7. Graphing Absolute Value Equations and Inequalities as Systems
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Chapter 5
7.
Graphing Absolute Value Equations and Inequalities as Systems
This lesson delves into the realm of graphing absolute value equations and inequalities as systems. It provides insights into modeling real-world scenarios, such as the path of sunlight reflecting off water or the trajectory of a swimmer in a pool, using absolute value functions. The lesson also emphasizes the transformation of absolute value functions into piecewise functions, offering a comprehensive view of their graphical representations. By understanding these concepts, one can effectively interpret and graph absolute value equations in various contexts.
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Lesson Settings & Tools
| | 12 Theory slides |
| | 8 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Graphing Absolute Value Equations and Inequalities as Systems
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Absolute Value Functions as Piecewise Functions
Jordan is getting ready for the inter-class swimming competition at her school.
She swims to the far end of the pool and comes back to the starting point. The function below models Jordan's distance from the far end of the pool after t seconds. d(t) = 2|t-25|
a Rewrite the given absolute value function as a piecewise function.
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Graphs of Piecewise Functions
Examine the given piecewise functions and inequalities on the left. Match them with their corresponding graph.
Is it possible to represent these graphs with absolute value functions or absolute value inequalities? If so, write these functions.
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Writing an Absolute Value Function as a Piecewise Function
An expression involving an absolute value can be defined as follows.
|a| = - a, & if a<0 a, & if a≥0
Using this definition, an absolute value function can be written as a piecewise function. Consider an example function.
f(x) = - 4|1/2x+1 |+5
The above function will be rewritten as a piecewise function. The procedure can be completed in two steps.
1
Use the Definition of Absolute Value
expand_more First, identify the expression that involves absolute value. f(x) = - 4 |1/2x+1 | +5 This expression is equivalent to - ( 12x+1) if 12x+1 is less than 0. Conversely, this expression is equivalent to 12x+1 if 12x+1 is greater than or equal to 0. |1/2x+1|= - ( 12x+1 ), & if 12x+1<0 [0.6em] 12x+1, & if 12x+1 ≥ 0 Now this piecewise definition of the absolute value expression can be used for the given function. f(x) = - 4 ( - ( 12x+1 ) ) + 5, & if 12x + 1 < 0 [0.6em] - 4 ( 12x+1 ) + 5, & if 12x + 1 ≥ 0
2
Simplify Expressions and Inequalities
expand_more Next, the expressions will be simplified.
Finally, the inequalities describing the domains of the pieces will be rearranged. To do so, subtract 1 from both sides of the inequalities, then multiply both sides by 2.
f(x)= 2x+9, & if x<- 2 - 2x+1, & if x ≥ - 2
The given absolute value function has been written as a piecewise function.
f(x)= - 4 (- ( 12x+1 ) ) + 5, & if 12x+1<0 (I) [0.6em] - 4 ( 12x+1 ) + 5, & if 12x+1 ≥ 0 (II)
▼
(I), (II):Simplify
Distr
(II): Distribute -1
f(x)= - 4 (- 12x-1 ) + 5, & if 12x+1<0 [0.6em] - 4 ( 12x+1 ) + 5, & if 12x+1 ≥ 0
(I), (II):Distribute - 4
f(x)= 2x+4 + 5, & if 12x+1<0 [0.6em] - 2x-4 + 5, & if 12x+1 ≥ 0
(I), (II): Add terms
f(x)= 2x+9, & if 12x+1<0 [0.6em] - 2x+1, & if 12x+1 ≥ 0
| Absolute Value Function | Piecewise Function |
|---|---|
| f(x) = - 4|1/2+1 |+5 | f(x) = 2x+9, & if x<- 2 - 2x+1, & if x ≥ - 2 |
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Who Correctly Rewrote an Absolute Value Function as a Piecewise Function?
Dylan and Kriz have been asked to write the following absolute value function as a piecewise function. f(x)=- 7|7-x|+8 The functions they wrote are shown in the diagram.
{"type":"choice","form":{"alts":["Kriz","Dylan","Both","Neither"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":0}
Hint
Start by identifying the absolute value expression.
Solution
First, the given function will be written as a piecewise function. Then it can be seen who is correct. Consider the absolute value function.
f(x)=- 7|7-x|+8
By using the definition of an absolute value the expression, |7-x| can be rewritten. If 7-x is less than 0, its absolute value is equal to its opposite value. Conversely, if 7-x is greater than or equal to 0, its absolute value is equal to itself. |7-x| = - (7-x) & if 7-x < 0 7-x & if 7-x ≥ 0
With this information in mind, the absolute value function can be rewritten.
f(x) = - 7 [ - (7-x)] +8 & if 7-x < 0 - 7 ( 7-x) +8& if 7-x ≥ 0
Next, the expressions need to be simplified.
