How do you write $∣ ∣ x^{2} + x - 12 ∣ ∣$ $abs(x^2 +x -12)$ as a piecewise function?

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Answer 1 · 2017-06-25T15:16:34

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 $y = |x^2 +x -12|$

Desmos.com

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:

Desmos.com

How do you write abs(x^2 +x -12)∣∣x2+x−12∣∣abs(x^2 +x -12) as a piecewise function?