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)(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)
Of course! I'd be happy to help you with that. To find the equation of the axis of symmetry of a parabola, you can use the formula "x = -b / (2a)". In this case, the equation of your parabola is y = 3x^2 + 6x + 9.
From the equation, we can identify the coefficients as follows:
a = 3
b = 6
To find the equation of the axis of symmetry, substitute these values into the formula:
x = -b / (2a)
x = -6 / (2*3)
x = -6 / 6
x = -1
So, the equation of the axis of symmetry is x = -1.
To find the coordinates of the vertex, substitute the value of x back into the original equation to get the corresponding y-coordinate.
Using the value of x = -1, substitute it into the equation y = 3x^2 + 6x + 9:
y = 3(-1)^2 + 6(-1) + 9
y = 3(1) - 6 + 9
y = 3 - 6 + 9
y = 6
Therefore, the vertex of the parabola is (-1, 6).
I hope this helps! Let me know if you have any further questions.