What is the vertex form of #y=x^2 + 12x + 36#?

2 Answers
Nov 24, 2017

#y=(x+6)^2#

Explanation:

#"the equation of a parabola in "color(blue)"vertex form"# is.

#color(red)(bar(ul(|color(white)(2/2)color(black)(a(x-h)^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 x-term")^2#
#"to "x^2+12x#

#x^2+2(6)xcolor(red)(+36)color(red)(-36)+36#

#=(x+6)^2+0larrcolor(red)"in vertex form"#

Nov 24, 2017

#y=(x-0)^2 -6#

Explanation:

YOUR EQUATION: #f(x) = ax^2 + bx + c#
VERTEX FORM: #f(x) = a(x - h)^2 + k#

  1. Find the vertex #(h,k)#
    Number 2-3 tells you how to find the vertex
    Remember #a=1#

  2. 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.

  3. 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#

  4. Plug the answers into vertex form
    #y=1(x-0)^2 -6#
    #y=(x-0)^2 -6#