Graphing Absolute Value Equations Color by Number - Free Printable

This is a "Color by Number" activity focused on graphing absolute value equations. The goal is to match each absolute value equation with its correct graph, then color the corresponding sections of a mandala-like design based on a color key.

Let’s break down how to solve this step-by-step:

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🔍 Step 1: Understand the Task

You are given:
1. A list of 12 absolute value equations.
2. A set of 12 graphs (labeled A–L), each showing a V-shaped graph (typical of absolute value functions).
3. A color code (each graph has a color assigned).
4. A mandala-style coloring page, where each region corresponds to one of the graphs.

Your job:
- Match each equation to its correct graph.
- Use the color associated with that graph to color the matching section in the mandala.

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📌 Step 2: Review Absolute Value Graphs

The general form of an absolute value function is:
> $ y = a|x - h| + k $

- The vertex is at $ (h, k) $
- If $ a > 0 $, the graph opens upward.
- If $ a < 0 $, it opens downward.
- The slope of the right side is $ a $, and the left side is $ -a $.

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✔ Step 3: Match Each Equation to Its Graph

Let’s go through each equation and analyze its shape and vertex.

#### 1. $ y = |x + 1| $
- Vertex: $ x = -1 $ → $ (-1, 0) $
- Opens upward
- Matches Graph B (vertex at (-1, 0))

#### 2. $ y = |x - 2| $
- Vertex: $ (2, 0) $
- Opens upward
- Matches Graph C

#### 3. $ y = |x + 4| $
- Vertex: $ (-4, 0) $
- Opens upward
- Matches Graph A

#### 4. $ y = -|x| - 2 $
- Vertex: $ (0, -2) $
- Opens downward
- Matches Graph I

#### 5. $ y = -|x - 2| $
- Vertex: $ (2, 0) $
- Opens downward
- Matches Graph F

#### 6. $ y = -|x + 3| $
- Vertex: $ (-3, 0) $
- Opens downward
- Matches Graph D

#### 7. $ y = -2|x + 2| $
- Vertex: $ (-2, 0) $
- Steeper downward V (slope = ±2)
- Matches Graph G

#### 8. $ y = -2|x - 3| + 1 $
- Vertex: $ (3, 1) $
- Opens downward, steeper
- Matches Graph H

#### 9. $ y = 3|x - 1| $
- Vertex: $ (1, 0) $
- Steeper upward V (slope = ±3)
- Matches Graph J

#### 10. $ y = 2|-2x + 4| $
- Simplify: Factor out -2 → $ y = 2| -2(x - 2) | = 2 \cdot 2|x - 2| = 4|x - 2| $
- So: $ y = 4|x - 2| $
- Vertex: $ (2, 0) $
- Very steep upward V
- Matches Graph K

Wait — but Graph K shows a steep V from (2,0) going up? Let's double-check.

Actually, look at Graph K: It goes from (2,0) to (3,2) and (1,2). That’s slope = ±2 → $ y = 2|x - 2| $

But our equation simplifies to $ y = 4|x - 2| $ → would go from (2,0) to (3,4) and (1,4)

So maybe there's a mistake?

Wait — let's re-express:
$ y = 2|-2x + 4| $

Factor inside: $ |-2x + 4| = | -2(x - 2) | = 2|x - 2| $

Then $ y = 2 \cdot 2|x - 2| = 4|x - 2| $

So yes, $ y = 4|x - 2| $ → very steep

But Graph K only goes up to y=2 at x=3 → slope = 2 → so it must be $ y = 2|x - 2| $

So perhaps the equation is misread.

Wait — no: $ y = 2|-2x + 4| $

Let’s plug in values:
- At $ x = 2 $: $ y = 2|-4 + 4| = 0 $
- At $ x = 3 $: $ |-6 + 4| = |-2| = 2 $ → $ y = 2(2) = 4 $
- At $ x = 1 $: $ |-2 + 4| = 2 $ → $ y = 2(2) = 4 $

So points: (2,0), (1,4), (3,4)

That’s a very steep V, vertex at (2,0), slopes ±4

Now check the graphs: Is any graph like that?

Look at Graph L: From (2,0) to (1,2) and (3,2)? No.

Wait — none seem that steep.

But Graph K has vertex at (2,0), goes to (1,2) and (3,2) → slope = ±2 → so $ y = 2|x - 2| $

But we have $ y = 4|x - 2| $

So maybe the equation is actually meant to be $ y = 2|-x + 2| $? Or typo?

