How do you find the max or minimum of #f(x)=4x^2+12x+9#?

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Answer 1 · 2016-11-28T17:44:45

it is a minimum and value is #""0#

1) complete the square.

#f(x)=4x^2+12x+9#

#f(x)=4(x^2+3x)+9#

#f(x)=4(x^2+3x+(3/2)^2)-9/4)+9#

#f(x)=4(x+3/2)^2-9+9#

#f(x)=4(x+3/2)^2#

2) identify max/min and cords/
.

since the eqn is a #""+x^2""#graph the value will be a minimum.

This occurs when the bracket equals #0#

so min at #""x=-3/2#

and this gives us #""f(x)=0#
#:.""#min value is #""0#
graph{4x^2+12x+9 [-10, 10, -5, 5]}