How do you rewrite #f(x)=x^2+12x+20# in vertex form?

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Answer 1 · 2018-08-12T17:09:52

#x^2+12x+20=(x+6)^2-16# in vertex form

#f(x)=x^2+12x+20#

#f'(x)=2x+12#

A critical point of #f(x)# is :

#2x+12=0<=> x=-6#

so:

#f(x)=(x+6)^2-16#
And you can see on the graph below that the vertex form is respected.
graph{ y=x^2+12x+20 [-15,15,-20,20]}

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