Math 2–3 Honors
Integrated Math 2–3 Honors · Lesson Review · August 31, 2026

Graphing Absolute-Value Inequalities

Graph the transformed V-shaped boundary, decide whether it is solid or dashed, test one ordered pair, and shade the complete two-variable solution region.

6-minute edited lesson6 chaptersCaptions + transcript

01 · Watch

Lesson video

VTT captionsSRT captions
0:0001Graphing Absolute-Value Inequalities0:0502What the Solution Set Represents0:3303Example 1: Graph a Solid V Boundary2:3004Use a Test Point and Shade3:3105Example 2: Graph a Dashed V Boundary4:3406Test the Second Region

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02 · Understand

Learning targets

03 · Vocabulary

Key definitions

Connect the inequality to its boundary and shaded region.

01Boundary equation
Replace the inequality symbol with == to graph the curve that separates the plane.
02Solid boundary
Use a solid curve for ≤\le or ≥\ge because equality includes boundary points.
03Dashed boundary
Use a dashed curve for << or >> because boundary points are not solutions.
04Test point
An ordered pair not on the boundary that determines which side makes the inequality true.
05Solution region
Every shaded ordered pair that makes the original inequality true.

04 · Apply

Worked examples

These are the two examples developed in the lesson.

Solid boundary

y≤2∣x−3∣−4y\le2|x-3|-4

  1. Graph the boundary y=2∣x−3∣−4y=2|x-3|-4.
  2. The vertex is (3,−4)(3,-4), and the two arms have slopes 22 and −2-2.
  3. Use a solid V because ≤\le includes equality.
Test the origin

(0,0)(0,0)

0≤2∣0−3∣−4=20\le2|0-3|-4=2

The statement is true, so shade the side of the solid V containing the origin.

Dashed boundary

y>−12∣x+1∣+1y>-\frac12|x+1|+1

  1. Graph the boundary y=−12∣x+1∣+1y=-\frac12|x+1|+1.
  2. The vertex is (−1,1)(-1,1). The negative coefficient reflects the V, and the arm slopes have magnitude 12\frac12.
  3. Use a dashed V because >> is strict.
Reject the origin

(0,0)(0,0)

0>−12∣0+1∣+1=120>-\frac12|0+1|+1=\frac12

The statement is false, so shade the region on the opposite side of the dashed V.

05 · Avoid

Common mistakes

Graphing the inequality before its boundary.

Replace the symbol with equality first.

Using the wrong line style.

Equality means solid; a strict inequality means dashed.

Reading the vertex sign literally.

In ∣x−h∣|x-h|, the horizontal coordinate is hh.

Choosing a point on the boundary.

Pick any ordered pair that is clearly off the V.

Testing the boundary equation.

Substitute into the original inequality, not the equation.

Shading from the symbol alone.

Use a test point; transformations can move or reflect the V.

06 · Check yourself

Quick check

Predict the boundary before opening each answer.

1For y≥∣x+2∣−3y\ge|x+2|-3, identify the vertex and line style.

Vertex (−2,−3)(-2,-3); solid V. The symbol includes equality.

2For y<−∣x−4∣+2y<-|x-4|+2, identify the vertex, orientation, and line style.

Vertex (4,2)(4,2); upside-down dashed V.

3Does (0,0)(0,0) satisfy y≥∣x∣−1y\ge|x|-1?

Yes. 0≥∣0∣−1=−10\ge|0|-1=-1 is true, so shade the side containing the origin.

4Why is one test point enough?

The continuous boundary separates the plane into regions. Every point in one region gives the same true-or-false result until the boundary is crossed.

07 · Revisit

Transcript and assigned practice

Search the corrected transcript for either example, then complete the eKadence worksheet Absolutevaluefunctioninequalities.pdf.

Download corrected transcript