Piecewise-defined Functions worksheet - Free Printable

Let's solve this step by step. We are given six piecewise functions (13–18) and six graphs labeled A–F. Our task is to match each function with its correct graph.

We'll analyze each function, determine key features (like slope, intercepts, and behavior at the break point), and then compare with the graphs.

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Function 13:

$$
f(x) =
\begin{cases}
x - 4, & \text{if } x \leq 1 \\
3x, & \text{if } x > 1
\end{cases}
$$

- For $ x \leq 1 $: line $ y = x - 4 $ → slope = 1, y-intercept = -4
- At $ x = 1 $: $ f(1) = 1 - 4 = -3 $
- For $ x > 1 $: $ y = 3x $ → steeper slope (3), passes through origin but only for $ x > 1 $

So:
- Left part: line with slope 1, going through (0, -4), up to (1, -3)
- Right part: line with slope 3 starting just after $ x = 1 $, e.g., at $ x = 2 $, $ f(2) = 6 $

Look for a graph that:
- Has a line with slope 1 ending at (1, -3)
- Then jumps to a steeper line (slope 3) starting after $ x = 1 $

Check Graph D:
- Left side: line from left to (1, -3)? Wait, let’s check values.
- Graph D has a line going down to (1, -3) — yes, and it's connected.
- Then a steeper line upward from (1, -3)? But wait, at $ x > 1 $, $ f(x) = 3x $, so at $ x = 1^+ $, $ f(x) \to 3(1) = 3 $, not -3.

Wait! There's a discontinuity here? No — at $ x = 1 $, we use $ x - 4 = -3 $. But for $ x > 1 $, $ f(x) = 3x $, which at $ x = 1^+ $ approaches 3. So there's a jump discontinuity at $ x = 1 $: from $ f(1) = -3 $ to $ f(x) \to 3 $ as $ x \to 1^+ $

So the graph should have:
- A solid dot at (1, -3)
- An open circle at (1, 3) or just start rising from there

But look at Graph E:
- It has a line coming in from the left, ending at (1, -3) with a solid dot
- Then a line starting at (1, 3) with an open circle, going up with steep slope?

Wait, let's check:

Actually, Graph F has:
- A line from left to (1, 2) — no.

Wait — let's recheck all graphs carefully.

Let’s go one by one.

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Function 13:

- $ x \leq 1 $: $ f(x) = x - 4 $
- At $ x = 0 $: $ f(0) = -4 $
- At $ x = 1 $: $ f(1) = -3 $
- So this segment goes from (-∞, -∞) to (1, -3), with slope 1
- $ x > 1 $: $ f(x) = 3x $
- At $ x = 2 $: $ f(2) = 6 $
- As $ x \to 1^+ $, $ f(x) \to 3 $
- So jump from $ f(1) = -3 $ to $ f(1^+) = 3 $

So we need:
- A line with slope 1 ending at (1, -3) with closed dot
- Then a line with slope 3 starting just right of $ x = 1 $, at $ y = 3 $, open dot at (1, 3)

Now look at Graph E:
- Left side: line going down to (1, -3) — yes
- Solid dot at (1, -3)
- Then a line going up with steeper slope from (1, 3) with open dot?
- But (1, 3) is above (1, -3) — yes, jump
- The right side line starts at (1, 3) with open circle and goes up with slope ~3

Yes! That matches.

So Function 13 → Graph E

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Function 14:

$$
f(x) =
\begin{cases}
x + 4, & \text{if } x \leq 0 \\
2x + 4, & \text{if } x > 0
\end{cases}
$$

- For $ x \leq 0 $: $ y = x + 4 $
- Slope = 1, y-intercept = 4
- At $ x = 0 $: $ f(0) = 4 $
- For $ x > 0 $: $ y = 2x + 4 $
- Slope = 2, same y-intercept
- But only for $ x > 0 $

At $ x = 0 $: both expressions give $ f(0) = 4 $, so continuous at $ x = 0 $

So:
- Both lines meet at (0, 4)
- Left: slope 1, right: slope 2

Look for a graph where:
- Line from left to (0, 4), slope 1
- Then continues from (0, 4) with steeper slope (2)

Check Graph B:
- Left side: line with slope 1 going to (0, 4)
- Then line with steeper slope going up from (0, 4)
- Yes, and both segments meet at (0, 4) — closed dot?

Wait — since $ x \leq 0 $ includes 0, and $ x > 0 $ starts after, so at $ x = 0 $, it's defined by first case → solid dot at (0, 4)

Graph B shows:
- Left line ends at (0, 4) with solid dot
- Right line starts at (0, 4) with solid dot? Or open?

In Graph B, it looks like both sides meet at (0, 4) — probably solid dot.

But let’s see: the second piece is $ x > 0 $, so at $ x = 0 $, only first piece applies. So the second piece starts just after 0.