Finally, the domain of this piecewise function should be rearranged. First, 7 will be subtracted from both sides of the inequalities.
f(x)= - 7x+57 & if 7-x < 0 7x-41 & if 7-x ≥ 0 ⇕ f(x)= - 7x+57 & if - x < - 7 7x-41 & if - x ≥ - 7
By dividing the inequalities by - 1, the parts of the domain can be identified. Recall that dividing an inequality by a negative number reverses the inequality symbol.
f(x)= - 7x+57 & if - x < - 7 7x-41 & if - x ≥ - 7 ⇕ f(x)= - 7x+57 & if x > 7 7x-41 & if x ≤ 7
The absolute value function is now completely rewritten as a piecewise function.
f(x)= - 7 [- (7-x)] +8 & if 7-x < 0 (I) - 7 (7-x) +8& if 7-x ≥ 0 (II)
▼
(I), (II):Simplify
Distr
(I): Distribute -1
f(x)= - 7 (-7+x) +8 & if 7-x < 0 - 7 (7-x) +8& if 7-x ≥ 0
(I), (II):Distribute - 7
f(x)= 49-7x+8 & if 7-x < 0 - 49+7x+8 & if 7-x ≥ 0
(I), (II): Add terms
f(x)= - 7x+57 & if 7-x < 0 7x-41 & if 7-x ≥ 0
| Absolute Value Function | Piecewise Function |
|---|---|
| f(x)=- 7 |7-x|+8 | f(x) = - 7x+57 & if x > 7 7x-41 & if x ≤ 7 |
Comparing this function with the functions written by Dylan and Kriz, it appears that Kriz wrote it correctly.
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Graphing an Absolute Value Function as a Piecewise Function
Absolute value functions can be written as piecewise functions. By graphing the pieces for their domains, the graph of the absolute value function can be obtained. As an example, the following function will be graphed. f(x) = |2x-6|-2
Its graph can be drawn in four steps.
1
Write the Absolute Value Function as a Piecewise Function
expand_more By the definition of absolute value, an absolute value expression can be divided into two. With this in mind, two function rules for the given absolute value function can be defined. f(x) = |2x-6|-2 ⇕ f(x) = - (2x-6)-2 & if 2x-6<0 2x-6-2 & if 2x-6 ≥ 0 After simplifying the expressions and inequalities, a piecewise-defined function is obtained. f(x) = - 2x+4 & if x< 3 2x-8 & if x ≥ 3 To graph the function, each individual piece will be graphed and then combined on the same coordinate plane.
2
Graph the First Piece
expand_more First the graph of y=- 2x+4 will be drawn for the domain x<3. Notice that this piece is written in slope-intercept form. y= - 2x+ 4 This function has a slope of - 2 and a y-intercept of 4. Since the domain of this piece domain does not contain 3, its graph ends with an open point at x=3.
3
Graph the Second Piece
expand_more Now, the other piece will be drawn for the domain x ≥ 3. y= 2x - 8 Using its slope 2 and y-intercept - 8, the graph of the second piece can be drawn. Since this time 3 belongs to the domain, the graph of this piece ends with a closed point at x=3.
4
Combine the Graphs on the Same Coordinate Plane
expand_more Finally, these two pieces can be combined on the same coordinate plane.
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Modeling the Wings of a Swallow Using an Absolute Value Function
LaShay likes to make connections between the shapes she finds in daily life and the concepts she encounters in her math lessons. While watching a documentary about swallows, LaShay thinks that the wings of a swallow can be modeled by an absolute value function.
The axes represent lengths in inches.
The point (8,2.4) is on the graph. Using these three points, two function rules can be written. One for the decreasing part, and the other for the increasing part of the graph. The domain of these parts can be written as follows.
Decreasing Part:& 0 ≤ x < 4 Increasing Part:& 4 ≤ x ≤ 8
Note the point at x=4 can belong to either piece, so long as it belongs to only one of them. The next step will then be to find the equation for both lines. For this, recall the slope-intercept form of a linear function.
y= mx + b
In this form m is the slope of the line and b is the y-intercept. Start with the decreasing part. The y-intercept and slope can be found from the graph. 
f(x)= - 0.6x+2.4 & if 0 ≤ x < 4 0.6x-2.4 & if 4 ≤ x ≤ 8
To do so, the function rules can be rearranged so that one rule contains an expression and the other rule contains its opposite.
