Determine the vertex and the axis of symmetry for the function below:

y = -648-05-01-00-00_files/i0110000.jpgx2 + 6x + 6

My crystal clairvoyance ball is a bit cloudy this morning.

I just can't see your graph

To determine the vertex and axis of symmetry for the given quadratic function, you need to use the formula for the vertex of a quadratic function:

x = -b / (2a)

Where "a" and "b" are the coefficients of the quadratic term and the linear term, respectively.

In the given function y = -648-05-01-00-00_files/i0110000.jpgx2 + 6x + 6, the coefficient of the quadratic term is -1, and the coefficient of the linear term is 6. Plugging these values into the formula, we have:

x = -6 / (2*(-1))

Simplifying further:

x = -6 / -2

x = 3

The x-coordinate of the vertex is 3. To find the y-coordinate, substitute this value back into the original function:

y = -648-05-01-00-00_files/i0110000.jpg(3)^2 + 6(3) + 6

y = -9 + 18 + 6

y = 15

Therefore, the vertex of the function is (3, 15).

To find the axis of symmetry, it is the vertical line passing through the vertex. In this case, the axis of symmetry is the line x = 3.