Math worksheet for practicing domain and range of radical functions with graphing exercises.

Let's solve the problem step by step. The worksheet asks you to:

1. Sketch the graph of two radical functions.
2. Identify the domain and range of eight radical functions.

We’ll go through each part carefully.

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🔷 Part 1: Sketch the Graph

#### 1) $ y = \sqrt{x} - 1 $

- This is a square root function shifted down by 1 unit.
- The parent function $ y = \sqrt{x} $ has:
- Domain: $ x \geq 0 $
- Range: $ y \geq 0 $
- Shifting down by 1:
- New range: $ y \geq -1 $
- Domain remains: $ x \geq 0 $

Graphing steps:
- Start at $ (0, -1) $ — this is the new "vertex".
- Plot points:
- $ x = 0 $ → $ y = \sqrt{0} - 1 = -1 $
- $ x = 1 $ → $ y = \sqrt{1} - 1 = 0 $
- $ x = 4 $ → $ y = \sqrt{4} - 1 = 2 - 1 = 1 $
- $ x = 9 $ → $ y = 3 - 1 = 2 $

Draw a smooth curve starting at $ (0, -1) $, increasing slowly.

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#### 2) $ y = 2\sqrt{x} + 2 $

- This is $ y = \sqrt{x} $ scaled vertically by 2 and shifted up by 2.
- Vertical stretch by factor 2 → steeper than original.
- Shift up by 2.

Domain: $ x \geq 0 $
Range: $ y \geq 2 $

Points:
- $ x = 0 $ → $ y = 2(0) + 2 = 2 $
- $ x = 1 $ → $ y = 2(1) + 2 = 4 $
- $ x = 4 $ → $ y = 2(2) + 2 = 6 $
- $ x = 9 $ → $ y = 2(3) + 2 = 8 $

Start at $ (0, 2) $, rise faster than standard square root.

> ✔ You would sketch both graphs on the provided grids accordingly.

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🔷 Part 2: Identify Domain and Range

We'll analyze each function using rules for radicals.

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#### 3) $ y = \sqrt{x - 5} + 3 $

- Square root: expression under root must be ≥ 0
- $ x - 5 \geq 0 $ → $ x \geq 5 $
- So, Domain: $ [5, \infty) $
- Minimum value of $ \sqrt{x - 5} $ is 0 when $ x = 5 $, so $ y = 0 + 3 = 3 $
- As $ x $ increases, $ y $ increases without bound
- Range: $ [3, \infty) $

✔ Answer:
- Domain: $ [5, \infty) $
- Range: $ [3, \infty) $

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#### 4) $ y = \sqrt{x + 1} - 2 $

- $ x + 1 \geq 0 $ → $ x \geq -1 $
- Domain: $ [-1, \infty) $
- When $ x = -1 $, $ y = \sqrt{0} - 2 = -2 $
- As $ x \to \infty $, $ y \to \infty $
- Range: $ [-2, \infty) $

✔ Answer:
- Domain: $ [-1, \infty) $
- Range: $ [-2, \infty) $

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#### 5) $ y = \sqrt[3]{x - 2} - 8 $

- Cube root! Defined for all real numbers.
- No restriction on input.
- Domain: $ (-\infty, \infty) $
- Cube root can output any real number → range is also all reals
- Then subtract 8 → still all reals

✔ Answer:
- Domain: $ (-\infty, \infty) $
- Range: $ (-\infty, \infty) $

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#### 6) $ y = \sqrt{3x - 9} + 6 $

- Inside radical: $ 3x - 9 \geq 0 $
- $ 3x \geq 9 $ → $ x \geq 3 $
- Domain: $ [3, \infty) $
- Minimum value when $ x = 3 $: $ \sqrt{0} + 6 = 6 $
- Increases as $ x $ increases
- Range: $ [6, \infty) $

✔ Answer:
- Domain: $ [3, \infty) $
- Range: $ [6, \infty) $

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#### 7) $ y = \sqrt{9x^2 - 9} $

- Simplify inside: $ 9x^2 - 9 = 9(x^2 - 1) $
- So $ y = \sqrt{9(x^2 - 1)} = 3\sqrt{x^2 - 1} $
- But we must have $ x^2 - 1 \geq 0 $ → $ x^2 \geq 1 $
- So $ x \leq -1 $ or $ x \geq 1 $
- Domain: $ (-\infty, -1] \cup [1, \infty) $

Now for range:
- $ \sqrt{x^2 - 1} \geq 0 $ → $ y \geq 0 $
- Minimum occurs at $ x = \pm1 $: $ y = 3\sqrt{1 - 1} = 0 $
- As $ |x| \to \infty $, $ y \to \infty $
- So range: $ [0, \infty) $

