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August 30, 2014

August 30, 2014

Posted by **Sam** on Wednesday, June 30, 2010 at 3:31pm.

Any help on this would be very much appreciated.

The graph y=3x^2+6x+9 is a parabola, I need to use algebra to find the equation of the axis of symmetry, I then need to use this information to find the coordinates of the vertex.

Any ideas? Many thanks

- Algebra -
**Damon**, Wednesday, June 30, 2010 at 4:56pm3x^2+6x+9 = y

x^2 + 2x + 3 = y/3

x^2 + 2 x = y/3 - 3

x^2 + 2x + 1 = y/3 - 2

(x+1)^2 = (1/3)(y-6)

symmetric about the vertical line x = -1 where (x+1) = 0

vertex at that point where x = -1 then y-6 = 0 and y = 6 so vertex at (-1,6)

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