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 ->
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
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(11628) (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
(