Question 309165
Convert to vertex form {{{y=a(x-h)^2+k}}} by completing the square.
Then the corrdinates of the vertex are (h,k) and the axis of symmetry is x=h.
{{{y=x^2-2x-5}}}
{{{y=(x^2-2x+1)-5-1}}}
{{{y=(x-1)^2-6}}}
Vertex : ({{{1}}},{{{-6}}})
Axis of Symmetry:{{{x=1}}}
{{{ drawing( 300, 300, -8, 8, -8, 8, grid(1),
circle(1,-6,.3),
line(1,-10,1,10),
graph( 300, 300, -8, 8, -8, 8, x^2-2x-5)) }}}
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{{{y=-x^2+4x-1}}}
{{{y=-(x^2-4x+4)-1+4}}}
{{{y=-(x-2)^2+3}}}
Vertex : ({{{2}}},{{{3}}})
Axis of Symmetry:{{{x=2}}}
{{{ drawing( 300, 300, -8, 8, -8, 8, grid(1),
circle(2,3,.3),
line(2,-10,2,10),
graph( 300, 300, -8, 8, -8, 8, -x^2+4x-1)) }}}  
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{{{y= 2x^2+4x-2 }}}
{{{y=2(x^2+2x+1)-2-2}}}
{{{y=2(x+1)^2-4}}}
Vertex : ({{{-1}}},{{{-4}}})
Axis of Symmetry:{{{x=-1}}}
{{{ drawing( 300, 300, -8, 8, -8, 8, grid(1),
circle(-1,-4,.3),
line(-1,-10,-1,10),
graph( 300, 300, -8, 8, -8, 8, 2x^2+4x-2)) }}}