Worksheet on sketching graphs of piecewise functions with four examples and their corresponding graphs.

The image you've uploaded is a worksheet from Kuta Software - Infinite Precalculus titled "Piecewise Functions", where the task is to sketch the graph of each given piecewise function. The worksheet contains four problems, and it appears that graphs have already been drawn (likely by a student or teacher), with some labeled as "Key," suggesting this might be an answer key.

Let’s go through each problem, explain how to sketch the graph step-by-step, and verify whether the provided graphs are correct.

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Problem 1:

$$
f(x) =
\begin{cases}
-2x - 1, & x \leq 2 \\
-x + 4, & x > 2
\end{cases}
$$

#### Step-by-step:
- First piece: $ f(x) = -2x - 1 $ for $ x \leq 2 $
- This is a line with slope $-2$, y-intercept $-1$
- At $ x = 2 $: $ f(2) = -2(2) - 1 = -5 $
- So we draw the line starting from left up to $ x = 2 $, including the point $ (2, -5) $ → closed circle
- Second piece: $ f(x) = -x + 4 $ for $ x > 2 $
- Slope $-1$, y-intercept $4$
- At $ x = 2 $: $ f(2) = -2 + 4 = 2 $, but since $ x > 2 $, we do not include this point → open circle at (2, 2)
- Draw the line to the right of $ x = 2 $

✔ Graph Check: The graph shows:
- A line with negative slope going down to $ (2, -5) $ → closed dot
- Then a different line starting just after $ x = 2 $, going downward with slope $-1$ → open dot at $ (2, 2) $
→ Correct

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Problem 2:

$$
f(x) =
\begin{cases}
-4, & x \leq -2 \\
x - 2, & -2 < x < 2 \\
-2x + 4, & x \geq 2
\end{cases}
$$

#### Step-by-step:
- First piece: $ f(x) = -4 $ for $ x \leq -2 $
- Horizontal line at $ y = -4 $, extending left to $ x = -2 $, including $ (-2, -4) $ → closed dot
- Second piece: $ f(x) = x - 2 $ for $ -2 < x < 2 $
- Line with slope 1, y-intercept $-2$
- At $ x = -2 $: $ f(-2) = -4 $, but excluded → open dot at (-2, -4)
- At $ x = 2 $: $ f(2) = 0 $, but excluded → open dot at (2, 0)
- Third piece: $ f(x) = -2x + 4 $ for $ x \geq 2 $
- Slope $-2$, y-intercept $4$
- At $ x = 2 $: $ f(2) = -4 + 4 = 0 $ → closed dot at (2, 0)

So:
- Left side: horizontal line at $ y = -4 $, ends at $ x = -2 $ with closed dot
- Middle: line from $ x = -2 $ (open) to $ x = 2 $ (open), rising
- Right: line starting at $ x = 2 $, $ y = 0 $, closed dot, going downward

✔ Graph Check: The graph shows:
- Horizontal line at $ y = -4 $ ending at $ x = -2 $ with closed dot
- Rising line from $ (-2, -4) $ to $ (2, 0) $, both ends open
- Then a descending line starting at $ (2, 0) $ with closed dot
→ Correct

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Problem 3:

$$
f(x) =
\begin{cases}
-2^x, & x < -4 \\
-|x|, & -4 \leq x \leq 0 \\
4 - x^2, & x > 0
\end{cases}
$$