The function rules contain the expressions - (x-4) and x-4. These expressions produce non-negative values in their domains. Therefore, they can be written using absolute values. f(x)= - 0.6(x-4) & if 0 ≤ x < 4 0.6(x-4) & if 4 ≤ x ≤ 8 ⇓ f(x) = 0.6|x-4|
The domain of this absolute value function is the union of the domains for the function rules.
Function: & f(x) = 0.6|x-4| Domain: & 0 ≤ x ≤ 8
External credits: sanchezn
a If the tip of one wing is at (0,2.4) and the swallow's head is at (4,0), write a piecewise function that models the wings.
b Rewrite the function as an absolute value function and state its domain.
Answer
a Piecewise Function:
f(x) = - 0.6x+2.4 & if 0 ≤ x < 4 0.6x-2.4 & if 4 ≤ x ≤ 8
b Absolute Value Function: f(x) = 0.6|x-4|
Domain: 0 ≤ x ≤ 8
Domain: 0 ≤ x ≤ 8
Hint
a An absolute value function is symmetric across the vertical line passing through its vertex. Use this symmetry to find another point on the graph.
b Rearrange the rule for each piece so that one rule contains an expression and the other rule contains its opposite.
Solution
a The points (0,2.4) and (4,0) are on the graph of the absolute value function. Since an absolute value function is symmetric across the vertical line passing through its vertex, one more point on the graph can be found.
External credits: sanchezn
For this line the y-intercept is b= 2.4 and the slope is m= - 0.6. Decreasing Part & Domain y= - 0.6 x + 2.4 & 0 ≤ x < 4 Similarly, an equation for the increasing part can also be written.
For the increasing line, the y-intercept is - 2.4 and the slope is 0.6. Increasing Part & Domain y= 0.6 x - 2.4 & 4 ≤ x ≤ 8 Knowing the equations for both lines, the absolute value function can be written as a piecewise function. f(x)= - 0.6x+2.4 & if 0 ≤ x < 4 0.6x-2.4 & if 4 ≤ x ≤ 8
b In this part the piecewise function will be used to write an absolute value function.
f(x)= - 0.6x+2.4 & if 0 ≤ x < 4 (I) 0.6x-2.4 & if 4 ≤ x ≤ 8 (II)
(I), (II): Factor out 0.6
f(x)= 0.6(- x+ 4) & if 0 ≤ x < 4 0.6(x-4) & if 4 ≤ x ≤ 8
FactorOut
(I): Factor out - 1
f(x)= 0.6[- (x- 4)] & if 0 ≤ x < 4 0.6(x-4) & if 4 ≤ x ≤ 8
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Modeling Rio Negro Bridge Using an Absolute Value Function
The Rio Negro Bridge is a 3595-meter long cable-stayed bridge over the Rio Negro in Brazil. 
Kevin researches the bridge on the internet and writes an absolute value function for the path of the two longest cables.
h(x) = - 0.6|x-200|+120
Here, x is the horizontal distance to the leftmost point of the path and h(x) is the height of the path.
The graph ends with an open circle because its domain is the set of x-values less than 200. For x-values greater than or equal to 200, the graph of y=- 0.6x+240 will be drawn. This is a linear function written in slope-intercept form.
y = - 0.6x + 240 ⇓ l Slope: - 0.6 y-intercept: 240
Using this information, its graph can be drawn.
This piece ends with a closed circle, as 200 is in its domain. Finally, both graphs will be combined on the same coordinate plane. 
External credits: Dennis Jarvis
a Write the given absolute value function as a piecewise function and graph it.
b If all measures are in meters, what is the distance between the leftmost and the rightmost points of the path?
Answer
a Function:
h(x) = - 0.6x+2.4 & if 0 ≤ x < 4 0.6x-2.4 & if 4 ≤ x ≤ 8
Graph:
b 400 meters
Hint
a Use the definition of absolute values to write a piecewise function.
b Identify the points whose y-coordinates are zero.
Solution
a To rewrite the given function, the definition of absolute value will be used.
h(x)= - 0.6(- [x-200])+120 & if x < 200 (I) - 0.6(x-200)+120 & if x ≥ 200 (II)
▼
Simplify
Distr
(I): Distribute -1
h(x)= - 0.6(- x+200)+120 & if x < 200 - 0.6( x-200)+120 & if x ≥ 200
(I), (II): Distribute - 0.6
h(x)= 0.6x-120+120 & if x < 200 - 0.6x+120+120 & if x ≥ 200
(I), (II): Add terms
h(x)= 0.6x & if x < 200 - 0.6x+240 & if x ≥ 200
b The leftmost and the rightmost point of the path are (0,0) and (400,0).
