Grade 8 Evaluating Piecewise Functions Worksheets 2024 - Free Printable

You’ve uploaded an answer key for a worksheet titled “Evaluating Piecewise Functions.” Since you asked to solve the problem and explain the solution, I’ll walk through each part of the worksheet, showing how each answer is derived — even though the answers are already filled in.

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## 🧩 PART A: Evaluate each function

1)

Function:
\[
f(x) =
\begin{cases}
-x - 4 & \text{if } x \leq 5 \\
2x^2 - 7 & \text{if } 5 < x \leq 10
\end{cases}
\]

#### i) \( f(-2) \)

- Since \(-2 \leq 5\), use the first piece: \( f(x) = -x - 4 \)
- Plug in \( x = -2 \):
\[
f(-2) = -(-2) - 4 = 2 - 4 = \boxed{-2}
\]

✔ Matches answer key.

#### ii) \( f(7) \)

- Since \( 5 < 7 \leq 10 \), use the second piece: \( f(x) = 2x^2 - 7 \)
- Plug in \( x = 7 \):
\[
f(7) = 2(7)^2 - 7 = 2(49) - 7 = 98 - 7 = \boxed{91}
\]

✔ Matches answer key.

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2)

Function:
\[
f(x) =
\begin{cases}
x^2 & \text{if } -15 \leq x \leq 0 \\
x - 5 & \text{if } 0 < x \leq 15
\end{cases}
\]

#### i) \( f(-5) \)

- Since \(-15 \leq -5 \leq 0\), use first piece: \( f(x) = x^2 \)
- \( f(-5) = (-5)^2 = \boxed{25} \)

✔ Matches.

#### ii) \( f(15) \)

- Since \( 0 < 15 \leq 15 \), use second piece: \( f(x) = x - 5 \)
- \( f(15) = 15 - 5 = \boxed{10} \)

✔ Matches.

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3)

Function:
\[
f(x) =
\begin{cases}
\frac{6}{x} - 1 & \text{if } x \neq 0 \\
3 & \text{if } x = 0
\end{cases}
\]

#### i) \( f(3) \)

- \( x = 3 \neq 0 \), so use first piece: \( \frac{6}{x} - 1 \)
- \( f(3) = \frac{6}{3} - 1 = 2 - 1 = \boxed{1} \)

✔ Matches.

#### ii) \( f(0) \)

- Defined directly as 3 when \( x = 0 \)
- So \( f(0) = \boxed{3} \)

✔ Matches.

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4)

Function:
\[
f(x) =
\begin{cases}
14 & \text{if } x \leq 0 \\
x^2 - 9x & \text{if } 0 < x < \infty
\end{cases}
\]

#### i) \( f(-5) \)

- Since \(-5 \leq 0\), use constant piece: \( f(x) = 14 \)
- So \( f(-5) = \boxed{14} \)

✔ Matches.

#### ii) \( f(1) \)

- Since \( 0 < 1 < \infty \), use second piece: \( x^2 - 9x \)
- \( f(1) = (1)^2 - 9(1) = 1 - 9 = \boxed{-8} \)

✔ Matches.

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## 🧩 PART B: Given \( f(x) = \begin{cases} x + 2 & \text{if } x \leq -7 \\ 8x & \text{if } -7 < x \leq 12 \end{cases} \)

We need to compute expressions using this function.

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1) \( \frac{7f(6)}{4f(-9)} \)

First, find \( f(6) \) and \( f(-9) \):

- \( f(6) \): Since \( -7 < 6 \leq 12 \), use \( f(x) = 8x \)
→ \( f(6) = 8 \cdot 6 = 48 \)

- \( f(-9) \): Since \( -9 \leq -7 \), use \( f(x) = x + 2 \)
→ \( f(-9) = -9 + 2 = -7 \)

Now plug into expression:
\[
\frac{7 \cdot 48}{4 \cdot (-7)} = \frac{336}{-28} = \boxed{-12}
\]

✔ Matches answer key.

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2) \( 3f(8) - f(-5) \)

- \( f(8) \): \( -7 < 8 \leq 12 \) → \( f(x) = 8x \) → \( f(8) = 64 \)

- \( f(-5) \): Is \(-5 \leq -7\)? ✘ No. So use second piece: \( f(x) = 8x \)
→ \( f(-5) = 8 \cdot (-5) = -40 \)

Wait — this is a common mistake!

Let’s double-check the domain:

> \( f(x) = \begin{cases} x + 2 & \text{if } x \leq -7 \\ 8x & \text{if } -7 < x \leq 12 \end{cases} \)

So for \( x = -5 \), since \( -7 < -5 \leq 12 \), we use \( 8x \).

Thus:
\[
3f(8) - f(-5) = 3(64) - (-40) = 192 + 40 = \boxed{232}
\]

✔ Matches answer key.

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3) \( -9f(-7) + f(0) \)

- \( f(-7) \): Since \( x = -7 \leq -7 \), use first piece: \( f(x) = x + 2 \)
→ \( f(-7) = -7 + 2 = -5 \)

- \( f(0) \): \( -7 < 0 \leq 12 \), so use \( 8x \)
→ \( f(0) = 8 \cdot 0 = 0 \)

Now compute:
\[
-9(-5) + 0 = 45 + 0 = \boxed{45}
\]

✔ Matches.

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4) \( f(-1) \times 2f(2) \)

- \( f(-1) \): \( -7 < -1 \leq 12 \) → use \( 8x \) → \( f(-1) = 8 \cdot (-1) = -8 \)

- \( f(2) \): same rule → \( f(2) = 8 \cdot 2 = 16 \)

Now compute:
\[
f(-1) \times 2f(2) = (-8) \times 2(16) = (-8) \times 32 = \boxed{-256}
\]

✔ Matches.

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## 🧩 PART C: If \( f(x) = \begin{cases} (x+4)^2 & \text{if } -20 \leq x \leq 0 \\ 3x^2 - x & \text{if } 0 < x \leq 20 \end{cases} \); what is \( f(-11) \)?

- Check domain: \( -20 \leq -11 \leq 0 \) → YES → use first piece: \( (x+4)^2 \)

Compute:
\[
f(-11) = (-11 + 4)^2 = (-7)^2 = \boxed{49}
\]

✔ Correct answer is iii) 49

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## ✔ FINAL SUMMARY

All answers in the key are correct, and here’s how they were derived:

| Problem | Answer | Reason |
|--------|--------|--------|
| A1i | -2 | Used \( -x - 4 \) at \( x = -2 \) |
| A1ii | 91 | Used \( 2x^2 - 7 \) at \( x = 7 \) |
| A2i | 25 | Used \( x^2 \) at \( x = -5 \) |
| A2ii | 10 | Used \( x - 5 \) at \( x = 15 \) |
| A3i | 1 | Used \( \frac{6}{x} - 1 \) at \( x = 3 \) |
| A3ii | 3 | Direct definition at \( x = 0 \) |
| A4i | 14 | Constant for \( x \leq 0 \) |
| A4ii | -8 | Used \( x^2 - 9x \) at \( x = 1 \) |
| B1 | -12 | Computed \( \frac{7 \cdot 48}{4 \cdot (-7)} \) |
| B2 | 232 | \( 3(64) - (-40) \) |
| B3 | 45 | \( -9(-5) + 0 \) |
| B4 | -256 | \( (-8) \times 2(16) \) |
| C | 49 | \( (-11 + 4)^2 = 49 \) |

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✔ You now understand how to evaluate piecewise functions — always check which piece of the function applies based on the input value’s domain, then plug into that expression.

Let me know if you’d like to practice similar problems!