Note: $ -2^x $ means $ -(2^x) $, not $ (-2)^x $

#### Step-by-step:
- First piece: $ f(x) = -2^x $ for $ x < -4 $
- Exponential decay, but reflected over x-axis
- As $ x \to -\infty $, $ 2^x \to 0 $, so $ -2^x \to 0^- $
- At $ x = -4 $: $ f(-4) = -2^{-4} = -1/16 \approx -0.0625 $
- But $ x < -4 $, so open dot at (-4, -1/16), and curve approaches $ y = 0 $ from below as $ x \to -\infty $
- Second piece: $ f(x) = -|x| $ for $ -4 \leq x \leq 0 $
- This is a V-shaped graph flipped upside down
- For $ x \in [-4, 0] $, $ |x| = -x $, so $ -|x| = -(-x) = x $? Wait: no!
- Actually: $ |x| = -x $ when $ x < 0 $, so $ -|x| = -(-x) = x $?
- No! Let's clarify:
- $ |x| = -x $ if $ x < 0 $
- So $ -|x| = -(-x) = x $ only if $ x < 0 $
- So $ -|x| = x $ for $ x < 0 $
- But at $ x = 0 $, $ -|0| = 0 $
- So on $ [-4, 0] $, $ f(x) = -|x| = -(-x) = x $ for $ x < 0 $, and $ 0 $ at $ x = 0 $
- Wait: actually $ -|x| $ for $ x \in [-4, 0] $:
- If $ x = -4 $: $ -|-4| = -4 $
- If $ x = -2 $: $ -2 $
- If $ x = 0 $: $ 0 $
- So this is a straight line from $ (-4, -4) $ to $ (0, 0) $
- Third piece: $ f(x) = 4 - x^2 $ for $ x > 0 $
- Parabola opening downward, vertex at $ (0, 4) $
- But $ x > 0 $, so start just after $ x = 0 $
- At $ x = 0 $: $ f(0) = 4 $, but not included → open dot at (0, 4)
- As $ x $ increases, it decreases

So putting together:
- Left: exponential decay approaching $ y = 0 $ from below, open dot at $ (-4, -1/16) $
- Middle: line from $ (-4, -4) $ to $ (0, 0) $, closed dots at both ends
- Right: parabola $ y = 4 - x^2 $ for $ x > 0 $, starting at open dot $ (0, 4) $, decreasing

Wait — but at $ x = -4 $, we have two definitions:
- From first piece: $ f(-4) $ is not defined because $ x < -4 $
- Second piece includes $ x = -4 $, so $ f(-4) = -|-4| = -4 $

So at $ x = -4 $: only second piece applies, so closed dot at $ (-4, -4) $

And for $ x < -4 $: $ f(x) = -2^x $, which is very close to 0 (negative), so near $ (-4, -0.06) $, open dot

Now check the graph:

✔ Graph Check: The graph shows:
- A small curve near $ x = -4 $, approaching from the left, open dot around $ (-4, -0.06) $
- A line from $ (-4, -4) $ to $ (0, 0) $, both ends closed
- Then a downward parabola starting at $ (0, 4) $ with open dot, going down

But wait: at $ x = 0 $, the second piece gives $ f(0) = -|0| = 0 $, so closed dot at (0, 0)
And third piece starts at $ x > 0 $, so open dot at (0, 4)? That doesn’t make sense — $ f(0) $ is defined by the middle piece as 0.

But in the graph, it shows a parabola starting at (0, 4) with open dot, which would suggest $ f(0) = 4 $, but that contradicts the definition.

✘ Error in graph?

Wait — let’s double-check:
For $ x > 0 $: $ f(x) = 4 - x^2 $
At $ x = 0^+ $, $ f(x) \to 4 $, but $ f(0) = -|0| = 0 $

So the graph should:
- Have a closed dot at (0, 0) from the middle piece
- And the parabola starts just to the right of $ x = 0 $, approaching $ y = 4 $, but does not include (0, 4) → open dot at (0, 4)? No — $ (0, 4) $ is not on the graph; the parabola approaches $ y = 4 $ as $ x \to 0^+ $, but never reaches it?

Wait: $ f(x) = 4 - x^2 $, so at $ x = 0 $, $ f(0) = 4 $, but x > 0, so we don't include $ x = 0 $. As $ x \to 0^+ $, $ f(x) \to 4 $, so we approach $ y = 4 $ from the right.

So the parabola should start near $ (0, 4) $ with open dot at (0, 4), and decrease.

But at $ x = 0 $, the value is $ f(0) = 0 $ from the middle piece.

So the graph must show:
- A closed dot at (0, 0) (from middle piece)
- A parabola starting just right of (0, 4) with open dot at (0, 4)

But in the given graph, it looks like the parabola starts at $ (0, 4) $ with open dot, and the line ends at $ (0, 0) $ — that’s fine.

But is there a jump? Yes — from $ (0, 0) $ to approaching $ (0, 4) $ — that’s okay.

