Piecewise Function - How to Graph? Examples, Evaluating - Free Printable

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 the graphs are already drawn (likely as part of an answer key). Below, I will:

1. Explain how to sketch each graph step-by-step.
2. Verify that the provided graphs match the functions.
3. Clarify any nuances in domain restrictions and continuity.

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🔷 Problem 1:

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

#### Step-by-step Graphing:
- First piece: $ f(x) = -2x - 1 $ for $ x \leq 2 $
- This is a line with slope $-2$, y-intercept $-1$.
- Plot points: when $ x = 2 $, $ f(2) = -2(2) - 1 = -5 $. So point $ (2, -5) $ is included (closed circle).
- Extend the line to the left (since $ x \leq 2 $).
- 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 $, this point is not included (open circle at $ (2, 2) $).
- Draw the line starting just after $ x = 2 $, going down to the right.

✔ The graph shows:
- A line decreasing steeply (slope -2) ending at $ (2, -5) $ with a closed dot.
- Another line (slope -1) starting at $ (2, 2) $ with an open dot, going downward.

✔️ 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 to the left.
- At $ x = -2 $, closed dot at $ (-2, -4) $.
- Second piece: $ f(x) = x - 2 $ for $ -2 < x < 2 $
- Line with slope 1, y-intercept $-2$.
- Open circle at $ x = -2 $: $ f(-2) = -2 - 2 = -4 $, but not included.
- Open circle at $ x = 2 $: $ f(2) = 0 $, but not included.
- So draw line segment between $ (-2, -4) $ and $ (2, 0) $, both ends open.
- Third piece: $ f(x) = -2x + 4 $ for $ x \geq 2 $
- Slope $-2$, y-intercept $4$.
- At $ x = 2 $, $ f(2) = -4 + 4 = 0 $, so closed dot at $ (2, 0) $.
- Continue the line downward to the right.

✔ The graph shows:
- Flat line at $ y = -4 $ up to $ x = -2 $ (closed dot).
- Diagonal line from $ (-2, -4) $ to $ (2, 0) $, both open circles.
- Then another diagonal line starting at $ (2, 0) $ (closed), going down.

✔️ Correct.

---

🔷 Problem 3:

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

#### Step-by-step:
- First piece: $ f(x) = -2^x $ for $ x < -4 $
- Exponential decay (but negative).
- As $ x \to -\infty $, $ 2^x \to 0 $, so $ -2^x \to 0^- $
- At $ x = -4 $, $ -2^{-4} = -1/16 \approx -0.0625 $
- But since $ x < -4 $, we have an open circle at $ (-4, -1/16) $
- The curve approaches zero from below as $ x \to -\infty $
- Second piece: $ f(x) = -|x| $ for $ -4 \leq x \leq 0 $
- This is a V-shape flipped upside-down.
- For $ x \in [-4, 0] $, $ |x| = -x $, so $ f(x) = -(-x) = x $? Wait — no!
- Actually: $ f(x) = -|x| $
- So for $ x \in [-4, 0] $, $ |x| = -x $, so $ f(x) = -(-x) = x $? No — wait:
- $ |x| = -x $ when $ x < 0 $, so $ f(x) = -(-x) = x $? That's incorrect.
- Let's clarify:
- If $ x = -2 $, $ |x| = 2 $, so $ f(x) = -2 $
- So $ f(x) = -|x| $ is a V-shape pointing downward, vertex at $ (0, 0) $
- From $ x = -4 $ to $ x = 0 $, it goes from $ f(-4) = -4 $ to $ f(0) = 0 $
- So it’s a straight line from $ (-4, -4) $ to $ (0, 0) $
- Closed dots at both ends because $ x \in [-4, 0] $
- Third piece: $ f(x) = 4 - x^2 $ for $ x > 0 $
- Parabola opening downward, vertex at $ (0, 4) $
- But only defined for $ x > 0 $, so open circle at $ (0, 4) $
- As $ x $ increases, it decreases: at $ x = 2 $, $ f(2) = 4 - 4 = 0 $; at $ x = 4 $, $ f(4) = 4 - 16 = -12 $

Now compare to the graph:
- Leftmost: exponential curve approaching $ y=0 $ from below, open at $ x=-4 $
- Middle: line from $ (-4, -4) $ to $ (0, 0) $, both closed
- Right: parabola from $ (0, 4) $ (open) downward

But wait — the graph shows:
- A point at $ (-4, -4) $, then a line to $ (0, 0) $
- Then a parabola starting at $ (0, 4) $, going down

But $ f(0) = -|0| = 0 $, so $ (0, 0) $ is correct.
And $ f(x) = 4 - x^2 $ for $ x > 0 $, so at $ x = 0^+ $, $ f(x) \to 4 $, but not including $ (0, 4) $

So the graph should show a jump discontinuity at $ x = 0 $:
- $ f(0) = 0 $ (from middle piece)
- But limit from right is $ 4 $

In the provided graph:
- There's a closed dot at $ (0, 0) $
- And a parabola starting at $ (0, 4) $ with open circle? Wait — actually, looking closely:
- The parabola starts at $ (0, 4) $ with no dot, suggesting open circle.
- But $ (0, 0) $ has a closed dot, which is correct.

So yes, the graph correctly shows:
- A gap between $ (0, 0) $ and $ (0, 4) $
- The parabola starts at $ (0, 4) $ with open circle

✔ Correct.

---

🔷 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 $
- Open circle at $ x = -2 $, since $ x < -2 $
- Second piece: $ g(x) = (x+1)^4 $ for $ x \geq -2 $
- Even-degree polynomial, always ≥ 0, minimum at $ x = -1 $
- At $ x = -2 $: $ (-2+1)^4 = (-1)^4 = 1 $
- So $ g(-2) = 1 $, so closed dot at $ (-2, 1) $
- At $ x = -1 $: $ (0)^4 = 0 $
- At $ x = 0 $: $ (1)^4 = 1 $
- At $ x = 1 $: $ (2)^4 = 16 $

So the graph:
- Horizontal line at $ y = -6 $ for $ x < -2 $, open at $ (-2, -6) $
- Then a smooth curve starting at $ (-2, 1) $, dipping to $ (-1, 0) $, rising to $ (0, 1) $, $ (1, 16) $, etc.

In the provided graph:
- Horizontal line at $ y = -6 $, open circle at $ (-2, -6) $
- Then a curve starting at $ (-2, 1) $ with closed dot
- It dips to $ (-1, 0) $, then rises sharply

✔ Correct.

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✔ Final Summary:

All four graphs are correctly sketched according to their piecewise definitions.

Key features checked:
- Domain intervals respected (open/closed circles)
- Function behavior in each interval
- Continuity or jumps properly shown
- Shape and direction of each piece (linear, exponential, absolute value, quadratic, power)

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📌 Tips for Sketching Piecewise Functions:

1. Identify domains for each piece.
2. Plot endpoints carefully: use closed circles for inclusive ($\leq, \geq$), open circles for exclusive ($<, >$).
3. Sketch each piece separately within its domain.
4. Check for continuity at boundary points.
5. Label key points like intercepts, vertices, asymptotes.

Let me know if you'd like these graphs redrawn or explained further!