algebra
posted by KR .
Find algebraically the equation of the axis of symmetry and the coordinates of the vertex of the parabola whose equation is y= 2x8x+3

2x^28x = y 3 (suspect you meant
x^2 + 4x = (1/2)y +3/2
x^2 + 4x + 4 = (1/2)y +11/2
(x+2)^2 = (1/2) (y11)
vertex at (2,11) axis is x = 2
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