How do you find the vertex of #y= 3x^2 +12x-6#?
1 Answer
Apr 24, 2018
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)#
