What is the vertex form of #y=x^2 + 12x + 36#?
2 Answers
Explanation:
#"the equation of a parabola in "color(blue)"vertex form"# is.
#color(red)(bar(ul(color(white)(2/2)color(black)(a(xh)^2+k)color(white)(2/2))))#
#"where "(h,k)" are the coordinates of the vertex and a"#
#"is a multiplier"#
#"to obtain this form use the method of"#
#color(blue)"completing the square"#
#• " ensure the coefficient of the "x^2" term is 1 which it is"#
#• " add/subtract "(1/2"coefficient of xterm")^2#
#"to "x^2+12x#
#x^2+2(6)xcolor(red)(+36)color(red)(36)+36#
#=(x+6)^2+0larrcolor(red)"in vertex form"#
Explanation:
YOUR EQUATION:
VERTEX FORM:

Find the vertex
#(h,k)#
Number 23 tells you how to find the vertex
Remember#a=1# 
Find b/2a (this is how to find
#h# )
In this equation b/2a would be 12/2(1)
The answer to 12/2(1) would be 6. 
Find
#k# by plugging in the answer for#h# into the equation.
#y=x^2 +12x +36#
#y=(6)^2 +12(6) +36#
#y=36 +36#
#y=0#
#h# would be#0# 
Plug the answers into vertex form
#y=1(x0)^2 6#
#y=(x0)^2 6#