Question

How do you write `f(x)= x^2 + 6x + 12` in vertex form?

Answer 1

`f(x)=(x+3)^2+3`

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.

`"Rearrange " f(x)=x^2+6x+12" into this form"`

`"using the method of "color(blue)" completing the square"`

`f(x)=(x^2+6xcolor(red)(+9))color(red)(-9)+12`

`color(white)(f(x))=(x+3)^2+3larrcolor(red)" in vertex form"`

Answer 2

`f(x)=(x+3)^2+3` (with some simplification, see below)

Explanation:

The general vertex form is
`color(white)("XXX")f(x)=color(green)m(x-color(red)a)^2+color(blue)b`
with vertex at `(color(red)a,color(blue)b)`

Given `f(x)=x^2+6x+12`

Extracting the "spread" factor `color(green)m`
`color(white)("XXX")f(x)=color(green)1(x^2+6x)+12`

Completing the square
`color(white)("XXX")f(x)=color(green)(x^2+6x+color(magenta)(3^2))+12 -color(green)1 * color(magenta)(3^2)`

Re-writing with a squared binomial and some simplification:
`color(white)("XXX")f(x)=color(green)1(x+3)^2+3`

...or in explicit vertex form:
`color(white)("XXX")f(x)=color(green)1(x-color(red)((-3)))^2+color(blue)3`
(note that I dropped the `color(green)1` and simplified the binomial in my "Answer" even though I consider this to be the more accurate version.

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In order to check that I didn't make any mechanical errors, I've plotted the original equation, and it does appear to have the vertex indicated by the derived equation: