Given:#" "y=-x^2+12x-4#
You can solve the traditional way or use a 'trick'
Just to give you an idea how useful the trick is:
By sight: #color(brown)("The axis of symmetry is "x=+6)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine axis of symmetry and "x_("vertex"))#
Consider the standard form of #y=ax^2+bx+c#
Write as: #y=a(x^2+b/a x)+c#
In your case #a=-1#
So #color(brown)(x_("vertex")=(-1/2)xx12/(-1) =+6)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine "y_("vertex"))#
Substitute #x=6# into original equation.
#y_("vertex")=-(6^2)+12(6)-4" "->" "y_("vertex")=32#
#color(white)(.)#
#color(magenta)("'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~")#
#color(magenta)("'~~~~~~~~~~~~Another method ~~~~~~~~~~~~~")#
#color(magenta)("'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~")#
#color(blue)("Completing the square "color(brown)(larr" not much detail given")#
#y=-(x^2-12x)-4+k#
#y=-(x-6)^2-4+k#
#But -36+k=0->k=36#
#y=-(x-6)^2 +32#
#x_("vertex")->(-1)xx(-6)=+6#
#y_("vertex")->32#
