#color(blue)("General introduction")#
Instead of a quadratic in #x# this is a quadratic in #y#
If the #y^2# term is positive then the general shape is #sub#
If the #y^2# term is negative then the general shape is #sup#
If you expand the brackets we end up with #-1/2y^2# which is negative. So the general shape is #sup#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Answering the question")#
I choose to opt for the 'completed square' form of equation
Expanding the brackets we have:
#x=-1/2(y^2-4y+4)-4y+12#
#x=-1/2y^2-2y+10#
#x=-1/2(y +2)^2+12" "......................Equation(1)#
.........................................................
Check
#x=-1/2y^2-2y-2+12" "->" "color(green)(x=-1/2y^2-2y+10)#
Original eqn: #x=-1/2(y-2)^2-4y+12#
#x=-1/2y^2+2y-2-4y+12#
#color(green)(x=-1/2y^2 -2y+10) color(red)(larr" Thay match")#
................................................................
From #Equation(1)#
#y_("vertex")=(-1)xx2=-2#
#x_("vertex")=+12#
Vertex #->(x,y)=(12,-2)#