Wait: but the middle piece goes to (0, 0), and the right piece starts near (0, 4), so there’s a gap between $ y = 0 $ and $ y = 4 $ at $ x = 0 $

But that’s correct — the function jumps from $ f(0) = 0 $ to $ f(x) \to 4 $ as $ x \to 0^+ $

So the graph should show:
- Closed dot at $ (0, 0) $
- Open dot at $ (0, 4) $ — but not connected to anything at $ x = 0 $

But in the given graph, it seems the parabola starts at $ (0, 4) $ with open dot, and the line ends at $ (0, 0) $ — that’s acceptable.

But wait — what about the left part? It shows a curve approaching $ (-4, -0.06) $ with open dot, and then a line from $ (-4, -4) $ to $ (0, 0) $

But $ f(-4) = -|-4| = -4 $, so closed dot at $ (-4, -4) $

And for $ x < -4 $, $ f(x) = -2^x $, which is negative and approaches 0 from below

So at $ x = -5 $: $ f(-5) = -2^{-5} = -1/32 \approx -0.031 $

So yes, the graph shows a curve approaching $ y = 0 $ from below, ending at $ x = -4 $ with open dot at $ (-4, -0.06) $, and then a line from $ (-4, -4) $ to $ (0, 0) $

But there is a discontinuity at $ x = -4 $ — one side has $ y \approx -0.06 $, other has $ y = -4 $

That’s correct.

So overall:
- Left: exponential decay to $ (-4, -0.06) $, open dot
- Middle: line from $ (-4, -4) $ to $ (0, 0) $, both closed
- Right: parabola $ y = 4 - x^2 $ for $ x > 0 $, starting near $ (0, 4) $, open dot at $ (0, 4) $

✔ Graph is correct

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Problem 4:

$$
g(x) =
\begin{cases}
-6, & x < -2 \\
(x+1)^4, & x \geq -2
\end{cases}
$$

#### Step-by-step:
- First piece: $ g(x) = -6 $ for $ x < -2 $
- Horizontal line at $ y = -6 $, all $ x < -2 $
- At $ x = -2 $: not included → open dot at (-2, -6)
- Second piece: $ g(x) = (x+1)^4 $ for $ x \geq -2 $
- This is a quartic function, always non-negative, symmetric about $ x = -1 $
- At $ x = -2 $: $ g(-2) = (-2+1)^4 = (-1)^4 = 1 $ → closed dot at (-2, 1)
- At $ x = -1 $: $ g(-1) = 0 $
- At $ x = 0 $: $ g(0) = (1)^4 = 1 $
- At $ x = 1 $: $ g(1) = (2)^4 = 16 $

So:
- Left: horizontal line at $ y = -6 $, ending at $ x = -2 $ with open dot
- Right: smooth curve starting at $ (-2, 1) $, going down to $ (-1, 0) $, then up to $ (0, 1) $, $ (1, 16) $, etc.

But in the graph:
- There is a horizontal line at $ y = -6 $ ending at $ x = -2 $ with open dot
- Then a curve starting at $ (-2, 1) $ with closed dot, going down to $ (-1, 0) $, then up
- But no connection between the two pieces — they are separated

✔ Graph is correct

However, note that at $ x = -2 $, the function jumps from $ y = -6 $ (not included) to $ y = 1 $ (included). So there is a jump discontinuity at $ x = -2 $

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✔ Final Summary:

All four graphs appear to be correctly sketched, matching the piecewise functions:

1. Correct: Two lines meeting at $ x = 2 $, one with closed dot, one with open dot
2. Correct: Horizontal line, then rising line, then falling line, with appropriate open/closed dots
3. Correct: Exponential decay, linear segment, parabola — with proper domain restrictions and open/closed dots
4. Correct: Constant line, then quartic function — jump at $ x = -2 $

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🔍 Key Points to Remember When Graphing Piecewise Functions:

1. Identify domains for each piece.
2. Evaluate endpoints to determine open or closed circles.
3. Sketch each piece separately within its domain.
4. Check continuity at boundaries — sometimes there are jumps.
5. Use open circles for strict inequalities ($<$, $>$), closed circles for inclusive ($\leq$, $\geq$).

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If you were asked to solve this problem (i.e., sketch the graphs), the provided sketches are accurate and well-executed. They correctly represent the behavior of each piecewise function.

Let me know if you'd like me to generate these graphs digitally or explain any specific part further!