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Absolute Value. Definition of absolute value of a real number: "The distance between the origin and the point representing the real number" (Algebra 2)
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1. Absolute Value Emily Benenhaley and Joe Gesualdi
I would like to adknowledge that most of this powerpoint presentation is the work of Elizabeth Stapel from www.purplemath.com and AlgebraLab at http://www.algebralab.org/lesson/lesson.aspx?file=Algebra_AbsoluteValueGraphing.xml
2. Absolute Value Definition of absolute value of a real number: "The distance between the origin and the point representing the real number" (Algebra 2)
The distance a number is from zero, thus being always positive (or zero).
The absolute value notation is bars, not parentheses or brackets.
It asks "how far?" not "in what direction?"
Example:
| 6 | = 6
|-6 | = 6
3. Absolute Value Inequalities Inserting Inequality signs:
| -2 |___| -5 |
Usually, -2 > -5, but with absolute value it changes:
| -2 | = 2
| -5 | = 5
Therefore...
| -2 | < | -5 |
4. Positive-Negative Property -If x>0, or if x is positive, then the value of the sign doesn't change:
For x = 0, | x | = x
-If x<0, or if x is negative, then the value of the sign does change:
For x = 0, | x | = -x
*The sign "-" indicates "the opposite sign" not necessarily a negative number
5. Solving Absolute Value Equations Absolute value of numbers are always positive or zero, whether input is negative or positive. This makes solving equations challenging.
| x | = 6
* This is easy because we know the answer: | 6 | = 6 and | -6 | = 6 but what if you don't know the answers?
*Use Positive/Negative Property of Absolute Value
-To remove absolute-value bars, you must isolate absolute value onto one side and split the equation into 2 possible cases.
6. Solving Absolute Value Equations cont. Example:
| x + 3 | = 8
(x+3) = 8 or -(x+3)= 8
x+3 = 8 -x - 3= 8
x = 5 -x = 11
x = -11
Solution: x = -11, 5
7. Examples:
1) | 2x - 3 | = 7
2) 2 | x + 4 | - 6 = 8
8. Solving Absolute Value Inequalities: Less Than Since | x | is the distance of x from zero, then | x | < 4 is all the points that are LESS THAN 4 units away from zero (without including four), so the answer is:
-4 < | x | < 4
To solve less than inequalities, you clear the absolute value bars and solve the linear inequality:
| 3x + 4 | < 8
-8 < 3x + 4 < 8
-8 - 4 < 3x < 8 - 4
-12 < 3x < 4
- 12/3 < x < 4/3
Solution: -4 < x< 4/3
9. Solving Absolute Value Inequalities: Greater Than | x | > 4 --> all points that are MORE than 4 units away from zero (without including exactly 4 units away), so the answer is:
| x | < -4 and | x | > 4
* Solutions for greater than inequalities are TWO inequalities, not one.
To solve greater than inequalities, you split the inequalities into 2 pieces: x < -a and x > a
10. Greater Than Examples:
Solve: | x - 4 | > 6
x - 4 < -6 and x - 4 > 6
Solution: x < -2 and x > 10
Solve: 2 | x + 3 | -3 > 8
11. Number Line Solution Sets Way of graphing answers to Absolute Value Inequalities using a number line.
Open Circle: shows that the number is NOT a solution
Filled-In Circle: shows that the number IS a solution
Shade in rest of number line that corresponds to inequality
To graph inequalities on number line you have to solve the inequality, finding the value(s) of x, and then graph the answer.
Greater-Than Example: | x | > 3 --------> solution: x > 3 and x < -3
<--------------------------------------------------------------->
Less-Than Example: | x | < 3 -------> solution: -3 < x < 3
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12. Graphs and Transformations Standard y=Mx+B graphs using absolute value form in a V-shape. When looking at the graph there is a clear distinction where the function seems to reflect like a mirror. The reason for this is that the negative values are changed and replaced with their distance from zero.
A parabolic shaped function being used with absolute value will look like a canyon with a small hill inside of it, unchanged, or reversed depending on the situation.
14. Absolute Value Graph y = | x | y = - | x |
x y
-3 3
-2 2
-1 1
0 0
1 1
2 2
3 3
If you have a negative sign in front of the absolute value, the graph will be reflected, or flipped, over the x-axis.
15. Transformations
When you have a function in the form y = |x + h| the graph will move h units to the left.
y = | x + 2 |
When you have a function in the form y = |x - h| the graph will move h units to the right.
16. Transformations When you have a function in the form y = |x| + k the graph will move up k units.
When you have a function in the form y = |x| - k the graph will move down k units.
y = | x | - 4
17. Graphs and Transformations When presented with an equation such as the following:
y=|x^2 -3x-10| + |x^2 -9|
the graph is represented as five different functions depending on the location of x. By factoring you can find that these sets have x-intercepts at -3, -2, 3, and 5. When x is less than -3, the function is (x^2 -3x-10) + (x^2 -9) which is simplified to 2x^2 -3x-19.
Between -3 and -2 the value changes because |x^2 -9| is reversed in this region. Here we use (x^2-3x-10) - (x^2 -9) which simplifies to -3x-1. That is the function that represents this region.
18. Graphs and Transformations Between -2 and 3 both values are reversed because both absolute values must be forced to become positive. Therefore, -(x^2-3x-10) - (x^2 -9) simplifies to -2x^2 -3x-19.
Between 3 and 5 |x^2 -9| becomes positive again and the new function is -(x^2 -3x-10) + (x^2 -9) which is simplified to -3x-19.
For any x-value greater than 5, the function returns to the initial function because both values are positive, so the equation is equal to 2x^2 -3x-19.