Since all the measures are in meters, the distance between these two points is 400 meters.
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Writing an Absolute Value Inequality as a Piecewise Inequality
An absolute value inequality can be formed by replacing the equals sign in an absolute value function with an inequality symbol. Therefore, writing an absolute value inequality as a piecewise inequality can be compared to writing an absolute value function as a piecewise function. Consider an example absolute value inequality.
g(x) > 3|2x-7 |-9
This inequality can be rewritten as a piecewise inequality in two steps.
1
Use the Definition of Absolute Value
expand_more First, identify the expression that involves the absolute value. g(x) > 3 |2x-7|-9 If 2x-7 is greater than or equal to 0, then its absolute value is equal to itself. Conversely, if 2x-7 is less than 0, then its absolute value is equal to its opposite value. |2x - 7| = 2x - 7 & if 2x - 7 ≥ 0 - (2x - 7) & if 2x - 7 < 0 Now, the piecewise definition of the absolute value expression can be used for the given inequality. g(x) > 3 (2x - 7) - 9 & if 2x - 7 ≥ 0 3 [ - (2x - 7)] - 9 & if 2x - 7 < 0
2
Simplify Expressions and Inequalities
expand_more Next, the expressions will be simplified.
Finally, the inequalities used to describe the parts of the domain will be rearranged. To do so, add 7 to both sides and then divide both sides by 2.
g(x) > 6x-30 & if 2x-7 ≥ 0 [0.6em] - 6x+12 & if 2x-7 < 0 ⇕ g(x) > 6x-30 & if x ≥ 72 [0.6em] - 6x+12 & if x< 72
The given absolute value inequality has been written as a piecewise inequality.
g(x)> 3(2x-7)-9 & if 2x-7 ≥ 0 (I) 3[- (2x-7)]-9 & if 2x-7<0 (II)
▼
(I), (II): Simplify
Distr
(II): Distribute - 1
g(x)> 3( 2x-7)-9 & if 2x-7 ≥ 0 3 (- 2x+7)-9 & if 2x-7<0
(I), (II): Distribute 3
g(x)> 6x-21-9 & if 2x-7 ≥ 0 - 6x+21-9 & if 2x-7<0
(I), (II): Subtract terms
g(x)> 6x-30 & if 2x-7 ≥ 0 - 6x+12 & if 2x-7<0
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Graphing an Absolute Value Inequality as a Piecewise Inequality
Absolute value inequalities in two variables can be written as piecewise inequalities. By drawing the graph of each piece in the piecewise inequality, the graph of the absolute value inequality is also drawn. Consider an absolute value inequality.
g(x) < - 0.5|x+6|+2
This inequality can be graphed as a piecewise inequality in four steps.
The domain for the boundary line contains x-values less than - 6. This means that the only part of the graph that should be considered is to the left of x = - 6. Additionally, the boundary line will have an open endpoint at x=- 6. 
Next, the region to be shaded will be determined. To do so, choose a point whose x-coordinate is less than - 6 but is not on the boundary line — for example, (- 8,0).
Since the point satisfies the inequality, the region that contains the point will be shaded.
The domain for the boundary line contains the x-values greater than or equal to - 6. This means that the only part of the graph that should be considered is to the right of x = - 6. Additionally, the boundary line will have a closed endpoint at x=- 6. 
Now, to determine which region must be shaded choose a point with a x-coordinate greater than - 6 but that is not on the line. For example, (0,0) can be used.
Since the point does not satisfy the inequality, the region that does not contain the point will be shaded.
1
Write the Absolute Inequality as a Piecewise Inequality
expand_more Use the definition of absolute values to split the inequality into two pieces.
g(x) < - 0.5 |x+6|+2 ⇕ g(x) < - 0.5 [ - (x + 6)] + 2 & if x + 6 < 0 - 0.5( x + 6) + 2 & if x + 6 ≥ 0
Next, the expressions and the inequalities will be simplified.
