# SOLUTION: Hi can you please help me with this question, graph the equation equal to: 2x^2-2x-2y-1=0 and find the vertix, focus point and axis of symmetry. Thank you all your help...

Algebra ->  Algebra  -> Quadratic-relations-and-conic-sections -> SOLUTION: Hi can you please help me with this question, graph the equation equal to: 2x^2-2x-2y-1=0 and find the vertix, focus point and axis of symmetry. Thank you all your help...      Log On

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 Question 409726: Hi can you please help me with this question, graph the equation equal to: 2x^2-2x-2y-1=0 and find the vertix, focus point and axis of symmetry. Thank you all your help...Answer by lwsshak3(6456)   (Show Source): You can put this solution on YOUR website!graph the equation equal to: 2x^2-2x-2y-1=0 and find the vertex, focus point and axis of symmetry. .. 2x^2-2x-2y-1=0 Standard form for parabola: (x-h)^2=4p(y-k),(h,k) being the (x,y) coordinates of the vertex,p=distance from vertex to focus or vertex to directrix(in the opposite direction) on the axis of symmetry. Completing the sqare: 2(x^2-x+1/4)-2y=1+1/2=3/2 2(x-1/2)^2=2y+3/2=2(y+3/4) equation in standard form: (x-1/2)^2=(y+3/4) axis of symmetry = x=1/2 vertex (1/2,-3/4) 4p=1 p=1/4 Directrix = y=-1 Focus (1/2,-1/2) .. y=(x-.5)^2-.75