What is the vertex of #y= 12x^2 - 18x - 6 #?

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Answer 1 · 2016-08-16T20:23:02

#P=(3/4,-51/4)#

#P=(h,k)" Vertex coordinates"#

#y=ax^2+bx+c#
#a=12" ; "b=-18" ; "c=-6#

#y=12x^2-18x-6#
#h=-b/(2a)#

#h=18/(2*12)=18/24=3/4#

#k=12*(3/4)^2-18*3/4-6#

#k=12*9/16-54/4-6#

#k=27/4-54/4-24/4=(27-78)/4=-51/4#

#P=(3/4,-51/4)#