19.1: The Absolute Value Function

Videos

Example: Graph Absolute Value Functions Using a Table

Example: Domain and Range of an Absolute Value Function

Example: Find the Equation of a Transformed Absolute Value Function From a Graph

Example: Find the Equation of a Transformed Absolute Value Function From a Graph

Definitions and Theorems

Definition: Absolute Value (as a number)

The absolute value (or modulus) of a real number \(x\), denoted \(|x|\), is its distance from zero on the real number line. It is defined by\[|x| = \begin{cases} x & \text{if } x \geq 0, \\ -x & \text{if } x \lt 0. \end{cases}\nonumber\]

Definition: Absolute Value (as a function)

The absolute value function is defined to be \( f(\blacksquare) = |\blacksquare| \), where\[ |\blacksquare| = \begin{cases}
\blacksquare, & \text{ if } \blacksquare \geq 0 \\[6pt]
-\blacksquare, & \text{ if } \blacksquare < 0 \\[6pt]
\end{cases} \nonumber \]Its domain is all real numbers, and its range is \([0,\infty)\). When written in its fundamental form, \(f(x) = |x|\), its graph is the V-shaped curve formed by the two rays \(y=x\) (for \(x\geq 0\)) and \(y=-x\) (for \(x\lt 0\)), meeting at a single sharp corner called the vertex. For the parent function the vertex is the origin, \((0,0)\).

Theorem: Radical Form of Absolute Value

For every real number \(x\),\[|x| = \sqrt{x^2}.\nonumber\]Equivalently, for every real number \(x\), \(\sqrt{x^2} = |x|\).

Caution: \(\sqrt{x^2}\) Is Not Always \(x\)

A common error is to simplify \(\sqrt{x^2}\) to \(x\). This is valid only when \(x \geq 0\). Because the radical symbol denotes the nonnegative square root, the correct simplification for every real number \(x\) is \(\sqrt{x^2} = |x|\). For instance, \(\sqrt{(-5)^2} = \sqrt{25} = 5 = |-5|\), not \(-5\).

Theorem: Properties of Absolute Value

For all real numbers \(x\) and \(y\), the absolute value satisfies\[|x| \geq 0, \qquad |-x| = |x|, \qquad |x|^2 = x^2.\nonumber\]The absolute value is multiplicative and respects quotients:\[|xy| = |x|\,|y| \qquad \text{and} \qquad \left|\dfrac{x}{y}\right| = \dfrac{|x|}{|y|} \quad (y \neq 0).\nonumber\]For every integer \(n \geq 0\),\[|x^n| = |x|^n.\nonumber\]

Figure \(\PageIndex{1}\): Graph of \(f(x)=|x|\). The two arms have slopes \(1\) and \(-1\).

The Sharp Vertex

The vertex of an absolute value graph is a corner, not a smooth turning point: the graph changes direction abruptly rather than curving. This corner is the lowest point of the graph when it opens upward and the highest point when it opens downward.

Definition: Transformed Absolute Value Function

A transformed absolute value function is a function of the form\[f(x)=a|x-h|+k, \nonumber\]where \(a\), \(h\), and \(k\) are constants and \(a\neq 0\). The point \((h,k)\) is the vertex of its graph.

Theorem: Graph of \(f(x)=a|x-h|+k\)

Let \(f(x)=a|x-h|+k\) with \(a\neq 0\). Then the graph of \(f\) is a V-shaped curve with the following features.

• The vertex is \((h,k)\).
• The graph opens upward if \(a\gt 0\) and downward if \(a\lt 0\).
• The right arm (for \(x\geq h\)) has slope \(a\), and the left arm (for \(x\leq h\)) has slope \(-a\).
• The larger \(|a|\) is, the narrower the graph; the smaller \(|a|\) is, the wider the graph.

Theorem: Piecewise Form of \(f(x)=a|x-h|+k\)

The function \(f(x)=a|x-h|+k\) can be written without absolute value bars as\[f(x) = \begin{cases} a(x-h)+k & \text{if } x \geq h \\ -a(x-h)+k & \text{if } x \lt h \end{cases} \nonumber\]

Caution: The Sign of the Horizontal Shift

The vertex occurs at the value of \(x\) that makes the expression inside the bars equal to zero. Because the form is \(x-h\), a function such as \(f(x)=|x-3|+1\) has vertex \((3,1)\), shifted to the right. To read the vertex of \(f(x)=|x+2|+1\), first rewrite it as \(f(x)=|x-(-2)|+1\); the vertex is \((-2,1)\), shifted to the left.

Examples

Example \(\PageIndex{1}\): Reading the Features from the Equation

For \(f(x)=2|x-3|-1\), state the vertex, the direction the graph opens, and the slope of each arm.

