posted by Sam on .
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)(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)