Question

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

Answer 1

`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"`

Answer 2

`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`