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

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Answer 1 · 2016-08-16T20:23:02

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

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

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

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

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

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

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

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

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

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