Post on 15-Mar-2022
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Absolute Value Functions
Properties of absolute values
Property Example |π₯| β₯ 0 |β7| = 7
and |5| = 5
|π₯| = |βπ₯| |β3| = |3|
3 = 3
|π₯ β π¦| = |π₯| β |π¦| |3 β β2| = |3| β |β2| |β6| = 3 β 2
6 = 6
|π₯
π¦| =
|π₯|
|π¦|
|β4
8| =
|β4|
|8|
|β0.5| =4
8
0.5 = 0.5
|π₯| + |π₯| = 2|π₯| |6| + |6| = 2|6| 6 + 6 = 2 β 6
12 = 12
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Absolute Value Function
The rule of an absolute value function is:
π(π₯) = π|π(π₯ β β)| + π vertex : (h, k)
We can remove the βbβ term (so π = 1) using properties of absolute values (which is especially useful
when we are trying to find the rule of an absolute value function):
π(π₯) = π|(π₯ β β)| + π
Ex: Write the following rule such that π = 1.
π(π₯) = 4|β9(π₯ + 7)| β 6
π(π₯) = 4 β |β9| β |(π₯ + 7)| β 6
π(π₯) = 4 β 9|(π₯ + 7)| β 6
π(π₯) = 36|(π₯ + 7)| β 6
Absolute Value Function Basics:
β’ An absolute value function is a β¨ or β§ graphically, formed by 2 linear rays meeting at the vertex.
β’ The graph is symmetrical with an axis of symmetry at π₯ = β.
β’ The slope of the rays is Β±π , provided π = 1. If π β 1, factor out the b term and remove is (as above) to determine the slope of the rays.
β’ If β1 < π < 1 and π β 0, the function is wider than the function when π = 1.
β’ If β1 < π or π > 1, the function is narrower than when π = 1.
β’ Domain: ]ββ, +β[
β’ Range: [π, +β[ if π > 0; ]ββ, π] if π < 0
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Graphing an Absolute Value Function
There are 4 steps to graphing an absolute value function
1) Determine (and plot) the vertex (β, π)
2) Determine whether the function is upright: β¨ (π > 0) or upside down: β§ (π < 0)
3) Determine the slope of each ray (Β±π) (and use slope to find a point on each ray)
4) Connect the vertex and the points with two linear rays.
Ex: Graph the following absolute value functions
a) π(π₯) = |π₯| Vertex: (0,0) Function is upright Right ray slope: 1 Left ray slope: β1
b) π(π₯) = β2|π₯| Vertex: (0,0) Function is upside down Right ray slope: β2 Left ray slope: 2
c) β(π₯) = 2|β2(π₯ β 3)| + 2 β(π₯) = 2|β2||(π₯ β 3)| + 2
β(π₯) = 4|(π₯ β 3)| + 2 Vertex: (3,2) Function is upright Right ray slope: 4 Left ray slope: β4
π(π₯)
π(π₯)
β(π₯)
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Finding the Rule
Recall the rule of an absolute value function is π(π₯) = π|π(π₯ β β)| + π, but the b value can be combined with the a value by using properties of absolute values. When finding the rule of an absolute value function, we use the simplified rule: π(π₯) = π|(π₯ β β)| + π (unless we are given the b value).
There are 4 basic cases (and variations of these) that might occur when asked to find the rule of an
absolute value function:
1) Given the vertex and a point
2) Given the vertex (without it being identified as a vertex) and two points (with the same y-values)
3) Given multiple points on each ray
4) Given multiple points on one ray and one on the other ray, where 2 of the points share y-values
Case 1
If we are given the vertex and a point, we can find the rule of an absolute value function by:
1) Write the general form of the rule
2) Substitute in (β, π) given the vertex
3) Substitute in (π₯, π¦) given the additional point
4) Solve for π
5) Write the rule (replacing π, β, πππ π)
Ex: Determine the rule of an absolute value function has a vertex of (3, 5) and passes through the point
(5, 8).
