Absolute Value Functions

An absolute value function has the form f(x) = a|x - h| + k. Its graph is a V shape with its vertex at (h, k). The V opens up when a \gt 0 and down when a \lt 0. The number a is the slope of the right side, and the left side has slope -a.

Before this lesson: You should know Transformations of Functions, Absolute Value Equations and Slope.

The Parent Graph and Its Changes

The parent function is f(x) = |x|. For x \ge 0 it equals x, and for x \lt 0 it equals -x. So its graph is two rays that meet at the origin: a V with slopes 1 and −1.

Every other absolute value function is a transformation of this V. In f(x) = a|x - h| + k:

• The vertex is (h, k). It is the lowest point if a \gt 0 and the highest point if a \lt 0.
• The axis of symmetry is the vertical line x = h. The two sides of the V are mirror images across it.
• The right side has slope a and the left side has slope -a. A bigger |a| makes a narrower V.
• The domain is all real numbers, (-\infty, \infty). The range is [k, \infty) if a \gt 0 and (-\infty, k] if a \lt 0.

How to Graph an Absolute Value Function

1. Read a, h and k. Rewrite |x + 3| as |x - (-3)|, so h = -3.
2. Plot the vertex (h, k).
3. From the vertex, move 1 unit right and a units up (or down if a \lt 0). Plot that point.
4. Plot the mirror point 1 unit left of the vertex at the same height.
5. Draw two rays from the vertex through the points.

To find the x-intercepts, set f(x) = 0 and solve the absolute value equation.

Worked Examples

Example 1: Graph and find intercepts

Describe g(x) = |x + 3| - 2. Give the vertex, the range and the x-intercepts.

1. Rewrite the inside: |x - (-3)|, so h = -3. With k = -2, the vertex is (−3, −2).
2. Here a = 1 \gt 0, so the V opens up with slopes 1 and −1.
3. The lowest y-value is −2, so the range is [-2, \infty).
4. Set g(x) = 0: |x + 3| - 2 = 0
5. Add 2 to both sides: |x + 3| = 2
6. Split into two equations: x + 3 = 2 or x + 3 = -2
7. Subtract 3 from both sides of each: x = -1 or x = -5

Answer: Vertex (−3, −2), range [-2, \infty), x-intercepts (−1, 0) and (−5, 0).

Example 2: A V that opens down

Describe g(x) = -2|x - 1| + 4. Give the vertex, the slopes, the range and the x-intercepts.

1. The vertex is (h, k) = (1, 4).
2. a = -2 \lt 0, so the V opens down. The right side has slope −2 and the left side has slope 2.
3. The highest y-value is 4, so the range is (-\infty, 4].
4. Set g(x) = 0: -2|x - 1| + 4 = 0
5. Subtract 4 from both sides: -2|x - 1| = -4
6. Divide both sides by −2: |x - 1| = 2
7. Split: x - 1 = 2 or x - 1 = -2
8. Add 1 to both sides of each: x = 3 or x = -1

Answer: Vertex (1, 4), slopes −2 and 2, range (-\infty, 4], x-intercepts (3, 0) and (−1, 0). This is the graph above.

Example 3: A wider V

Graph g(x) = \tfrac{1}{2}|x| - 3 by plotting points.

1. The vertex is (0, −3), and a = \tfrac{1}{2}, so the V opens up and is wider than |x|.
2. Right 2 from the vertex: g(2) = \tfrac{1}{2}(2) - 3 = 1 - 3 = -2. Point (2, −2).
3. Left 2 from the vertex: g(-2) = \tfrac{1}{2}(2) - 3 = -2. Point (−2, −2).
4. Right 6: g(6) = \tfrac{1}{2}(6) - 3 = 3 - 3 = 0. Point (6, 0), an x-intercept.
5. By symmetry, (−6, 0) is the other x-intercept.

Answer: Draw rays from (0, −3) through (2, −2) and (−2, −2). The x-intercepts are (6, 0) and (−6, 0).

Example 4: Write the equation from a graph

The vertex of a V-shaped graph is (2, −1), and the graph passes through (4, 5). Write its equation.

1. Start with the vertex: f(x) = a|x - 2| - 1
2. Substitute the point (4, 5): 5 = a|4 - 2| - 1
3. Simplify inside: 5 = a|2| - 1
4. Take the absolute value: 5 = 2a - 1
5. Add 1 to both sides: 6 = 2a
6. Divide both sides by 2: a = 3

Answer: f(x) = 3|x - 2| - 1

Example 5: Word problem with a train

A train moves at 60 mi/h on a straight track. It passes a station at t = 3 hours. Its distance from the station is d(t) = 60|t - 3|. At what times is the train 90 miles from the station?

1. Set the distance equal to 90: 60|t - 3| = 90
2. Divide both sides by 60: |t - 3| = 1.5
3. Split: t - 3 = 1.5 or t - 3 = -1.5
4. Add 3 to both sides of each: t = 4.5 or t = 1.5

Answer: At 1.5 hours (before it reaches the station) and at 4.5 hours (after it passes). The graph of d is a V with its vertex at (3, 0).

