what is the global maximum of the parabola
y= -x^2 -12x +4
you should recall that for
y = ax^2+bx+c
the vertex is at x = -b/2a
So, figure that for x, then evaluate y there. Since this parabola opens downward, its vertex is the maximum value.
y= -x^2 -12x +4
y = ax^2+bx+c
the vertex is at x = -b/2a
So, figure that for x, then evaluate y there. Since this parabola opens downward, its vertex is the maximum value.