Graphing Linear Piecewise Functions Worksheet - Free Printable

Let's solve each of the piecewise functions step by step and explain how to graph them. Since I can't draw on the image, I’ll guide you through the process clearly so you can accurately graph them on the provided coordinate grids.

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Problem 1:

$$
f(x) =
\begin{cases}
2x + 1 & \text{if } x < -2 \\
-x - 1 & \text{if } x \geq -2
\end{cases}
$$

#### Step 1: Analyze each piece

First Piece: $ f(x) = 2x + 1 $ for $ x < -2 $

- This is a linear function with slope = 2 and y-intercept = 1.
- But it's only defined for $ x < -2 $. So we will graph this line only to the left of $ x = -2 $.
- At $ x = -2 $, this piece does not include the point (since it’s strictly less than), so we use an open circle at $ x = -2 $.

Second Piece: $ f(x) = -x - 1 $ for $ x \geq -2 $

- This is a linear function with slope = -1 and y-intercept = -1.
- It's defined for $ x \geq -2 $, so includes $ x = -2 $, meaning we use a closed circle at $ x = -2 $.

#### Step 2: Find key points

For $ f(x) = 2x + 1 $, $ x < -2 $:

Pick values less than -2:
- $ x = -3 $: $ f(-3) = 2(-3) + 1 = -6 + 1 = -5 $ → point: (-3, -5)
- $ x = -4 $: $ f(-4) = 2(-4) + 1 = -8 + 1 = -7 $ → point: (-4, -7)

Now, check what happens as $ x $ approaches -2 from the left:
- $ f(-2^-) = 2(-2) + 1 = -4 + 1 = -3 $
- So, plot an open circle at $ (-2, -3) $

For $ f(x) = -x - 1 $, $ x \geq -2 $:

- At $ x = -2 $: $ f(-2) = -(-2) - 1 = 2 - 1 = 1 $ → point: (-2, 1) → closed circle
- $ x = 0 $: $ f(0) = -0 - 1 = -1 $ → (0, -1)
- $ x = 1 $: $ f(1) = -1 - 1 = -2 $ → (1, -2)

#### Step 3: Graphing

- Draw the line $ y = 2x + 1 $ for $ x < -2 $, ending with an open circle at $ (-2, -3) $
- Draw the line $ y = -x - 1 $ for $ x \geq -2 $, starting with a closed circle at $ (-2, 1) $
- Note: The two lines do not meet — they are separate at $ x = -2 $

> ⚠️ Important: Even though both expressions evaluate to something near $ x = -2 $, they have different values at that point:
> - Left side: $ f(-2^-) = -3 $ (but not included)
> - Right side: $ f(-2) = 1 $ (included)

So there’s a jump discontinuity at $ x = -2 $.

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✔ Final Graph Description for Problem 1:

- Line 1: $ y = 2x + 1 $, drawn only for $ x < -2 $, ends with open circle at (-2, -3)
- Line 2: $ y = -x - 1 $, drawn for $ x \geq -2 $, starts with closed circle at (-2, 1)

Plot the points and connect accordingly.

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Problem 2:

$$
f(x) =
\begin{cases}
\frac{1}{2}x + 2 & \text{if } x < 2 \\
2x - 3 & \text{if } x \geq 2
\end{cases}
$$

#### Step 1: Analyze each piece

First Piece: $ f(x) = \frac{1}{2}x + 2 $ for $ x < 2 $

- Slope = $ \frac{1}{2} $, y-intercept = 2
- Only valid for $ x < 2 $ → open circle at $ x = 2 $

Second Piece: $ f(x) = 2x - 3 $ for $ x \geq 2 $

- Slope = 2, y-intercept = -3
- Valid for $ x \geq 2 $ → closed circle at $ x = 2 $

#### Step 2: Find key points

For $ f(x) = \frac{1}{2}x + 2 $, $ x < 2 $:

- $ x = 0 $: $ f(0) = 0 + 2 = 2 $ → (0, 2)
- $ x = 1 $: $ f(1) = 0.5 + 2 = 2.5 $ → (1, 2.5)
- As $ x \to 2^- $: $ f(2^-) = \frac{1}{2}(2) + 2 = 1 + 2 = 3 $
→ Open circle at $ (2, 3) $

For $ f(x) = 2x - 3 $, $ x \geq 2 $:

- At $ x = 2 $: $ f(2) = 2(2) - 3 = 4 - 3 = 1 $ → (2, 1) → closed circle
- $ x = 3 $: $ f(3) = 6 - 3 = 3 $ → (3, 3)
- $ x = 4 $: $ f(4) = 8 - 3 = 5 $ → (4, 5)

#### Step 3: Graphing

- Draw $ y = \frac{1}{2}x + 2 $ for $ x < 2 $, ending with open circle at (2, 3)
- Draw $ y = 2x - 3 $ for $ x \geq 2 $, starting with closed circle at (2, 1)

> ⚠️ Note: There is a jump discontinuity at $ x = 2 $:
> - Left limit: $ f(2^-) = 3 $
> - Right value: $ f(2) = 1 $

So the graph jumps down from 3 to 1 at $ x = 2 $.

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✔ Final Graph Description for Problem 2:

- Line 1: $ y = \frac{1}{2}x + 2 $, drawn for $ x < 2 $, ends with open circle at (2, 3)
- Line 2: $ y = 2x - 3 $, drawn for $ x \geq 2 $, starts with closed circle at (2, 1)

Connect the points accordingly.

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📌 Summary of Key Steps for Graphing Piecewise Functions:

1. Identify domains for each piece.
2. Graph each piece separately over its domain.
3. Use open circles for endpoints where the inequality is strict (< or >).
4. Use closed circles for endpoints where the inequality includes equality (≤ or ≥).
5. Check for continuity or jumps at boundary points.
6. Make sure no part of the graph extends outside its domain.

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You can now carefully plot these on the provided grid using the points and rules above.

Let me know if you'd like a visual sketch description or help checking your work!