How do you write #abs(x^2 +x -12)# as a piecewise function?
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Use the definition:
#|A|={(A;A>=0),(-A;A<0):}#
Find the points where the quadratic is zero to simplify the restrictions and add pieces as needed.
Given #y = |x^2 +x -12|#
Use the definition, #|A|={(A;A>=0),(-A;A<0):}#:
#y = {(x^2 +x -12;x^2 +x -12>=0),(-(x^2 +x -12);x^2 +x -12<0):}#
Find the x values for #x^2 +x -12=0#:
Factor:
#(x-3)(x+4)=0#
#x = -4 and x = 3#
This means that #x^2 +x -12 >=0# for #x <= -4# and #x>=3#
Modify the restriction for the first piece to be #x <=-4# and add a third peace with the restriction #x>=3#:
#y = {(x^2 +x -12;x<=-4),(-(x^2 +x -12);x^2 +x -12 < 0),(x^2 +x -12;x>=3):}#
Modify the restriction for the middle piece to be #-4 < x < 3# and distribute the -1
#y = {(x^2 +x -12;x<=-4),(-x^2 -x +12;-4 < x < 3),(x^2 +x -12;x>=3):}#
Here is a graph of #y = |x^2 +x -12|#
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Here is a graph of the piece-wise function
#y = {(x^2 +x -12;x<=-4),(-x^2 -x +12;-4 < x < 3),(x^2 +x -12;x>=3):}#
with each piece in a different color:
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