This is a quadratic in #y# instead of #x#

So #y# is the independent variable. The consequence is that the graph of type #uu# is rotated clockwise by #90^o#

Being given this question suggests that your mathematical manipulation skills are more than basic. So not a lot of explanation is given.

#12x=-y^2-6y-33#

#color(red)("Correction:")# as #y^2# is negative so the graph type of #uu# would be rotated #90^o# anticlockwise giving shape #sup#

#x=-1/12y^2-1/2 y-11/4#

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

Instead of #-11/4# being the intercept of the y-axis it is the intercept of the x-axis

#color(blue)("Determine the y-intercept")#

The determinant of form

#b^2-4ac""->""(-1/2)^2-4(-1/12)(-11/4) = +1/4-11/12#

#=-2/3#

As the determinant is negative there is no y-intercept

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#color(blue)("Determine vertex")#

Write as:#" "x=-1/12(y^2 +12/2y)-11/4#

Consider the #12/2# from #12/2y#

#=>y_("vertex")=(-1/2)xx12/2 = -3#

Substituting #y=-3#

#x_("vertex")=-1/12(-3)^2-1/2(-3)-11/4 #

#x_("vertex")= -3/4 +3/2-11/4 = -2 #

#"Vertex" -> (x,y)=(-2,-3)#