Solution

Compare with \(f(x)=a|x-h|+k\): here \(a=2\), \(h=3\), and \(k=-1\). The vertex is \((h,k)=(3,-1)\). Since \(a=2\gt 0\), the graph opens upward. The right arm has slope \(a=2\), and the left arm has slope \(-a=-2\).

Example \(\PageIndex{2}\): A Downward-Opening Graph

For \(f(x)=-|x+4|+5\), state the vertex, the direction the graph opens, and the slope of each arm.

Solution

Rewrite the inside as \(x-h\): \(f(x)=-|x-(-4)|+5\), so \(a=-1\), \(h=-4\), and \(k=5\). The vertex is \((-4,5)\). Since \(a=-1\lt 0\), the graph opens downward. The right arm has slope \(a=-1\), and the left arm has slope \(-a=1\).

Example \(\PageIndex{3}\): A Fractional Value of \(a\)

For \(f(x)=\dfrac{1}{2}|x+2|-3\), state the vertex, the direction the graph opens, and the slope of each arm. Compare its width to that of the parent function.

Solution

Writing the inside as \(x-h\) gives \(f(x)=\dfrac{1}{2}|x-(-2)|-3\), so \(a=\dfrac{1}{2}\), \(h=-2\), and \(k=-3\). The vertex is \((-2,-3)\). Since \(a=\dfrac{1}{2}\gt 0\), the graph opens upward. The right arm has slope \(\dfrac{1}{2}\), and the left arm has slope \(-\dfrac{1}{2}\). Because \(|a|=\dfrac{1}{2}\lt 1\), the graph is wider than the parent function \(f(x)=|x|\).

Example \(\PageIndex{4}\): Graphing with a Table of Values

Graph \(f(x)=|x-2|-3\). Identify the vertex, the \(x\)-intercepts, and the \(y\)-intercept.

Solution

Here \(a=1\), \(h=2\), and \(k=-3\), so the vertex is \((2,-3)\) and the graph opens upward with arms of slope \(1\) and \(-1\). Building a table of values centered at the vertex gives the following points.

\(x\)	\(f(x)\)	\((x,f(x))\)
\(-1\)	\(0\)	\((-1,0)\)
\(0\)	\(-1\)	\((0,-1)\)
\(2\)	\(-3\)	\((2,-3)\)
\(4\)	\(-1\)	\((4,-1)\)
\(5\)	\(0\)	\((5,0)\)

The \(y\)-intercept is \((0,-1)\). To find the \(x\)-intercepts, set \(f(x)=0\):\[|x-2|-3=0 \quad\Longrightarrow\quad |x-2|=3 \quad\Longrightarrow\quad x-2=\pm 3. \nonumber\]Thus \(x=5\) or \(x=-1\), giving the \(x\)-intercepts \((-1,0)\) and \((5,0)\).

Figure \(\PageIndex{2}\): Graph of \(f(x)=|x-2|-3\).

Example \(\PageIndex{5}\): Writing the Function in Piecewise Form

Write \(f(x)=-2|x+1|+4\) as a piecewise-defined function.

Solution

Here \(a=-2\) and the vertex is at \(h=-1\), so the two pieces split at \(x=-1\). Using the piecewise form \(f(x)=a(x-h)+k\) for \(x\geq h\) and \(f(x)=-a(x-h)+k\) for \(x\lt h\):

For \(x\geq -1\):\[f(x)=-2(x+1)+4=-2x-2+4=-2x+2. \nonumber\]For \(x\lt -1\):\[f(x)=2(x+1)+4=2x+2+4=2x+6. \nonumber\]Therefore,\[f(x) = \begin{cases} -2x+2 & \text{if } x \geq -1 \\ 2x+6 & \text{if } x \lt -1 \end{cases} \nonumber\]As a check, both pieces give \(f(-1)=4\), which matches the vertex \((-1,4)\).

Example \(\PageIndex{6}\): Finding the Equation from a Graph

The graph below is an absolute value function that opens upward. Use its vertex and one other point to write the function in the form \(f(x)=a|x-h|+k\).

Figure \(\PageIndex{3}\): Graph of an upward-opening absolute value function.

Solution

Reading the graph, the vertex is \((-3,2)\), so \(h=-3\) and \(k=2\), which gives \(f(x)=a|x+3|+2\). The graph also passes through the marked point \((-1,6)\). Substitute this point to solve for \(a\):\[6=a|-1+3|+2 \quad\Longrightarrow\quad 6=a(2)+2 \quad\Longrightarrow\quad 2a=4 \quad\Longrightarrow\quad a=2. \nonumber\]Therefore \(f(x)=2|x+3|+2\).

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Sources

Several parts of this text use modifications from the following source:

• Wikipedia article: "Absolute value"

This source is released under the Creative Commons Attribution-Share-Alike License 4.0.