Algebra 2 Trig Unit 1 Lesson 3 Transformations Of Graphs - Fill ... - Free Printable

To solve the problem, we need to determine the domain and range of each given function without using technology. Let's analyze each function step by step.

---

1. \( f(x) = x - 2x^2 + 6 \)

#### Domain:
- This is a quadratic function in the form \( f(x) = ax^2 + bx + c \).
- Quadratic functions are defined for all real numbers.
- Therefore, the domain is:
\[
\text{Domain: } (-\infty, \infty)
\]

#### Range:
- The function can be rewritten as \( f(x) = -2x^2 + x + 6 \).
- Since the coefficient of \( x^2 \) is negative (\( -2 \)), the parabola opens downward.
- To find the maximum value, we use the vertex formula for a quadratic function \( f(x) = ax^2 + bx + c \):
\[
x = -\frac{b}{2a}
\]
Here, \( a = -2 \) and \( b = 1 \):
\[
x = -\frac{1}{2(-2)} = \frac{1}{4}
\]
- Substitute \( x = \frac{1}{4} \) into the function to find the maximum value:
\[
f\left(\frac{1}{4}\right) = -2\left(\frac{1}{4}\right)^2 + \frac{1}{4} + 6
\]
\[
= -2\left(\frac{1}{16}\right) + \frac{1}{4} + 6
\]
\[
= -\frac{2}{16} + \frac{1}{4} + 6
\]
\[
= -\frac{1}{8} + \frac{2}{8} + 6
\]
\[
= \frac{1}{8} + 6 = \frac{1}{8} + \frac{48}{8} = \frac{49}{8}
\]
- The maximum value is \( \frac{49}{8} \), and since the parabola opens downward, the range is:
\[
\text{Range: } \left(-\infty, \frac{49}{8}\right]
\]

---

2. \( f(x) = x + \sqrt{x^2 - 1} \)

#### Domain:
- The term \( \sqrt{x^2 - 1} \) requires that the expression inside the square root be non-negative:
\[
x^2 - 1 \geq 0
\]
\[
(x - 1)(x + 1) \geq 0
\]
- Solving this inequality, we find:
\[
x \leq -1 \quad \text{or} \quad x \geq 1
\]
- Therefore, the domain is:
\[
\text{Domain: } (-\infty, -1] \cup [1, \infty)
\]

#### Range:
- For \( x \leq -1 \):
- As \( x \to -\infty \), \( \sqrt{x^2 - 1} \to |x| \), so \( f(x) \to -\infty \).
- At \( x = -1 \):
\[
f(-1) = -1 + \sqrt{(-1)^2 - 1} = -1 + \sqrt{0} = -1
\]
- For \( x \geq 1 \):
- As \( x \to \infty \), \( \sqrt{x^2 - 1} \to x \), so \( f(x) \to 2x \to \infty \).
- At \( x = 1 \):
\[
f(1) = 1 + \sqrt{1^2 - 1} = 1 + \sqrt{0} = 1
\]
- Combining these observations, the range is:
\[
\text{Range: } (-\infty, -1] \cup [1, \infty)
\]

---

3. \( f(x) = x + \sqrt{x + 1} + 6 \)

#### Domain:
- The term \( \sqrt{x + 1} \) requires that the expression inside the square root be non-negative:
\[
x + 1 \geq 0
\]
\[
x \geq -1
\]
- Therefore, the domain is:
\[
\text{Domain: } [-1, \infty)
\]

#### Range:
- As \( x \to -1 \):
\[
f(-1) = -1 + \sqrt{-1 + 1} + 6 = -1 + 0 + 6 = 5
\]
- As \( x \to \infty \):
- Both \( x \) and \( \sqrt{x + 1} \) grow without bound, so \( f(x) \to \infty \).
- The function is increasing for \( x \geq -1 \), so the minimum value is at \( x = -1 \).
- Therefore, the range is:
\[
\text{Range: } [5, \infty)
\]

