How do you write the quadratic function #y=x^2+12x+6# in vertex form?
1 Answer
May 28, 2017
Explanation:
#"the equation of a parabola in "color(blue)"vertex form"# is.
#color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))#
where ( h , k ) are the coordinates of the vertex and a is a constant.
#"for a quadratic in standard form " ax^2+bx+c#
#x_(color(red)"vertex")=-b/(2a)#
#x^2+12x+6" is in this form"#
#"with " a=1,b=12,c=6#
#rArrx_(color(red)"vertex")=-12/2=-6#
#"substitute this value into the equation for y"#
#y_(color(red)"vertex")=(-6)^2+(12xx-6)+6=-30#
#rArrcolor(magenta)"vertex "=(-6,-30)#
#rArry=(x+6)^2-30larrcolor(red)" in vertex form"#