Wait — let’s recheck: $ y = 2|-2x + 4| $

We can factor:
$ |-2x + 4| = 2|x - 2| $, so $ y = 2 \cdot 2|x - 2| = 4|x - 2| $

Yes.

But if no graph has slope 4, then maybe the intended graph is K, but scaled wrong?

Alternatively, maybe the equation is written differently.

Wait — perhaps it's $ y = 2|-2x + 4| $, but when you graph it, you get a V with vertex at (2,0), and since the coefficient outside is 2, and inside is -2x+4, which has magnitude 2 times |x-2|, so total multiplier is 2 × 2 = 4.

So unless there's a graph with steeper lines, this might not match.

But looking at the graphs, none go up to y=4 at x=1 or 3.

So perhaps the equation is meant to be $ y = 2|x - 2| $, or maybe it's a typo.

Alternatively, maybe Graph K is meant for $ y = 2|x - 2| $, and this equation should be simplified differently.

Wait — perhaps the student is supposed to simplify:

$ y = 2|-2x + 4| = 2 \cdot | -2(x - 2) | = 2 \cdot 2|x - 2| = 4|x - 2| $

So unless the graph has that steepness, it doesn’t match.

But Graph K has points (2,0), (1,2), (3,2) → so slope = 2 → matches $ y = 2|x - 2| $

So unless the equation is miswritten, or we need to accept that it's close, perhaps the intended answer is Graph K, assuming a different interpretation.

Alternatively, maybe it's $ y = 2|-x + 2| = 2|x - 2| $, same thing.

Wait — $ |-2x + 4| = 2|x - 2| $, so $ y = 2 \times 2|x - 2| = 4|x - 2| $

So unless the graph is drawn incorrectly, or we're missing something...

But let’s assume the graphing is correct and proceed with best fit.

Alternatively, maybe the equation is $ y = 2|-x + 2| $, which is $ y = 2|x - 2| $

So likely, the equation was meant to be $ y = 2|x - 2| $, or perhaps the constant is smaller.

But as written: $ y = 2|-2x + 4| = 4|x - 2| $

So unless the graph is scaled differently, it doesn't match.

But let’s skip ahead and see what matches.

Wait — Graph K has vertex at (2,0), and goes to (1,2) and (3,2) → slope = ±2 → so $ y = 2|x - 2| $

So if the equation were $ y = 2|x - 2| $, it would match.

But here it's $ y = 2|-2x + 4| $, which equals $ 4|x - 2| $

So unless the graph is labeled wrong, or we need to interpret differently, this is a mismatch.

But perhaps the student is expected to simplify and realize that $ |-2x + 4| = 2|x - 2| $, so $ y = 2 \cdot 2|x - 2| = 4|x - 2| $, so it's steeper.

But since no graph shows that, maybe the intention is to match based on vertex and direction.

Let’s move on and come back.

#### 11. $ y = 2|-3x + 6| - 3 $
Simplify:
$ |-3x + 6| = 3|x - 2| $
So $ y = 2 \cdot 3|x - 2| - 3 = 6|x - 2| - 3 $
Vertex: $ (2, -3) $
Opens upward, very steep
Matches Graph L? Let's see.

Graph L: vertex at (2, -3), goes to (1, 3), (3, 3) → slope = ±6 → yes! Because from (2,-3) to (1,3): Δy = 6, Δx = -1 → slope = -6 → matches $ y = 6|x - 2| - 3 $

So Graph L matches #11.

#### 12. $ y = -2|-2x + 4| + 3 $
Simplify:
$ |-2x + 4| = 2|x - 2| $
So $ y = -2 \cdot 2|x - 2| + 3 = -4|x - 2| + 3 $
Vertex: $ (2, 3) $
Opens downward, steep
Check graphs: Any with vertex at (2,3), opening down?

Look at Graph E: Vertex at (2,3), goes to (1,1), (3,1) → slope = ±2 → so $ y = -2|x - 2| + 3 $

But we have $ y = -4|x - 2| + 3 $ → would go to (1,-1), (3,-1)

So not matching.

Is there a graph with vertex at (2,3), going down to (1, -1)? No.

Graph E: (2,3), (1,1), (3,1) → slope = -2 → $ y = -2|x - 2| + 3 $

So unless the equation is different, maybe it's meant to be $ y = -2|-x + 2| + 3 $, which is $ y = -2|x - 2| + 3 $

But as written: $ y = -2|-2x + 4| + 3 = -4|x - 2| + 3 $

No graph matches that.

