Question 754965
write the parabola in standard form and graph: 
x^2-4y-28=0
4y=x^2-28
y=(1/4)x^2-7
standard form: y=A(x-h)^2+k, (h,k)=(x,y) coordinates of vertex.
see graph below:
{{{ graph( 300, 300, -10, 10, -10, 10,  x^2/4-7) }}}
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write the ellipse in standard form and graph 
2x^2+y^2-4x-4y+4=0
2x^2-4x+y^2-4y+4=0
complete the square:
2(x^2-2x+1)+(y^2-4y+4)=-4+2+4
2(x-1)^2+(y-2)^2=2
(x-1)^2+(y-2)^2/2=1
see graph below:
y=(2-2(x-1)^2)^.5+2
{{{ graph( 300, 300, -4, 4, -4, 4,(2-2(x-1)^2)^.5+2,-(2-2(x-1)^2)^.5+2) }}}
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write the hyperbola in standard form and graph 
x^2-8x+11=-y^2
x^2-8x+y^2=-11
complete the square:
(x^2-8x+16)+y^2=-11+16
(x-4)^2+y^2=5
(x-4)^2+y^2=5
This is an equation of a circle, not a hyperbola.
see graph below:
y=(5-(x-4)^2)^.5
{{{ graph( 300, 300, -10, 10, -10, 10,(5-(x-4)^2)^.5,-(5-(x-4)^2)^.5) }}}