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

2 Answers
Apr 14, 2017

#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"#

Apr 14, 2017

#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:
enter image source here