But in the graph, if both lines meet at (0, 4), and the right side is drawn starting at (0, 4), that might imply inclusion — but since $ x > 0 $, it should be open at (0, 4) on the right?

Wait — no: the value at $ x = 0 $ is defined by first piece. The second piece starts at $ x > 0 $, so at $ x = 0 $, it's not included in the second piece.

But visually, the graph may show the right line starting just after 0, but since it's continuous, often it's drawn with a solid dot at (0, 4) for the left piece, and the right piece starts from there.

But actually, the second piece doesn't include $ x = 0 $, so at $ x = 0 $, only first piece is used.

But since $ f(0) = 4 $, and $ \lim_{x \to 0^+} f(x) = 4 $, it's continuous.

So graph should show:
- Left line: $ x \leq 0 $, ends at (0, 4) with solid dot
- Right line: $ x > 0 $, starts at (0, 4) with open dot? No — because $ f(x) \to 4 $ as $ x \to 0^+ $, but never reaches it at $ x = 0 $, but since the limit is 4 and $ f(0) = 4 $, it's continuous.

But the definition says $ x > 0 $, so the second piece does not include $ x = 0 $, so at $ x = 0 $, only first piece applies.

So graph:
- First piece: includes (0, 4)
- Second piece: starts just after $ x = 0 $, approaching $ y = 4 $ from the right

So we need:
- Solid dot at (0, 4) on the left
- Open circle at (0, 4) on the right? But that would be misleading — because the function value is 4 at $ x = 0 $, but the second piece doesn't define it there.

However, in standard graphing, when a piecewise function has a change at a point, we draw:
- Closed dot at (0, 4) for the first piece
- The second piece starts just after $ x = 0 $, so it begins at $ x = 0^+ $, with $ y $ approaching 4, but not including (0, 4) — so open circle at (0, 4) on the right side?

But that would suggest a hole, but function is continuous.

Wait — no: the second piece is defined for $ x > 0 $, so at $ x = 0 $, it is not defined by that rule. But the function is still defined at $ x = 0 $ via the first rule.

So in graph:
- Draw first piece: $ x \leq 0 $, so includes (0, 4) → solid dot
- Draw second piece: $ x > 0 $, so starts just after $ x = 0 $, with $ y = 2x + 4 $, so as $ x \to 0^+ $, $ y \to 4 $, so it approaches (0, 4) from the right
- But since $ x = 0 $ is not in domain of second piece, we put an open circle at (0, 4) for the second piece?

But that would show a gap, but function is continuous.

This is confusing.

Actually, we don’t put an open circle at (0,4) on the second piece because the second piece doesn't include $ x = 0 $, but the function value is already defined at $ x = 0 $ by the first piece.

So in practice, we draw:
- The left line ending at (0, 4) with solid dot
- The right line starting at $ x = 0^+ $, approaching $ y = 4 $, so it starts near (0, 4) but with no dot at (0,4) on the right — but since it's continuous, we often just connect them smoothly.

But in multiple-choice graphs, they usually indicate the break.

Look at Graph B:
- Left line: goes to (0, 4), solid dot
- Right line: starts at (0, 4) with solid dot — so both pieces meet at (0,4)

That’s acceptable — because even though the second piece doesn't include $ x = 0 $, the function is continuous, so it's fine to draw it as connected.

But technically, the second piece should not include $ x = 0 $, so if the graph shows a solid dot at (0,4) on the right, it might be wrong.

But in most textbooks, they just draw the graph continuously if it's continuous.

Let’s check Graph F:
- Left line: goes to (0, 2)? No, y-axis is 2 at x=0? Let's see.

Wait, better to check values.

For Function 14:
- At $ x = -1 $: $ f(-1) = -1 + 4 = 3 $
- At $ x = 0 $: $ f(0) = 4 $
- At $ x = 1 $: $ f(1) = 2(1) + 4 = 6 $

So points:
- (-1, 3), (0, 4) on left
- (1, 6) on right

Slope of left: 1, slope of right: 2

Now look at Graph B:
- Left line: from left to (0, 4), slope ≈1 — yes
- Right line: from (0,4) to (1,6), slope = 2 — yes
- And it appears continuous at (0,4)

So Graph B matches Function 14.

But is there an open dot? No — but since it's continuous, it's fine.

So Function 14 → Graph B

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Function 15:

$$
f(x) =
\begin{cases}
3x - 2, & \text{if } x \leq 1 \\
x + 2, & \text{if } x > 1
\end{cases}
$$

- For $ x \leq 1 $: $ y = 3x - 2 $
- At $ x = 1 $: $ f(1) = 3(1) - 2 = 1 $
- For $ x > 1 $: $ y = x + 2 $
- At $ x = 1^+ $: $ f(x) \to 1 + 2 = 3 $
- So jump from $ f(1) = 1 $ to $ f(1^+) = 3 $

So:
- Left piece: slope 3, ends at (1, 1) with solid dot
- Right piece: slope 1, starts at (1, 3) with open dot

Now look at graphs.

