Posted by **:)** on Friday, June 4, 2010 at 9:12pm.

find algebraically the equation of the axis of symmetry and the coordinates of the vertex of the parabola whose equation is y=-2x^2-8x+3. explain the steps please. :)

- algebra -
**MathMate**, Friday, June 4, 2010 at 9:56pm
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=c-b²/4a

Thus the vertex of the parabola

y=ax²+bx+c is at

(-b/2a, c-b²/4a)

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.

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