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Analyzing Graphs of Absolute Value Functions


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)=xf(x)=|x| will always lie on or above the xx-axis. Note that -5=5and5=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=axh+k y=a|x-h|+k

where a,a, h,h, and kk are real numbers and a0.a\neq0.

Graph the absolute value function f(x)=x23f(x)=|x-2|-3 using a table of values.

Show 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(-3)=-323f({\color{#0000FF}{\text{-} 3}})=|{\color{#0000FF}{\text{-} 3}}-2|-3
f(-3)=-53f(\text{-} 3)=|\text{-} 5|-3
f(-3)=53f(\text{-} 3)= 5-3
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 x23|x-2|-3 f(x)f(x)
-2{\color{#0000FF}{\text{-} 2}} -223|{\color{#0000FF}{\text{-} 2}}-2|-3 11
-1{\color{#0000FF}{\text{-} 1}} -123|{\color{#0000FF}{\text{-} 1}}-2|-3 00
0{\color{#0000FF}{0}} 023|{\color{#0000FF}{0}}-2|-3 -1\text{-} 1
1{\color{#0000FF}{1}} 123|{\color{#0000FF}{1}}-2|-3 -2\text{-} 2
2{\color{#0000FF}{2}} 223|{\color{#0000FF}{2}}-2|-3 -3\text{-} 3
3{\color{#0000FF}{3}} 323|{\color{#0000FF}{3}}-2|-3 -2\text{-} 2
4{\color{#0000FF}{4}} 423|{\color{#0000FF}{4}}-2|-3 -1\text{-} 1
5{\color{#0000FF}{5}} 523|{\color{#0000FF}{5}}-2|-3 00
6{\color{#0000FF}{6}} 623|{\color{#0000FF}{6}}-2|-3 11
7{\color{#0000FF}{7}} 723|{\color{#0000FF}{7}}-2|-3 22

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


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 2x28=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.

Solve the equation graphically. 2x53=0 |2x-5|-3=0

Show Solution

When solving an equation graphically, the first step is to rearrange the equation so that the constant term is alone on one side. 2x53=02x5=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)=2x5.f(x)=|2x-5|. We'll draw the graph of f.f.

The solutions to the equation y=2x5=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.
215=?3|2\cdot {\color{#0000FF}{1}}-5|\stackrel{?}{=}3
-3=?3|\text{-} 3|\stackrel{?}{=}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=33=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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