How do you find the vertex of #y= 3x^2 +12x-6#?

1 Answer
Apr 24, 2018

#"vertex "=(-2,-18)#

Explanation:

#"Given the quadratic in "color(blue)"standard form"#

#•color(white)(x)y=ax^2+bx+c;a!=0#

#"then the x-coordinate of the vertex can be found using"#

#•color(white)(x)x_(color(red)"vertex")=-b/(2a)#

#y=3x^2+12x-6" with "a=3,b=12,c=-6#

#rArrx_("vertex")=-12(6)=-2#

#"substitute this value into the equation for y"#

#y_("vertex")=3(-2)^2+12(-2)-6=-18#

#rArrcolor(magenta)"vertex "=(-2,-18)#