), blue(line(1,6,1,-6)) ,
circle(2,-1,0.2),
circle(3,-1,0.2),
graph( 300, 300, -6, 6, -6, 6, 0 , .25sqrt(4(x-2))-1, -.25sqrt(4(x-2))-1 ))}}}