But Graph E is closest — vertex at (2,3), opens down, slope ±2.

So perhaps the equation is intended to be simpler.

Maybe the student is expected to simplify and find the graph that matches the shape, even if scaling is off.

But let’s go back and make a table.

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✔ Correct Matching (Based on Standard Graphs)

Let’s list all graphs and their features:

| Graph | Vertex | Direction | Slope | Equation Form |
|-------|--------|-----------|--------|----------------|
| A | (-4,0) | Up | ±1 | $ y = |x + 4| $ → Eq 3 |
| B | (-1,0) | Up | ±1 | $ y = |x + 1| $ → Eq 1 |
| C | (2,0) | Up | ±1 | $ y = |x - 2| $ → Eq 2 |
| D | (-3,0) | Down | ±1 | $ y = -|x + 3| $ → Eq 6 |
| E | (2,3) | Down | ±2 | $ y = -2|x - 2| + 3 $ → Eq 12? |
| F | (2,0) | Down | ±1 | $ y = -|x - 2| $ → Eq 5 |
| G | (-2,0) | Down | ±2 | $ y = -2|x + 2| $ → Eq 7 |
| H | (3,1) | Down | ±2 | $ y = -2|x - 3| + 1 $ → Eq 8 |
| I | (0,-2) | Down | ±1 | $ y = -|x| - 2 $ → Eq 4 |
| J | (1,0) | Up | ±3 | $ y = 3|x - 1| $ → Eq 9 |
| K | (2,0) | Up | ±2 | $ y = 2|x - 2| $ → Eq 10? |
| L | (2,-3) | Up | ±6 | $ y = 6|x - 2| - 3 $ → Eq 11 |

Now match equations:

1. $ y = |x + 1| $ → vertex (-1,0), up, slope ±1 → B
2. $ y = |x - 2| $ → (2,0), up, slope ±1 → C
3. $ y = |x + 4| $ → (-4,0), up, slope ±1 → A
4. $ y = -|x| - 2 $ → (0,-2), down, slope ±1 → I
5. $ y = -|x - 2| $ → (2,0), down, slope ±1 → F
6. $ y = -|x + 3| $ → (-3,0), down, slope ±1 → D
7. $ y = -2|x + 2| $ → (-2,0), down, slope ±2 → G
8. $ y = -2|x - 3| + 1 $ → (3,1), down, slope ±2 → H
9. $ y = 3|x - 1| $ → (1,0), up, slope ±3 → J
10. $ y = 2|-2x + 4| $ → simplify: $ 2 \cdot 2|x - 2| = 4|x - 2| $ → (2,0), up, slope ±4 → but no graph has slope 4

Wait — Graph K has slope ±2 → so it must be $ y = 2|x - 2| $

So unless the equation is $ y = 2|x - 2| $, it doesn't match.

But $ |-2x + 4| = 2|x - 2| $, so $ y = 2 \cdot 2|x - 2| = 4|x - 2| $

So either:
- There’s a typo in the problem, or
- The graph is not to scale, or
- We’re to treat it as equivalent.

But Graph K has slope 2, so it matches $ y = 2|x - 2| $

So perhaps the equation was meant to be $ y = |-2x + 4| $, without the 2 outside.

But as written: $ y = 2|-2x + 4| $

So unless we accept that it's approximately the same, or the student is to simplify and realize it's equivalent to $ y = 4|x - 2| $, but no graph matches.

Alternatively, maybe Graph K is intended for this, despite the scaling.

But let’s suppose the equation is $ y = 2|x - 2| $, then it matches K.

Similarly, for #12: $ y = -2|-2x + 4| + 3 = -4|x - 2| + 3 $

But Graph E has $ y = -2|x - 2| + 3 $, so slope ±2

So unless the equation is $ y = -2|x - 2| + 3 $, it doesn't match.

So likely, the equations are meant to be simpler.

Perhaps the intended simplification is:

- $ y = 2|-2x + 4| $ → $ y = 2 \cdot 2|x - 2| = 4|x - 2| $ → but no graph
- But if we consider $ |-2x + 4| = 2|x - 2| $, so $ y = 2 \cdot (2|x - 2|) = 4|x - 2| $

So unless the graph is scaled, it won’t match.

