To find the quadratic of the form ax^2+bx+c=y set up a system of equations with the points provided...
a+b+c=4, 0a+0b+c=3, a-b+c=-6 getting differences:
a+b=1, -a+b=9 and getting difference again:
2a=-8, so a=-4, making a+b=1 become:
-4+b=1, so b=5, making a+b+c=4 become:
-4+5+c=4, so c=3
thus the quadratic that produces those three points is:
y=-4x^2+5x+3
...
The vertex occurs when the velocity is equal to zero, the derivative of the quadratic...
dy/dx=-8x+5, dy/dx=0 when x=5/8=0.625
Then you can find y(5/8)=4.5625
So the vertex is (0.625, 4.5625)
Or you may wish to remember that the vertex will always be:
(-b/(2a), (4ac-b^2)/(4a)) (this can be derived from conservation of energy under constant acceleration or the vertex form of the quadratic)
The axis of symmetry will be the vertical line about the x-coordinate of the vertex, in this case:
x=0.625