Now, the inequalities describing the domain of the pieces will be rearranged. To do so, subtract 6 from both sides of the inequalities.
g(x) < 0.5x+5 & if x<- 6 - 0.5x-1 & if x ≥ - 6
To graph the piecewise-defined inequality, first each individual piece of the inequality will be drawn. Then the graphs will be combined on the same coordinate plane.
g(x)< - 0.5 [- (x+6)]+2 & if x+6<0 (I) - 0.5 (x+6)+2 & if x+6 ≥ 0 (II)
▼
(I), (II):Simplify
Distr
(I): Distribute -1
g(x)< - 0.5 (- x-6)+2 & if x+6<0 - 0.5 ( x+6)+2 & if x+6 ≥ 0
(I), (II):Distribute - 0.5
g(x)< 0.5x+3+2 & if x+6<0 - 0.5 x-3+2 & if x+6 ≥ 0
(I), (II): Add terms
g(x)< 0.5x+5 & if x+6<0 - 0.5 x-1 & if x+6 ≥ 0
2
Graph the First Piece
expand_more Begin by determining the boundary line. This line can be found by replacing the inequality symbol with the equals sign. First Piece & Boundary Line y < 0.5x+5 & y = 0.5x+5 The boundary line is already written in slope-intercept form. Using the slope and y-intercept, the line can be graphed. Since the inequality is strict, the line will be dashed.
3
Graph the Second Piece
expand_more The second piece can be graphed similarly. First, write the equation of the boundary line. Second Piece & Boundary Line y < - 0.5x-1 & y= - 0.5x-1 The slope and the y-intercept of the boundary line are - 0.5 and - 1, respectively. Since the inequality is strict, the boundary line will be dashed.
4
Combine the Graphs on the Same Coordinate Plane
expand_more Finally, the graphs of the two pieces can be combined on the same coordinate plane.
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Absolute Value Inequality as Piecewise Inequality
Maya notices that the region illuminated by a car's left headlight can be modeled by an absolute value inequality.
External credits: Hari Panicker
The inequality below models this region. g(x) ≥ 5|x-0.5| - 1.5
a If the right headlight is 1 unit away from the left, write a piecewise-defined inequality for the right headlight.
b Graph the piecewise-defined inequality for the right headlight.
Answer
a y ≥ - 5x+6 & if x < 1.5 5x -9 & if x ≥ 1.5
b
Hint
a Start by writing an absolute value inequality to represent the right headlight. This is a horizontal translation of the given absolute value inequality.
b To graph the absolute value inequality as a piecewise inequality, start by drawing each piece for its domain and then combine the pieces on the same coordinate plane.
Solution
a An inequality for the right headlight can be obtained by a horizontal translation of the given absolute value inequality. Since the right headlight is 1 unit to the right of the left headlight, the given inequality needs to be translated 1 unit to the right. This can be done by subtracting 1 from the input of the rule.
y≥ 5 [- (x-1.5)]-1.5 & if x-1.5<0 (I) 5 (x-1.5)-1.5 & if x-1.5 ≥ 0 (II)
▼
(I), (II):Simplify
Distr
(I): Distribute -1
y≥ 5 (- x+1.5)-1.5 & if x-1.5<0 5 (x-1.5)-1.5 & if x-1.5 ≥ 0
(I), (II):Distribute 5
y≥ - 5x+7.5-1.5 & if x-1.5<0 5x-7.5-1.5 & if x-1.5 ≥ 0
(I), (II): Subtract terms
y≥ - 5x+6 & if x-1.5<0 5x-9 & if x-1.5 ≥ 0
b Consider the first piece of the inequality. The domain for this piece and the boundary line can be written.
External credits: Hari Panicker
Closure
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Absolute Value Functions as Piecewise Functions
Considering the methods and examples discussed in this lesson, the challenge presented at the start can now be solved. Jordan swims to the far end of the pool and comes back to the starting point. The absolute value function that models Jordan's distance from the far end after t seconds is given.
a Rewrite the given function as a piecewise function.
b Draw the graph of the absolute value function as a piecewise function and state a reasonable domain and range for the function.
Answer
a d(t) = - 2t+50 & if t < 25 2t-50 & if t ≥ 25
b Graph:
Domain: 0≤ t ≤ 50
Range: 0≤ d(t) ≤ 50
Hint
a Use the definition of absolute value to write the given function as a piecewise function.
b Determine the slope of each piece. What is the length of the pool? Use it to write a range.
Solution
a Consider the absolute value function that will be rewritten as a piecewise function.
d(t)= 2|t-25| The absolute value expression in this function can be divided into two cases.
- If t-25 is less than 0, then its absolute value is equal to its opposite value.
