How do you find the vertex and intercepts for #x =-1/2(y-2)^2-4#?

Instead of an equation in #x# we have an equation in #y#. What that does is 'rotate' the graph #90^o# to the right

So instead the general shape being #nn# it is #sup#

Also the format of the given equation is in Vertex form.

#color(blue)("Determine the vertex")#
'..................................................................
#color(red)("If this had been an equation in "x) color(magenta)(larr" For comparison only")#

Then #color(red)(x_("vertex")=(-1)xx(-2) = +2)#

#color(red)(y_("vertex")=-4#
'.......................................................................

But our equation is in #y# so we need to reverse them in that

#color(green)(y_("vertex")=(-1)xx(-2)=+2)#
#color(green)(x_("vertex")=-4)#

#color(blue)("Vertex "->(x,y)->(-4,2))#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(blue)("Determine x intercept")#

Set #y=0# giving

#color(brown)(x=-1/2(y-2)^2-4)color(green)(" "->" "x=-1/2(0-2)^2-4#

#color(blue)(x_("intercept")=-6)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine y intercept")#

Set #x=0#

#color(brown)(x=-1/2(y-2)^2-4)color(green)(" "->" "0=-1/2(y-2)^2-4#

Add 4 to both sides

#4=-1/2(y-2)^2#

Multiply both sides by (-2)

#-8=(y-2)^2#

Square root both sides

#sqrt(-8)=y-2#

#y=2+-sqrt(-8)#

As you have square root of a negative number the graph does not cross the y axis.

So the only solution for the values of y are in the complex numbers set of values

#y=2+-(2sqrt(2)) i#