simply way to find domain and range of f(x) = square root of 2+3x-5x^2

To find the domain and range of the function \( f(x) = \sqrt{2 + 3x - 5x^2} \), follow these steps:

### Domain:
The domain of a function is the set of all possible input values (x-values) for which the function is defined. For the square root function, the expression inside the square root must be non-negative (greater than or equal to zero) because the square root of a negative number is not a real number.

1. **Set the expression inside the square root to be greater than or equal to zero:**
   \[
   2 + 3x - 5x^2 \geq 0
   \]

2. **Rewrite the quadratic inequality:**
   \[
   -5x^2 + 3x + 2 \geq 0
   \]

3. **Solve the inequality:**
   First, find the roots of the quadratic equation by setting the expression equal to zero:
   \[
   -5x^2 + 3x + 2 = 0
   \]
   Use the quadratic formula:
   \[
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   where \( a = -5 \), \( b = 3 \), and \( c = 2 \).

   \[
   x = \frac{-3 \pm \sqrt{3^2 - 4(-5)(2)}}{2(-5)} = \frac{-3 \pm \sqrt{9 + 40}}{-10} = \frac{-3 \pm \sqrt{49}}{-10} = \frac{-3 \pm 7}{-10}
   \]
   So, the roots are:
   \[
   x = \frac{4}{-10} = -0.4 \quad \text{and} \quad x = \frac{-10}{-10} = 1
   \]

   The roots are \( x = -0.4 \) and \( x = 1 \).

4. **Determine the intervals where the inequality holds:**
   Test the sign of the expression in each interval determined by the roots \( x = -0.4 \) and \( x = 1 \):
   - For \( x < -0.4 \), the expression is negative.
   - For \( -0.4 \leq x \leq 1 \), the expression is non-negative.
   - For \( x > 1 \), the expression is negative.

   Therefore, the domain is:
   \[
   -0.4 \leq x \leq 1
   \]

### Range:
The range of a function is the set of all possible output values (y-values).

1. **Evaluate the function at the endpoints of the domain:**
   \[
   f(-0.4) = \sqrt{2 + 3(-0.4) - 5(-0.4)^2} = \sqrt{2 - 1.2 - 0.8} = \sqrt{0} = 0
   \]
   \[
   f(1) = \sqrt{2 + 3(1) - 5(1)^2} = \sqrt{2 + 3 - 5} = \sqrt{0} = 0
   \]

2. **Check the vertex of the parabola (maximum value within the domain):**
   The vertex occurs at \( x = -\frac{b}{2a} = \frac{-3}{-10} = 0.3 \).
   \[
   f(0.3) = \sqrt{2 + 3(0.3) - 5(0.3)^2} = \sqrt{2 + 0.9 - 0.45} = \sqrt{2.45} \approx 1.565
   \]

   Since \( f(x) \) is a square root function and has a maximum at \( x = 0.3 \), the range is:
   \[
   0 \leq y \leq \sqrt{2.45}
   \]

### Final Answer:
- **Domain:** \( -0.4 \leq x \leq 1 \)
- **Range:** \( 0 \leq y \leq \sqrt{2.45} \)

Would you like further details or explanations?

### Related Questions:
1. How do you solve quadratic inequalities in general?
2. What are the steps to determine the vertex of a quadratic function?
3. How do you find the maximum or minimum value of a quadratic function?
4. What is the significance of the discriminant in finding roots?
5. How can you verify the domain and range of a function graphically?

### Tip:
When solving quadratic inequalities, always consider testing intervals around the roots to determine where the function is positive or negative.

Simply way to find domain and range of f(x) = square root of 2+3x-5x^2

Learn how to find the domain and range of the function f(x) = √(2 + 3x - 5x^2) using quadratic inequalities and the quadratic formula. This detailed solution is ideal for students in Grades 9-12 and covers key algebraic concepts.

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