Domain and Range of Radical Functions worksheets - Free Printable

Problem Overview:

The task involves sketching the graphs of radical functions and identifying their domains and ranges. Let's solve each part step by step.

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Part 1: Sketch the graph of each function

#### 1. \( y = \sqrt{x} - 1 \)

- Domain: The square root function \( \sqrt{x} \) is defined for \( x \geq 0 \). Therefore, the domain is \( [0, \infty) \).
- Range: The minimum value of \( \sqrt{x} \) is 0 (when \( x = 0 \)), so the minimum value of \( y = \sqrt{x} - 1 \) is \( -1 \). As \( x \to \infty \), \( y \to \infty \). Thus, the range is \( [-1, \infty) \).
- Graph: Start with the graph of \( y = \sqrt{x} \), then shift it downward by 1 unit.

#### 2. \( y = 2\sqrt{x} + 2 \)

- Domain: Similar to the previous function, \( \sqrt{x} \) is defined for \( x \geq 0 \). Therefore, the domain is \( [0, \infty) \).
- Range: The minimum value of \( \sqrt{x} \) is 0, so the minimum value of \( y = 2\sqrt{x} + 2 \) is 2. As \( x \to \infty \), \( y \to \infty \). Thus, the range is \( [2, \infty) \).
- Graph: Start with the graph of \( y = \sqrt{x} \), stretch it vertically by a factor of 2, and then shift it upward by 2 units.

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Part 2: Identify the domain and range of each function

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

- Domain: The expression inside the square root, \( x - 5 \), must be non-negative: \( x - 5 \geq 0 \). Therefore, \( x \geq 5 \). The domain is \( [5, \infty) \).
- Range: The minimum value of \( \sqrt{x - 5} \) is 0 (when \( x = 5 \)), so the minimum value of \( y = \sqrt{x - 5} + 3 \) is 3. As \( x \to \infty \), \( y \to \infty \). Thus, the range is \( [3, \infty) \).

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

- Domain: The expression inside the square root, \( x + 1 \), must be non-negative: \( x + 1 \geq 0 \). Therefore, \( x \geq -1 \). The domain is \( [-1, \infty) \).
- Range: The minimum value of \( \sqrt{x + 1} \) is 0 (when \( x = -1 \)), so the minimum value of \( y = \sqrt{x + 1} - 2 \) is \( -2 \). As \( x \to \infty \), \( y \to \infty \). Thus, the range is \( [-2, \infty) \).

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

- Domain: The cube root function \( \sqrt[3]{x} \) is defined for all real numbers. Therefore, the domain is \( (-\infty, \infty) \).
- Range: The cube root function can take any real value, so \( \sqrt[3]{x - 2} \) can take any real value. Shifting it downward by 8 does not change this property. Thus, the range is \( (-\infty, \infty) \).

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

- Domain: The expression inside the square root, \( 3x - 9 \), must be non-negative: \( 3x - 9 \geq 0 \). Solving for \( x \), we get \( x \geq 3 \). The domain is \( [3, \infty) \).
- Range: The minimum value of \( \sqrt{3x - 9} \) is 0 (when \( x = 3 \)), so the minimum value of \( y = \sqrt{3x - 9} + 6 \) is 6. As \( x \to \infty \), \( y \to \infty \). Thus, the range is \( [6, \infty) \).

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

- Domain: The expression inside the square root, \( 9x^2 - 9 \), must be non-negative: \( 9x^2 - 9 \geq 0 \). Factoring, we get \( 9(x^2 - 1) \geq 0 \), or \( x^2 - 1 \geq 0 \). This inequality holds when \( x \leq -1 \) or \( x \geq 1 \). Therefore, the domain is \( (-\infty, -1] \cup [1, \infty) \).
- Range: The minimum value of \( 9x^2 - 9 \) is 0 (when \( x = \pm 1 \)), so the minimum value of \( y = \sqrt{9x^2 - 9} \) is 0. As \( x \to \infty \) or \( x \to -\infty \), \( y \to \infty \). Thus, the range is \( [0, \infty) \).

