Algebra
posted by Sam .
Hi,
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

3x^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)(y6)
symmetric about the vertical line x = 1 where (x+1) = 0
vertex at that point where x = 1 then y6 = 0 and y = 6 so vertex at (1,6)
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