Common Mistakes

Mistake	Example of the error	Fix
Using the wrong sign for the vertex	|x + 3| gives vertex at x = 3	Inside, the sign flips: x = -3.
Forgetting that a controls direction	Saying -2|x - 1| + 4 opens up	A negative a opens the V down.
Giving one slope only	“The slope is 3”	The sides have slopes a and -a: 3 and −3.
Solving only one case	|x - 1| = 2 gives only x = 3	Also solve x - 1 = -2, so x = -1.
Wrong range for a downward V	[4, \infty) for -2|x - 1| + 4	The vertex is the highest point: (-\infty, 4].

Practice Problems

1. Give the vertex of f(x) = |x - 4|.
2. Give the vertex and range of f(x) = |x| + 6.
3. For f(x) = -|x + 2| + 5, give the vertex, direction, range and x-intercepts.
4. What are the slopes of the two sides of f(x) = 3|x|?
5. Find the x-intercepts and y-intercept of f(x) = 2|x - 3| - 8.
6. For f(x) = \tfrac{1}{4}|x + 4|, give the vertex and two other points.
7. Write the equation of the V with vertex (−1, 3) that passes through (1, −1).
8. Write the equation of the V with vertex (0, −4) that opens up with slopes 2 and −2.
9. Does f(x) = |x - 1| + 2 have any x-intercepts? Explain.
10. A town’s average temperature is modeled by T(m) = -3|m - 6| + 80, where m is the month number. What is the highest average temperature, and in which months is it 71°?

Show answers

1. h = 4, k = 0: vertex (4, 0)
2. Vertex (0, 6). It opens up, so the range is [6, \infty).
3. Rewrite: -|x - (-2)| + 5, so the vertex is (−2, 5)
   a = -1 \lt 0: opens down
   Range: (-\infty, 5]
   Set to 0: -|x + 2| + 5 = 0
   Subtract 5 from both sides: -|x + 2| = -5
   Multiply both sides by −1: |x + 2| = 5
   Split: x + 2 = 5 or x + 2 = -5
   Subtract 2: x = 3 or x = -7
4. The right side has slope 3 and the left side has slope −3.
5. Set to 0: 2|x - 3| - 8 = 0
   Add 8 to both sides: 2|x - 3| = 8
   Divide both sides by 2: |x - 3| = 4
   Split: x - 3 = 4 or x - 3 = -4
   Add 3: x = 7 or x = -1
   y-intercept: f(0) = 2|0 - 3| - 8 = 2(3) - 8 = 6 - 8 = -2, so (0, −2)
6. Vertex (−4, 0)
   Right 4: f(0) = \tfrac{1}{4}|4| = 1, point (0, 1)
   Left 4: f(-8) = \tfrac{1}{4}|-4| = 1, point (−8, 1)
7. Start: f(x) = a|x + 1| + 3
   Substitute (1, −1): -1 = a|1 + 1| + 3
   Simplify: -1 = 2a + 3
   Subtract 3 from both sides: -4 = 2a
   Divide by 2: a = -2
   Equation: f(x) = -2|x + 1| + 3
8. Vertex (0, −4) and a = 2: f(x) = 2|x| - 4
9. No. The V opens up and its lowest point is the vertex (1, 2). Since 2 is above 0, the graph never reaches the x-axis.
10. The highest value is at the vertex, m = 6: T(6) = 80 degrees
    Set T(m) = 71: -3|m - 6| + 80 = 71
    Subtract 80 from both sides: -3|m - 6| = -9
    Divide both sides by −3: |m - 6| = 3
    Split: m - 6 = 3 or m - 6 = -3
    Add 6: m = 9 or m = 3. The months are March and September.

FAQ

How do you find the vertex of an absolute value function?

Write the function as f(x) = a|x - h| + k. The vertex is (h, k). For |x + 3| - 2, the vertex is (−3, −2).

What is the range of an absolute value function?

If a \gt 0, the range is [k, \infty). If a \lt 0, the range is (-\infty, k]. The domain is always all real numbers.

What does the a value do in an absolute value function?

Its sign tells you whether the V opens up or down. Its size sets the slopes of the sides, a and -a, so a larger |a| makes a narrower V.

How is an absolute value function different from an absolute value equation?

An equation like |x - 1| = 2 has a few solutions. A function like f(x) = |x - 1| has a whole graph. Solving f(x) = 0 gives the graph’s x-intercepts.

Summary

• f(x) = a|x - h| + k has a V-shaped graph with vertex (h, k) and axis of symmetry x = h.
• It opens up if a \gt 0 and down if a \lt 0. The sides have slopes a and -a.
• Domain (-\infty, \infty). Range [k, \infty) or (-\infty, k].
• Find x-intercepts by solving f(x) = 0 as two cases.

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