What is the the vertex of #x =-1/2(y-2)^2-4y+12 #?

1 Answer
Aug 30, 2017

Vertex #->(x,y)=(12,-2)#

Explanation:

#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#
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#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)#
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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")#
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From #Equation(1)#

#y_("vertex")=(-1)xx2=-2#

#x_("vertex")=+12#

Vertex #->(x,y)=(12,-2)#

Tony B