---

4. \( f(x) = \sqrt{x^2 - 1} + 3 \)

#### Domain:
- The term \( \sqrt{x^2 - 1} \) requires that the expression inside the square root be non-negative:
\[
x^2 - 1 \geq 0
\]
\[
(x - 1)(x + 1) \geq 0
\]
- Solving this inequality, we find:
\[
x \leq -1 \quad \text{or} \quad x \geq 1
\]
- Therefore, the domain is:
\[
\text{Domain: } (-\infty, -1] \cup [1, \infty)
\]

#### Range:
- For \( x \leq -1 \) or \( x \geq 1 \):
- The minimum value of \( \sqrt{x^2 - 1} \) is 0, which occurs when \( x = \pm 1 \).
- Therefore, the minimum value of \( f(x) \) is:
\[
f(x) = \sqrt{x^2 - 1} + 3 = 0 + 3 = 3
\]
- As \( x \to \pm\infty \), \( \sqrt{x^2 - 1} \to |x| \), so \( f(x) \to \infty \).
- Therefore, the range is:
\[
\text{Range: } [3, \infty)
\]

---

5. \( f(x) = \sqrt{x} - 2 \)

#### Domain:
- The term \( \sqrt{x} \) requires that \( x \geq 0 \).
- Therefore, the domain is:
\[
\text{Domain: } [0, \infty)
\]

#### Range:
- As \( x \to 0 \):
\[
f(0) = \sqrt{0} - 2 = 0 - 2 = -2
\]
- As \( x \to \infty \):
- \( \sqrt{x} \to \infty \), so \( f(x) \to \infty \).
- The function is increasing for \( x \geq 0 \), so the minimum value is at \( x = 0 \).
- Therefore, the range is:
\[
\text{Range: } [-2, \infty)
\]

---

6. \( f(x) = 2^x + 3 \)

#### Domain:
- The function \( 2^x \) is defined for all real numbers.
- Therefore, the domain is:
\[
\text{Domain: } (-\infty, \infty)
\]

#### Range:
- The function \( 2^x \) is always positive and approaches 0 as \( x \to -\infty \).
- Adding 3 shifts the entire graph up by 3 units.
- Therefore, the range is:
\[
\text{Range: } (3, \infty)
\]

---

Final Answers:

1. \( f(x) = x - 2x^2 + 6 \)
- Domain: \( (-\infty, \infty) \)
- Range: \( \left(-\infty, \frac{49}{8}\right] \)

2. \( f(x) = x + \sqrt{x^2 - 1} \)
- Domain: \( (-\infty, -1] \cup [1, \infty) \)
- Range: \( (-\infty, -1] \cup [1, \infty) \)

3. \( f(x) = x + \sqrt{x + 1} + 6 \)
- Domain: \( [-1, \infty) \)
- Range: \( [5, \infty) \)

4. \( f(x) = \sqrt{x^2 - 1} + 3 \)
- Domain: \( (-\infty, -1] \cup [1, \infty) \)
- Range: \( [3, \infty) \)

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

6. \( f(x) = 2^x + 3 \)
- Domain: \( (-\infty, \infty) \)
- Range: \( (3, \infty) \)

\boxed{
\begin{array}{ll}
1. & \text{Domain: } (-\infty, \infty), \text{ Range: } \left(-\infty, \frac{49}{8}\right] \\
2. & \text{Domain: } (-\infty, -1] \cup [1, \infty), \text{ Range: } (-\infty, -1] \cup [1, \infty) \\
3. & \text{Domain: } [-1, \infty), \text{ Range: } [5, \infty) \\
4. & \text{Domain: } (-\infty, -1] \cup [1, \infty), \text{ Range: } [3, \infty) \\
5. & \text{Domain: } [0, \infty), \text{ Range: } [-2, \infty) \\
6. & \text{Domain: } (-\infty, \infty), \text{ Range: } (3, \infty) \\
\end{array}
}