π(π₯) = π|(π₯ β β)| + π
π(π₯) = π|(π₯ β 3)| + 5
8 = π|(5 β 3)| + 5
3 = π|(2)|
3 = 2π
3
2= π
β΄ π‘βπ ππ’ππ ππ π‘βπ ππ’πππ‘πππ ππ π(π₯) =3
2|(π₯ β 3)| + 5
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Case 2
If we are given the vertex (without it being labeled as a vertex) and two points with the same y-value,
we can find the rule of an absolute value function by:
1) Using the two points with the same y-value to determine β.
2) Check that the third point is the vertex (does π₯ = β?)
3) Write the general form of the rule
4) Substitute in (β, π) given the vertex
5) Substitute in (π₯, π¦) given the additional point
6) Solve for π
7) Write the rule (replacing π, β, πππ π)
Ex: Determine the rule of an absolute value function that passes through the points:
(4, 2), (β2, 2), and (1, β10).
Notice that the points (4, 2) and (β2, 2) have the same y-values. Because absolute value functions
have an axis of symmetry at π₯ = β, we know h must be half way between the two x values of these
points.
β =π₯1 + π₯2
2
β =4 β 2
2
β =2
2
β = 1
π₯ = 1 for the third point we are given, therefore, the point (1, β10) must be the vertex of the function.
So we can use that point and either other point to determine the rule
π(π₯) = π|(π₯ β β)| + π
π(π₯) = π|(π₯ β 1)| β 10
2 = π|(4 β 1)| β 10
12 = π|(3)|
12 = 3π
4 = π
β΄ π‘βπ ππ’ππ ππ π‘βπ ππ’πππ‘πππ ππ π(π₯) = 4|(π₯ β 1)| β 10
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Case 3
If we are given at least two points on each ray, we can find the rule of an absolute value function by:
1) Make a sketch, plotting all 4 points and drawing two lines
2) Use two points on one of the lines to determine the rule of that line
3) Use the negative slope of the one found in step 2 and an additional point, to find the rule for the second line.
4) Find the intersection point of the two lines (this is the vertex)
5) Determine whether a is positive or negative (using the sketch)
6) Write the rule (replacing π, β, πππ π)
Ex: Determine the rule of an absolute value function passing through the points (β1, 4), (β3, β2), (4, 2), πππ (6, β4).
Ray 1:
Ray 2: π = β3 π¦ = ππ₯ + π
π¦ = β3π₯ + π
2 = (β3)(4) + π
2 = β12 + π
14 = π
β΄ π¦ = β3π₯ + 14
Point of Intersection 3π₯ + 7 = β3π₯ + 14
6π₯ + 7 = 14 6π₯ = 7
π₯ =7
6= β
π¦ = 3π₯ + 7
π¦ = 3 (7
6) + 7
π¦ =7
2+ 7
π¦ =21
2= π
Write the Rule Upside down, so π is negative π = β3
Vertex: (7
6,
21
2)
β΄ π(π₯) = β3 |π₯ β7
6| +
21
2
π =π¦2 β π¦1
π₯2 β π₯1
π =4 β (β2)
β1 β (β3)
π =6
2
π = 3
π¦ = ππ₯ + π
π¦ = 3π₯ + π
4 = (3)(β1) + π
4 = β3 + π
7 = π
β΄ π¦ = 3π₯ + 7
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Case 4
If we are given two points on one ray and one point on the other ray (that shares a y-value with one of
the points on the other ray), we can:
1. Determine h
2. Determine a
3. Find the rule of one ray
4. Use h and the rule of one ray to solve for k
5. Write the rule (substituting in a, h, and k)
Ex: Determine the rule of an absolute value function that passes through the points
(β6, β7), (β8, β11) on one ray, and (0, β11) on the other ray.
Note that there are two points with a y-value of β11.