✔ Answer:
- Domain: $ (-\infty, -1] \cup [1, \infty) $
- Range: $ [0, \infty) $

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#### 8) $ y = \sqrt{x^2 + 9} - 4 $

- $ x^2 + 9 \geq 0 $ always true (since $ x^2 \geq 0 $)
- So domain: $ (-\infty, \infty) $
- $ x^2 + 9 \geq 9 $ → $ \sqrt{x^2 + 9} \geq \sqrt{9} = 3 $
- So $ y = \sqrt{x^2 + 9} - 4 \geq 3 - 4 = -1 $
- As $ |x| \to \infty $, $ y \to \infty $
- Range: $ [-1, \infty) $

✔ Answer:
- Domain: $ (-\infty, \infty) $
- Range: $ [-1, \infty) $

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#### 9) $ y = 7\sqrt[5]{2x^4 + 5x^3 - x^2 - x + 1} + 7 $

- Fifth root: odd root → defined for all real numbers, even if inside is negative.
- So no restrictions on $ x $
- Domain: $ (-\infty, \infty) $
- The fifth root function outputs all real numbers → $ \sqrt[5]{\text{anything}} \in \mathbb{R} $
- Multiply by 7 and add 7 → still covers all reals
- So range: $ (-\infty, \infty) $

✔ Answer:
- Domain: $ (-\infty, \infty) $
- Range: $ (-\infty, \infty) $

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#### 10) $ y = 2x\sqrt[4]{x^4 + 1} + 1 $

- Fourth root: even root → requires non-negative radicand
- $ x^4 + 1 \geq 0 $ → always true (since $ x^4 \geq 0 $, so $ x^4 + 1 \geq 1 > 0 $)
- So domain: $ (-\infty, \infty) $
- Now check range:

Let’s analyze behavior:
- $ \sqrt[4]{x^4 + 1} $ is always positive, slightly larger than $ |x| $
- $ x $ can be positive or negative
- So $ 2x \cdot \sqrt[4]{x^4 + 1} $ can be positive or negative depending on sign of $ x $
- As $ x \to \infty $: $ \sqrt[4]{x^4 + 1} \approx |x| = x $ → $ 2x \cdot x = 2x^2 $ → $ y \to \infty $
- As $ x \to -\infty $: $ x < 0 $, $ \sqrt[4]{x^4 + 1} \approx |x| = -x $, so $ 2x \cdot (-x) = -2x^2 $ → $ y \to -\infty $
- At $ x = 0 $: $ y = 0 \cdot \sqrt[4]{1} + 1 = 1 $

So $ y $ can take any real value — goes from $ -\infty $ to $ \infty $

✔ Answer:
- Domain: $ (-\infty, \infty) $
- Range: $ (-\infty, \infty) $

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✔ Final Answers Summary:

| Problem | Function | Domain | Range |
|--------|---------|--------|-------|
| 1 | $ y = \sqrt{x} - 1 $ | $ [0, \infty) $ | $ [-1, \infty) $ |
| 2 | $ y = 2\sqrt{x} + 2 $ | $ [0, \infty) $ | $ [2, \infty) $ |
| 3 | $ y = \sqrt{x - 5} + 3 $ | $ [5, \infty) $ | $ [3, \infty) $ |
| 4 | $ y = \sqrt{x + 1} - 2 $ | $ [-1, \infty) $ | $ [-2, \infty) $ |
| 5 | $ y = \sqrt[3]{x - 2} - 8 $ | $ (-\infty, \infty) $ | $ (-\infty, \infty) $ |
| 6 | $ y = \sqrt{3x - 9} + 6 $ | $ [3, \infty) $ | $ [6, \infty) $ |
| 7 | $ y = \sqrt{9x^2 - 9} $ | $ (-\infty, -1] \cup [1, \infty) $ | $ [0, \infty) $ |
| 8 | $ y = \sqrt{x^2 + 9} - 4 $ | $ (-\infty, \infty) $ | $ [-1, \infty) $ |
| 9 | $ y = 7\sqrt[5]{...} + 7 $ | $ (-\infty, \infty) $ | $ (-\infty, \infty) $ |
|10 | $ y = 2x\sqrt[4]{x^4 + 1} + 1 $ | $ (-\infty, \infty) $ | $ (-\infty, \infty) $ |

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📌 Key Concepts Used:

- Square roots: require $ \text{radicand} \geq 0 $
- Odd roots (cube, fifth): defined for all real numbers
- Even roots (square, fourth): require $ \text{radicand} \geq 0 $
- Transformations: shifts, stretches affect domain/range accordingly
- For ranges, consider minimum/maximum values and end behavior

Let me know if you'd like the graph sketches drawn digitally or explained in more detail!