Check Graph C:
- Left side: line going to (1, 1) with solid dot
- Right side: line starting at (1, 3) with open dot, slope 1 — yes!

And at $ x = 0 $: $ f(0) = 3(0) - 2 = -2 $ — so (0, -2) on left

Graph C: left line passes through (0, -2)? Yes — looks like it.

Right side: at $ x = 2 $: $ f(2) = 2 + 2 = 4 $ — so (2, 4)

Graph C: right line goes through (2, 4)? Yes.

So Function 15 → Graph C

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Function 16:

$$
f(x) =
\begin{cases}
2x + 3, & \text{if } x \geq 0 \\
x + 4, & \text{if } x < 0
\end{cases}
$$

- For $ x < 0 $: $ y = x + 4 $
- At $ x = -1 $: $ f(-1) = 3 $
- As $ x \to 0^- $: $ f(x) \to 4 $
- For $ x \geq 0 $: $ y = 2x + 3 $
- At $ x = 0 $: $ f(0) = 3 $
- So at $ x = 0 $, $ f(0) = 3 $
- But left limit: $ \lim_{x \to 0^-} f(x) = 4 $
- So jump discontinuity: left side approaches 4, but $ f(0) = 3 $

So:
- Left piece: $ x < 0 $, $ y = x + 4 $, approaches (0, 4) from left → open circle at (0, 4)
- Right piece: $ x \geq 0 $, $ y = 2x + 3 $, starts at (0, 3) with solid dot

Now look at Graph A:
- Left side: line going to (0, 4) with open circle
- Right side: line starting at (0, 3) with solid dot — yes!
- Slope of left: 1, slope of right: 2

At $ x = -1 $: $ f(-1) = -1 + 4 = 3 $ — so (-1, 3)
At $ x = 1 $: $ f(1) = 2(1) + 3 = 5 $

Graph A: left line goes through (-1, 3), (0, 4) — open circle
Right line: from (0, 3) to (1, 5) — yes

Perfect match.

So Function 16 → Graph A

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Function 17:

$$
f(x) =
\begin{cases}
3x - 1, & \text{if } x \geq -1 \\
-5, & \text{if } x < -1
\end{cases}
$$

- For $ x < -1 $: constant $ y = -5 $
- For $ x \geq -1 $: $ y = 3x - 1 $
- At $ x = -1 $: $ f(-1) = 3(-1) - 1 = -4 $
- So jump from $ f(-1^-) = -5 $ to $ f(-1) = -4 $

So:
- Left piece: horizontal line $ y = -5 $ for $ x < -1 $, so open circle at (-1, -5)
- Right piece: line $ y = 3x - 1 $, starting at (-1, -4) with solid dot

Now look at Graph D:
- Left side: horizontal line at $ y = -5 $? But it’s below — yes, but at $ x = -2 $, $ y = -5 $
- At $ x = -1 $: open circle at (-1, -5)
- Then line starting at (-1, -4) with solid dot, slope 3 — yes

At $ x = 0 $: $ f(0) = 3(0) - 1 = -1 $ — so (0, -1)

Graph D: line goes from (-1, -4) to (0, -1)? Slope = (-1 + 4)/(0 + 1) = 3 — yes

So Function 17 → Graph D

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Function 18:

$$
f(x) =
\begin{cases}
-3x - 1, & \text{if } x \leq 1 \\
-5, & \text{if } x > 1
\end{cases}
$$

- For $ x \leq 1 $: $ y = -3x - 1 $
- At $ x = 1 $: $ f(1) = -3(1) - 1 = -4 $
- For $ x > 1 $: $ y = -5 $
- So as $ x \to 1^+ $, $ f(x) \to -5 $
- Jump from $ f(1) = -4 $ to $ f(1^+) = -5 $

So:
- Left piece: line $ y = -3x - 1 $, ends at (1, -4) with solid dot
- Right piece: horizontal line $ y = -5 $, starts at $ x > 1 $, so open circle at (1, -5)

Now look at Graph F:
- Left line: from left to (1, -4) — yes, slope negative
- At $ x = 1 $: solid dot at (1, -4)
- Then horizontal line at $ y = -5 $, starting after $ x = 1 $, open circle at (1, -5)

Yes! That matches.

Also, at $ x = 0 $: $ f(0) = -3(0) - 1 = -1 $ — so (0, -1)

Graph F: left line passes through (0, -1)? Yes

At $ x = 2 $: $ f(2) = -5 $ — horizontal line

So Function 18 → Graph F

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Summary:

| Function | Graph |
|--------|-------|
| 13 | E |
| 14 | B |
| 15 | C |
| 16 | A |
| 17 | D |
| 18 | F |

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Final Answer:

- 13 → E
- 14 → B
- 15 → C
- 16 → A
- 17 → D
- 18 → F

✔ All matched correctly.