Find h
β =π₯1 + π₯2
2
β =β8 + 0
2
β = β4
Determine slope for ray
π =π¦2 β π¦1
π₯2 β π₯1
π =β11 β (β7)
β8 β (β6)
π =β11 β (β7)
β8 β (β6)
π =β4
β2
π = 2
Determine sign of a
For the absolute value function, π = β2 because the function is upside down.
Determine rule of ray π¦ = ππ₯ + π
β11 = 2(β8) + π β11 = β16 + π
5 = π
Determine k π¦ = 2π₯ + 5
π¦ = 2(β4) + 5 π¦ = β8 + 5
π¦ = β3 so π = β3
Write rule π¦ = π|π₯ β β| + π
π¦ = β2|π₯ + 4| β 3
β΄ π¦ = β2|π₯ + 4| β 3
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Solving Absolute Value Functions
To solve for y, plug in x and solve (remember to the absolute value of a number is always positive).
To solve for x:
1) Isolate the absolute value
2) Create two cases, one where the | | = + and one where the | | = β
3) Solve for x in each case
4) Check for extraneous answers (plug solution into equation)
Ex: Solve 2|π₯ + 4| β 2 = 0
2|π₯ + 4| β 2 = 0
2|π₯ + 4| = 2
|π₯ + 4| = 1
Check π₯ = β3
2|π₯ + 4| β 2 = 0
2|β3 + 4| β 2 = 0
2|1| β 2 = 0
2 β 2 = 0
0 = 0
True
Check π₯ = β5
2|π₯ + 4| β 2 = 0
2|β5 + 4| β 2 = 0
2|β1| β 2 = 0
2 β 2 = 0
0 = 0
True
π₯ + 4 = 1 π₯ = β3
π₯ + 4 = β1 π₯ = β5
β΄ 2|π₯ + 4| β 2 = 0 when
π₯ = β3 πππ π₯ = β5
Ex: Solve 4|π₯ + 2| + 6 = 1
4|π₯ + 2| + 6 = 1
4|π₯ + 2| = β5
|π₯ + 2| = β5
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No solutions (because the absolute value of a term can never be negative).
Recall
If: |π₯| = 2
Then π₯ = 2 and π₯ = β2
So if: |π₯ + 6| = 2
Then π₯ + 6 = 2 and π₯ + 6 = β2
And π₯ = β4 and π₯ = β8
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Ex: Given π(π₯) = |π₯ β 2| β 1 and π(π₯) = β2π₯ , solve π(π₯) = π(π₯)
π(π₯) = π(π₯)
|π₯ β 2| β 1 = β2π₯
|π₯ β 2| = β2π₯ + 1
Check π₯ = 1 |1 β 2| β 1 = β2(1)
|β1| β 1 = β2
1 β 1 = β2
0 = β2
False (so not a solution)
Check π₯ = β1 |β1 β 2| β 1 = β2(β1)
|β3| β 1 = 2
3 β 1 = 2
2 = 2
True
π₯ β 2 = β2π₯ + 1
3π₯ = 3
π₯ = 1
π₯ β 2 = β(β2π₯ + 1)
π₯ β 2 = 2π₯ β 1)
βπ₯ = 1
π₯ = β1
β΄ π(π₯) = π(π₯) π€βππ π₯ = β1
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Solving Absolute Value Inequalities
Recall that |π₯| > 3 means that π₯ (whatever is inside the absolute value) is more than 3 units away from
0. On a number line, this means:
So |π₯| > 3 means π₯ > 3 and x < β3 (note the change of direction in the inequality)
We will use this when solving inequalities for absolute value functions.
