Absolute Value Notes and Worksheets - Lindsay Bowden - Free Printable

Let’s solve each problem step by step. We’ll look at the function, think about what it does to the basic absolute value graph (which is a V-shape), and then find all the characteristics.

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Problem 1: f(x) = –|x| – 3

This starts with |x|, which is a V opening upward with vertex at (0,0).

- The negative sign in front flips it upside down → now it opens downward.
- Then we subtract 3 → shift the whole graph down by 3 units.

So vertex is at (0, -3).

Now let’s find each characteristic:

• Domain: Absolute value functions are defined for all real numbers → Domain: All real numbers or (-∞, ∞)

• Range: Since it opens downward and the highest point is y = -3, everything else is below that → Range: y ≤ -3 or (-∞, -3]

• Transformations:
- Reflected over x-axis (because of the negative)
- Shifted down 3 units

• Axis of symmetry: For any absolute value function, it’s the vertical line through the vertex → x = 0

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

• Zero(s): Set f(x) = 0 → –|x| – 3 = 0 → –|x| = 3 → |x| = –3 → But absolute value can’t be negative! So no solution → No zeros

✔ Final Answer for Problem 1:
- domain: all real numbers
- range: y ≤ -3
- transformations: reflected over x-axis, shifted down 3
- axis of symmetry: x = 0
- y-intercept: (0, -3)
- zero(s): none

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Problem 2: f(x) = 2|x – 1|

Starts with |x|, then:

- Inside the absolute value: x – 1 → shift right 1 unit
- Multiply by 2 outside → stretch vertically by factor of 2 (makes it steeper)

Vertex is at (1, 0)

• Domain: Still all real numbers → (-∞, ∞)

• Range: Opens upward, lowest point is y=0 → y ≥ 0 or [0, ∞)

• Transformations:
- Shifted right 1 unit
- Vertically stretched by factor of 2

• Axis of symmetry: Through vertex → x = 1

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

• Zero(s): Set f(x)=0 → 2|x–1|=0 → |x–1|=0 → x–1=0 → x=1 → One zero at x=1

✔ Final Answer for Problem 2:
- domain: all real numbers
- range: y ≥ 0
- transformations: shifted right 1, vertical stretch by 2
- axis of symmetry: x = 1
- y-intercept: (0, 2)
- zero(s): x = 1

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Problem 3: f(x) = |x + 2|

Basic |x|, but inside we have x + 2 → that means shift LEFT 2 units.

Vertex at (-2, 0)

• Domain: All real numbers

• Range: Opens up, minimum y=0 → y ≥ 0

• Transformations: Shifted left 2 units

• Axis of symmetry: x = -2

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

• Zero(s): Set f(x)=0 → |x+2|=0 → x+2=0 → x=-2 → One zero at x=-2

✔ Final Answer for Problem 3:
- domain: all real numbers
- range: y ≥ 0
- transformations: shifted left 2
- axis of symmetry: x = -2
- y-intercept: (0, 2)
- zero(s): x = -2

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Problem 4: f(x) = –|x| + 4

Start with |x|, flip it (negative sign) → opens down, then add 4 → shift up 4 units.

Vertex at (0, 4)

• Domain: All real numbers

• Range: Opens down, max y=4 → y ≤ 4

• Transformations: Reflected over x-axis, shifted up 4

• Axis of symmetry: x = 0

• y-intercept: x=0 → f(0)=–|0|+4=4 → (0,4)

• Zero(s): Set f(x)=0 → –|x| + 4 = 0 → –|x| = –4 → |x| = 4 → x = 4 or x = -4 → Two zeros

✔ Final Answer for Problem 4:
- domain: all real numbers
- range: y ≤ 4
- transformations: reflected over x-axis, shifted up 4
- axis of symmetry: x = 0
- y-intercept: (0, 4)
- zero(s): x = -4 and x = 4

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Final Answer:

Problem 1:
- domain: all real numbers
- range: y ≤ -3
- transformations: reflected over x-axis, shifted down 3
- axis of symmetry: x = 0
- y-intercept: (0, -3)
- zero(s): none

Problem 2:
- domain: all real numbers
- range: y ≥ 0
- transformations: shifted right 1, vertical stretch by 2
- axis of symmetry: x = 1
- y-intercept: (0, 2)
- zero(s): x = 1

Problem 3:
- domain: all real numbers
- range: y ≥ 0
- transformations: shifted left 2
- axis of symmetry: x = -2
- y-intercept: (0, 2)
- zero(s): x = -2

Problem 4:
- domain: all real numbers
- range: y ≤ 4
- transformations: reflected over x-axis, shifted up 4
- axis of symmetry: x = 0
- y-intercept: (0, 4)
- zero(s): x = -4 and x = 4