Question 156941
Find the vertex, line of symmetry, the maximum or minimum value of the quadratic equation, and graph the function. f(x)= -2x^2+2x+6
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set f(x) = 0 to find where it crosses the x-axis.
-2x^2+2x+6=0
x^2 - x - 3 = 0
*[invoke solve_quadratic_equation 1,-1,-3]
The vertex is at the minimum:  set the 1st derivative to 0
2x-1=0
x = 1/2
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sub for x in the eqn
y = (1/4) - 1/2 - 3
y = -3 1/4 = -13/4
So the vertex is at (1/2,-13/4)

Since there's no xy term, the axis of symmetry is parallel to the y-axis.  It goes thru the vertex, so it's:
x = 1/2