#y= -3x^2 +12x -8#
#= -3(x^2 +4x)-8#
#=-3 (x^2 +4x +4 -4) -8#
#= - 3(x^2 +4x +4) +12-8#
#=-3 (x+2)^2 +4#, is the required vertex form.
This quadratic function represents a vertical parabola opening down.
The vertex is at #(-2,4)#.
Axis of symmetry is #x= -2#
#y#- intercept is given by #x=0#. So it is #-8#
#x#- intercept is given by #y=0#.
#x#-intercepts would be given by the quadratic equation:
#-3x^2 +12x-8=0#,
or, #3x^2 -12x +8=0#,
#x=(12+- sqrt(144-96))/6" "# using the quadratic formula.:
#x=(12+-sqrt48)/6#
= #2+-sqrt48/6#
#x = 3.15 and x = 0.85" "# (to 2 d.p.)
