Wize High School Algebra I Textbook (Common Core) > Absolute Value Functions

Basics of Absolute Value

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Basics of Absolute Value
The absolute value of a number xxx is denoted ∣x∣|x|∣x∣, and it is the "positive version" of that number.

It keeps positive numbers the same, but turns negative numbers into positive numbers.


Examples
∣7∣=7⟶|7|=7 \quad \longrightarrow∣7∣=7⟶  the absolute value does not change positive numbers
∣−7∣=7⟶|-7|=7 \quad \longrightarrow∣−7∣=7⟶  the absolute value of a negative number is the same value, except positive!
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Magnitude
The magnitude of a quantity xxx is simply the absolute value of the quantity, ∣x∣|x|∣x∣.

The magnitude is the size, and it does not depend on whether the quantity is positive or negative.

Wize Tip
The magnitude ∣x∣|x|∣x∣ represents the distance of the number xxx from 000 on the number line.
Ex. ∣−4∣|-4|∣−4∣ is 4 steps away from 0.
  


Example

Which quantity has a larger magnitude, −4-4−4 or 333?

Note that −4<3-4<3−4<3, but this just means that, on a number line, -4 is to the left of 3.

However, the magnitude is not concerned with positive or negative!
∣−4∣=4∣3∣=3|-4| = 4\qquad |3|=3∣−4∣=4∣3∣=3
Since ∣−4∣>∣3∣|-4|>|3|∣−4∣>∣3∣, the magnitude of -4 is larger than the magnitude of 3.

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Distance
We can calculate the distance between two numbers by finding the absolute value of the difference:

distance(x, y)=∣x−y∣\boxed{\quad
\text{distance(\textit x, \textit y)} = |x-y|
\quad}distance(x, y)=∣x−y∣​

Example

How far apart are the integers -3 and 2?

The distance between the numbers is ∣−3−2∣=∣−5∣=5|-3-2|=|-5|=\boxed{5}∣−3−2∣=∣−5∣=5​.

We can see this on a number line:

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Example: Absolute Value
Evaluate the following expressions:

a) ∣−7∣ − ∣−2∣|-7|\,-\,|-2|∣−7∣−∣−2∣

∣−7∣−∣−2∣=7−2=5\begin{array}{rcl}
&|-7|&-&|-2| \\[0.5em]
=& 7&-&2  \\[0.5em]
=& \boxed{5}
\end{array}==​∣−7∣75​​−−​∣−2∣2​

b) −2∣3∣−3∣2−5∣-2|3|-3|2-5|−2∣3∣−3∣2−5∣

−2∣3∣−3∣2−5∣=−2(3)−3∣−3∣=−6−3(3)=−15\begin{array}{rcl}
&-2|3| &-& 3|2-5| \\[0.5em]
=&-2(3) &-& 3|-3| \\[0.5em]
=&-6 &-& 3(3) \\[0.5em]
=& \boxed{-15}
\end{array}===​−2∣3∣−2(3)−6−15​​−−−​3∣2−5∣3∣−3∣3(3)​

c) −3∣2(−4)−8∣-3|2(-4)-8|−3∣2(−4)−8∣

−3∣2(−4)−8∣=−3∣−16∣=−3(16)=−48\begin{array}{rcl}
&-3|2(-4)-8| \\[0.5em]
=& -3|-16|  \\[0.5em]
=& -3(16)  \\[0.5em]
=& \boxed{-48}
\end{array}===​−3∣2(−4)−8∣−3∣−16∣−3(16)−48​​

Practice: Absolute Value
Match each expression with the correct value.

A.

6

B.

0

C.

-6

D.

-3

∣−3−3∣|-3-3|∣−3−3∣

∣3−3∣|3-3|∣3−3∣

−3∣1−2∣-3|1-2|−3∣1−2∣

−2∣−2−1∣-2|-2-1|−2∣−2−1∣

I don't know

Practice: Absolute Value
Find the distance between the following pairs of numbers. Sketch a number line to visualize the distances.

a) −1.5-1.5−1.5 and −3.2-3.2−3.2

b) −38-\dfrac{3}{8}−83​ and 222

b) Answer as an exact improper fraction

I don't know

Practice: Absolute Value
A grocery store has 3 apples left to sell. The apples have the following weights (in grams): {100,122,90}\{ 100, 122, 90 \}{100,122,90}

a) Find the average weight of the apples.

b) Find the deviation (the distance: the absolute value of the difference) of each apple's weight from the mean weight. What is the sum of these values?

c) What is the average deviation from the mean?

a) average weight of the apples (in grams):

b) Sum of the deviations from the mean (in grams):

c) Average deviation from the mean (in grams):

I don't know