Graphing Absolute Value Functions: Vertex, Points, and Intercepts

Find the vertex before plotting points

The parent graph y = |x| has its corner at (0, 0). For example, both x = −2 and x = 2 give y = 2. Its right arm has slope 1 and its left arm has slope −1.

For y = a|x − h| + k, the expression inside the bars is zero when x = h. Substituting gives y = k, so the vertex is (h, k), and the line of symmetry is x = h.

• Horizontal shift: |x − 3| moves the corner right 3 units. |x + 3| means |x − (−3)|, so the corner moves left 3 units.
• Vertical shift: adding k moves every output up or down by k.
• Direction: a > 0 opens upward; a < 0 opens downward.
• Steepness: |a| > 1 makes the arms steeper than y = |x|. If 0 < |a| < 1, they are less steep. The right arm has slope a; the left arm has slope −a.

Worked example: graph y = −2|x − 3| + 4

1. Locate the corner. Here h = 3 and k = 4, so plot (3, 4).
2. Choose the direction. Since a = −2, the graph opens downward. Moving 1 unit away from x = 3 lowers y by 2.
3. Calculate points on both sides. At x = 1, y = −2|1 − 3| + 4 = −2(2) + 4 = 0. Use the same substitution for the other x-values.
4. Draw straight rays. Join the points to the corner, then extend both arms. Do not round the corner into a curve.

Check the symmetry: (2, 2) and (4, 2) are equally far from x = 3 and have the same height. A mismatch usually means a sign or substitution error.

Find the intercepts, domain, and range

The y-intercept: substitute x = 0. In the worked example, y = −2|0 − 3| + 4 = −2, giving (0, −2).

The x-intercepts: set y = 0. Then −2|x − 3| + 4 = 0, so |x − 3| = 2. Solve x − 3 = 2 or x − 3 = −2 to get x = 5 or x = 1. The intercepts are (5, 0) and (1, 0).

The domain is all real numbers for y = a|x − h| + k. The range is y ≥ k when a > 0, and y ≤ k when a < 0. Our downward graph has a maximum height of 4, so its range is y ≤ 4.

A V can have two, one, or no x-intercepts. For example, y = |x| + 2 never reaches y = 0. If solving for an intercept gives an absolute value equal to a negative number, there is no real solution.

What if a coefficient is inside the absolute value?

For y = |2x + 6| − 1, do not read the vertex as (−6, −1). First factor the expression: |2x + 6| = |2(x + 3)| = 2|x + 3|. The function becomes y = 2|x + 3| − 1.

Now the vertex is (−3, −1), and the graph opens upward. At x = −4 or −2, y = 1; these two points sit on opposite sides of the vertex.

Another way to locate the corner is to solve 2x + 6 = 0. This method works for y = |mx + b| + k when m ≠ 0. After finding the corner, substitute nearby x-values to check the steepness.

Try three graphs, then check your work

For each function, find the vertex, two matching points on opposite sides, and the range. Sketch the rays before opening the answer.

1. y = |x + 2| − 3
2. y = −|x − 1| + 2
3. y = |3x − 6| + 1

For more practice solving the equations that locate x-intercepts, use the absolute value equations section of our Algebra 1 worksheets. Those sheets cover equations and inequalities; the three problems above provide graphing practice.

For a broader study plan, visit the Algebra 1 Online Center for lessons, worksheets, and book choices.