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

- Domain: The expression inside the square root, \( x^2 + 9 \), is always positive for all real \( x \). Therefore, the domain is \( (-\infty, \infty) \).
- Range: The minimum value of \( x^2 + 9 \) is 9 (when \( x = 0 \)), so the minimum value of \( y = \sqrt{x^2 + 9} - 4 \) is \( \sqrt{9} - 4 = 3 - 4 = -1 \). As \( x \to \infty \) or \( x \to -\infty \), \( y \to \infty \). Thus, the range is \( [-1, \infty) \).

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

- Domain: The fifth root function \( \sqrt[5]{x} \) is defined for all real numbers. Therefore, the domain is \( (-\infty, \infty) \).
- Range: The fifth root function can take any real value, so \( \sqrt[5]{2x^4 + 5x^3 - x^2 - x + 1} \) can take any real value. Multiplying by 7 and adding 7 does not change the fact that \( y \) can take any real value. Thus, the range is \( (-\infty, \infty) \).

#### 10. \( y = 2x^4\sqrt{x^4 + 1} + 1 \)

- Domain: The expression inside the square root, \( x^4 + 1 \), is always positive for all real \( x \). Therefore, the domain is \( (-\infty, \infty) \).
- Range: The minimum value of \( x^4 + 1 \) is 1 (when \( x = 0 \)), so the minimum value of \( \sqrt{x^4 + 1} \) is 1. When \( x = 0 \), \( y = 2(0)^4\sqrt{(0)^4 + 1} + 1 = 1 \). As \( x \to \infty \) or \( x \to -\infty \), both \( x^4 \) and \( \sqrt{x^4 + 1} \) grow without bound, so \( y \to \infty \). Thus, the range is \( [1, \infty) \).

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Final Answers:

1. \( y = \sqrt{x} - 1 \)
- Domain: \( [0, \infty) \)
- Range: \( [-1, \infty) \)

2. \( y = 2\sqrt{x} + 2 \)
- Domain: \( [0, \infty) \)
- Range: \( [2, \infty) \)

3. \( y = \sqrt{x - 5} + 3 \)
- Domain: \( [5, \infty) \)
- Range: \( [3, \infty) \)

4. \( y = \sqrt{x + 1} - 2 \)
- Domain: \( [-1, \infty) \)
- Range: \( [-2, \infty) \)

5. \( y = \sqrt[3]{x - 2} - 8 \)
- Domain: \( (-\infty, \infty) \)
- Range: \( (-\infty, \infty) \)

6. \( y = \sqrt{3x - 9} + 6 \)
- Domain: \( [3, \infty) \)
- Range: \( [6, \infty) \)

7. \( y = \sqrt{9x^2 - 9} \)
- Domain: \( (-\infty, -1] \cup [1, \infty) \)
- Range: \( [0, \infty) \)

8. \( y = \sqrt{x^2 + 9} - 4 \)
- Domain: \( (-\infty, \infty) \)
- Range: \( [-1, \infty) \)

9. \( y = 7\sqrt[5]{2x^4 + 5x^3 - x^2 - x + 1} + 7 \)
- Domain: \( (-\infty, \infty) \)
- Range: \( (-\infty, \infty) \)

10. \( y = 2x^4\sqrt{x^4 + 1} + 1 \)
- Domain: \( (-\infty, \infty) \)
- Range: \( [1, \infty) \)

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Boxed Final Answer:

\[
\boxed{
\begin{array}{ll}
1. & \text{Domain: } [0, \infty), \text{ Range: } [-1, \infty) \\
2. & \text{Domain: } [0, \infty), \text{ Range: } [2, \infty) \\
3. & \text{Domain: } [5, \infty), \text{ Range: } [3, \infty) \\
4. & \text{Domain: } [-1, \infty), \text{ Range: } [-2, \infty) \\
5. & \text{Domain: } (-\infty, \infty), \text{ Range: } (-\infty, \infty) \\
6. & \text{Domain: } [3, \infty), \text{ Range: } [6, \infty) \\
7. & \text{Domain: } (-\infty, -1] \cup [1, \infty), \text{ Range: } [0, \infty) \\
8. & \text{Domain: } (-\infty, \infty), \text{ Range: } [-1, \infty) \\
9. & \text{Domain: } (-\infty, \infty), \text{ Range: } (-\infty, \infty) \\
10. & \text{Domain: } (-\infty, \infty), \text{ Range: } [1, \infty) \\
\end{array}
}
\]