algebra
posted by :) .
find algebraically the equation of the axis of symmetry and the coordinates of the vertex of the parabola whose equation is y=2x^28x+3. explain the steps please. :)

The axis of symmetry of a parabola of the form y=ax²+bx+c is always vertical.
If a > 0, the parabola is concave upwards, and if a<0, the parabola is concave downwards.
The vertex of a parabola can be obtained by completing the squares, namely
y=a((x+(b/2a)²+c/a(b/2a)²)
or at
x= b/2a, at which point
y=cb²/4a
Thus the vertex of the parabola
y=ax²+bx+c is at
(b/2a, cb²/4a)
The vertical axis of symmetry passes through the vertex, thus has the equation:
x=b/2a
Apply the above the equation to the given problem and post the answer for verification if required.