Absolute Value Notes and Worksheets - Lindsay Bowden - Free Printable

Let's solve each problem step by step, analyzing the absolute value functions and their characteristics.

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Problem 1:

Function: $ f(x) = -|x| - 3 $

#### Step-by-step Analysis:

- Domain: The domain of an absolute value function is all real numbers since there are no restrictions on $ x $.
→ Domain: $ (-\infty, \infty) $

- Range: The basic $ |x| $ has a minimum at 0. Here, we have $ -|x| $, which flips the V-shape downward, and then subtract 3, shifting it down. So the maximum value is $ -3 $.
→ Range: $ (-\infty, -3] $

- Transformations:
- Reflection over the x-axis (due to the negative sign in front of $ |x| $)
- Vertical shift down by 3 units
→ Transformations: Reflected over x-axis, shifted down 3 units

- Axis of Symmetry: Absolute value functions are symmetric about the vertex. For $ f(x) = -|x| - 3 $, the vertex is at $ x = 0 $.
→ Axis of symmetry: $ x = 0 $

- y-intercept: Plug in $ x = 0 $:
$$
f(0) = -|0| - 3 = -3
$$
→ y-intercept: $ (0, -3) $

- Zero(s): Solve $ f(x) = 0 $
$$
-|x| - 3 = 0 \Rightarrow -|x| = 3 \Rightarrow |x| = -3
$$
This is impossible because absolute value can't be negative.
→ No zeros (or none)

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✔ Answer for Problem 1:

- Domain: $ (-\infty, \infty) $
- Range: $ (-\infty, -3] $
- Transformations: Reflection over x-axis, vertical shift down 3 units
- Axis of symmetry: $ x = 0 $
- y-intercept: $ (0, -3) $
- Zero(s): None

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Problem 2:

Function: $ f(x) = 2|x - 1| $

#### Step-by-step Analysis:

- Domain: All real numbers
→ Domain: $ (-\infty, \infty) $

- Range: The expression $ |x - 1| $ has a minimum of 0 at $ x = 1 $. Multiplying by 2 makes it steeper, but still minimum is 0. So $ f(x) \geq 0 $
→ Range: $ [0, \infty) $

- Transformations:
- Horizontal shift right by 1 unit (because $ x - 1 $)
- Vertical stretch by factor of 2 (coefficient 2 in front)
→ Transformations: Shift right 1 unit, vertical stretch by 2

- Axis of Symmetry: The vertex occurs at $ x = 1 $, so symmetry line is:
→ Axis of symmetry: $ x = 1 $

- y-intercept: Plug in $ x = 0 $:
$$
f(0) = 2|0 - 1| = 2(1) = 2
$$
→ y-intercept: $ (0, 2) $

- Zero(s): Solve $ f(x) = 0 $
$$
2|x - 1| = 0 \Rightarrow |x - 1| = 0 \Rightarrow x = 1
$$
→ Zero(s): $ x = 1 $

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✔ Answer for Problem 2:

- Domain: $ (-\infty, \infty) $
- Range: $ [0, \infty) $
- Transformations: Shift right 1 unit, vertical stretch by factor of 2
- Axis of symmetry: $ x = 1 $
- y-intercept: $ (0, 2) $
- Zero(s): $ x = 1 $

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Problem 3:

Function: $ f(x) = |x + 2| $

We'll graph using transformations and find characteristics.

#### Step-by-step Analysis:

- Base Function: $ f(x) = |x| $ — V-shaped, vertex at origin.

- Transformation: $ |x + 2| = |x - (-2)| $ → shift left by 2 units.

- Graph: Start with $ |x| $, move vertex from $ (0, 0) $ to $ (-2, 0) $. Two rays go up from that point.

- Domain: All real numbers
→ Domain: $ (-\infty, \infty) $

- Range: Minimum value is 0 (at vertex), goes up infinitely
→ Range: $ [0, \infty) $

- Transformations: Horizontal shift left 2 units

- Axis of Symmetry: Vertex at $ x = -2 $
→ Axis of symmetry: $ x = -2 $

- y-intercept: Plug $ x = 0 $:
$$
f(0) = |0 + 2| = 2
$$
→ y-intercept: $ (0, 2) $

- Zero(s): Solve $ |x + 2| = 0 $
$$
x + 2 = 0 \Rightarrow x = -2
$$
→ Zero(s): $ x = -2 $

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✔ Answer for Problem 3:

- Domain: $ (-\infty, \infty) $
- Range: $ [0, \infty) $
- Transformations: Shift left 2 units
- Axis of symmetry: $ x = -2 $
- y-intercept: $ (0, 2) $
- Zero(s): $ x = -2 $

> 🔹 Graph: Draw a V-shape with vertex at $ (-2, 0) $, going up to the right and left. Passes through $ (0, 2) $ and $ (-4, 2) $.

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Problem 4:

Function: $ f(x) = -|x| + 4 $

#### Step-by-step Analysis:

- Transformations:
- $ -|x| $: Reflects the V-shape over x-axis (opens downward)
- $ +4 $: Shifts up 4 units

- Vertex: Original vertex at $ (0, 0) $; after reflection and shift → $ (0, 4) $

- Domain: All real numbers
→ Domain: $ (-\infty, \infty) $

- Range: Maximum at $ y = 4 $, goes down infinitely
→ Range: $ (-\infty, 4] $

- Transformations: Reflection over x-axis, vertical shift up 4 units

- Axis of Symmetry: Still symmetric about $ x = 0 $
→ Axis of symmetry: $ x = 0 $

- y-intercept: $ f(0) = -|0| + 4 = 4 $
→ y-intercept: $ (0, 4) $

- Zero(s): Solve $ f(x) = 0 $
$$
-|x| + 4 = 0 \Rightarrow |x| = 4 \Rightarrow x = \pm 4
$$
→ Zero(s): $ x = -4 $ and $ x = 4 $

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✔ Answer for Problem 4:

- Domain: $ (-\infty, \infty) $
- Range: $ (-\infty, 4] $
- Transformations: Reflection over x-axis, shift up 4 units
- Axis of symmetry: $ x = 0 $
- y-intercept: $ (0, 4) $
- Zero(s): $ x = -4 $ and $ x = 4 $

> 🔹 Graph: V-shape opening downward, vertex at $ (0, 4) $, crossing x-axis at $ (-4, 0) $ and $ (4, 0) $

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✔ Final Summary:

| Problem | Function | Domain | Range | Transformations | Axis of Symmetry | y-intercept | Zeros |
|--------|---------|--------|-------|------------------|--------------------|-------------|--------|
| 1 | $ f(x) = -|x| - 3 $ | $ (-\infty, \infty) $ | $ (-\infty, -3] $ | Reflect over x-axis, shift down 3 | $ x = 0 $ | $ (0, -3) $ | None |
| 2 | $ f(x) = 2|x - 1| $ | $ (-\infty, \infty) $ | $ [0, \infty) $ | Shift right 1, stretch by 2 | $ x = 1 $ | $ (0, 2) $ | $ x = 1 $ |
| 3 | $ f(x) = |x + 2| $ | $ (-\infty, \infty) $ | $ [0, \infty) $ | Shift left 2 | $ x = -2 $ | $ (0, 2) $ | $ x = -2 $ |
| 4 | $ f(x) = -|x| + 4 $ | $ (-\infty, \infty) $ | $ (-\infty, 4] $ | Reflect over x-axis, shift up 4 | $ x = 0 $ | $ (0, 4) $ | $ x = \pm 4 $ |

Let me know if you'd like the graphs sketched or explained further!