Piecewise Functions Graphing Worksheet - Free Printable

Let's solve each problem step by step.

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Problem 1:

Given the piecewise function:

$$
f(x) =
\begin{cases}
x + 5 & \text{if } x < -2 \\
-2x - 1 & \text{if } x \geq -2
\end{cases}
$$

We are to:
1. Graph the function.
2. Determine if it is a function.
3. Evaluate $ f(3), f(-4), f(-2) $.

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#### Step 1: Evaluate the function at given points

a) $ f(3) $:

Since $ 3 \geq -2 $, use the second rule:
$$
f(3) = -2(3) - 1 = -6 - 1 = -7
$$

✔ $ f(3) = -7 $

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b) $ f(-4) $:

Since $ -4 < -2 $, use the first rule:
$$
f(-4) = (-4) + 5 = 1
$$

✔ $ f(-4) = 1 $

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c) $ f(-2) $:

Since $ x = -2 $ satisfies $ x \geq -2 $, use the second rule:
$$
f(-2) = -2(-2) - 1 = 4 - 1 = 3
$$

✔ $ f(-2) = 3 $

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#### Step 2: Graph the function

We graph two parts:

1. For $ x < -2 $: $ f(x) = x + 5 $
This is a straight line with slope 1 and y-intercept 5.
But only for $ x < -2 $.
At $ x = -2 $, we do not include this part — so draw an open circle at $ x = -2 $.

Let’s find a few points:
- $ x = -3 $: $ f(-3) = -3 + 5 = 2 $
- $ x = -4 $: $ f(-4) = 1 $ (as above)
- $ x = -5 $: $ f(-5) = 0 $

So plot points like $ (-5,0), (-4,1), (-3,2) $, and draw a ray going left from $ (-2,3) $ but not including $ (-2,3) $ on this side.

2. For $ x \geq -2 $: $ f(x) = -2x - 1 $
This is a line with slope $-2$, y-intercept $-1$.
Include $ x = -2 $: $ f(-2) = 3 $ → closed circle at $ (-2,3) $

Find a few more points:
- $ x = -1 $: $ f(-1) = -2(-1) - 1 = 2 - 1 = 1 $
- $ x = 0 $: $ f(0) = -1 $
- $ x = 1 $: $ f(1) = -2(1) - 1 = -3 $
- $ x = 2 $: $ f(2) = -5 $

Plot these and draw a ray starting at $ (-2,3) $ going right.

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#### Step 3: Is it a function?

Yes — it is a function because every input $ x $ has exactly one output.
Even though it's defined in pieces, each $ x $ falls into exactly one case, and there are no vertical lines that cross the graph more than once.

✔ It passes the Vertical Line Test.

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✔ Summary for Problem 1:

- $ f(3) = -7 $
- $ f(-4) = 1 $
- $ f(-2) = 3 $
- Graph: Two rays meeting at $ x = -2 $, with open circle on left side, closed on right.
- It is a function.

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Problem 2:

Given the piecewise function:

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

Evaluate:
- $ f(-2) $
- $ f(6) $
- $ f(1) $

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#### Step 1: Evaluate

a) $ f(-2) $:

Since $ -2 < 1 $, use the second rule:
$$
f(-2) = \frac{-2}{2} - 3 = -1 - 3 = -4
$$

✔ $ f(-2) = -4 $

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b) $ f(6) $:

Since $ 6 \geq 1 $, use the first rule:
$$
f(6) = 2(6) + 1 = 12 + 1 = 13
$$

✔ $ f(6) = 13 $

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c) $ f(1) $:

Since $ x = 1 \geq 1 $, use the first rule:
$$
f(1) = 2(1) + 1 = 2 + 1 = 3
$$

✔ $ f(1) = 3 $

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#### Step 2: Graph the function

1. For $ x < 1 $: $ f(x) = \frac{x}{2} - 3 $
This is a line with slope $ \frac{1}{2} $, y-intercept $ -3 $.
But only for $ x < 1 $.
At $ x = 1 $, we do not include this part → open circle at $ x = 1 $.

Points:
- $ x = 0 $: $ f(0) = 0/2 - 3 = -3 $
- $ x = -2 $: $ f(-2) = -1 - 3 = -4 $ (as above)
- $ x = -4 $: $ f(-4) = -2 - 3 = -5 $

Draw a ray going left from $ x = 1 $, ending with an open circle at $ (1, f(1)) $ from this side.

2. For $ x \geq 1 $: $ f(x) = 2x + 1 $
Slope 2, y-intercept 1.
Includes $ x = 1 $: $ f(1) = 3 $ → closed circle at $ (1,3) $

Points:
- $ x = 1 $: $ f(1) = 3 $
- $ x = 2 $: $ f(2) = 5 $
- $ x = 3 $: $ f(3) = 7 $

Draw a ray starting at $ (1,3) $ going right.

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#### Step 3: Is it a function?

Yes — every $ x $ maps to exactly one $ y $.
No overlaps or contradictions.

✔ It is a function.

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✔ Summary for Problem 2:

- $ f(-2) = -4 $
- $ f(6) = 13 $
- $ f(1) = 3 $
- Graph: Two rays meeting at $ x = 1 $, with open circle on left side, closed on right.
- It is a function.

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Final Answers:

#### Problem 1:
- $ f(3) = \boxed{-7} $
- $ f(-4) = \boxed{1} $
- $ f(-2) = \boxed{3} $
- Is a function? Yes

#### Problem 2:
- $ f(-2) = \boxed{-4} $
- $ f(6) = \boxed{13} $
- $ f(1) = \boxed{3} $
- Is a function? Yes

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Let me know if you'd like a sketch of the graphs!