How do you write the quadratic function #y=x^2+12x+6# in vertex form?

1 Answer
Jim G.
May 28, 2017

#y=(x+6)^2-30#

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"#

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