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

Jun 25, 2017

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.

#### Explanation:

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:

$\left(x - 3\right) \left(x + 4\right) = 0$

$x = - 4 \mathmr{and} x = 3$

This means that ${x}^{2} + x - 12 \ge 0$ for $x \le - 4$ and $x \ge 3$

Modify the restriction for the first piece to be $x \le - 4$ and add a third peace with the restriction $x \ge 3$:

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 $y = | {x}^{2} + x - 12 |$ 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: 