02Fabsval

toc = Absolute Value (Modulus) Function =

The function y = | f(x) | is callled the ** absolute value ** function (or the ** modulus ** function).

The symbol | x | represents the __**magnitude**__ of x (magnitude means size).
 * So | x | = x (ignore the sign)


 * eg | +5 | = 5
 * eg | –5 | = 5

More technically, this can be written as a __**hybrid function**__:

math . \qquad |x|=\left\{ \begin{matrix} +x,\text{ if } x \geqslant 0 \\ \\ -x, \text{ if } x < 0 \\ \end{matrix} \right. math

The graph of y = | f(x)| starts off looking like y = f(x) except that all negative y-values are reflected up above the x-axis.



Graph of Absolute Value
... ... y = | x |

... ... Domain: x Î R

... ... Range: y __>__ 0

The standard absolute value graph has a ** cusp ** (sharp minimum point) at ( 0, 0 )

Transformations

 * We can apply the standard transformations such as dilations, translations and reflections to the graph y = |x|.

Advanced Functions

 * If there is a more advanced function within the absolute value signs, | f(x ) |
 * Sketch the function first, then everything below the x-axis is reflected across the x-axis


 * More formally, write the absolute value function as a hybrid function
 * with f(x) for the parts of the graph where f(x) __>__ 0
 * and –f(x) for the parts of the graph where f(x) < 0

** Example 1 **


... ... y = | x 2 – 2 |


 * Solution:**

... ... This is the graph of y = x 2 – 2 with all negative y-values reflected above the x-axis.

... ... More technically, this can be written as a __**hybrid function**__.

math . \qquad f(x)=\left\{ \begin{matrix} +(x^2-2),\text{ if } \{ x \leqslant -\sqrt{2} \} \cup \{ x \geqslant \sqrt{2} \} \\ \\ -(x^2-2), \text{ if } x \in \left( -\sqrt{2}, \sqrt{2} \right) \qquad \quad \; \; \\ \end{matrix} \right. math

... ... Domain: x Î R

... ... Range: y __>__ 0

... ... Local Maximum (0, 2)
 * Turning Points:**

math . \qquad \text{Local minimums at } \; \left( -\sqrt{2}, 0 \right) \; \; \left( \sqrt{2}, 0 \right) math
 * Cusps** (sharp points)

We could apply further standard transformations such as dilations, translations and reflections to the graph y = |x 2 – 2|.

Example 2
Find the x-intercepts for y = 2|x + 2| – 6


 * Solution:**

... ... ... 2|x + 2| – 6 = 0 ... ... ... ... 2|x + 2| = 6 ... ... ... ..... |x + 2| = 3 ... ... ... ...... x + 2 = ±3 ... ... ... ...... x + 2 = +3 ... ... or ... ... x + 2 = -3 ... ... ... .......... x = +1 .... ... or ..... ... x = -5

** Questions **
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