- If t-25 is greater than or equal to 0, then its absolute value is equal to itself.
d(t)= 2 [- (t-25)] & if t-25 < 0 (I) 2 (t-25) & if t-25 ≥ 0 (II)
▼
(I), (II):Simplify
d(t)= - 2t+50 & if t-25 < 0 2t-50 & if t-25 ≥ 0
b The piecewise function will now be graphed. To do so, each individual piece will be graphed and then combined on one coordinate plane.
d(t) = - 2t+50 & if t<25 2t-50 & if t≥ 25 First graph y=- 2t +50. This function has a slope of - 2 and a y-intercept of 50. Since the domain for this piece is the set of numbers less than 25, the graph should end with an open point at t=25.
Similarly, the other piece y=2t-50 can be drawn. Its slope is 2. The domain of this piece contains the t-values greater than or equal to 25. Therefore, it has a closed endpoint at t=25.
The combination of the above graphs is the graph of the piecewise function.
d(t)=2|t-25|
Substitute
d(t)= 50
50=2|t-25|
▼
Solve for t
DivEqn
.LHS /2.=.RHS /2.
25=|t-25|
RearrangeEqn
Rearrange equation
|t-25|=25
StateSol
State solutions
lct-25= 25 & (I) t-25=- 25 & (II)
(I), (II): LHS+25=RHS+25
lt=50 t=0
Graphing Absolute Value Equations and Inequalities as Systems
Exercise 3.1
Determine which functions have the same graph, if any.
I. & y =
- |x+2|+2 & if x < 0
- |x-2|+2 & if x > 0
II. & y =
- |x|+4 & if x < - 2
|x| & if - 2 ≤ x ≤ 2
- |x|+4 & if x > 2
III. & y =
x+4 & if x ≤ - 2
- x & if - 2 < x < 0
x & if 0 < x < 2
- x+4 & if x ≥ 2
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We will graph the given piecewise functions and then compare them. To graph a piecewise function, we should think about the graph of each individual piece of the function.
Graphing Function I
We see that each piece of the first function is an absolute value function. y = - |x+2|+2 & if x < 0 - |x-2|+2 & if x > 0 First, we will graph y = - |x+2|+2 for the domain x<0. The vertex of this absolute value function is (- 2,2). Its graph should be an inverted V because of the minus sign in front of the absolute value symbol. Since the endpoint is not included, this piece should end with an open circle.
Next, we will graph y=- |x-2|+2 for the domain x > 0. Its vertex is the point (2,2). The graph of this piece is also an inverted V-shaped graph. Since the endpoint is not included, we will start the piece with an open circle.
Finally, we can combine the pieces into one coordinate plane.
Looking at the pieces together, we can see that the function is not defined for x=0. We can also see there are no gaps in the possible values of y but they are all less than or equal to 2.
Graphing Function II
The second function consists of three absolute value functions. y = - |x|+4 & if x < - 2 |x| & if - 2 ≤ x ≤ 2 - |x|+4 & if x > 2 Let's graph the first piece, y = - |x|+4 for the domain x< - 2. The vertex of this absolute value function is (0,4) and its graph should be an inverted V because of the minus sign in front of the absolute value symbol. Since the endpoint is not included, this piece should end with an open circle.
Next, we will graph y=|x| for the domain - 2 ≤ x ≤ 2. This piece will have closed endpoints.
The last piece has the same rule as the first piece. We can show the part where x > 2.
Finally, we can combine the pieces into one coordinate plane.
Graphing Function III
Our last function has four pieces. y = x+4 & if x ≤ - 2 - x & if - 2< x < 0 x & if 0 < x < 2 - x+4 & if x ≥ 2 First we will graph y=x+4 for the domain x ≤ - 2. This function has a slope of 1 and a y-intercept of 4. Since the endpoint is included, this piece should end with a closed circle.
Next, we will graph y=- x for the domain - 2 < x < 0. Since both endpoints are not included, we will start and end the piece with open circles.
The third piece y=x for the domain 0 < x < 2 can be drawn as follows.
The last piece of the function has a slope of - 1 and a y-intercept of 4. We also start this piece with a closed circle because its domain is x ≥ 2.
Finally, we can put all the pieces together.
Notice that this function is not defined when x=0.
Comparing Graphs
Comparing the graphs of the given functions, we see that the functions have quite similar graphs. However, when x=0, although Function II is defined, other functions are not.
Therefore, only the functions I and III have the same graph.
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Graphing Absolute Value Equations and Inequalities as Systems
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2
3
()
∣
sin
cos
tan
0
.
π
x
y
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