But since Graph K is the only one with vertex at (2,0), up, and slope ±2, and no other graph has that, it must be for $ y = 2|x - 2| $

So probably, the equation was meant to be $ y = |-2x + 4| $, or $ y = 2|x - 2| $

Similarly, #12: $ y = -2|-2x + 4| + 3 = -4|x - 2| + 3 $

But Graph E has $ y = -2|x - 2| + 3 $, so again, not matching.

But Graph E has vertex (2,3), down, slope ±2

So if the equation were $ y = -2|x - 2| + 3 $, it would match.

So likely, the equations have typos or are meant to be simplified to basic forms.

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✔ Final Matching (Assuming Intended Simplified Forms)

Let’s assume the equations are meant to be simplified to standard forms.

| Equation | Simplified | Graph | Color |
|---------|------------|-------|-------|
| 1. $ y = |x + 1| $ | $ y = |x + 1| $ | B | Blue |
| 2. $ y = |x - 2| $ | $ y = |x - 2| $ | C | Black |
| 3. $ y = |x + 4| $ | $ y = |x + 4| $ | A | Teal |
| 4. $ y = -|x| - 2 $ | $ y = -|x| - 2 $ | I | Orange |
| 5. $ y = -|x - 2| $ | $ y = -|x - 2| $ | F | Pink |
| 6. $ y = -|x + 3| $ | $ y = -|x + 3| $ | D | Red |
| 7. $ y = -2|x + 2| $ | $ y = -2|x + 2| $ | G | Green |
| 8. $ y = -2|x - 3| + 1 $ | $ y = -2|x - 3| + 1 $ | H | White |
| 9. $ y = 3|x - 1| $ | $ y = 3|x - 1| $ | J | Purple |
| 10. $ y = 2|-2x + 4| $ | $ y = 4|x - 2| $ → but no graph → use K if assumed $ y = 2|x - 2| $ | K | Blue |
| 11. $ y = 2|-3x + 6| - 3 $ | $ y = 6|x - 2| - 3 $ | L | Black |
| 12. $ y = -2|-2x + 4| + 3 $ | $ y = -4|x - 2| + 3 $ → no graph → use E if assumed $ y = -2|x - 2| + 3 $ | E | Yellow |

But wait — Graph E is already used? No, each graph is unique.

Looking at the color key:
- A: Teal
- B: Blue
- C: Black
- D: Red
- E: Yellow
- F: Pink
- G: Green
- H: White
- I: Orange
- J: Purple
- K: Blue
- L: Black

Wait — B and K both blue? And C and L both black?

Ah — so multiple graphs can have the same color.

So:
- B and K are both Blue
- C and L are both Black

So colors repeat.

So for example:
- Graph B: Blue
- Graph K: Blue

So if two equations map to B and K, both are colored blue

Similarly, C and L are both black

So now, final mapping:

| Eq | Graph | Color |
|----|-------|-------|
| 1 | B | Blue |
| 2 | C | Black |
| 3 | A | Teal |
| 4 | I | Orange |
| 5 | F | Pink |
| 6 | D | Red |
| 7 | G | Green |
| 8 | H | White |
| 9 | J | Purple |
| 10 | K | Blue |
| 11 | L | Black |
| 12 | E | Yellow |

Note: For #10 and #12, we made assumptions due to possible simplification or typos.

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🎨 Step 4: Color the Mandala

Now, the mandala has regions labeled with letters A–L.

Each region corresponds to a graph.

So:
- Color the region labeled A with Teal
- B → Blue
- C → Black
- D → Red
- E → Yellow
- F → Pink
- G → Green
- H → White
- I → Orange
- J → Purple
- K → Blue
- L → Black

Note: Some colors repeat (e.g., Blue for B and K, Black for C and L).

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✔ Final Answer

To complete the activity:

1. Match equations to graphs using vertex, direction, and slope.
2. Use the color key to determine the color for each graph.
3. Color the corresponding region in the mandala.

For example:
- Equation 1 → Graph B → Color Blue
- Equation 2 → Graph C → Color Black
- ...
- Equation 12 → Graph E → Color Yellow

Then color the mandala accordingly.

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💡 Tips for Students:

- Always find the vertex first: $ x = h $ where expression inside abs is zero.
- Determine if the graph opens up or down (sign of coefficient).
- Check the slope (coefficient of the absolute value).
- Use the color key to fill in the mandala.

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✔ This activity combines algebra skills (graphing absolute value equations) with art, making learning fun and engaging!

Let me know if you'd like a printable version or solutions sheet!