To solve an absolute value inequality:
1) Isolate the absolute value
2) Create two cases (remember to flip the inequality when you take the negative in the second
case)
3) Solve (remember to change the inequality if you multiply or divide by a negative number)
4) Check using equality
5) State solution
Ex: Solve |π₯ β 7| β 4 < β2
|π β π| β π < βπ
|π₯ β 7| < 2
Check π < π (so use π = π) |π₯ β 7| β 4 = β2
|9 β 7| β 4 = β2
|2| β 4 = β2
2 β 4 = β2
β2 = β2
True
Check π > π (so use (π = π) |π₯ β 7| β 4 = β2
|5 β 7| β 4 = β2
|β2| β 4 = β2
2 β 4 = β2
β2 = β2
True
π₯ β 7 < 2
π₯ < 9
π₯ β 7 > β2
π₯ > 5
So |π₯ β 7| β 4 < β2 when π₯ is greater than 5 and less than 9
β΄ πππ ππππππππ ππ ]π, π[
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Ex: Solve |2π₯| β 4 β₯ 0
|ππ| β π β₯ π
|2π₯| β₯ 4
Check π β₯ π (so use π = π) |2π₯| β 4 = 0
|2 β 2| β 4 = 0
|4| β 4 = 0
4 β 4 = 0
0 = 0
True
Check π β€ βπ (so use π = βπ) |2π₯| β 4 = 0
|2 β β2| β 4 = 0
|β4| β 4 = 0
4 β 4 = 0
0 = 0
True
2π₯ β₯ 4
π₯ β₯ 2
2π₯ β€ β4
π₯ β€ β2
So |2π₯| β 4 β₯ 0 when π₯ is greater than or equal to 2 and less than or equal to β2
β΄ πππ ππππππππ ππ ]ββ, βπ] βͺ [π, +β[
Ex: Given π(π₯) = |π₯ β 2| and π(π₯) = 4π₯ + 8, determine the interval over which π(π₯) β₯ π(π₯)
π(π) β₯ π(π)
|π₯ β 2| β₯ 4π₯ + 8
Check π β€ βππ
π
|β10
3β 2| = 4 (β
10
3) + 8
|β16
3| = β
40
3+ 8
16
3= β
16
3
False (so not a solution)
Check π β€ βπ
π
|β6
5β 2| = 4 (β
6
5) + 8
|β16
5| = β
24
5+ 8
16
5=
16
5
True
π₯ β 2 β₯ 4π₯ + 8
β3π₯ β₯ 10
π₯ β€ β10
3
π₯ β 2 β€ β(4π₯ + 8)
π₯ β 2 β€ β4π₯ β 8
5π₯ β€ β6
π₯ β€ β6
5
So π(π₯) β₯ π(π₯) when π₯ is less
than or equal to β6
5
β΄ πππ ππππππππ ππ ]ββ, βπ
π]
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Finding the Inverse of an Absolute Value Function
Important note: Remember that taking the inverse of a parabola gives us a sideways parabola that is not
a function (and specifically not a square root function). When we take the inverse of an absolute value
function, we will also get something that is not a function (and looks like a sideways absolute value
function).
When we find the inverse of an absolute value function, we will get what look like two linear functions,
and we must state the domain of the πβ1(π₯), which was the range of π(π₯).
Ex: Find the inverse of π(π₯) = 2|π₯ β 2| + 1 and sketch both π(π₯) and πβ1(π₯).
π(π₯) = 2|π₯ β 2| + 1 Domain: ]ββ, +β[ Range: [1, +β[
Inverse
π₯ = 2|π¦ β 2| + 1
π₯ β 1 = 2|π¦ β 2|
π₯ β 1
2= |π¦ β 2|
π₯ β 1
2= π¦ β 2
π₯ β 1
2+ 2 = π¦
π¦ =1
2π₯ β
1
2+ 2
π¦ =1
2π₯ +
3
2
β (π₯ β 1
2) = π¦ β 2
β (π₯ β 1
2) + 2 = π¦
π¦ = β1
2π₯ +
1
2+ 2
π¦ = β1
2π₯ +
5
2
β΄ the inverse of π(π₯) = 2|π₯ β 2| + 1 is π¦ =1
2π₯ +
3
2 and π¦ = β
1
2π₯ +
5
2 with a domain of [1, +β[
π(π₯)
πβ1(π₯)