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Analyzing Graphs of Absolute Value Functions

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Algebra 2
Miscellaneous Functions
Concept

Absolute Value Function

An absolute value function is any function that contains the absolute value of a variable expression. These functions can also be described as any function that is a transformation of the absolute value parent function, f(x)=∣x∣. f(x)=|x|. Since the absolute value of an expression is never negative, the graph of the function f(x)=∣x∣f(x)=|x| will always lie on or above the xx-axis. Note that ∣-5∣=5and∣5∣=5. |\text{-} 5| = 5 \quad \text{and} \quad |5|=5. This means that the points (-5,5)(\text{-} 5,5) and (5,5)(5,5) both lie on f(x)=∣x∣.f(x)=|x|. As it turns out, these points lie directly across from each other. In fact, this symmetry exists for all inverse input values. Thus, absolute value graphs have a distinct V-shape.

Any function belonging to the absolute value function family can be written using the equation y=a∣x−h∣+k y=a|x-h|+k

where a,a, h,h, and kk are real numbers and a≠0.a\neq0.
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Exercise

Graph the absolute value function f(x)=∣x−2∣−3f(x)=|x-2|-3 using a table of values.

Show Solution
Solution
To graph the given function using a table of values, we can substitute various xx-values into the rule and solve for the corresponding yy-values. The absolute value of a number is always the positive value of that number. For instance, ∣-2∣=2.|\text{-} 2| = 2. Let's first calculate the yy-value that corresponds with x=-3.x=\text{-} 3.
f(x)=∣x−2∣−3f(x)=|x-2|-3
Substitutex=-3x={\color{#0000FF}{\text{-} 3}}
f(-3)=∣-3−2∣−3f({\color{#0000FF}{\text{-} 3}})=|{\color{#0000FF}{\text{-} 3}}-2|-3
SubTermSubtract term
f(-3)=∣-5∣−3f(\text{-} 3)=|\text{-} 5|-3
AbsNeg∣-5∣=5 \left|\text{-}5\right|=5
f(-3)=5−3f(\text{-} 3)= 5-3
SubTermSubtract term
f(-3)=2f(\text{-} 3)= 2
We have found that f(-3)=2.f(\text{-} 3)= 2. Thus, the point (-3,2)(\text{-} 3,2) lies on the graph. We can find other points on the graph in the same way.
xx ∣x−2∣−3|x-2|-3 f(x)f(x)
-2{\color{#0000FF}{\text{-} 2}} ∣-2−2∣−3|{\color{#0000FF}{\text{-} 2}}-2|-3 11
-1{\color{#0000FF}{\text{-} 1}} ∣-1−2∣−3|{\color{#0000FF}{\text{-} 1}}-2|-3 00
0{\color{#0000FF}{0}} ∣0−2∣−3|{\color{#0000FF}{0}}-2|-3 -1\text{-} 1
1{\color{#0000FF}{1}} ∣1−2∣−3|{\color{#0000FF}{1}}-2|-3 -2\text{-} 2
2{\color{#0000FF}{2}} ∣2−2∣−3|{\color{#0000FF}{2}}-2|-3 -3\text{-} 3
3{\color{#0000FF}{3}} ∣3−2∣−3|{\color{#0000FF}{3}}-2|-3 -2\text{-} 2
4{\color{#0000FF}{4}} ∣4−2∣−3|{\color{#0000FF}{4}}-2|-3 -1\text{-} 1
5{\color{#0000FF}{5}} ∣5−2∣−3|{\color{#0000FF}{5}}-2|-3 00
6{\color{#0000FF}{6}} ∣6−2∣−3|{\color{#0000FF}{6}}-2|-3 11
7{\color{#0000FF}{7}} ∣7−2∣−3|{\color{#0000FF}{7}}-2|-3 22

To draw the graph, we can plot these points, then connect them with a V-shaped curve.

Method

Solving Absolute Value Equations Graphically

An absolute value equation is an equation that contains the absolute value of a variable expression. An example of this kind of equation is ∣2x2−8∣=5. \left| 2x^2-8 \right|=5.

As is the case with most equations, these can be solved graphically. This is done by moving all terms except the constant term to one side. The function, which is the non-constant side of the equation, is then graphed and the solution(s) to the equation are found as the point(s) on the graph having the yy-coordinate that equals the constant.
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Exercise

Solve the equation graphically. ∣2x−5∣−3=0 |2x-5|-3=0

Show Solution
Solution

When solving an equation graphically, the first step is to rearrange the equation so that the constant term is alone on one side. ∣2x−5∣−3=0⇔∣2x−5∣=3 |2x-5|-3=0 \quad \Leftrightarrow \quad |2x-5|=3 The left-hand side of the equation can now be expressed as a function, f(x)=∣2x−5∣.f(x)=|2x-5|. We'll draw the graph of f.f.

The solutions to the equation y=∣2x−5∣=3 y=|2x-5|=3 are the xx-coordinates of the points on the graph that have the yy-coordinate 3.3.

The equation has the solutions x=1x=1 and x=4.x=4. We can verify the solutions by testing them in the equation. We'll start with x=1.x=1.
∣2x−5∣=3|2x-5|=3
Substitutex=1x={\color{#0000FF}{1}}
∣2⋅1−5∣=?3|2\cdot {\color{#0000FF}{1}}-5|\stackrel{?}{=}3
MultiplyMultiply
∣2−5∣=?3|2-5|\stackrel{?}{=}3
SubTermSubtract term
∣-3∣=?3|\text{-} 3|\stackrel{?}{=}3
AbsNeg∣-3∣=3 \left|\text{-}3\right|=3
3=33=3
Since x=1x=1 makes a true statement, it is a solution to the equation. Next, we'll verify x=4x=4 in the same way. ∣2(4)−5∣=3⇔∣3∣=3 |2({\color{#0000FF}{4}})-5|=3 \quad \Leftrightarrow \quad |3|=3

Since x=4x=4 makes a true statement, it is also a solution. Thus, the equation has the solutions x=1andx=4. x=1 \quad \text{and} \quad x=4.

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